tag:blogger.com,1999:blog-36285423351682311612017-08-20T00:32:55.458-07:00GM Jackson Physics and MathematicsGM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.comBlogger92125tag:blogger.com,1999:blog-3628542335168231161.post-30288285243179780672017-08-17T12:22:00.000-07:002017-08-17T12:22:06.606-07:00Deriving the Gravitational Constant G<p><iframe width="560" height="315" src="https://www.youtube.com/embed/jwPc0kK9VHU" frameborder="0" allowfullscreen></iframe></p><p>Today we will derive the gravitational constant G, also known as Newton's constant. Here are the variables we will be working with: </p><p><a href="https://2.bp.blogspot.com/-NjjCfqOSFiw/WZXk0DolbpI/AAAAAAAADDk/4hjC7E5r4ooBa53wYbrca9iSDLl4m-wxgCLcBGAs/s1600/1.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-NjjCfqOSFiw/WZXk0DolbpI/AAAAAAAADDk/4hjC7E5r4ooBa53wYbrca9iSDLl4m-wxgCLcBGAs/s1600/1.png" data-original-width="484" data-original-height="841" /></a></p><p>Below is a crude diagram of a satellite orbiting a star or planet at velocity v, at a distance of radius r. According to its clock, the proper time is t'. The star or planet has a mass of m. </p><p><a href="https://3.bp.blogspot.com/-MTMmu7nfSUs/WZXsNNUguGI/AAAAAAAADEM/QwkSWiMBeuQyvSelOJh8y_DcFWugy2eAACLcBGAs/s1600/2.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-MTMmu7nfSUs/WZXsNNUguGI/AAAAAAAADEM/QwkSWiMBeuQyvSelOJh8y_DcFWugy2eAACLcBGAs/s1600/2.png" data-original-width="615" data-original-height="844" /></a></p><p>Our starting point shall be the Lorentz equation, courtesy of Einstein's theory of special relativity: </p><p><a href="https://1.bp.blogspot.com/-sandknoN4V4/WZXk9Ye2crI/AAAAAAAADDs/MpT4bvdVax0zsy5eJT_IgLtOopr8Yy_MQCLcBGAs/s1600/3.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-sandknoN4V4/WZXk9Ye2crI/AAAAAAAADDs/MpT4bvdVax0zsy5eJT_IgLtOopr8Yy_MQCLcBGAs/s1600/3.png" data-original-width="521" data-original-height="279" /></a></p><p>By doing some algebra we can derive equation 7 below: </p><p><a href="https://3.bp.blogspot.com/-DZC4nR0_1Ts/WZXlCA2HbzI/AAAAAAAADDw/Oupu99NbCdgUXpI_nf-pW3tXNA47tJh8QCLcBGAs/s1600/4.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-DZC4nR0_1Ts/WZXlCA2HbzI/AAAAAAAADDw/Oupu99NbCdgUXpI_nf-pW3tXNA47tJh8QCLcBGAs/s1600/4.png" data-original-width="533" data-original-height="857" /></a></p><p>Equation 7's right side expresses v^2 in terms c^2 and a time ratio. The bigger the time ratio, the faster the velocity and vice versa. </p><p>We manipulate Einstein's energy equation to get equation 8: </p><p><a href="https://1.bp.blogspot.com/-XJM9jrCb5NM/WZXlGJwJicI/AAAAAAAADD0/k2PFlIHvmacR3s7QOWd4vulAh4FYkV-bwCLcBGAs/s1600/5.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-XJM9jrCb5NM/WZXlGJwJicI/AAAAAAAADD0/k2PFlIHvmacR3s7QOWd4vulAh4FYkV-bwCLcBGAs/s1600/5.png" data-original-width="307" data-original-height="155" /></a></p><p>We make a substitution, then do some more algebra until we derive G at equation 15: </p><p><a href="https://1.bp.blogspot.com/-1oXa27nTtVQ/WZXlK8XmU1I/AAAAAAAADD4/O6ndaY1bt20k-BlmzNEMlmn6crIhumifQCLcBGAs/s1600/6.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-1oXa27nTtVQ/WZXlK8XmU1I/AAAAAAAADD4/O6ndaY1bt20k-BlmzNEMlmn6crIhumifQCLcBGAs/s1600/6.png" data-original-width="656" data-original-height="858" /></a></p><p> <a href="https://2.bp.blogspot.com/-Qw8DfMnnCxk/WZXlO0kBszI/AAAAAAAADD8/87_YaJB0SAs3QVTYFf3VjYExaj9q6GqpgCLcBGAs/s1600/7.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-Qw8DfMnnCxk/WZXlO0kBszI/AAAAAAAADD8/87_YaJB0SAs3QVTYFf3VjYExaj9q6GqpgCLcBGAs/s1600/7.png" data-original-width="586" data-original-height="222" /></a></p><p>We can now see why G is the constant it is: Any change in velocity (v^2) is offset by a change of the radius-mass ratio. Any change in radius-mass ratio is offset by a change in the time ratio. </p><p> </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com2tag:blogger.com,1999:blog-3628542335168231161.post-59480097270381213172017-08-12T15:43:00.000-07:002017-08-12T15:43:39.254-07:00How to Derive a Black Hole From Einstein's Field Equations<p><iframe width="560" height="315" src="https://www.youtube.com/embed/uiATpTX4VJs" frameborder="0" allowfullscreen></iframe></p><p>According to Stephen Hawking, if we start with a volume of space, say, a public library, and add books and more books, eventually the total number of books will become so massive they will collapse into a black hole. This is a little difficult to verify experimentally, but we can derive a black hole from Einstein's field equations. Below are the variables we will need: </p><p><a href="https://1.bp.blogspot.com/-Yfu6M4r5tys/WY9pRAAAVcI/AAAAAAAADCg/eGqlbSUffVwtkE4xPQHrQHywB-12Ib2ZQCLcBGAs/s1600/1.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-Yfu6M4r5tys/WY9pRAAAVcI/AAAAAAAADCg/eGqlbSUffVwtkE4xPQHrQHywB-12Ib2ZQCLcBGAs/s1600/1.png" data-original-width="441" data-original-height="649" /></a></p><p>In the diagram below, the blue circle represents the compressed mass (library books); the yellow and blue circle represent the mass's initial volume (library shelf space); The largest circle has a Scharzschild radius. As more and more mass is added, the blue circle shrinks and the singularity radius (r) shrinks as well. </p><p><a href="https://4.bp.blogspot.com/-cJzjPKd7mK4/WY9pZUZpi_I/AAAAAAAADCk/r11NoaGAYQENIDjKRrniE2TkyURlEGHqQCLcBGAs/s1600/2.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-cJzjPKd7mK4/WY9pZUZpi_I/AAAAAAAADCk/r11NoaGAYQENIDjKRrniE2TkyURlEGHqQCLcBGAs/s1600/2.png" data-original-width="570" data-original-height="623" /></a></p><p>Let's begin the derivation with equation 1: </p><p><a href="https://2.bp.blogspot.com/-b3YJ48TPECg/WY9pdLqtdiI/AAAAAAAADCo/8V-DjZc2tzYaFfpfSpQa3cV5nJdG5ZviQCLcBGAs/s1600/3.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-b3YJ48TPECg/WY9pdLqtdiI/AAAAAAAADCo/8V-DjZc2tzYaFfpfSpQa3cV5nJdG5ZviQCLcBGAs/s1600/3.png" data-original-width="564" data-original-height="191" /></a></p><p>Equation 1 has second-order tensors. We want to convert these to easy-to-work-with scalars (aka: invariant zero-order tensors). We can do this by contracting the indices. At equation 3 we pull the metric tensor (g) out of the Ricci tensor (R). At equation 4, the contravariant and covariant indices (i) cancel and vanish. </p><p><a href="https://1.bp.blogspot.com/-7N81ynirokg/WY9phgPqi2I/AAAAAAAADCs/6ikml3rvO6Mxhlj05LXYzWjat42heTWXACLcBGAs/s1600/4.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-7N81ynirokg/WY9phgPqi2I/AAAAAAAADCs/6ikml3rvO6Mxhlj05LXYzWjat42heTWXACLcBGAs/s1600/4.png" data-original-width="548" data-original-height="435" /></a></p><p>Now let's do the subtraction at equation 4 to get equation 5: </p><p><a href="https://4.bp.blogspot.com/-0dZu3LiEJxQ/WY9pl11pdhI/AAAAAAAADCw/L3T4pwTomQ47Pxmz04PhUzSEg2EuZzjvACLcBGAs/s1600/5.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-0dZu3LiEJxQ/WY9pl11pdhI/AAAAAAAADCw/L3T4pwTomQ47Pxmz04PhUzSEg2EuZzjvACLcBGAs/s1600/5.png" data-original-width="443" data-original-height="157" /></a></p><p>At equation 6 we set g equal to 8pi, so we can divide both sides of equation 5 by 8pi to get equation 7: </p><p><a href="https://3.bp.blogspot.com/-2qmMy5anyfw/WY9pqINWyOI/AAAAAAAADC0/qgKF4o2bSBU2Zpl8pRPnAdru9Bh4x-IGgCLcBGAs/s1600/6.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-2qmMy5anyfw/WY9pqINWyOI/AAAAAAAADC0/qgKF4o2bSBU2Zpl8pRPnAdru9Bh4x-IGgCLcBGAs/s1600/6.png" data-original-width="470" data-original-height="320" /></a></p><p>At 8 we set R equal to 1/r^2 and make a substitution to get equation 9: </p><p><a href="https://2.bp.blogspot.com/-7-rmlD8Ewsk/WY9pt2ymBMI/AAAAAAAADC4/4zfs6TwJqoYmIedkeCFtcjYLkFLFihiqwCLcBGAs/s1600/7.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-7-rmlD8Ewsk/WY9pt2ymBMI/AAAAAAAADC4/4zfs6TwJqoYmIedkeCFtcjYLkFLFihiqwCLcBGAs/s1600/7.png" data-original-width="469" data-original-height="311" /></a></p><p>The energy-stress tensor (T) has units of energy density. We do what we must to convert energy density to mass density (equations 10 and 11). We do a little algebra at 12 and 13 to get the Scharzschild radius (equation 13). </p><p><a href="https://3.bp.blogspot.com/-MGUj9SCK7H0/WY9pxoe7tqI/AAAAAAAADC8/zlmFqSWybDUDaiL6WVVkZGUtAUPSYTqBACLcBGAs/s1600/8.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-MGUj9SCK7H0/WY9pxoe7tqI/AAAAAAAADC8/zlmFqSWybDUDaiL6WVVkZGUtAUPSYTqBACLcBGAs/s1600/8.png" data-original-width="447" data-original-height="608" /></a></p><p>Taking equation 12 and applying limits gives the black-hole equation 14: </p><p><a href="https://3.bp.blogspot.com/-8E1r9gO5QNY/WY9p1VYDJ8I/AAAAAAAADDA/zc3afg56wrQzTNpKWmjIyK9AfChjaTB3QCLcBGAs/s1600/9.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-8E1r9gO5QNY/WY9p1VYDJ8I/AAAAAAAADDA/zc3afg56wrQzTNpKWmjIyK9AfChjaTB3QCLcBGAs/s1600/9.png" data-original-width="528" data-original-height="210" /></a></p><p>Just as Stephen Hawking said: If we keep adding library books (mass), the library's radius (r) shrinks and collapses into a black-hole singularity. </p><p>Below are bonus equations that show the maximum potential velocity is light speed and the maximum rest-mass energy is still mc^2. </p><p><a href="https://3.bp.blogspot.com/-r9Or1fbKT3c/WY9p5farH-I/AAAAAAAADDE/gvIJy0KcFD4XTJ5MkOADmwddNgRJfH7egCLcBGAs/s1600/10.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-r9Or1fbKT3c/WY9p5farH-I/AAAAAAAADDE/gvIJy0KcFD4XTJ5MkOADmwddNgRJfH7egCLcBGAs/s1600/10.png" data-original-width="475" data-original-height="548" /></a></p><p> </p><p> </p><p> </p><p> </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-75830080181070046332017-08-07T14:28:00.001-07:002017-08-07T14:43:45.399-07:00How to Conserve Dark Energy and the Rest<p><iframe width="560" height="315" src="https://www.youtube.com/embed/GHCc9b2phn0" frameborder="0" allowfullscreen></iframe> </p><p>In the above video the Physics Girl discusses how the expanding universe causes galaxies to move apart, and in turn causes photon wavelengths to stretch out. As photon wavelengths grow, they lose energy. "Where does the energy go?" she asks. </p><p>Other physicists, including myself, have a different question: "Where does dark energy come from?" As the universe expands, there is apparently more dark energy and less photon energy? Perhaps energy is conserved after all. If nothing else, it can be mathematically demonstrated. First, let's define the variables: </p><p><a href="https://1.bp.blogspot.com/-cIboK6f_WAw/WYjP3py2JPI/AAAAAAAADA4/XSX1MuQB6KUp_K5trRwFGDePzI-o3OuRgCLcBGAs/s1600/1.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-cIboK6f_WAw/WYjP3py2JPI/AAAAAAAADA4/XSX1MuQB6KUp_K5trRwFGDePzI-o3OuRgCLcBGAs/s1600/1.png" data-original-width="447" data-original-height="842" /></a></p><p>Equation 1 below shows how photon energy (Ep) is a function of its wavelength (lambda). The bigger lambda gets, the smaller the photon energy. </p><p><a href="https://1.bp.blogspot.com/-vfKu25L0w5s/WYjQqhrLctI/AAAAAAAADBE/7kegWLuLkQAICWtZjo1xGdKm5iETY9TqACLcBGAs/s1600/2.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-vfKu25L0w5s/WYjQqhrLctI/AAAAAAAADBE/7kegWLuLkQAICWtZjo1xGdKm5iETY9TqACLcBGAs/s1600/2.png" data-original-width="445" data-original-height="304" /></a></p><p>Equation 2 is dark energy (Ed)--a function of energy density (pd) times volume (V). As volume gets bigger, so does dark energy. </p><p>Equation 3 below shows the universe's radius (r) depends on how much dark energy there is. Equation 4 shows photon wavelength depends on how little photon energy there is: </p><p><a href="https://1.bp.blogspot.com/-jbO_a-XoqWE/WYjSXnCsmaI/AAAAAAAADBQ/r7jM6HTyS6UX74eWaoitG8omzmgHmaMewCLcBGAs/s1600/3.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-jbO_a-XoqWE/WYjSXnCsmaI/AAAAAAAADBQ/r7jM6HTyS6UX74eWaoitG8omzmgHmaMewCLcBGAs/s1600/3.png" data-original-width="419" data-original-height="384" /></a></p><p>Consider the universe's history. It started out with little or no space (dark energy) and it was very hot (photon energy). Over time space grew and the universe cooled (more dark energy, less photon energy). One way to conserve energy is to multiply photon energy and dark energy together. This creates a constant: as one energy grows, the other shrinks, but their product is always constant. Below we do a little algebra to get the product of the two energies: </p><p><a href="https://3.bp.blogspot.com/-d2KGQSbtWhA/WYjUUGuqm0I/AAAAAAAADBc/gO8-3cPWieEDein26oTvM2Yilmkn60QVQCLcBGAs/s1600/4.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-d2KGQSbtWhA/WYjUUGuqm0I/AAAAAAAADBc/gO8-3cPWieEDein26oTvM2Yilmkn60QVQCLcBGAs/s1600/4.png" data-original-width="569" data-original-height="413" /></a></p><p>Now, one thing we note is both energies are motion energies. Neither is at rest. Given the fact both energies have momentum (p) (due to mass or mass equivalence) we can make a substitution and derive equation 7 below: </p><p><a href="https://2.bp.blogspot.com/-0mSNZ3YpW3Y/WYjWAmaWYII/AAAAAAAADBo/XlYTEkCqv5gKLSIV7sL0gVlizhArYZ7cgCLcBGAs/s1600/5.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-0mSNZ3YpW3Y/WYjWAmaWYII/AAAAAAAADBo/XlYTEkCqv5gKLSIV7sL0gVlizhArYZ7cgCLcBGAs/s1600/5.png" data-original-width="440" data-original-height="186" /></a></p><p>You might recognize the momentum-energy term on equation 7's left side. It appears in this famous equation: </p><p><a href="https://2.bp.blogspot.com/-XbXjqBpFv8A/WYjYJwr6NNI/AAAAAAAADB0/jPUM9F0Ap0kI9ihN4GxV1vNj3LKAde3qgCLcBGAs/s1600/6.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-XbXjqBpFv8A/WYjYJwr6NNI/AAAAAAAADB0/jPUM9F0Ap0kI9ihN4GxV1vNj3LKAde3qgCLcBGAs/s1600/6.png" data-original-width="441" data-original-height="177" /></a></p><p>Einstein's energy equation, in this instance, shall represent the universe's total momentum and rest-mass energy. If we make one more substitution we get this: </p><p><a href="https://3.bp.blogspot.com/-blOHo3DY17g/WYjZQvPXuTI/AAAAAAAADCA/wHVv4JLvd5Eu-AfbOIKWnIiqX7iophd5wCLcBGAs/s1600/7.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-blOHo3DY17g/WYjZQvPXuTI/AAAAAAAADCA/wHVv4JLvd5Eu-AfbOIKWnIiqX7iophd5wCLcBGAs/s1600/7.png" data-original-width="499" data-original-height="174" /></a></p><p>Equation 9 above says the universe's conserved energy is the square root of dark energy times boson energy plus rest-mass energy squared. It includes all matter, radiation and vacuum energy. </p><p> </p><p> </p><p> GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-13955577397116641832017-08-01T15:36:00.001-07:002017-08-01T16:19:51.166-07:00Adding Gravity to the Standard Model Lagrangian<p>Here is one version of the Standard Model Lagrangian. Click on the image below to learn more details. Lagrangian L = <p><a href="http://www.symmetrymagazine.org/article/the-deconstructed-standard-model-equation" target="blank"><img src="https://www.symmetrymagazine.org/sites/default/files/images/standard/sml.png" height="1000" width="500"></a></p><p><a href="http://www.einstein-schrodinger.com/Standard_Model.pdf" target="blank">To read a pretty good outline re: the Standard Model Lagrangian, click here</a>. </p><p>Now you have probably been told a gazillion times that the Standard Model does not include gravity, that Einstein's field equations are incompatible with the big messy equation you see above. Let's have a look at the field equations: </p><p><a href="https://2.bp.blogspot.com/-sH0Ln_x1AVA/WYD5NWf-lrI/AAAAAAAAC_k/7CiuH5uwRM4XUhwaRCkTAkMYMYbcuhMVgCLcBGAs/s1600/1.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-sH0Ln_x1AVA/WYD5NWf-lrI/AAAAAAAAC_k/7CiuH5uwRM4XUhwaRCkTAkMYMYbcuhMVgCLcBGAs/s1600/1.png" data-original-width="660" data-original-height="774" /></a></p><p>The energy-stress tensor (Tij) provides a clue to how we can unify gravity with the Standard Model. Its units are energy density or energy over a volume (V). The Lagrangian has units of energy. Hmmmm ... if we contract the tensor indices and do a little algebra, we get equation 5 below: </p><p><a href="https://2.bp.blogspot.com/-_NAhInkP7L4/WYD8QLg9l9I/AAAAAAAAC_0/9G2KDDKsjeIoHnYG5hvE8eI3fM8fJHB-wCLcBGAs/s1600/2.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-_NAhInkP7L4/WYD8QLg9l9I/AAAAAAAAC_0/9G2KDDKsjeIoHnYG5hvE8eI3fM8fJHB-wCLcBGAs/s1600/2.png" data-original-width="703" data-original-height="629" /></a></p><p>Equation 5 shows that the scalar or zero-order tensor T is equivalent to the Lagrangian (L) divided by a volume (V). That leads us to equation 6 below: </p><p><a href="https://1.bp.blogspot.com/-kpZZD645rwE/WYD-ZS2OZLI/AAAAAAAAC_8/T2RgE3mER-wRM_oPMm1TTkNdRo_n0ypbQCLcBGAs/s1600/3.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-kpZZD645rwE/WYD-ZS2OZLI/AAAAAAAAC_8/T2RgE3mER-wRM_oPMm1TTkNdRo_n0ypbQCLcBGAs/s1600/3.png" data-original-width="647" data-original-height="159" /></a></p><p>We can now see the relationship between gravity and the other forces. If we do a little more algebra, we get an interesting result: </p><p><a href="https://1.bp.blogspot.com/-i2Diiz1wYKM/WYEABqX-KBI/AAAAAAAADAI/k_mLhEVvU3A5WFGkEayKFOns4kCbtRtQgCLcBGAs/s1600/4.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-i2Diiz1wYKM/WYEABqX-KBI/AAAAAAAADAI/k_mLhEVvU3A5WFGkEayKFOns4kCbtRtQgCLcBGAs/s1600/4.png" data-original-width="839" data-original-height="493" /></a></p><p>At equation 9 we add the newly-formed gravity Lagrangian to the Standard Model Lagrangian. Doing this yields the ground-state vacuum energy--and this quantity is consistent with the Wilkinson Microwave Anisotropy Probe measurement. So if we add gravity to the Standard Model as illustrated above, we not only get a result that is mathematically consistent, but also consistent with observations. </p><p> </p><p> </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com2tag:blogger.com,1999:blog-3628542335168231161.post-2067200188739815442017-07-24T12:11:00.000-07:002017-08-18T19:14:58.805-07:00Relativizing Hubble's Constant<p><iframe width="560" height="315" src="https://www.youtube.com/embed/ZjyFENuVuSo" frameborder="0" allowfullscreen></iframe></p><p><a href="https://1.bp.blogspot.com/-99Bpt9SVny8/WXYt6oCIBLI/AAAAAAAAC98/B7kAZnEv7b4mSBvwVoXTJ0Qo-UeOoX74gCLcBGAs/s1600/0.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-99Bpt9SVny8/WXYt6oCIBLI/AAAAAAAAC98/B7kAZnEv7b4mSBvwVoXTJ0Qo-UeOoX74gCLcBGAs/s1600/0.png" data-original-width="499" data-original-height="841" /></a></p><p>Albert Einstein had good reasons to believe nothing could go faster than light in a vacuum, including the vacuum itself (aka: dark energy). Consider his famous equation along with equation 2: </p><p><a href="https://3.bp.blogspot.com/-Nq9ijn8sKPk/WXYqhSZwAYI/AAAAAAAAC9w/dkJUk2hCD3E7NZotywk6_YCx1BcqfNG8wCLcBGAs/s1600/1.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-Nq9ijn8sKPk/WXYqhSZwAYI/AAAAAAAAC9w/dkJUk2hCD3E7NZotywk6_YCx1BcqfNG8wCLcBGAs/s1600/1.png" data-original-width="306" data-original-height="235" /></a></p><p>If the total energy (E) equals mass (m) times c^2, then c is the maximum potential velocity. If a greater velocity is possible, then Einstein was wrong. Equation 2 above shows the folly of trying to increase the velocity beyond light speed. To have more speed requires more energy, but more energy has more mass (or mass equivalence), since the equation represents mass energy. This is why c^2 is a constant. </p><p>Consider equation 3 below. <a href="http://gmjacksonphysics.blogspot.com/2017/07/how-to-derive-dark-energy-etc-from.html" target="blank">(To see how it was derived, click here.)</a> It shows how spacetime expansion velocity (Hr) relates to the gravitational constant G. The term on the right side has the cosmological constant in the denominator, but you'll notice there is also c^2. At equation 4 we set the expansion velocity at the cosmological horizon (Hr[u]) equal to c. After a little algebra, we derive equation 7. The idea is to test whether the expansion velocity can exceed c. </p><p><a href="https://4.bp.blogspot.com/-pLYIB9e-ddQ/WXYvFFdi94I/AAAAAAAAC-E/_q6RhSdaQvE67B5er28VVheEYSdCt1vfgCLcBGAs/s1600/2.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-pLYIB9e-ddQ/WXYvFFdi94I/AAAAAAAAC-E/_q6RhSdaQvE67B5er28VVheEYSdCt1vfgCLcBGAs/s1600/2.png" data-original-width="483" data-original-height="836" /></a></p><p>If we try to increase the left side of equation 7, the right side must also increase, but how? If we increase mass density (p), then spacetime curvature must also increase; i.e., the cosmological constant must increase. If we try reducing the cosmological constant (i.e. spacetime curvature), then mass density will decrease. In other words, we can't make the universe expand faster than c. </p><p>But surely if the universe continues to expand, the radius (r) should grow and the velocity (Hr) should also grow beyond c. But if we use all the energy the universe has to offer, the maximum potential is mc^2, not m(infinite velocity)^2. Perhaps the following crude diagram can assist us: </p><p><a href="https://3.bp.blogspot.com/-_QWWhc3A-ok/WXY2uGuPdGI/AAAAAAAAC-U/h845zUDXvlImEFKL4GSsOu5igtZ9WAcsgCLcBGAs/s1600/3.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-_QWWhc3A-ok/WXY2uGuPdGI/AAAAAAAAC-U/h845zUDXvlImEFKL4GSsOu5igtZ9WAcsgCLcBGAs/s1600/3.png" data-original-width="548" data-original-height="617" /></a></p><p>The outer circle represents the cosmological horizon; the inner circle is an arbitrary distance an observer may be looking into space (Do). If HDo is the expansion velocity observed, then the remaining universe must be expanding at a rate of HDr. The sum of these velocities is c or Hru. With this information we can write the following mathematical proof: </p><p><a href="https://4.bp.blogspot.com/-OhdNcDjjBH4/WXY5TUQnLmI/AAAAAAAAC-g/GWQJiYxcpRgnu5RLvZnDJmhafKVLHJd_QCLcBGAs/s1600/4.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-OhdNcDjjBH4/WXY5TUQnLmI/AAAAAAAAC-g/GWQJiYxcpRgnu5RLvZnDJmhafKVLHJd_QCLcBGAs/s1600/4.png" data-original-width="653" data-original-height="842" /></a></p><p><a href="https://4.bp.blogspot.com/-OiJhtZX9Fws/WXY56JW6hHI/AAAAAAAAC-k/WXMsiMgCdNE765bCvrWm3e7l5IN_xKR3QCLcBGAs/s1600/5.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-OiJhtZX9Fws/WXY56JW6hHI/AAAAAAAAC-k/WXMsiMgCdNE765bCvrWm3e7l5IN_xKR3QCLcBGAs/s1600/5.png" data-original-width="586" data-original-height="825" /></a></p><p>The result is a system of equations at 21 above. Any observer will see the universe expanding at an accelerated rate up to the cosmological horizon. Before that point, Hubble's constant remains reasonably constant (perhaps due to the fact that the mass density of spacetime, which could alter the value of H, is fairly consistent throughout the universe). At the horizon and beyond, Hubble's constant shrinks to accommodate the maximum potential velocity c. </p><p>Given this information we can hypothesize that any point in spacetime moves at a maximum potential velocity of c relative to any point that is separated by a distance of Do + Dr. Hence we have c^2 as a maximum potential. If we couple it with a mass, we have mc^2. We can also show the following relationships between gravity and dark energy: </p><p><a href="https://3.bp.blogspot.com/-DIaQVCNHitw/WXY_i28jBQI/AAAAAAAAC-w/CEAQYE4FavYljlNQ9oREMsZo2oPSbe-zgCLcBGAs/s1600/6.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-DIaQVCNHitw/WXY_i28jBQI/AAAAAAAAC-w/CEAQYE4FavYljlNQ9oREMsZo2oPSbe-zgCLcBGAs/s1600/6.png" data-original-width="583" data-original-height="426" /></a></p><p>Equations 22 through 24 show that when mass (m) is added to an expanding universe, proper time (t') shrinks. This would slow the rate of expansion if it were not for gravity making up the difference. <a href="http://gmjacksonphysics.blogspot.com/2017/06/dark-energy-in-dark-energy-out-gravity.html" target="blank">For more details click here.</a> </p><p>Update: Here is another mathematical proof showing that Hubble's constant shrinks beyond the cosmological horizon. Take special note of equation 31. It shows how the constant c^2 is maintained. </p><p><a href="https://3.bp.blogspot.com/-4FNzgH8q9JY/WX5KQciRhJI/AAAAAAAAC_M/eoNggMo_9DgSNKdyVOdTXHWrgXCzgXrmACLcBGAs/s1600/8.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-4FNzgH8q9JY/WX5KQciRhJI/AAAAAAAAC_M/eoNggMo_9DgSNKdyVOdTXHWrgXCzgXrmACLcBGAs/s1600/8.png" data-original-width="626" data-original-height="827" /></a></p><p><a href="https://1.bp.blogspot.com/-Go89dFR8Yrs/WX5KV3x6D2I/AAAAAAAAC_Q/n-pRfBwvgFwXJSk9ferEMo7mQ5K5w0dPACLcBGAs/s1600/9.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-Go89dFR8Yrs/WX5KV3x6D2I/AAAAAAAAC_Q/n-pRfBwvgFwXJSk9ferEMo7mQ5K5w0dPACLcBGAs/s1600/9.png" data-original-width="599" data-original-height="765" /></a></p><p><a href="https://1.bp.blogspot.com/-WDWIB7JSYwQ/WX5KZdPo32I/AAAAAAAAC_U/1OjVcx7zR1YsBfFsxk4dKTmHO7YhhbR2ACLcBGAs/s1600/10.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-WDWIB7JSYwQ/WX5KZdPo32I/AAAAAAAAC_U/1OjVcx7zR1YsBfFsxk4dKTmHO7YhhbR2ACLcBGAs/s1600/10.png" data-original-width="833" data-original-height="771" /></a></p><p> </p><p> </p><p> </p><p> GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-2529084668200794652017-07-07T16:45:00.000-07:002017-07-11T20:37:02.399-07:00How to Derive Dark Energy, Etc. From Heisenberg's Uncertainty Principle<p><iframe width="560" height="315" src="https://www.youtube.com/embed/LPoV_v7xL2w" frameborder="0" allowfullscreen></iframe></p><p>This post is a sequel to <a href= "http://gmjacksonphysics.blogspot.com/2017/06/dark-energy-in-dark-energy-out-gravity.html" target="blank">"Dark Energy In - Dark Energy Out = Gravity."</a> Today we are going to find the relationship between Hubble's observations (i.e. Hubble's constant), dark energy and gravity--and we are going to derive it from Heisenberg's uncertainty principle. Let's kick things off with defining the variables: </p><p><a href="https://1.bp.blogspot.com/-ArG77Rsx8RY/WWFA9AY1rsI/AAAAAAAAC7k/z5x6oR8kejgdsYs3Hd5GOHLk1ivz9J70ACLcBGAs/s1600/D1.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-ArG77Rsx8RY/WWFA9AY1rsI/AAAAAAAAC7k/z5x6oR8kejgdsYs3Hd5GOHLk1ivz9J70ACLcBGAs/s1600/D1.png" data-original-width="492" data-original-height="818" /></a></p><p>Equation(and inequality) 1 below is the energy-time version of Heisenberg's uncertainty principle: </p><p><a href="https://2.bp.blogspot.com/-tct6IdehtjI/WV_mI9jhmtI/AAAAAAAAC5g/YW-8-bgviDY_89Kc45aPwKWR0hKkYT5QQCLcBGAs/s1600/D2.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-tct6IdehtjI/WV_mI9jhmtI/AAAAAAAAC5g/YW-8-bgviDY_89Kc45aPwKWR0hKkYT5QQCLcBGAs/s1600/D2.png" data-original-width="405" data-original-height="221" /></a></p><p>The idea here is to build an expanding universe by taking a bottom-up approach. We build the very large by starting with something very small. We derive a simple energy equation (see equation 4 below). </p><p><a href="https://3.bp.blogspot.com/-R6fS3tGhhAk/WV_m55zDGII/AAAAAAAAC5k/FV25-V1ezd4QjabQTsgEr2ghVpJFaXZIgCLcBGAs/s1600/D3.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-R6fS3tGhhAk/WV_m55zDGII/AAAAAAAAC5k/FV25-V1ezd4QjabQTsgEr2ghVpJFaXZIgCLcBGAs/s1600/D3.png" data-original-width="467" data-original-height="488" /></a></p><p>Note the change in energy or energy difference variable on the left side of equation 4. We can substitute some arbitrary energy (E) minus the ground state (epsilon * E): </p><p><a href="https://3.bp.blogspot.com/-r3qErtcQK7w/WV_osUhJ1kI/AAAAAAAAC5o/Wt1oj4BthowiFGWltHRy2s4T7BJn9gbjgCLcBGAs/s1600/D4.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-r3qErtcQK7w/WV_osUhJ1kI/AAAAAAAAC5o/Wt1oj4BthowiFGWltHRy2s4T7BJn9gbjgCLcBGAs/s1600/D4.png" data-original-width="467" data-original-height="175" /></a></p><p>Let's bring all the terms to the left side and derive equation 10 below: </p><p><a href="https://1.bp.blogspot.com/-JL2VbO7TD9M/WWWWZTHPQVI/AAAAAAAAC8A/lOgw50lujzIip-DoAbSQnW02EfaR8aBuACLcBGAs/s1600/D5.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-JL2VbO7TD9M/WWWWZTHPQVI/AAAAAAAAC8A/lOgw50lujzIip-DoAbSQnW02EfaR8aBuACLcBGAs/s1600/D5.png" data-original-width="615" data-original-height="822" /></a></p><p>We now have an energy squared minus another energy squared minus the ground state squared equals the final energy (Ef) squared. We get equations 11 and 12 below by using Planck's reduced constant (h-bar), the wave number (k), and the light-speed constant (c)--and making substitutions. </p><p><a href="https://1.bp.blogspot.com/-XodK5pDoCdE/WWWYVfVoxzI/AAAAAAAAC8M/IVPrgDPK7SEKYM3BFFJlHmtu3mSd70FXQCLcBGAs/s1600/D6.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-XodK5pDoCdE/WWWYVfVoxzI/AAAAAAAAC8M/IVPrgDPK7SEKYM3BFFJlHmtu3mSd70FXQCLcBGAs/s1600/D6.png" data-original-width="782" data-original-height="431" /></a></p><p>Checking the units, we find equation 12 to be eerily similar to Einstein's field equations. Not a bad thing, by the way. It allows us to rewrite equation 12 to get 13: </p><p><a href="https://3.bp.blogspot.com/-YN1ZoPl8SHo/WWWYcuZInfI/AAAAAAAAC8Q/1QXx5nLaBJ00kZvnSwkux-J-aG4G7m3QgCLcBGAs/s1600/D7.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-YN1ZoPl8SHo/WWWYcuZInfI/AAAAAAAAC8Q/1QXx5nLaBJ00kZvnSwkux-J-aG4G7m3QgCLcBGAs/s1600/D7.png" data-original-width="578" data-original-height="186" /></a></p><p>Multiply both sides of 13 by the volume (D^3) to get 14 and 15: </p><p><a href="https://3.bp.blogspot.com/-h1aJ_fEnX7I/WWWYmRqtSTI/AAAAAAAAC8U/JfAcun2GP6Y2CVXg22iSMIMcymJ4M4lZQCLcBGAs/s1600/D8.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-h1aJ_fEnX7I/WWWYmRqtSTI/AAAAAAAAC8U/JfAcun2GP6Y2CVXg22iSMIMcymJ4M4lZQCLcBGAs/s1600/D8.png" data-original-width="586" data-original-height="334" /></a></p><p>Multiply both sides by Hubble's constant (H): </p><p><a href="https://4.bp.blogspot.com/-eDT7YSxlUeg/WWWYxPfXIsI/AAAAAAAAC8Y/ugNi6U5OQgsmN63KzXbf49u7GFZdO12dACLcBGAs/s1600/D9.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-eDT7YSxlUeg/WWWYxPfXIsI/AAAAAAAAC8Y/ugNi6U5OQgsmN63KzXbf49u7GFZdO12dACLcBGAs/s1600/D9.png" data-original-width="677" data-original-height="188" /></a></p><p>Multiply both sides by c^2/D: </p><p><a href="https://1.bp.blogspot.com/-6hk-qVc4Dk8/WWWY4Ia3jYI/AAAAAAAAC8c/WuOzQOiJQo43Y4ufrgfRZxEQNq94Ow8KACLcBGAs/s1600/D10.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-6hk-qVc4Dk8/WWWY4Ia3jYI/AAAAAAAAC8c/WuOzQOiJQo43Y4ufrgfRZxEQNq94Ow8KACLcBGAs/s1600/D10.png" data-original-width="598" data-original-height="171" /></a></p><p>From here we can derive 21 below: </p><p><a href="https://1.bp.blogspot.com/-1YmS2oEBXf8/WWWZBr8kmnI/AAAAAAAAC8g/ipAj9HzZNtoi_fg9T8yi92xXdnOgcwTCgCLcBGAs/s1600/D11.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-1YmS2oEBXf8/WWWZBr8kmnI/AAAAAAAAC8g/ipAj9HzZNtoi_fg9T8yi92xXdnOgcwTCgCLcBGAs/s1600/D11.png" data-original-width="709" data-original-height="686" /></a></p><p>Equation 21 is a power equation that has two components: the force of gravity, and the velocity the universe is expanding at distance D. However, this is only part of the story. Equation 21 does not take into account the mass density of spacetime or vacuum. A more complete equation is 22: </p><p><a href="https://2.bp.blogspot.com/-YjYwNbHmAoM/WWFOt5QBbiI/AAAAAAAAC7s/WcS5tcXITOsJQ1ouMTC_U5vhrGCslBfOACLcBGAs/s1600/D12.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-YjYwNbHmAoM/WWFOt5QBbiI/AAAAAAAAC7s/WcS5tcXITOsJQ1ouMTC_U5vhrGCslBfOACLcBGAs/s1600/D12.png" data-original-width="590" data-original-height="768" /></a></p><p>Note that as distance D increases, Volume V increases. The vacuum-mass-density gravity grows(dark matter effect) while classical Newton's gravity shrinks. If we utilize the cosmological constant, we can see a more precise relation between gravity and dark energy. First, we need to go back a few steps and work the cosmological constant into the math. Let's start with equation 13 and work forward: </p><p><a href="https://3.bp.blogspot.com/-9ccgdPVhc64/WWWZSllpwjI/AAAAAAAAC8k/QRyqTT0Q2JgRhioRt7wnHyMybCZr6cnegCLcBGAs/s1600/D13.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-9ccgdPVhc64/WWWZSllpwjI/AAAAAAAAC8k/QRyqTT0Q2JgRhioRt7wnHyMybCZr6cnegCLcBGAs/s1600/D13.png" data-original-width="653" data-original-height="773" /></a></p><p>A note re: equation 23. We want epsilon/D^2 to represent a ground state and a ground state does not increase or decrease, so epsilon must be proportionate to D^2. Thus we can set the term equal to the cosmological constant. </p><p><a href="https://2.bp.blogspot.com/-4yMIy7NK-q4/WWAOxzGKxFI/AAAAAAAAC60/NrDv99PpUoQFqU6tImLypgEptq4bJpITgCLcBGAs/s1600/D14.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-4yMIy7NK-q4/WWAOxzGKxFI/AAAAAAAAC60/NrDv99PpUoQFqU6tImLypgEptq4bJpITgCLcBGAs/s1600/D14.png" data-original-width="764" data-original-height="391" /></a></p><p>Equation 29 reveals something interesting: the instant velocity of expansion (HD) appears to be unaffected by gravity. The gravity in the numerator seems to be proportionate to the gravity in the denominator. This suggests the big crunch ain't gonna happen. But wait! It gets better. Suppose the universe expands to a point where Newtonian gravity (GM/r^2) is insignificant? We can drop it and get equations 30 and 31: </p><p><a href="https://1.bp.blogspot.com/-sgiAM1hw9Dw/WWARyL_BO2I/AAAAAAAAC68/9YcrW7kFotkvB94maDj1Q1nF_ypq9d6KACLcBGAs/s1600/D15.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-sgiAM1hw9Dw/WWARyL_BO2I/AAAAAAAAC68/9YcrW7kFotkvB94maDj1Q1nF_ypq9d6KACLcBGAs/s1600/D15.png" data-original-width="491" data-original-height="374" /></a></p><p>Look at equation 31. The only variable that isn't a constant is distance D. Now here's the awesome part: When D increases, so does gravity and so does the rate of expansion. The expansion rate (HD) is a function of gravity ... or is it dark energy? They both appear to be two sides of the same coin. And why not? They are both components of vacuum power (P). </p><p>Update: We can take equation 31 and derive the value of the cosmological constant:</p><p> <a href="https://2.bp.blogspot.com/-M5GrmbeMGQw/WWErZ_iOVHI/AAAAAAAAC7Q/UedF8YYLW4QTDg5UzhhFX-TDFbdpYpuUwCLcBGAs/s1600/D16.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-M5GrmbeMGQw/WWErZ_iOVHI/AAAAAAAAC7Q/UedF8YYLW4QTDg5UzhhFX-TDFbdpYpuUwCLcBGAs/s1600/D16.png" data-original-width="650" data-original-height="583" /></a></p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-47746793071913209262017-07-05T16:48:00.000-07:002017-07-08T15:05:13.458-07:00The Expanding Universe Conundrum<p><iframe width="560" height="315" src="https://www.youtube.com/embed/MoTNGmlOO2g" frameborder="0" allowfullscreen></iframe></p><p>How fast is the universe expanding? To figure this out, we could take the radius of the known universe and divide it by the universe's age. That should give us the average velocity the universe has expanded since the beginning of time: </p><p><a href="https://2.bp.blogspot.com/-ZdMUssQCJo0/WV1hXTHjCkI/AAAAAAAAC4I/C_AP7g_EXqgEL1mVQ554RaAWmk_cK3e3ACLcBGAs/s1600/1.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-ZdMUssQCJo0/WV1hXTHjCkI/AAAAAAAAC4I/C_AP7g_EXqgEL1mVQ554RaAWmk_cK3e3ACLcBGAs/s1600/1.png" data-original-width="526" data-original-height="797" /></a></p><p>Wow! It appears the universe has been expanding at the speed of light since day one! If that's true, if the galaxies, for example, have always been moving apart from each other at light speed, how did their photons reach us? How can we see them? If the space between galaxies were expanding at light speed, the light from the neighboring galaxies would never have made up the increasing distance. As soon as a photon, say, from NGC-2419 covered a light year, another light year of distance would have been added. The result? There would have been no discovery of good old NGC-2419 and more distant galaxies. </p><p>But what if the science is wrong? What if we are at rest and space does not expand--and those distant galaxies are moving away from us through space? Their light would eventually reach us no matter how fast they're moving away. Unfortunately this is not the case. If it were, the most distant galaxies would become dimmer and dimmer as they move further and further away. To see stuff beyond our known universe, all we would need is a more powerful telescope. If the science is correct (and it is), there should be a point in the distant cosmos, a cosmological horizon, beyond which we can see nothing (the light can't reach us due to expanding space). </p><p>So then how did the light from distant galaxies reach us? Imagine the universe is a sphere. Let's pretend we can draw a small circle anywhere we want on that sphere. That small circle encircles some galaxies (including the Milky Way). The sphere, represented by the 2D larger circle below, expands at light speed (c). The smaller circle within also expands over the same time period (t): </p><p><a href="https://4.bp.blogspot.com/-zNGYIX4MaCk/WV1sP7kXZAI/AAAAAAAAC4Y/gApPVhd-lQYlGxxxF-M6V2SJJwDtuR3sQCLcBGAs/s1600/2.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-zNGYIX4MaCk/WV1sP7kXZAI/AAAAAAAAC4Y/gApPVhd-lQYlGxxxF-M6V2SJJwDtuR3sQCLcBGAs/s1600/2.png" data-original-width="526" data-original-height="783" /></a></p><p>Now, let's double-check the velocity of the circles: </p><p><a href="https://4.bp.blogspot.com/-chL-mly2b8k/WV1wdhxwd7I/AAAAAAAAC4k/NtyyGaQK2FA64EZ7Sr7tT00t9H8fiDHHQCLcBGAs/s1600/3.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-chL-mly2b8k/WV1wdhxwd7I/AAAAAAAAC4k/NtyyGaQK2FA64EZ7Sr7tT00t9H8fiDHHQCLcBGAs/s1600/3.png" data-original-width="543" data-original-height="836" /></a></p><p>The larger circle expanded at velocity c as expected, but the smaller circle expanded less over the same time period; its velocity is only v or Hr. Therefore, the observable universe has indeed expanded at the speed of light (or more), but galaxies nearer to us have not. Their photons were able to reach us and we can see them. </p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com2tag:blogger.com,1999:blog-3628542335168231161.post-9949354903817860342017-07-02T14:46:00.000-07:002017-07-02T15:12:51.799-07:00Deriving Hubble's Constant, Etc. <p><iframe width="560" height="315" src="https://www.youtube.com/embed/5NFowhTBwI0" frameborder="0" allowfullscreen></iframe></p><p>Here is one way to derive Hubble's constant. First, let's define the variables: </p> <p><a href="https://4.bp.blogspot.com/-Lmy6A20j69o/WVla_gY5gmI/AAAAAAAAC2M/Tavk959lJfYmvcY_J5UFUYSGbuTXso0TwCLcBGAs/s1600/0.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-Lmy6A20j69o/WVla_gY5gmI/AAAAAAAAC2M/Tavk959lJfYmvcY_J5UFUYSGbuTXso0TwCLcBGAs/s1600/0.png" data-original-width="439" data-original-height="784" /></a></p><p>We start with an arbitrary total distance (D) which is the product of a small coefficient (a) and distance (x). As x grows, so does D, and vice versa. At equation 2 below, we find the time derivative of D (which equals V). </p><p><a href="https://3.bp.blogspot.com/-HN1CP-qQTXQ/WVlo6RYB-FI/AAAAAAAAC2w/ga-5qCJ1gSAGgPygUOd9Dx4vqFPshWKigCLcBGAs/s1600/1.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-HN1CP-qQTXQ/WVlo6RYB-FI/AAAAAAAAC2w/ga-5qCJ1gSAGgPygUOd9Dx4vqFPshWKigCLcBGAs/s1600/1.png" data-original-width="422" data-original-height="838" /></a></p><p>At equation 3 above, we multiply the left side by x/x. At equations 4 through 7 we do some algebraic slight of hand to derive Hubble's constant (H). Equation 8 shows the approximate numerical value of Hubble's constant. Its units are s^-1--the reciprocal of time. </p><p>Next, let's see what its relationship is to proper or relativistic time. To accomplish this we use 1/H in lieu of time (t). </p><p><a href="https://4.bp.blogspot.com/-3u-ZK0SAo6A/WVlddPw_A2I/AAAAAAAAC2Y/NIKlrkto25I0ovBrQQnLZnC1s6rnGwZpACLcBGAs/s1600/2.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-3u-ZK0SAo6A/WVlddPw_A2I/AAAAAAAAC2Y/NIKlrkto25I0ovBrQQnLZnC1s6rnGwZpACLcBGAs/s1600/2.png" data-original-width="518" data-original-height="858" /></a></p><p>Equation 14 shows how Hubble's constant stays constant. If proper time t' changes, so does variable u. The two are proportionate. Equation 15 below is the velocity the universe expands at distance D. When D increases, so does t'. You would think this amounts to a constant velocity for any distance D, But variable u also increases and offsets t'. As a result, velocity increases as D increases. </p><p><a href="https://2.bp.blogspot.com/-ocln9CwnvKU/WVldhRYfXYI/AAAAAAAAC2c/Vjr6PJMZrXwfJsxLmCoovywsptmjRW6aACLcBGAs/s1600/3.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-ocln9CwnvKU/WVldhRYfXYI/AAAAAAAAC2c/Vjr6PJMZrXwfJsxLmCoovywsptmjRW6aACLcBGAs/s1600/3.png" data-original-width="420" data-original-height="697" /></a></p><p>Equations 16 through 20 express distance D in terms of time (ctD). When examining the relativistic consequences of Hubble's constant, it is important to recognize that the time that makes up distance (ct) is a function of D only. Whereas t' is a function of D, mass and energy. High mass, for example, will contract time t to time t', and will contract length from ct to ct'. But distance D equals ct or act'; i.e., it can be measured out even if the units shrink due to relativity. </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-52193274157937424382017-06-03T16:05:00.000-07:002017-07-24T09:25:54.072-07:00Dark Energy In - Dark Energy Out = Gravity<iframe width="560" height="315" src="https://www.youtube.com/embed/9kgPjSlLnVw" frameborder="0" allowfullscreen></iframe> </p><p>In the diagram below there is a body falling to earth. After a distance of dr, what is its instant velocity? </p><p><a href="https://4.bp.blogspot.com/-IBFvMq3Ngz0/WS9AKxL2cII/AAAAAAAACrE/eQirdoQICTI2GCVV3OzqnxuTxhzoqcLDACLcB/s1600/a1.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-IBFvMq3Ngz0/WS9AKxL2cII/AAAAAAAACrE/eQirdoQICTI2GCVV3OzqnxuTxhzoqcLDACLcB/s1600/a1.png" data-original-width="779" data-original-height="518" /></a></p><p><a href="https://1.bp.blogspot.com/-HMlhDohPjnE/WS9ASugBNqI/AAAAAAAACrI/vsdxHmBfZw0sr_cI_bhFogMFwgWXwqdzgCLcB/s1600/a2.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-HMlhDohPjnE/WS9ASugBNqI/AAAAAAAACrI/vsdxHmBfZw0sr_cI_bhFogMFwgWXwqdzgCLcB/s1600/a2.png" data-original-width="469" data-original-height="628" /></a></p><p>Let's derive a formula that can give us the instant velocity: </p><p><a href="https://3.bp.blogspot.com/-KgLNfHJW5xw/WS9AXM7oSAI/AAAAAAAACrM/1o2XV8hK_cYcQjdr-8nmpUmIVyJZilfogCLcB/s1600/a3.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-KgLNfHJW5xw/WS9AXM7oSAI/AAAAAAAACrM/1o2XV8hK_cYcQjdr-8nmpUmIVyJZilfogCLcB/s1600/a3.png" data-original-width="565" data-original-height="832" /></a></p><p><a href="https://3.bp.blogspot.com/-3FF7P12hQC0/WS9Aa9GpP2I/AAAAAAAACrQ/bu9oc6Cb2NkZ7aT_O-HFT6cZdgMVJgGQgCLcB/s1600/a4.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-3FF7P12hQC0/WS9Aa9GpP2I/AAAAAAAACrQ/bu9oc6Cb2NkZ7aT_O-HFT6cZdgMVJgGQgCLcB/s1600/a4.png" data-original-width="527" data-original-height="596" /></a></p><p>Equations 8 and 9 give us the square of the instant velocity. Equation 7 shows that the squared velocity is also equal to gdr. Equation 8 shows how gdr is a function of the different time rates. At the top of dr (see diagram above), time is t; at the bottom, time is t'. Such time differences can be tested with atomic clocks. </p><p> At equation 9, notice how gdr or the instant velocity squared is the difference between light speed squared (c^2) and a lesser velocity squared (u^2). We can derive this difference again, starting with Einstein's field equations: </p><p><a href="https://4.bp.blogspot.com/-e3jplgYr0jQ/WS9IuK_LQbI/AAAAAAAACrg/WBSyfBMIKKUsA_O7VJ0HrNsvq-HPXXG5ACLcB/s1600/a5.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-e3jplgYr0jQ/WS9IuK_LQbI/AAAAAAAACrg/WBSyfBMIKKUsA_O7VJ0HrNsvq-HPXXG5ACLcB/s1600/a5.png" data-original-width="442" data-original-height="596" /></a></p><p><a href="https://1.bp.blogspot.com/-aoOUkpJWmvc/WS9IzDvaIwI/AAAAAAAACrk/x3zTStLZcRUNySMmctQigeepDwj2dMG-gCLcB/s1600/a6.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-aoOUkpJWmvc/WS9IzDvaIwI/AAAAAAAACrk/x3zTStLZcRUNySMmctQigeepDwj2dMG-gCLcB/s1600/a6.png" data-original-width="748" data-original-height="803" /></a></p><p><a href="https://4.bp.blogspot.com/-TCVmMrxjcWg/WS9I3qFWWWI/AAAAAAAACro/ZjFp1ZWScFU5N6vfTNYuBU48pq_6GT71gCLcB/s1600/a7.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-TCVmMrxjcWg/WS9I3qFWWWI/AAAAAAAACro/ZjFp1ZWScFU5N6vfTNYuBU48pq_6GT71gCLcB/s1600/a7.png" data-original-width="581" data-original-height="824" /></a></p><p>Equation 20 shows us again the instant velocity is the square root of the difference between two squared velocities. What exactly are these velocities? Starting with Einstein's energy equation, we can derive them yet again: </p><p><a href="https://4.bp.blogspot.com/-w3UCN1Apumk/WS9Yy8PX1aI/AAAAAAAACr4/_QKh9udWUhwVMbYnaooGz_krX7bjfiBYACLcB/s1600/a8.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-w3UCN1Apumk/WS9Yy8PX1aI/AAAAAAAACr4/_QKh9udWUhwVMbYnaooGz_krX7bjfiBYACLcB/s1600/a8.png" data-original-width="463" data-original-height="510" /></a></p><p> <a href="https://2.bp.blogspot.com/-LJcjSwpm8x4/WS9Y3ao1zRI/AAAAAAAACr8/pwemxrmYMgk1j5x46cOPiN9Qc0AuGx9bgCLcB/s1600/a9.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-LJcjSwpm8x4/WS9Y3ao1zRI/AAAAAAAACr8/pwemxrmYMgk1j5x46cOPiN9Qc0AuGx9bgCLcB/s1600/a9.png" data-original-width="533" data-original-height="770" /></a></p><p> <a href="https://2.bp.blogspot.com/-vs5NJ_PipTo/WS9Y7Lt54sI/AAAAAAAACsA/7SY_CPw4f6k2kzFsWYR0FMY0uwKKrTxegCLcB/s1600/a10.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-vs5NJ_PipTo/WS9Y7Lt54sI/AAAAAAAACsA/7SY_CPw4f6k2kzFsWYR0FMY0uwKKrTxegCLcB/s1600/a10.png" data-original-width="640" data-original-height="657" /></a></p><p>Equations 30 and 31 are all too familiar. One interesting property they seem to have is momentum is not conserved. It looks as though we could couple the velocities with any mass and they would remain constant? </p><p><a href="https://4.bp.blogspot.com/-DkkT2Tw0bYM/WTBIeNLm-EI/AAAAAAAACsg/AVElpK_OPQwXvG9562bg-quZo3R6Kmy-gCLcB/s1600/a11.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-DkkT2Tw0bYM/WTBIeNLm-EI/AAAAAAAACsg/AVElpK_OPQwXvG9562bg-quZo3R6Kmy-gCLcB/s1600/a11.png" data-original-width="545" data-original-height="156" /></a></p><p>Momentum not being conserved is consistent with any mass falling in a gravitational field--all masses fall at the same rate. By contrast, momentum is conserved when an electromagnetic force is applied to different masses. (See equations below.) </p><p><a href="https://3.bp.blogspot.com/--_gzOrjxVb4/WTBVmSX_CkI/AAAAAAAACtc/TNpgHhJgt_grNn5b3lyK2Z4If6JTfDT6QCLcB/s1600/a11b.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/--_gzOrjxVb4/WTBVmSX_CkI/AAAAAAAACtc/TNpgHhJgt_grNn5b3lyK2Z4If6JTfDT6QCLcB/s1600/a11b.png" data-original-width="444" data-original-height="521" /></a></p><p><a href="https://1.bp.blogspot.com/-EIiIk8jt4Ls/WTBJnQuVu1I/AAAAAAAACss/ayDaryOWqSYeAVQ0eXA5HQmdfj7snLetQCLcB/s1600/a12.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-EIiIk8jt4Ls/WTBJnQuVu1I/AAAAAAAACss/ayDaryOWqSYeAVQ0eXA5HQmdfj7snLetQCLcB/s1600/a12.png" data-original-width="423" data-original-height="556" /></a></p><p>Equations 33 and 34 are classical force equations. Normally, when mass (m) increases, acceleration (a) decreases--force is conserved. The time derivative of conserved force is conserved momentum, so the instant velocity decreases as well. When mass is decreased, the acceleration and velocity increase. Equation 35 fits the norm; however, equation 36 does not. Note that electromagnetism (EM, equation 35) is not enhanced by mass, but charges (q' and q). By contrast, gravity (equation 36) has mass (m) in both the numerator and denominator, so it cancels itself. </p><p>The mathematical proof below confirms the squared velocities we derived above are unaffected by different masses. We start with equation 27: </p><p><a href="https://2.bp.blogspot.com/-poPn2kGK-8Y/WTBMGYyr5lI/AAAAAAAACs4/p32_y5p3fMM7IzQk7HCUCY9EbZ9rsvZTQCLcB/s1600/a13.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-poPn2kGK-8Y/WTBMGYyr5lI/AAAAAAAACs4/p32_y5p3fMM7IzQk7HCUCY9EbZ9rsvZTQCLcB/s1600/a13.png" data-original-width="556" data-original-height="511" /></a></p><p>At equation 38, momentum is not conserved. If we increase the mass (ps/c) and assume velocity (v) drops (making the term constant), ps on the equation's right also increases, but there is no v there to offset it. A similar problem occurs if we reduce ps. Armed with this information we derive equation 41 below: </p><p><a href="https://3.bp.blogspot.com/-Iutd-FuxQ6k/WTiFuDAXUrI/AAAAAAAACwk/-_zLNrDizfku3nCwOf_uDLSkMCNAaUcUwCLcB/s1600/a14.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-Iutd-FuxQ6k/WTiFuDAXUrI/AAAAAAAACwk/-_zLNrDizfku3nCwOf_uDLSkMCNAaUcUwCLcB/s1600/a14.png" data-original-width="570" data-original-height="593" /></a></p><p>Equation 41 works if we hold the velocities constant, but allow any mass. What we end up with is a momentum squared equals the difference between two squared momenta. Now look at what we can derive from this: </p><p><a href="https://1.bp.blogspot.com/-GVWjzj9-Lq4/WTBRltqzW0I/AAAAAAAACtQ/YAcdCzj3PiMoiXsOVnFyWLWrpoWJyHkgwCLcB/s1600/a15.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-GVWjzj9-Lq4/WTBRltqzW0I/AAAAAAAACtQ/YAcdCzj3PiMoiXsOVnFyWLWrpoWJyHkgwCLcB/s1600/a15.png" data-original-width="564" data-original-height="659" /></a></p><p>Equation 46 shows that curved spacetime is the difference between wave numbers (wave cycles per distance). The larger wave number (ks) corresponds to more wave cycles per distance, more space expansion pressure, and faster time. The smaller wave number (ke) corresponds to fewer wave cycles per distance, less space expansion pressure, and slower time (see proof at the update below). The net wave number (k) is the net spacetime expansion or gravitational field. </p><p>Now, take a close look at equation 22: </p><p><a href="https://3.bp.blogspot.com/-6E6Y0lDESpc/WTCObvycmtI/AAAAAAAACts/zZphlXTaNzQFyCeGcA2VYQ-056bWuhgRwCLcB/s1600/a16.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-6E6Y0lDESpc/WTCObvycmtI/AAAAAAAACts/zZphlXTaNzQFyCeGcA2VYQ-056bWuhgRwCLcB/s1600/a16.png" data-original-width="460" data-original-height="166" /></a></p><p>The intent of equation 22 was to isolate the momentum energy (pc). That's easy enough. It's just the total energy (E) minus the rest-mass energy (mc^2), all squared, of course. But something anti-gravity happens when you assume the total energy is conserved and increase the rest-mass energy: the momentum energy decreases! If equation 22 is a gravity equation, the momentum should increase when the rest-mass increases. </p><p>Suppose we assume there are two energy fields instead of one? Each has its own momentum and rest-mass energies. Spacetime would have a low rest-mass energy compared to, say, earth, but its momentum energy (dark energy?) could be higher than earth's energy field's. So imagine high-momentum spacetime expansion pressing against the earth and lower-momentum spacetime expansion pressing away from earth. The difference is a net inward force we call gravity. Equations 23 through 25 below concur: </p><p><a href="https://1.bp.blogspot.com/-iG0oGxE8OTc/WTCWoaPgUgI/AAAAAAAACt4/wHjXu95P_K0A7DewaOjPCP7eQg3d6OKMgCLcB/s1600/a17.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-iG0oGxE8OTc/WTCWoaPgUgI/AAAAAAAACt4/wHjXu95P_K0A7DewaOjPCP7eQg3d6OKMgCLcB/s1600/a17.png" data-original-width="509" data-original-height="409" /></a></p><p>If we take the above three equations together, we get a gravitational relationship between the net momentum (p) and earth's rest-mass energy (me*c^2). The greater the potential energy or rest-mass energy, the greater the net momentum towards earth. This is consistent with greater energy density causing more spacetime curvature, i.e., gravity. Where there's more energy density, there is bound to be more rest-mass energy. </p><p>Now, let's take a look at dark energy. Let's define it as vacuum energy. According to the Wilkinson Microwave Anisotropy probe, the energy density of the vacuum is around e-10 Joules per cubic meter. That would give it a mass or mass-equivalent density of approximately e-27 kilograms per cubic meter. Energy density, of course, is equivalent to pressure, so the vacuum has a pressure of e-27c^2 Pascals. Equations 47 and 48 below reveal a key difference between typical pressure and vacuum pressure. </p><p> <a href="https://3.bp.blogspot.com/-O9WDTOPug6I/WTH5MmjglSI/AAAAAAAACuU/ZjLYCMDCNyUbIAXSJqrmajvY_F0a8EIQACLcB/s1600/a18.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-O9WDTOPug6I/WTH5MmjglSI/AAAAAAAACuU/ZjLYCMDCNyUbIAXSJqrmajvY_F0a8EIQACLcB/s1600/a18.png" data-original-width="484" data-original-height="525" /></a></p><p> <a href="https://2.bp.blogspot.com/-H_kfx11moXY/WTiKVlrsgsI/AAAAAAAACww/CguX8gjF6TIs-CEakOHMaQcy36vJD42dQCLcB/s1600/a19.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-H_kfx11moXY/WTiKVlrsgsI/AAAAAAAACww/CguX8gjF6TIs-CEakOHMaQcy36vJD42dQCLcB/s1600/a19.png" data-original-width="676" data-original-height="855" /></a></p><p>At equation 47, expanding volume (V) relieves pressure, and, one would guess that the pressure eventually drops to zero and expansion stops. Equation 48 shows this is not the case. An increase in volume increases the vacuum energy, so the pressure remains constant and expansion continues. Equation 50, using the vacuum mass density, calculates the net pressure we interpret as gravity. At equation 51 we can increase the net pressure by adding a falling mass (m) to the vacuum mass. Equation 52 shows that the instant falling velocity is unaffected by the additional mass, since the mass's inertia cancels the mass's pressure contribution. </p><p>The diagram below is a one-dimensional representation of how gravity emerges from the expanding vacuum: </p><p><a href="https://1.bp.blogspot.com/-SNykck1_O24/WTH5bPSHJRI/AAAAAAAACuc/6KwciMgpgXsHDOcFthmEdGw_LwLiP1CDQCLcB/s1600/a20.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-SNykck1_O24/WTH5bPSHJRI/AAAAAAAACuc/6KwciMgpgXsHDOcFthmEdGw_LwLiP1CDQCLcB/s1600/a20.png" data-original-width="652" data-original-height="850" /></a></p><p>The red rectangle shows the difference between the incoming pressure (m'c^2/V) and the outgoing pressure (m'u^2/V). Notice how the falling mass (m') slows vacuum expansion to velocity w. One would think this would affect velocity v, but the velocity w goes in all directions and cancels itself. Equation 53 confirms this and confirms that mass m' inertia cancels its pressure contribution: m'/m'. </p><p> </p><p>The next diagram demonstrates how mass m' causes a tidal effect: </p><p><a href="https://3.bp.blogspot.com/-Qmsh9Yu8bDc/WTMRYLZoRJI/AAAAAAAACvg/n43odF2EIi0Cc6Ej00CYa_Di4N56wTsWACLcB/s1600/a21.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-Qmsh9Yu8bDc/WTMRYLZoRJI/AAAAAAAACvg/n43odF2EIi0Cc6Ej00CYa_Di4N56wTsWACLcB/s1600/a21.png" data-original-width="507" data-original-height="876" /></a></p><p>In the above diagram, gravity is weaker above mass m and stronger below. The next diagram demonstrates the gravity of mass m'. (<a href="http://gmjacksonphysics.blogspot.com/2016/11/how-fast-is-universe-really-expanding.html" target="blank">Notice in all the diagrams how the spacetime vacuum expands at the outskirts at a maximum rate of c^2. This appears as acceleration when Hubble's constant is used. For more details on the rate of expansion, click here.</a>) </p><p><a href="https://4.bp.blogspot.com/-m-s2dCrfLes/WTNgGnHT4gI/AAAAAAAACwU/eh0njF_q6QgLeh3VaVWdtdwph3T3rtiHwCLcB/s1600/a22.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-m-s2dCrfLes/WTNgGnHT4gI/AAAAAAAACwU/eh0njF_q6QgLeh3VaVWdtdwph3T3rtiHwCLcB/s1600/a22.png" data-original-width="502" data-original-height="533" /></a> </p><p><a href="https://2.bp.blogspot.com/-ZEMNwvOuzFU/WTH50xo8VDI/AAAAAAAACuo/D84vKQ7YjYM7KNgXUEjO_HvHTTqkK6CpwCLcB/s1600/a23.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-ZEMNwvOuzFU/WTH50xo8VDI/AAAAAAAACuo/D84vKQ7YjYM7KNgXUEjO_HvHTTqkK6CpwCLcB/s1600/a23.png" data-original-width="658" data-original-height="863" /></a></p><p>Of course we know when masses fall, they accelerate, so velocity v must increase, but how does this work exactly? Let's zoom in and have a look: </p><p><a href="https://4.bp.blogspot.com/-i21aztr0MDM/WTLxg4ruqII/AAAAAAAACvA/bmggYlsf1H0rQqroHqpOp1gbSwsGBjDKgCLcB/s1600/a24.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-i21aztr0MDM/WTLxg4ruqII/AAAAAAAACvA/bmggYlsf1H0rQqroHqpOp1gbSwsGBjDKgCLcB/s1600/a24.png" data-original-width="507" data-original-height="844" /></a></p><p><a href="https://4.bp.blogspot.com/-04x4u_BzUWE/WTLwdZL1mmI/AAAAAAAACu8/WbyA2fE4R5gVY95jOeJauwjy_4GQbYZQQCLcB/s1600/a24b.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-04x4u_BzUWE/WTLwdZL1mmI/AAAAAAAACu8/WbyA2fE4R5gVY95jOeJauwjy_4GQbYZQQCLcB/s1600/a24b.png" data-original-width="459" data-original-height="194" /></a></p><p>As a body falls closer to the planet's surface, overall energy density increases (more rest-mass energy in the given space). The vacuum expansion is slower and slower (progressively shorter double arrows). As a result, velocity v grows. Equations 56 and 56b sum all the little changes in velocity during a body's descent. </p><p>Using equation 57 below we can calculate the velocity a mass will accelerate to in a gravitational field. Massless particles, such as photons, don't accelerate, they gain frequency (see equation 58). </p><p><a href="https://3.bp.blogspot.com/-xuC1DQMiv0M/WTH5_zCulJI/AAAAAAAACuw/TW_-kOsUm1sSXHUdcZ3qR00qdwY2OkB7ACLcB/s1600/a25.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-xuC1DQMiv0M/WTH5_zCulJI/AAAAAAAACuw/TW_-kOsUm1sSXHUdcZ3qR00qdwY2OkB7ACLcB/s1600/a25.png" data-original-width="702" data-original-height="373" /></a></p><p>On a cosmological scale, galaxies are moving apart but they also have gravity. The diagram below shows how this possible: </p><p><a href="https://2.bp.blogspot.com/-kcLMYgk9QAY/WTNM5ywvrRI/AAAAAAAACv8/4f8q5NTVa1gpjPdSqpbNjmpSL_oMrsfsQCLcB/s1600/a26.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-kcLMYgk9QAY/WTNM5ywvrRI/AAAAAAAACv8/4f8q5NTVa1gpjPdSqpbNjmpSL_oMrsfsQCLcB/s1600/a26.png" data-original-width="366" data-original-height="837" /></a></p><p>In the above diagram, if we sum up all the vector arrows within the red rectangles, we get a net gravitational velocity (v) of zero. As a result, spacetime is free to expand. If, however, we zoom in closer to one of the galaxies (red dots), we encounter gravity (blue rectangles: c^2 - u^2 = v^2). Ironically, this gravity has no impact on the overall expansion. If two galaxies (blue dots below) are sufficiently close together, they can slow the expansion enough to create a gravitational attraction between them, but spacetime on the outskirts will still expand (see diagram below). </p><p><a href="https://4.bp.blogspot.com/-vfnW8tnUc58/WTNNEBAk7QI/AAAAAAAACwA/6VHWarOaITkr4OCy9Kyg9CRKHpSORPdzgCLcB/s1600/a27.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-vfnW8tnUc58/WTNNEBAk7QI/AAAAAAAACwA/6VHWarOaITkr4OCy9Kyg9CRKHpSORPdzgCLcB/s1600/a27.png" data-original-width="374" data-original-height="818" /></a></p><p>At the quantum scale, it seems highly probable that gravity boson(s) originate from the vacuum. The mysterious "dark energy" and graviton could be one and the same. It's also possible that any boson or combination of bosons can do the job. </p><p> Imagine a black hole with gravity so strong that not even gravitons or dark energy can escape. In other words, the huge mass of the black hole prevents an outflow of expanding spacetime. How does a particle floating in space know there's a black hole nearby? It knows from the massive inflow of vacuum pressure! In general, particles know how to move due to the net flow of the vacuum pressure--and whatever boson(s) it carries with it. </p><p>Update: Here's a mathematical proof showing the correlation between wave number and time rate:</p><p><a href="https://2.bp.blogspot.com/-a4Y9YTsmYOg/WUhtREmMzqI/AAAAAAAAC0w/1DkzoDdR3j8FjegsiryllWIWEi3cwOgRACLcBGAs/s1600/k1.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-a4Y9YTsmYOg/WUhtREmMzqI/AAAAAAAAC0w/1DkzoDdR3j8FjegsiryllWIWEi3cwOgRACLcBGAs/s1600/k1.png" data-original-width="380" data-original-height="826" /></a> </p><p><a href="https://2.bp.blogspot.com/-sqWSqq144Js/WUhtfB1OFUI/AAAAAAAAC00/puQEHdi8q8U_nbJ8fRZtOjHCC4hJeF00ACLcBGAs/s1600/k2.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-sqWSqq144Js/WUhtfB1OFUI/AAAAAAAAC00/puQEHdi8q8U_nbJ8fRZtOjHCC4hJeF00ACLcBGAs/s1600/k2.png" data-original-width="459" data-original-height="725" /></a></p><p>Update: Below are dark energy-gravity equations that take into account Hubble's observations. First we define the variables: </p><p><a href="https://2.bp.blogspot.com/-BnTFETc7KVA/WVWTPJ_zGBI/AAAAAAAAC1g/gxZavalekn8ox1sV1ubk7rQU55uOeQCEQCLcBGAs/s1600/0.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-BnTFETc7KVA/WVWTPJ_zGBI/AAAAAAAAC1g/gxZavalekn8ox1sV1ubk7rQU55uOeQCEQCLcBGAs/s1600/0.png" data-original-width="476" data-original-height="835" /></a></p><p>The first equation below is power per area at any location in spacetime. Check out what can be derived from this equation. </p><p><a href="https://2.bp.blogspot.com/-0n3IW79j1aI/WVWTUZeY5vI/AAAAAAAAC1k/TT2xU2a9GlgfYLih4c28A0gxd1Y-RE-wwCLcBGAs/s1600/1.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-0n3IW79j1aI/WVWTUZeY5vI/AAAAAAAAC1k/TT2xU2a9GlgfYLih4c28A0gxd1Y-RE-wwCLcBGAs/s1600/1.png" data-original-width="724" data-original-height="834" /></a></p><p>Equation 3 above is the orbital velocity of a satellite. Equation 4 is the familiar Newtonian g-force. These equations were derived by dividing both sides of equation 1 by the dark energy velocity (Hr) and the mass density (ps). Below we derive the dark energy velocity by dividing both sides of equation 1 by the gravitational velocity and mass density. (See equation 7.) </p><p><a href="https://2.bp.blogspot.com/-sHlxPWeoSzw/WVWTYkIo7tI/AAAAAAAAC1o/Txes-feh0M0hPsL0xmmOVteFIQ_Cjv_sQCLcBGAs/s1600/2.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-sHlxPWeoSzw/WVWTYkIo7tI/AAAAAAAAC1o/Txes-feh0M0hPsL0xmmOVteFIQ_Cjv_sQCLcBGAs/s1600/2.png" data-original-width="810" data-original-height="829" /></a></p><p>We use the reciprocal of Hubble's constant as a benchmark time. Relative or proper time is 1/H'. If we take the limit of H' to infinity, we get the dark energy velocity at half the Schwarzschild radius. (See equation 8.) This shows that there is always some expansion velocity even when gravity is as powerful as a black hole's. </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-19951824155846068292017-05-18T18:05:00.001-07:002017-05-19T13:46:29.970-07:00Derive the Big Bang From Einstein's Field Equations<p><iframe width="560" height="315" src="https://www.youtube.com/embed/ePkPYA4AQ3o" frameborder="0" allowfullscreen></iframe></p><p>Was the universe once a singularity? If so, did it have infinite gravity? What forces overcame infinite gravity so our universe could expand to its current size? We can get some answers from Einstein's field equations. </p><p>Equation 1 below is the typical textbook rendition of the field equations. At equation 2, on the left side, I changed a plus sign to a minus sign. The third term contains the cosmological constant and shall represent spacetime expansion which is the opposite of gravity, so a minus sign is placed in front of it. We shall treat gravity as positive and expansion as negative. </p><p><a href="https://4.bp.blogspot.com/-vF2BdfyhtPY/WR4wqua_ivI/AAAAAAAACo8/Kz1wM2dPcTUsrr_U8a1OLgKPsj-Z6vwPACLcB/s1600/1.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-vF2BdfyhtPY/WR4wqua_ivI/AAAAAAAACo8/Kz1wM2dPcTUsrr_U8a1OLgKPsj-Z6vwPACLcB/s1600/1.png" /></a></p><p>Normally the first two terms on left side are huge compared to the third term, so the left side of the equation has a positive value. Albeit, something surprising happens if we reduce the universe's radius (r) and hold all other values constant. To make this more obvious, let's change up the variables a bit: </p><p><a href="https://4.bp.blogspot.com/-MBFL8jhI5sA/WR4wvMmRjKI/AAAAAAAACpA/_yJg3aaURnUHMXRDKFvmDhOYsWuZ_pEEQCLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-MBFL8jhI5sA/WR4wvMmRjKI/AAAAAAAACpA/_yJg3aaURnUHMXRDKFvmDhOYsWuZ_pEEQCLcB/s1600/2.png" /></a></p><p>We now reduce the radius to zero: </p><p><a href="https://4.bp.blogspot.com/-8IZ5CzaAOHA/WR4wzKL60dI/AAAAAAAACpE/ypIVv_peDc81lS8ljy237XEEh-REqCaMwCLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-8IZ5CzaAOHA/WR4wzKL60dI/AAAAAAAACpE/ypIVv_peDc81lS8ljy237XEEh-REqCaMwCLcB/s1600/3.png" /></a></p><p>Equation 13 above shows we derive infinite negative pressure or expansion pressure--and no gravity to hold it back! How is that possible? Well, if we start with the assumption that there was infinite gravity, we are stuck trying to find a force that overcame it. On the other hand, if there was infinite outward pressure and no gravity, it's pretty obvious why the universe grew to its present size. </p><p>Gravity must have emerged from the rapid expansion that resulted from so much pressure. To have gravity, there needs to be more than just a point of spacetime. The first two terms of the field equations show that there must be a difference of spacetime curvature between two points in space. In other words, there needs to be more than one reference frame. If the universe was just a single point, there was only one reference frame. Because they represent the same point of spacetime, the first term has a value of infinity, the second term also has a value of infinity. Infinity - infinity = zero spacetime curvature. As soon as there was some distance between, say, point A and point B, gravity emerged. </p><p>But then the question becomes, "What caused the infinite pressure at a single point?" The answer is nothing. If the infinite pressure was the beginning, nothing happened before it; there was no time or space. A "cause" is an event, and an event is a function of time and space. So, in the beginning there was infinite outward pressure from a single point ... </p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com1tag:blogger.com,1999:blog-3628542335168231161.post-1497721733240636342017-04-25T15:32:00.000-07:002017-04-25T15:32:51.117-07:00Why Gravity is Weak<p><iframe width="560" height="315" src="https://www.youtube.com/embed/4yyb_RNJWUM" frameborder="0" allowfullscreen></iframe></p><p>Perhaps you are familiar with some of the more exotic and far-out theories that attempt to explain why gravity is so weak compared to, say, electromagnetism. One of my favorites involves gravitons hiding in extra dimensions. The logic is as follows: if gravitons weren't hiding in extra dimensions, gravity would be much stronger. It's analogous to someone screaming at the top of his lungs. The intensity of the sound is very strong until you hide that person away in a broom closet and lock the door. </p><p>As much as I like this theory, it has a problem that can't be hidden in a closet or extra dimension: there is no empirical evidence of extra dimensions. That begs the question: is it possible to explain gravity's weakness without using extra dimensions? Yes! When doing physics, I prefer taking an Occam's razor approach and using concepts and ideas that are testable and/or have been tested and confirmed. The end result is not as exciting as extra dimensions--but more likely to be true. </p><p>Let's begin with Einstein's field equations: </p><p><a href="https://1.bp.blogspot.com/-F2wEOnR8ib8/WP-qgYciIKI/AAAAAAAAClw/9Mq1YxO79q8JcGjvpwxFDZLf9d7it9eiwCLcB/s1600/1.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-F2wEOnR8ib8/WP-qgYciIKI/AAAAAAAAClw/9Mq1YxO79q8JcGjvpwxFDZLf9d7it9eiwCLcB/s1600/1.png" /></a></p><p>To simplify matters we equate the left side of equation 1 to K: </p><p><a href="https://4.bp.blogspot.com/-kQWUYGZ5nwQ/WP-mxh4WrPI/AAAAAAAAClc/f08xT5cdV082yILnR9ZDcgE9yCrbixCEgCLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-kQWUYGZ5nwQ/WP-mxh4WrPI/AAAAAAAAClc/f08xT5cdV082yILnR9ZDcgE9yCrbixCEgCLcB/s1600/2.png" /></a></p><p>In equation 3 above, we have energy density (T44) multiplied by a constant that consists of G and c. Why do we need such a constant? Why can't we get by using equations 4 and 5 below? </p><p><a href="https://2.bp.blogspot.com/-16Dx5Hiq_9M/WP-nXHFjPmI/AAAAAAAAClk/wzLBRoBsJfQXFJmMdu_9qLlE_s2IIoXtACLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-16Dx5Hiq_9M/WP-nXHFjPmI/AAAAAAAAClk/wzLBRoBsJfQXFJmMdu_9qLlE_s2IIoXtACLcB/s1600/3.png" /></a></p><p>The constant in question is a very small number and substantially reduces the value of K, the spacetime curvature. Perhaps equations 6 through 8 can shed some light on the subject: </p><p><a href="https://3.bp.blogspot.com/-1OyUKPCS4Rc/WP-qs_8CE4I/AAAAAAAACl0/4KkgGWsc2O8lohVEYPsw6zGCg8O6pApDwCLcB/s1600/4.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-1OyUKPCS4Rc/WP-qs_8CE4I/AAAAAAAACl0/4KkgGWsc2O8lohVEYPsw6zGCg8O6pApDwCLcB/s1600/4.png" /></a></p><p>Imagine energy (E) interacting with a cubic meter of spacetime. Equation 6 shows that spacetime energy density (Ts) is a tiny number. Since curved spacetime causes gravity, and since spactime's energy is so weak, is it any wonder that gravity is weak? Let's see if the math agrees. Equations 7 and 8 are the average radius of a nucleus and the nucleus volume respectively. Why are these numbers important? Imagine energy (E) added to a cubic meter of spacetime. To see how that energy interacts with spacetime, it's easier to take all spacetime's energy and reduce it to a particle. We do the same with the added energy. So we have a particle interacting with another particle within a cubic meter. Spacetime's total energy is equivalent to a proton's. We can imagine it having an approximate radius of e-15 meters and a volume of e-45 cubic meters. Our energy particle will have the same dimensions. </p><p>To provide a simple visual demonstration of the interaction between matter and spacetime, let's pretend that the volume of our spacetime proton is .25 cubic meters and that the energy particle we add has a .25 probability of interacting with it. </p><p><a href="https://2.bp.blogspot.com/-L10fDZ69iPE/WP-9zmUdS5I/AAAAAAAACmE/4UjW_HRgXYc3CnFeukQHXemuNLS8b9qJgCLcB/s1600/5.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-L10fDZ69iPE/WP-9zmUdS5I/AAAAAAAACmE/4UjW_HRgXYc3CnFeukQHXemuNLS8b9qJgCLcB/s1600/5.png" /></a></p><p>Because the probability of interaction is .25, the average energy ends up being .25 the total energy. To get the proper value for K, we would need to multiply E/m^3 by a constant of .25. So K is weaker than E/m^3. To illustrate the point further, let's distribute the energy evenly throughout the cubic meter. This time it will be a field of isotropic energy interacting with our spacetime proton: </p><p><a href="https://2.bp.blogspot.com/-Qguq4Q26ADo/WP_BZOpXjpI/AAAAAAAACmQ/pfpMJQ01zz8OS6-gZmEhRgVrQ42aiBYPQCLcB/s1600/6.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-Qguq4Q26ADo/WP_BZOpXjpI/AAAAAAAACmQ/pfpMJQ01zz8OS6-gZmEhRgVrQ42aiBYPQCLcB/s1600/6.png" /></a></p><p>At any given time, only 25% of the energy field interacts with the spacetime proton. The remaining 75% contributes nothing to the value of K. To drive this point completely home, let's cut up the spacetime proton and evenly distribute it within the cubic meter: </p><p><a href="https://1.bp.blogspot.com/-iyaqofcM_zY/WP_E8rKbt9I/AAAAAAAACmc/VTVEjyAui4Us7UfGrc_0MxQ6CVKGrUHhgCLcB/s1600/7.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-iyaqofcM_zY/WP_E8rKbt9I/AAAAAAAACmc/VTVEjyAui4Us7UfGrc_0MxQ6CVKGrUHhgCLcB/s1600/7.png" /></a></p><p>As you can see, it matters not how we distribute the matter energy and spacetime energy. In the case above, only 25% of the total matter energy interacts with the spacetime energy. Our constant of .25 remains constant. Now, let's stop pretending and let's replace the .25 constant with the volume of a nucleus, which is approximately e-45 m^3. </p><p><a href="https://1.bp.blogspot.com/-3heSKMtp9zc/WP_HbwvTLSI/AAAAAAAACmo/W5427Xesqeoly3LaZekTivQi9pM50hn-gCLcB/s1600/8.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-3heSKMtp9zc/WP_HbwvTLSI/AAAAAAAACmo/W5427Xesqeoly3LaZekTivQi9pM50hn-gCLcB/s1600/8.png" /></a></p><p>If e-45 is the right constant, then that means only e-45 of the energy contained in T44 contributes to K. Let's check it: </p><p><a href="https://3.bp.blogspot.com/-NtjOPzQp-GE/WP_MA-DduxI/AAAAAAAACm0/apcV7dGDsnQkwO_7mymS0UeGHDGNfnVKwCLcB/s1600/9.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-NtjOPzQp-GE/WP_MA-DduxI/AAAAAAAACm0/apcV7dGDsnQkwO_7mymS0UeGHDGNfnVKwCLcB/s1600/9.png" /></a></p><p>It looks as though it's in the ballpark of the actual constant used in Einstein's field equations. So it's highly plausible that gravity is weak due to energy interacting with very weak spacetime energy. </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com2tag:blogger.com,1999:blog-3628542335168231161.post-43891312717672814522017-04-20T13:09:00.000-07:002017-04-22T10:01:33.871-07:00Deriving the Schwarzschild Solution to Einstein's Field Equations<p><iframe width="560" height="315" src="https://www.youtube.com/embed/LQe6hLdPHl8" frameborder="0" allowfullscreen></iframe></p><p>Step one: Beginning with Einstein's field equations, derive the Scharzschild radius (equation 13 below): </p><p><a href="https://1.bp.blogspot.com/-_xYaLOACljk/WPfqSKV5Q1I/AAAAAAAACgo/PiC4kMYwatkKrwkFIa3BGxRVunAQO3WkQCLcB/s1600/1.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-_xYaLOACljk/WPfqSKV5Q1I/AAAAAAAACgo/PiC4kMYwatkKrwkFIa3BGxRVunAQO3WkQCLcB/s1600/1.png" /></a></p><p><a href="https://3.bp.blogspot.com/-SNDIR9BZ18w/WPkCPh-vZRI/AAAAAAAACkI/9KBCiY-NKAwPZS0adqt2EVE72GC6JZCtwCLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-SNDIR9BZ18w/WPkCPh-vZRI/AAAAAAAACkI/9KBCiY-NKAwPZS0adqt2EVE72GC6JZCtwCLcB/s1600/2.png" /></a></p><p><a href="https://3.bp.blogspot.com/-00mlkcQTw9Y/WPkDUHCwNnI/AAAAAAAACkU/DXzZ0UUiLM03Arjrw5275luzLxi1hkpdQCLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-00mlkcQTw9Y/WPkDUHCwNnI/AAAAAAAACkU/DXzZ0UUiLM03Arjrw5275luzLxi1hkpdQCLcB/s1600/3.png" /></a></p><p><a href="https://2.bp.blogspot.com/-lYSoIprmehU/WPfqpP7JbGI/AAAAAAAACg0/Mvxa-W6i0ggPY_5jjwhmc8VrqPm_GXsLACLcB/s1600/4.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-lYSoIprmehU/WPfqpP7JbGI/AAAAAAAACg0/Mvxa-W6i0ggPY_5jjwhmc8VrqPm_GXsLACLcB/s1600/4.png" /></a></p><p>Next, we call on Pythagoras and a right triangle to derive a basic metric equation (equation 15 below): </p><p><a href="https://4.bp.blogspot.com/-PDyBhjjEX6I/WPuMYbJ_4xI/AAAAAAAAClA/H05TUySqXsQwMbPAMt2vd_1LcBtADgrBwCLcB/s1600/5.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-PDyBhjjEX6I/WPuMYbJ_4xI/AAAAAAAAClA/H05TUySqXsQwMbPAMt2vd_1LcBtADgrBwCLcB/s1600/5.png" /></a></p><p>Using the same right triangle we derive the Lorentz factor (equation 19 below): </p><p><a href="https://2.bp.blogspot.com/-jf427yuvLyM/WPf3A1pauNI/AAAAAAAAChQ/FdziLBD2R04RfSDMFfA-eleGsH5Av8GxgCLcB/s1600/6.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-jf427yuvLyM/WPf3A1pauNI/AAAAAAAAChQ/FdziLBD2R04RfSDMFfA-eleGsH5Av8GxgCLcB/s1600/6.png" /></a></p><p>Now check out equation 20: </p><p><a href="https://4.bp.blogspot.com/-Lw4QkCfT9NA/WPf3-ig5oTI/AAAAAAAAChc/CSpounRs-xIdP81Yq5by0B3Sr1_5G8ARgCLcB/s1600/7.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-Lw4QkCfT9NA/WPf3-ig5oTI/AAAAAAAAChc/CSpounRs-xIdP81Yq5by0B3Sr1_5G8ARgCLcB/s1600/7.png" /></a></p><p>Because of equation 20, we can make a substitution and derive equations 22 and 23: </p><p><a href="https://3.bp.blogspot.com/-qhU39KRp3Ok/WPf4-_RdDrI/AAAAAAAACho/knFFARxfMP4i5Wxa5cfx8FwXC6LwVrx3gCLcB/s1600/8.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-qhU39KRp3Ok/WPf4-_RdDrI/AAAAAAAACho/knFFARxfMP4i5Wxa5cfx8FwXC6LwVrx3gCLcB/s1600/8.png" /></a></p><p>Equations 22 and 23 allow us to make more substitutions. The result is something that resembles the Schwarzschild metric (equation 24): </p><p><a href="https://2.bp.blogspot.com/-IyCH93mAPOA/WPf5-zwnCsI/AAAAAAAACh0/lfpKApS3xbg3MzltzckJpBS5Lt0Lg5a4ACLcB/s1600/10.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-IyCH93mAPOA/WPf5-zwnCsI/AAAAAAAACh0/lfpKApS3xbg3MzltzckJpBS5Lt0Lg5a4ACLcB/s1600/10.png" /></a></p><p>Here's the actual Scharzschild metric: </p><p><a href="https://2.bp.blogspot.com/-bCjULFZuGHQ/WPf88NPBBEI/AAAAAAAACiA/-ZgFUFu97GMT3ObH0-0kB-L9Wkia73KrQCLcB/s1600/11.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-bCjULFZuGHQ/WPf88NPBBEI/AAAAAAAACiA/-ZgFUFu97GMT3ObH0-0kB-L9Wkia73KrQCLcB/s1600/11.png" /></a></p><p>We can replace equation 24's cdt' with dr (differential radius) to get the following: </p><p><a href="https://4.bp.blogspot.com/-PXVbb3RSt9k/WPf_oK5hKvI/AAAAAAAACiM/YpoWXQDXzWMXIR49rgJdVdyrzp4NLq3_ACLcB/s1600/12.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-PXVbb3RSt9k/WPf_oK5hKvI/AAAAAAAACiM/YpoWXQDXzWMXIR49rgJdVdyrzp4NLq3_ACLcB/s1600/12.png" /></a></p><p>It would be great if the middle term (vdt)^2 had a plus sign instead of a minus sign in front of it. With some trigonometric slight of hand we change the minus sign to a plus. The result is equation 29: </p><p><a href="https://4.bp.blogspot.com/-8Fxx4Gx-MMc/WPgDG9yi6gI/AAAAAAAACiY/16TYCx4X2CsQyuyCYilnF_17gnIb6zABwCLcB/s1600/13.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-8Fxx4Gx-MMc/WPgDG9yi6gI/AAAAAAAACiY/16TYCx4X2CsQyuyCYilnF_17gnIb6zABwCLcB/s1600/13.png" /></a></p><p>So far we've used a triangle with only two space dimensions. We are one dimension short, but we can fix that: </p><p><a href="https://3.bp.blogspot.com/-UQu7KzHHacA/WPgHsEdF1NI/AAAAAAAACik/bUr52tGFrI4aPTVJNjwnb-S-sTbqrBMIACLcB/s1600/14.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-UQu7KzHHacA/WPgHsEdF1NI/AAAAAAAACik/bUr52tGFrI4aPTVJNjwnb-S-sTbqrBMIACLcB/s1600/14.png" /></a></p><p>Each dimension in space is a hypotenuse of a right triangle with two other dimensions which can replace the hypotenuse. We make a final substitution and we get the Schwarzschild metric (equation 32). Schwarzschild used spherical coordinates. For clarity and to help you visualize this type of coordinate system, I provide the diagrams below. The first two diagrams show the front and side view of a sphere of spacetime with a mass in the center. The position in spacetime is given by the radius (r), the first angle (top diagram), and a shorter radius (rsin[first angle]) and the second angle (bottom diagram). </p><p><a href="https://2.bp.blogspot.com/-Kj3z3Cr588A/WPjx4mxZr1I/AAAAAAAACi0/C4QlM_f2GncRRbby9VnYN1p1KUPa441CgCLcB/s1600/15.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-Kj3z3Cr588A/WPjx4mxZr1I/AAAAAAAACi0/C4QlM_f2GncRRbby9VnYN1p1KUPa441CgCLcB/s1600/15.png" /></a></p><p>The variables used in the Schwarzschild metric, however, are differential--a tiny piece of the radius and each angle. The value of each space variable is indicated in red below: </p><p><a href="https://4.bp.blogspot.com/-8AfVjy0_kS0/WPjy-YEatMI/AAAAAAAACjA/n3Xp8hgiJT415mMjvFiwVI_IF40CAhaiwCLcB/s1600/16.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-8AfVjy0_kS0/WPjy-YEatMI/AAAAAAAACjA/n3Xp8hgiJT415mMjvFiwVI_IF40CAhaiwCLcB/s1600/16.png" /></a></p><p>If we take the limit of these variables we get a point in spacetime indicated by the red dot in the diagram below: </p><p><a href="https://4.bp.blogspot.com/-KpabwEfbaBU/WPkUyfx1SVI/AAAAAAAACkw/gaDn7BcAywYMP1c7bNGvXAEwkvytXrXbQCLcB/s1600/17.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-KpabwEfbaBU/WPkUyfx1SVI/AAAAAAAACkw/gaDn7BcAywYMP1c7bNGvXAEwkvytXrXbQCLcB/s1600/17.png" /></a></p><p>The objective is to figure out the spacetime curvature in that tiny (red) region of space. To solve the field equations, we need to know the metric tensor components; i.e., the g's. </p><p><a href="https://1.bp.blogspot.com/-avbmH2jIRdE/WPj4LR6DXjI/AAAAAAAACjk/k0UbhsT2GKIucsPMZzR5ijJnW7wUZQATgCLcB/s1600/18.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-avbmH2jIRdE/WPj4LR6DXjI/AAAAAAAACjk/k0UbhsT2GKIucsPMZzR5ijJnW7wUZQATgCLcB/s1600/18.png" /></a></p><p>We can find the value of each g component within the Scharzschild metric: </p><p><a href="https://3.bp.blogspot.com/-Ew1CCxLzDtE/WPj4bHh60nI/AAAAAAAACjo/Bt3XS1zTpW4yT-zXOJ_LuxMSRhpsCMLuQCLcB/s1600/19.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-Ew1CCxLzDtE/WPj4bHh60nI/AAAAAAAACjo/Bt3XS1zTpW4yT-zXOJ_LuxMSRhpsCMLuQCLcB/s1600/19.png" /></a></p><p>Thus the metric tensor is as follows: </p><p><a href="https://3.bp.blogspot.com/-hUc4bcHpkWc/WPj8KLegNbI/AAAAAAAACj0/pbHpiMFmJJ0Q_w0ooKQ3tfuhnrPzvtfmQCLcB/s1600/20.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-hUc4bcHpkWc/WPj8KLegNbI/AAAAAAAACj0/pbHpiMFmJJ0Q_w0ooKQ3tfuhnrPzvtfmQCLcB/s1600/20.png" /></a></p><p><a href="https://4.bp.blogspot.com/-X1GUH-fsDhY/WPj8QKKeCOI/AAAAAAAACj4/CcUH2qPVvvUm95j0ViVZ2LY1ep8ukSwBQCLcB/s1600/21.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-X1GUH-fsDhY/WPj8QKKeCOI/AAAAAAAACj4/CcUH2qPVvvUm95j0ViVZ2LY1ep8ukSwBQCLcB/s1600/21.png" /></a></p><p>With the information we have, we can derive equation 42 below. </p><p><a href="https://1.bp.blogspot.com/-TfxaYVHXdGc/WPkRRmpX3ZI/AAAAAAAACkk/asmw01YG5_QfdNrPzQJouk6KsPhpBYm3gCLcB/s1600/22.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-TfxaYVHXdGc/WPkRRmpX3ZI/AAAAAAAACkk/asmw01YG5_QfdNrPzQJouk6KsPhpBYm3gCLcB/s1600/22.png" /></a></p><p>We can solve for R44--the spacetime curvature--by plugging in the mass (m) of a star, planet or black hole; the volume (V) of the space, mass and energy within an imaginary sphere with radius (r); and radius (r). </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-29792084682293489472017-04-04T15:59:00.000-07:002017-04-17T10:46:15.125-07:00Debunking the Equivalence Principle Thought Experiments<p><iframe width="560" height="315" src="https://www.youtube.com/embed/6APfzIHZ6tk" frameborder="0" allowfullscreen></iframe></p><p>According to the textbooks, the principle of equivalence asserts that gravity and inertia are one and the same--not similar, proportionate or related to each other, but the same. According to the principle, you could kidnap some scientists by smothering each of them with an ether rag, place them in a rocket ship with an acceleration of 9.8 meters per second per second. When they regain consciousness, they won't be able to tell if the gravity they feel is from the earth or the rocket ship--there are no windows for them to look outside. </p><p>As the video above says, they can drop a pen and it will fall to the floor like a pen on earth. They are free to do whatever experiment they wish--and they won't be able to distinguish earth's gravity from the rocket's acceleration? Hmmm ... let's take a closer look at how a rocket propels itself: </p><p><a href="https://3.bp.blogspot.com/-VJO4TZaLxcU/WOVo28BpIEI/AAAAAAAACcQ/BbM9ZCe8btUi2CQH68mzRPo59YqHmF5qQCLcB/s1600/1.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-VJO4TZaLxcU/WOVo28BpIEI/AAAAAAAACcQ/BbM9ZCe8btUi2CQH68mzRPo59YqHmF5qQCLcB/s1600/1.png" /></a></p><p>According to equation 1) above, rocket fuel and oxygen enter the nozzle at a certain mass-flow rate and velocity, then exit the nozzle at a higher velocity and final mass-flow rate. Add to that the pressure difference. The total is the thrust force. Recalling Newton's third law: this is the action. The equal and opposite reaction is the rocket accelerating. </p><p><a href="https://1.bp.blogspot.com/-ELMv6YHjAIo/WOVro5lvAiI/AAAAAAAACcU/YAFtw3fh7qMj9VnBGyIqZv7KQ6C1ICVjQCLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-ELMv6YHjAIo/WOVro5lvAiI/AAAAAAAACcU/YAFtw3fh7qMj9VnBGyIqZv7KQ6C1ICVjQCLcB/s1600/2.png" /></a></p><p>Equation 2) shows that the perceived gravity inside the rocket is acceleration (g). Since action and reaction are equal and opposite, we can set the thrust force equal to the total mass of the rocket (passengers, supplies, fuel included) times the acceleration g: </p><p><a href="https://2.bp.blogspot.com/-pH4GXD9i3L0/WOVtTG_71lI/AAAAAAAACcY/jOdw2r0QUt0ntB1rxcVu6UdcrC3wtTLCACLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-pH4GXD9i3L0/WOVtTG_71lI/AAAAAAAACcY/jOdw2r0QUt0ntB1rxcVu6UdcrC3wtTLCACLcB/s1600/3.png" /></a></p><p>Doing a bit of algebra gives us equation 4)--the function g. Let's compare equation 4) to equation 5) below: </p><p><a href="https://4.bp.blogspot.com/-0bQsfuKHhZ8/WOVuNAC6VTI/AAAAAAAACcc/Wdk6Noc5a8kHNoUmSyMu2iunUYoUbGCMgCLcB/s1600/4.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-0bQsfuKHhZ8/WOVuNAC6VTI/AAAAAAAACcc/Wdk6Noc5a8kHNoUmSyMu2iunUYoUbGCMgCLcB/s1600/4.png" /></a></p><p>Equation 5), of course, is Newton's law of gravitation. Let's assume the scientists aboard the rocket are aware of equations 4) and 5). They're also aware of equation 1) that shows fuel mass flowing out the rocket's nozzle. How hard would it be for them to figure out that they are in a rocket ship and not in a sealed box on earth? </p><p>Not hard at all! In fact, all they have to do is weigh themselves repeatedly over time. The rocket's total mass (m) is decreasing thanks to the burning fuel exploding out the rocket's back end. This would cause the rocket's acceleration (g) to increase. So when the scientists weigh themselves they notice over time they are gaining weight (scientists' mass * increased g). Either their Weight Watchers diet is failing or they are in an accelerating rocket. </p><p>Unlike the rocket, earth's mass is virtually constant. If the scientists weighed themselves there, they would notice little or no change (assuming they didn't quit their diet). </p><p>Now, take another look at equations 4) and 5). Notice how mass (m) is in equation 4's denominator, but equation 5) has a mass variable in its numerator. Any change in mass would yield opposite results when comparing the two equations. If mass increases, the acceleration (g) in equation 4) goes down, but equation 5's acceleration (g) increases--and vice versa! </p><p>To claim the g in the rocket is indistinguishable from the g on earth requires us to ignore the physical processes involved. Prior to any mass loss, if one drops a pen in a rocket accelerating at 9.8 meters per second per second, the pen's inertia keeps it in place and allows the floor of the rocket to accelerate up to it. A casual observer aboard the rocket will most likely get the impression the pen dropped to the floor. Is this really sufficient evidence to support the claim that inertia and gravity are one and the same? If so, one could also claim that gravity is not a genuine force; it's a byproduct of another force such as the rocket's thrust force. </p><p>If gravity and inertia are the same, it should be possible to fool the kidnapped scientists aboard the rocket. Our first attempt failed--the rocket lost some of its mass causing g to increase. Suppose we install some computer software that changes the thrust force in proportion to how much mass is lost or gained. This would allow the rocket's acceleration to remain constant. Perhaps we can rub our hands together in glee: how will the scientists ever figure out they are in a rocket ship and not on earth? Ah, but alas, there's equation 6): </p><p><a href="https://2.bp.blogspot.com/-Nkdy3X4G_zA/WOWEGQOHpVI/AAAAAAAACcs/nmi7MBNsWmsgT8K6U5GFV-6nXKqYOU-lACLcB/s1600/5.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-Nkdy3X4G_zA/WOWEGQOHpVI/AAAAAAAACcs/nmi7MBNsWmsgT8K6U5GFV-6nXKqYOU-lACLcB/s1600/5.png" /></a></p><p>And there's the inequality 7). Equation 6) and inequality 7) show what happens over time (t) if the rocket accelerates at a constant g. The rocket's velocity (v) increases. This causes the the rocket and every person and thing inside the rocket to have more mass than anticipated. When the scientists weigh themselves they once again discover a weight gain. Each starts out with a weight of (m * g) and end up with (m' * g). Compare this result to what happens on earth: </p><p><a href="https://3.bp.blogspot.com/-jxp_AM6RQ3w/WOWLb1M9slI/AAAAAAAACc0/XTc6dzrGJEsi-2U_csE2FRXd-6AKeIFUQCLcB/s1600/6.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-jxp_AM6RQ3w/WOWLb1M9slI/AAAAAAAACc0/XTc6dzrGJEsi-2U_csE2FRXd-6AKeIFUQCLcB/s1600/6.png" /></a></p><p><a href="https://3.bp.blogspot.com/-CzDm_JxRUUQ/WOWLfs81aqI/AAAAAAAACc4/yPENaue5fOYNDRjfZb4I3Qhbqnh7QXX4ACLcB/s1600/7.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-CzDm_JxRUUQ/WOWLfs81aqI/AAAAAAAACc4/yPENaue5fOYNDRjfZb4I3Qhbqnh7QXX4ACLcB/s1600/7.png" /></a></p><p>Equations 8) and 9) show that both the initial earth velocity and final earth velocity are pretty much the same. The right side of each equation is the total tangential earth velocity (which includes earth's spin and orbit around the sun) relative to the galaxy and universe's motion. Cutting to the chase, the earth's velocity relative to itself is zero. As a result, equation 12) shows that if the scientists were on earth, their respective weights would not increase over time through no fault of their own. By contrast, the rocket's velocity increases relative to the earth's--and so must the scientists' respective weights. </p><p>OK, strike two. We failed to fool the scientists once again. Time for desperate measures. We smother the scientists with ether rags and place them in a rotating, donut-shaped space station. When they regain consciousness, it is hoped they will fail to distinguish between the angular acceleration (g) and the earth's gravity. </p><p><a href="https://2.bp.blogspot.com/-77KqIGvJWqc/WOWT1y3HyaI/AAAAAAAACdI/tjRN7_PaPU4mg3ly4-86Gtp60QiNjaDcQCLcB/s1600/8.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-77KqIGvJWqc/WOWT1y3HyaI/AAAAAAAACdI/tjRN7_PaPU4mg3ly4-86Gtp60QiNjaDcQCLcB/s1600/8.png" /></a></p><p>The advantage of this new arrangement and equation 13) is the velocity does not change, so relativistic mass increases are avoided. We avoid a mass decrease, due to fuel consumption, by continually topping off the fuel tank (using a maintenance spacecraft). These measures should provide an artificial gravity indistinguishable from earth' gravity, right? But then something very routine happens: the scientists place their garbage and waste in the disposal chute. The waste is jettisoned into space. The scientists are not aware of this. For all they know, the garbage goes to a dumpster here on earth. But once again, when they weigh themselves, their respective weights increase. </p><p>Curses! It's that bloody conservation-of-momentum rule! When overall mass decreases due to garbage disposal, the velocity increases, so does the angular acceleration. If the scientists were on earth, jettisoning waste would still decrease the space station's mass, but the overall mass of earth would remain virtually the same, since the garbage ends up in a dumpster on earth. As a consequence, earth's gravity maintains its value. Strike three. </p><p>Then again, if the garbage stays on the space station, the perceived gravity there should be the same as earth's, and, we should be able to say with confidence, "Inertia and gravity are one and the same." But before we get too cocky, let's put this claim to another test. Imagine the earth and a red meteor coming together. We can interpret this event in one of two ways: the meteor is falling to earth, or, the meteor is at rest and the earth is moving toward it. </p><p><a href="https://2.bp.blogspot.com/-8IsfizALyUA/WOZwJGJJ5pI/AAAAAAAACdY/Q8lYrRujGEYRdTboeund-sUGyXDC-g7jACLcB/s1600/9.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-8IsfizALyUA/WOZwJGJJ5pI/AAAAAAAACdY/Q8lYrRujGEYRdTboeund-sUGyXDC-g7jACLcB/s1600/9.png" /></a></p><p>If inertia and gravity are the same, then it should be possible for a blue meteor to be at rest on the other side of the earth. The earth should move toward both meteors: </p><p><a href="https://3.bp.blogspot.com/-u5ofArDHBAo/WOZxcdWGwNI/AAAAAAAACdk/rTxwJ1i6vewbEX3l1v56URGlR9bPlPxiwCLcB/s1600/10.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-u5ofArDHBAo/WOZxcdWGwNI/AAAAAAAACdk/rTxwJ1i6vewbEX3l1v56URGlR9bPlPxiwCLcB/s1600/10.png" /></a></p><p>No force acts on the meteors--they are at rest, floating in space. The force, whatever it may be, is moving the earth in opposite directions to meet the meteors. Then again, maybe it doesn't really work that way. Could it be that the meteors are really falling to earth? </p><p><a href="https://2.bp.blogspot.com/-2yfzcLT00L4/WOZzyPQDmPI/AAAAAAAACdw/IdfUMfZa38MtlsUJtxWe90lQatumP1-MwCLcB/s1600/11.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-2yfzcLT00L4/WOZzyPQDmPI/AAAAAAAACdw/IdfUMfZa38MtlsUJtxWe90lQatumP1-MwCLcB/s1600/11.png" /></a></p><p>Assuming the meteors are falling to earth, a force must be causing them to accelerate. We call that force gravity. It seems clear at this point that gravity and inertia are not one and the same but are only similar to a casual observer who doesn't ask too many questions. </p><p>There is another key difference between inertia and gravity that involves time ... the time it takes a boson, say, a graviton to notify matter of a change in status in other matter. To properly distinguish between gravity and inertia, we need to make Newton's equation and Einstein's field equations time dependent. The same goes for the rocket-thrust equation. </p><p><a href="https://4.bp.blogspot.com/-Gp_LUBNrCpo/WOaEzZaHDII/AAAAAAAACeA/pj8nU70HJQozzd1S7_tV2JNJYymhGBmvgCLcB/s1600/13.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-Gp_LUBNrCpo/WOaEzZaHDII/AAAAAAAACeA/pj8nU70HJQozzd1S7_tV2JNJYymhGBmvgCLcB/s1600/13.png" /></a></p><p><a href="https://4.bp.blogspot.com/-EDX_Yqn99dU/WOaE3qmwCcI/AAAAAAAACeE/zMy52QMuUNI3_x1iwCEdBwiOwuIdtcV0QCLcB/s1600/12.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-EDX_Yqn99dU/WOaE3qmwCcI/AAAAAAAACeE/zMy52QMuUNI3_x1iwCEdBwiOwuIdtcV0QCLcB/s1600/12.png" /></a></p><p>Take a look at equation 15) above. If the sun's mass (m) were to drastically change in the current time (t), it would take about eight minutes (r/c) for us here on earth to feel the gravitational effects. The change will be felt by us at time t+r/c. Equation 16) concurs. Any change in spacetime curvature won't happen instantaneously. It takes time for gravitons to move through space from the sun to the earth. </p><p>Equation 17) tells a different story. If the thrust force changes in the rocket ship we discussed earlier, the acceleration (or perceived gravity) changes instantaneously--there is no need for gravitons to notify distant objects of the change in status, and to tell them to accelerate. For it is the rocket floor that is accelerating to objects at rest. Thus, when inertia mimics gravity, it takes less time. </p><p>The diagrams below illustrate the time difference between gravity and inertia: </p><p><a href="https://4.bp.blogspot.com/-K0r56IqskxI/WOaSG4rCJ7I/AAAAAAAACew/qjPonIFoesk3drWkV3lsfnCxByaNvffbgCLcB/s1600/14.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-K0r56IqskxI/WOaSG4rCJ7I/AAAAAAAACew/qjPonIFoesk3drWkV3lsfnCxByaNvffbgCLcB/s1600/14.png" /></a></p><p><a href="https://1.bp.blogspot.com/-RfYTOc4-FCg/WOaNFL2INkI/AAAAAAAACec/xy_82essAskir84-oqjujoHKoMJH8uIUwCLcB/s1600/15.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-RfYTOc4-FCg/WOaNFL2INkI/AAAAAAAACec/xy_82essAskir84-oqjujoHKoMJH8uIUwCLcB/s1600/15.png" /></a></p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-47487014195574784162017-03-27T12:16:00.000-07:002017-07-12T08:42:08.488-07:00Using Quantum Physics to Solve Relativity Conundrums<p><iframe width="560" height="315" src="https://www.youtube.com/embed/ttZCKAMpcAo" frameborder="0" allowfullscreen></iframe></p><p>Imagine you are driving down a lonesome highway. According to relativity theory, you could look at the situation in two ways: the highway is at rest and your car is moving; or, your car is at rest and the highway is moving, so are all the trees lining the highway, so are the buildings and the sky above--the whole earth is moving while you and your car sit still and watch it all go by. </p><p>But which way of looking at the situation is true? Could they both be equally valid? No. Here's why: the amount of energy needed to move the entire earth beneath your car far exceeds the amount of energy that is in your gas tank. Thus, conservation of energy is violated if we take seriously the notion that your car is at rest and it is the earth that is moving. Therefore, it is your car that is moving relative to the earth. </p><p>Now, consider Bob and Alice. Each are inside a separate spaceship. The two spaceships are at rest relatively speaking. Bob takes some measurements inside his spaceship; Alice does likewise. Both drink a potion that puts them to sleep. While they sleep, one of their spaceships is accelerated to speed v; the other remains at rest relative to the other. When Alice and Bob wake up, they must figure out which of their spaceships is going speed v. According to relativity theory, they won't be able to tell. If there is no friction and if there are no windows or portholes, both will perceive they are still at rest. </p><p>Both Alice and Bob experience the same laws of physics but ... not the same measurements. As I mentioned earlier, both took some measurements when both ships were at rest. One measurement was a momentum, given their respective masses. They each fired themselves out of a small cannon. They each measured the distance they flew and the time it took; i.e., the velocity. They calculated the momentum by multiplying mass and velocity. Using the same method, Alice now measures her velocity again. She notices no change. Bob, however, is in for a surprise. His velocity has slowed down. The momentum is conserved, so that means his mass has increased! Thus Alice and Bob deduce, pursuant to equation 1) below, that Alice's spaceship is at rest relative to Bob's. Bob's ship is the one moving at velocity v. </p><p><a href="https://4.bp.blogspot.com/-gcXxRQo0VfA/WNbkJCyU14I/AAAAAAAACUY/s1xm9ZtszykR4L7lpDKK_V4A9GLpEeJhgCLcB/s1600/1.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-gcXxRQo0VfA/WNbkJCyU14I/AAAAAAAACUY/s1xm9ZtszykR4L7lpDKK_V4A9GLpEeJhgCLcB/s1600/1.png" /></a></p><p>Suppose Alice and Bob can see inside each other's ship as Bob's ship passes by Alice's. According to relativity theory, Alice sees Bob's clock running slower and the length of Bob's ship has shrunk. Bob, however, sees Alice's clock running slower and it is her ship that has shrunk. Is Bob right? Or is Alice? Are they both right? Are they both wrong? Did one or both of their ships shrink? Is time really running slower for Bob and/or Alice? </p><p>Since his ship is moving faster than Alice's, Chances are it's Bob's ship that is experiencing time dilation and length contraction. But how can we be sure? </p><p>Let's assume Bob and Alice each have a box that tells time by firing two photons vertically, one up and one down. The photons have a velocity c and take time t to traverse the box. The vertical distance is ct: </p><p><a href="https://4.bp.blogspot.com/-7tlx75FbHg0/WNbrsDs2NTI/AAAAAAAACUo/jpX2N__5zhA51FqzWDdaz2lINtrV41KIACLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-7tlx75FbHg0/WNbrsDs2NTI/AAAAAAAACUo/jpX2N__5zhA51FqzWDdaz2lINtrV41KIACLcB/s1600/2.png" /></a></p><p>When Bob's ship passes Alice's here's how they see each other's clock: </p><p><a href="https://1.bp.blogspot.com/-NvQ7Xmtgf-I/WNbs7oTC4dI/AAAAAAAACU0/9_ABnua-JKQ6_jl1nUvKvUWx5zGZ70xkQCLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-NvQ7Xmtgf-I/WNbs7oTC4dI/AAAAAAAACU0/9_ABnua-JKQ6_jl1nUvKvUWx5zGZ70xkQCLcB/s1600/3.png" /></a></p><p>They each see the other's clock lagging behind their own. When Alice notices her photons have completed the distance (ct), she looks over at Bob's and notices his photons have not. (His have gone ct'.) Bob, of course, sees the opposite. </p><p><a href="https://2.bp.blogspot.com/-IXsJq3oKDlo/WNbu_yNMknI/AAAAAAAACVA/9sifSPFfKVMYSaVdSXbk_9fYuBc7htPAQCLcB/s1600/4.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-IXsJq3oKDlo/WNbu_yNMknI/AAAAAAAACVA/9sifSPFfKVMYSaVdSXbk_9fYuBc7htPAQCLcB/s1600/4.png" /></a></p><p>Alice and Bob rotate their clocks 90 degrees, so that the photons go horizontally. </p><p><a href="https://1.bp.blogspot.com/-uPbw1IPrAHQ/WNbwrV0XsXI/AAAAAAAACVM/zeSAacTJVRk6DLelYBM0Ft_PyjJfCkwvwCLcB/s1600/5.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-uPbw1IPrAHQ/WNbwrV0XsXI/AAAAAAAACVM/zeSAacTJVRk6DLelYBM0Ft_PyjJfCkwvwCLcB/s1600/5.png" /></a></p><p>Again, they see their clocks going a distance of ct. When they look at the other's clock, they see this: </p><p><a href="https://2.bp.blogspot.com/-wM50IAsSLvI/WNbx3e4eC7I/AAAAAAAACVY/pRAJXaWG6OA8XU3Qo79jyQr2xonrrQFWQCLcB/s1600/6.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-wM50IAsSLvI/WNbx3e4eC7I/AAAAAAAACVY/pRAJXaWG6OA8XU3Qo79jyQr2xonrrQFWQCLcB/s1600/6.png" /></a></p><p>The above diagram indicates that time appears distorted. Going from left to right, the distance is longer and time rate is faster, but going right to left, distance is shorter and time rate appears slower. Both Alice and Bob calculate the average time and distance. They both notice the photons in both clocks, on average, complete the distance simultaneously. Since Alice and Bob have concluded the other's clock is slower, there is only one way the photons could complete their respective distances simultaneously: the other's horizontal distance has shrunk to length ct'. (In the diagram below, each assumes the other is in motion while he or she is at rest.) </p><p><a href="https://4.bp.blogspot.com/-Bg_xjJVjxjk/WNb14L5opoI/AAAAAAAACVk/_9Rvlf-BrUoq2gi_D5Xo5X_SJw0iWWbVACLcB/s1600/7.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-Bg_xjJVjxjk/WNb14L5opoI/AAAAAAAACVk/_9Rvlf-BrUoq2gi_D5Xo5X_SJw0iWWbVACLcB/s1600/7.png" /></a></p><p>However, what Alice and Bob see and conclude in the above thought experiment depends on one's confirmation bias. Instead of seeing time slow down and length contract, Alice and Bob could see the other's clock (vertical space) grow taller and the other's time in sync with their own: </p><p><a href="https://3.bp.blogspot.com/-vl6j9ClEtTA/WNb4lXTF6lI/AAAAAAAACVo/ZM-K0osd2cEFVCjfro9PA0d-WaFUX1gzgCLcB/s1600/8.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-vl6j9ClEtTA/WNb4lXTF6lI/AAAAAAAACVo/ZM-K0osd2cEFVCjfro9PA0d-WaFUX1gzgCLcB/s1600/8.png" /></a></p><p>If the vertical space grows taller, that explains why the other's photons take longer to complete the distance. If the horizontal space does not contract for either ship, that explains why both sets of photons completed the distance simultaneously. Also the fact that both Alice and Bob see the other's ship having the relativistic effects suggests that what they see in the other's ship is some sort of optical illusion. It ain't really happening--or is it? To know the truth about relativity we must delve deeper. Quantum physics can assist us. </p><p>Let's address whether a spaceship in motion experiences horizontal contraction or vertical growth. We can think of the particles that make up a spaceship as waves. Waves have wavelength and amplitude. We can imagine each particle's wavelength contracting. This may cause the macroscopic ship length to contract. Or, we can imagine each wave's amplitude increasing--possibly causing the ship's height to grow. We begin our investigation by defining some variables: </p><p><a href="https://1.bp.blogspot.com/-WRi7SW0K0-A/WNmLREY3B6I/AAAAAAAACZA/Jwu0n5FlF4UkJt3esLmaF3MXogIY0VnFwCLcB/s1600/9.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-WRi7SW0K0-A/WNmLREY3B6I/AAAAAAAACZA/Jwu0n5FlF4UkJt3esLmaF3MXogIY0VnFwCLcB/s1600/9.png" /></a></p><p>Here's the math that can help us figure out what is going on: </p><p><a href="https://4.bp.blogspot.com/-v8B5m5NuKlI/WNld2RUdZwI/AAAAAAAACWY/u8IBss02nI8iFoXBlp2rVod5rdGlQUMQACLcB/s1600/10.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-v8B5m5NuKlI/WNld2RUdZwI/AAAAAAAACWY/u8IBss02nI8iFoXBlp2rVod5rdGlQUMQACLcB/s1600/10.png" /></a></p><p><a href="https://4.bp.blogspot.com/-_n3d_QKftYU/WNleqy8iVZI/AAAAAAAACWk/ohmyjeHNNT4n0a7j3J1vYK-bfAgiUfyXwCLcB/s1600/11.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-_n3d_QKftYU/WNleqy8iVZI/AAAAAAAACWk/ohmyjeHNNT4n0a7j3J1vYK-bfAgiUfyXwCLcB/s1600/11.png" /></a></p><p>From Hooke's law we derive equation 13) above. We mentioned earlier that Bob's mass increased. Equation 13) shows that an increase in mass is a function of increased amplitude and/or decreased wavelength. So Bob's spaceship may have really shrunk horizontally and/or grown vertically. By contrast, Alice measured no increase in her mass. We can assume when Bob sees her spaceship change shape, he's seeing an illusion, like a guy looking out a train window, watching the train station go by--or, maybe he sees no change in Alice's ship. </p><p>If equation 13) is valid, we should be able to derive from it the Lorentz equation. Let's give it a go: </p><p><a href="https://1.bp.blogspot.com/-ft8-kIqcDfY/WNlj7eqXx_I/AAAAAAAACW0/5mSdh2RzIzshlRWmHCKLOIdqcSnP4CfagCLcB/s1600/12.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-ft8-kIqcDfY/WNlj7eqXx_I/AAAAAAAACW0/5mSdh2RzIzshlRWmHCKLOIdqcSnP4CfagCLcB/s1600/12.png" /></a></p><p><a href="https://2.bp.blogspot.com/-rrr8GBhtoQc/WNllB_9wZRI/AAAAAAAACXA/bB6ZiXZ7JGog8RSstzv-FQ-F01TyBqTLQCLcB/s1600/13.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-rrr8GBhtoQc/WNllB_9wZRI/AAAAAAAACXA/bB6ZiXZ7JGog8RSstzv-FQ-F01TyBqTLQCLcB/s1600/13.png" /></a></p><p>Equation 28) above confirms equation 13's validity. However, we still can't be sure what Alice sees when Bob's ship whizzes by. Bob's ship could have shorter length and/or taller height. Perhaps combining the Lorentz factor and quantum mechanic's wave function can help: </p><p><a href="https://4.bp.blogspot.com/-_KRbrPTJgmQ/WNlrMhEPD2I/AAAAAAAACXY/fvCj1vcdUoMYV0upPvX1REl8CZGK6VqZQCLcB/s1600/14.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-_KRbrPTJgmQ/WNlrMhEPD2I/AAAAAAAACXY/fvCj1vcdUoMYV0upPvX1REl8CZGK6VqZQCLcB/s1600/14.png" /></a></p><p><a href="https://3.bp.blogspot.com/-nwXvsv2G3j0/WNlsoQbpZyI/AAAAAAAACXk/AMQvclvQzCMZ_sNOnknzf31D20TTbUa2gCLcB/s1600/15.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-nwXvsv2G3j0/WNlsoQbpZyI/AAAAAAAACXk/AMQvclvQzCMZ_sNOnknzf31D20TTbUa2gCLcB/s1600/15.png" /></a></p><p>Take a close look at equation 37) above. Notice that the right side's numerator is cosine squared plus sine squared. This value always equals one. It is constant. This constant value represents the amplitude or height, so when velocity (v) increases, the amplitudes of all the particles in Bob's ship don't change. It's possible then that the height of Bob's ship remains constant. Only the denominator (cosine squared) can change. This represents the wavelength. So when Alice sees Bob's ship whiz by, she probably sees a shorter length and a constant height. Such a scenario is consistent with the following wavelength equation: </p><p><a href="https://4.bp.blogspot.com/-6ylNsYPjvIM/WOvTg-hLqNI/AAAAAAAACfA/wMu-1oJv3WYCdqsvXH0ERGDrdgKDwc7pQCLcB/s1600/28.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-6ylNsYPjvIM/WOvTg-hLqNI/AAAAAAAACfA/wMu-1oJv3WYCdqsvXH0ERGDrdgKDwc7pQCLcB/s1600/28.png" /></a></p><p>Equation 37b) makes no mention of wave amplitude--which implies the amplitude does not change when there's a change in momentum (p). Only the wavelength changes. Below we derive equation 41). If you check it against the Lorentz (equation 42) you will find it yields the same results--the wavelength shortens when velocity (v) increases. </p><p><a href="https://4.bp.blogspot.com/-GCP-Dpmk4Z4/WNlxRlFYE7I/AAAAAAAACXw/Ekn9xRN3kAkt7Gaiv3KdzzlYgWULSgMiQCLcB/s1600/16.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-GCP-Dpmk4Z4/WNlxRlFYE7I/AAAAAAAACXw/Ekn9xRN3kAkt7Gaiv3KdzzlYgWULSgMiQCLcB/s1600/16.png" /></a></p><p>The above math seems to have solved the riddle of what Alice really sees when Bob's ship flies by. However, the following math suggests that all the dimensions of Bob's ship shrink--not just the length. If we make the Lorentz equation a function of mass, the y and z dimensions shrink. </p><p><a href="https://4.bp.blogspot.com/-_7sBWMqYJ1I/WNl0VzlJFBI/AAAAAAAACX8/nzImdp6qk0UvUSKVG90jNB_7Ac1n1QgbQCLcB/s1600/17.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-_7sBWMqYJ1I/WNl0VzlJFBI/AAAAAAAACX8/nzImdp6qk0UvUSKVG90jNB_7Ac1n1QgbQCLcB/s1600/17.png" /></a></p><p><a href="https://2.bp.blogspot.com/-8RyerDgxliU/WNl1MVfkB4I/AAAAAAAACYI/SpQw54Lz5KEpzuZS_6-DmIKSMf0-knRVACLcB/s1600/18.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-8RyerDgxliU/WNl1MVfkB4I/AAAAAAAACYI/SpQw54Lz5KEpzuZS_6-DmIKSMf0-knRVACLcB/s1600/18.png" /></a></p><p><a href="https://2.bp.blogspot.com/-4PBfhQqC8oI/WNl2RNpVUoI/AAAAAAAACYY/s7kmqAnGYPYjL6C0_PeZbiLMMOV947-MwCLcB/s1600/19.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-4PBfhQqC8oI/WNl2RNpVUoI/AAAAAAAACYY/s7kmqAnGYPYjL6C0_PeZbiLMMOV947-MwCLcB/s1600/19.png" /></a></p><p><a href="https://2.bp.blogspot.com/-oongxb4W_1w/WNl3OesAHcI/AAAAAAAACYk/VOWnCe14zoMrfx0BwUBt9LmLXB5hhn_3gCLcB/s1600/20.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-oongxb4W_1w/WNl3OesAHcI/AAAAAAAACYk/VOWnCe14zoMrfx0BwUBt9LmLXB5hhn_3gCLcB/s1600/20.png" /></a></p><p>Could that be how it really works? Mass increases due to high velocity, and all three dimensions shrink? Not according to the definition of a wavelength. A wavelength is the distance along the direction of propagation ... If Bob's ship is moving along the x-axis, then we can imagine all the spaceship's particle-waves doing likewise. Thus their shortening wavelengths are along x--the ship's length. So the length of Bob's ship will likely be ct'. The height, depending on which clock is used, will be nct' or ct (nct'=ct; n is a coefficient). The width will also be some factor of ct'. </p><p>The diagrams below show the relationship between particle-waves and Bob's square clock. As velocity (v) increases, Bob's ship propagates further as expected, but the wavelengths of his ship's particles shorten, so his ship's length shortens. You can think of Bob's clock and ship as macroscopic particle waves propagating from left to right. The photon (see wavy diagonal line) is moving vertically, but its shorter wavelengths have no effect on the height, since it is the only particle moving vertically. </p><p><a href="https://2.bp.blogspot.com/-sjHM-ka0rCk/WNmG3V2AyLI/AAAAAAAACY0/3ByQiPlzKSMgWmPXXOlCrZ4tC5uF-yrpQCLcB/s1600/21.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-sjHM-ka0rCk/WNmG3V2AyLI/AAAAAAAACY0/3ByQiPlzKSMgWmPXXOlCrZ4tC5uF-yrpQCLcB/s1600/21.png" /></a></p><p>Equation 59) below expresses the Lorentz in terms of frequency, wavelength, mass, and time: </p><p><a href="https://2.bp.blogspot.com/-8o2FOlVQ-UQ/WNmYgROU3OI/AAAAAAAACZQ/g6UyGmhJkd8cB9maPLWaLo116GQfzZkqgCLcB/s1600/22.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-8o2FOlVQ-UQ/WNmYgROU3OI/AAAAAAAACZQ/g6UyGmhJkd8cB9maPLWaLo116GQfzZkqgCLcB/s1600/22.png" /></a></p><p>When energy is added, frequency increases from fo to f'. The wavelength shortens. The light-speed term (second term) stays constant, since wavelength and frequency inversely offset each other. The last term is the velocity v term. Here, mass increases, frequency increases to inversely offset the mass increase, so what's left to cause higher velocity? A shorter wavelength. </p><p>To further understand what's happening at the quantum level, we rely on a Hamiltonian (H) equation expressed in terms of mass, wavelength, amplitude, and time: </p><p><a href="https://1.bp.blogspot.com/-BSJXeQt32go/WNmds8k3DZI/AAAAAAAACZg/ehgxrQhLT2cu-b74v0FhMHqvsUvLK662gCLcB/s1600/23.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-BSJXeQt32go/WNmds8k3DZI/AAAAAAAACZg/ehgxrQhLT2cu-b74v0FhMHqvsUvLK662gCLcB/s1600/23.png" /></a></p><p><a href="https://3.bp.blogspot.com/-Vb3De4UBzJ0/WNmeuehKS_I/AAAAAAAACZs/oX5hakawWVQ_ke7Dl7Bdcd-W1vSg0MIjgCLcB/s1600/24.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-Vb3De4UBzJ0/WNmeuehKS_I/AAAAAAAACZs/oX5hakawWVQ_ke7Dl7Bdcd-W1vSg0MIjgCLcB/s1600/24.png" /></a></p><p>Equation 65) shows what happens to a typical particle-wave. Velocity is seen by Alice, not Bob. Bob can treat his spaceship's particles as though they are at rest. This is why the velocity term or kinetic energy term has time (t), Alice's time. Amplitude stays constant and, once again, velocity and kinetic energy are determined by wavelength. The last term is rest-mass energy. Here, when Bob's time (t') reduces, so must the wavelength (lambda): otherwise, light speed (c) would not remain constant. </p><p>After making the case above, can we conclude that Bob's spaceship actually contracts when whizzing by Alice? No. We have to take into account thermal expansion (see equations below). High velocity implies high kinetic energy; high kinetic energy implies high temperatures; high temperatures cause thermal expansion which could more than offset length contraction. At very high speeds Bob might notice he's sweating profusely. If his velocity is great enough, Alice might see a fireball going by. If she sees anything, she might presume she's at rest, since anyone moving at such high speeds might be dead. </p><p><a href="https://1.bp.blogspot.com/-jI5amSbBT0s/WNmoDjCtGlI/AAAAAAAACZ8/oCWMNWu1Eg8xp3q0yVt7Qto15glcAxc8gCLcB/s1600/25.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-jI5amSbBT0s/WNmoDjCtGlI/AAAAAAAACZ8/oCWMNWu1Eg8xp3q0yVt7Qto15glcAxc8gCLcB/s1600/25.png" /></a></p><p>Then again, thermal transfer to the cold surrounding space might provide Bob some relief. Equation 70) below gives us the rate of heat loss. We combine equation 70) with equations 68) and 69) to get equation 71) which tells us the thermal expansion of length L, taking into account the temperature decrease due to heat loss as well as any temperature gain due to high speeds. </p><p><a href="https://1.bp.blogspot.com/-ENdkRpIVgzA/WNr2pNPwN0I/AAAAAAAACa8/Zmb_tN9ENwgY5rf5zicVymchUIKIzyA1wCLcB/s1600/26.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-ENdkRpIVgzA/WNr2pNPwN0I/AAAAAAAACa8/Zmb_tN9ENwgY5rf5zicVymchUIKIzyA1wCLcB/s1600/26.png" /></a></p><p>Keeping relativity in mind, thermal expansion occurs when the molecules that make up Bob's spaceship move at higher velocities relative to each other. If the molecules only move in the direction Bob's ship is moving, their relative velocity to each other is zero--no thermal expansion occurs. If, on the other hand, they move in many directions, their relative velocity to each other is greater than zero--and thermal expansion could offset relativity's length contraction. </p><p>What Alice sees when Bob's spaceship flies by can vary depending on the circumstances: </p><p><a href="https://4.bp.blogspot.com/-G-6tEGZkqYM/WNv4Zm7G-UI/AAAAAAAACbc/t0bKWBrR99sE41pjN-FIqC2Gtwehif_XgCLcB/s1600/27.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-G-6tEGZkqYM/WNv4Zm7G-UI/AAAAAAAACbc/t0bKWBrR99sE41pjN-FIqC2Gtwehif_XgCLcB/s1600/27.png" /></a></p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-67161836351279873572017-03-12T15:10:00.001-07:002017-03-16T18:47:58.102-07:00The Essence of Time<p><iframe width="560" height="315" src="https://www.youtube.com/embed/G-R8LGy-OVs" frameborder="0" allowfullscreen></iframe></p><p>What exactly is time? Is it just an abstract idea? Or does it exist independently of human imagination and perception? Atomic clocks reveal that the rate of time appears to run slower on the earth's surface than way out in space. Assuming time is a real entity, what is it made of? What is its essence? We start our investigation by defining some variables: </p><p><a href="https://3.bp.blogspot.com/-mtXZBc5zUkU/WMWi5xQv9eI/AAAAAAAACNo/_d-fxphayFgTZ-9UUPskIV1RsJXkbaJAACLcB/s1600/1.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-mtXZBc5zUkU/WMWi5xQv9eI/AAAAAAAACNo/_d-fxphayFgTZ-9UUPskIV1RsJXkbaJAACLcB/s1600/1.png" /></a></p><p>We know that nothing goes faster than light in a vacuum. If we add velocity (v) to velocity (c) we still get the speed of light (c). </p><p><a href="https://4.bp.blogspot.com/-_hlLZvH9-Tk/WMWi-Yv0kqI/AAAAAAAACNs/L8_qtwa5oB8hTQulZeHdmbrNski2WMymACLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-_hlLZvH9-Tk/WMWi-Yv0kqI/AAAAAAAACNs/L8_qtwa5oB8hTQulZeHdmbrNski2WMymACLcB/s1600/2.png" /></a></p><p>Equation 1) above seems absurd. When we combine velocities, we should get a higher velocity than c ... unless ... the rate of time (t') shrinks. Equation 2) below works: </p><p><a href="https://4.bp.blogspot.com/-JXwTdNbuuqw/WMWjDranTbI/AAAAAAAACNw/wMfSmCXoY8IE4xqcb4BgHdbZKPoBNaHVgCLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-JXwTdNbuuqw/WMWjDranTbI/AAAAAAAACNw/wMfSmCXoY8IE4xqcb4BgHdbZKPoBNaHVgCLcB/s1600/3.png" /></a></p><p>And from equation 2) we can derive the famous Lorentz equation: </p><p><a href="https://2.bp.blogspot.com/-XLP7VHROFYQ/WMWjJWQPBRI/AAAAAAAACN0/hYuNf8_lXEkKBypN53xrvkUT44xLU9_SwCLcB/s1600/4.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-XLP7VHROFYQ/WMWjJWQPBRI/AAAAAAAACN0/hYuNf8_lXEkKBypN53xrvkUT44xLU9_SwCLcB/s1600/4.png" /></a></p><p>When velocity (v) increases, time (t') shrinks, but something else happens that's also strange: mass (m') increases. We can verify this if we start with Einstein's energy equation below. </p><p><a href="https://2.bp.blogspot.com/-3CTEyO4o_s0/WMWjODN3qPI/AAAAAAAACN4/EQsIQd8CWAkEXhjGqBtLO9ZXbHy4e8YNQCLcB/s1600/5.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-3CTEyO4o_s0/WMWjODN3qPI/AAAAAAAACN4/EQsIQd8CWAkEXhjGqBtLO9ZXbHy4e8YNQCLcB/s1600/5.png" /></a></p><p>From equation 6) we can derive the relative mass equation: </p><p><a href="https://2.bp.blogspot.com/-DN-uybZG04E/WMWjSfKxBbI/AAAAAAAACN8/LBzcp9HJXfU1wdVKJ_WiZ8s8ya5eDwlQQCLcB/s1600/6.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-DN-uybZG04E/WMWjSfKxBbI/AAAAAAAACN8/LBzcp9HJXfU1wdVKJ_WiZ8s8ya5eDwlQQCLcB/s1600/6.png" /></a></p><p>Equation 13) confirms that when velocity (v) is increased, mass (m') increases. Now, let's take equation 10) and derive equation 14) below: </p><p><a href="https://3.bp.blogspot.com/-OnBGBxx8PJk/WMWjWRThUAI/AAAAAAAACOA/eE-uG-hCypg6FC6tNvZZM4vuYXxSsDRrQCLcB/s1600/7.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-OnBGBxx8PJk/WMWjWRThUAI/AAAAAAAACOA/eE-uG-hCypg6FC6tNvZZM4vuYXxSsDRrQCLcB/s1600/7.png" /></a></p><p>Equation 14) shows why increased velocity increases mass. When the velocity of a system increases, velocity (u) decreases. To conserve momentum, mass (m') must increase. We start with momentum (mc) and end up with (m'u). Of course m'u must always equal mc. But what exactly is this velocity u? I call it the velocity of time. </p><p>The rate of time is the relative speed (u) of a photon or (c^2-v^2)^.5. If a system is moving at velocity (v) and we assume that system is at rest, then the photons in that system may still appear to be moving at c, but relative to v they have slowed to velocity u. Since photons are bosons, their relative speed (how fast they carry force) will determine how fast or how slow the system evolves. If the system is your watch, your watch will noticeably slow down if it moves at a significant fraction of light speed. This suggests that time is real and not just a concept. After all, the original concept of time was that the rate of time is fixed. </p><p>Below is a Feynman diagram where velocity v is zero, so velocity u equals velocity c. At the beginning of time (t), two electrons colide. Next, a photon is emitted, then the electrons fly apart. </p><p><a href="https://1.bp.blogspot.com/-ZiFZ9sdgUsI/WMWzO1cWrCI/AAAAAAAACOQ/WsUGyzQ13rgp3qYkqCY2IDr0PngmrLZygCLcB/s1600/8.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-ZiFZ9sdgUsI/WMWzO1cWrCI/AAAAAAAACOQ/WsUGyzQ13rgp3qYkqCY2IDr0PngmrLZygCLcB/s1600/8.png" /></a></p><p>In the next diagram imagine that the entire diagram is moving through space at velocity v. If we assume the diagram is at rest, the emitted photon will be relatively slower and the measure of time will also be slower: </p><p><a href="https://1.bp.blogspot.com/-pQ8HAlYY39s/WMW0jZzlPJI/AAAAAAAACOc/NQpI4tmWUH4YftPlNXHA_yQhZF7F96u9ACLcB/s1600/9.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-pQ8HAlYY39s/WMW0jZzlPJI/AAAAAAAACOc/NQpI4tmWUH4YftPlNXHA_yQhZF7F96u9ACLcB/s1600/9.png" /></a></p><p>The electrons in the above diagram are moving faster, but photons can't increase their speed, so, relative to the electrons, the photon is slower. This seemingly slow moving photon is the time we measure or proportionate to the time we measure. Now, just for fun, what happens if the electrons go faster than light? </p><p><a href="https://3.bp.blogspot.com/-TpgkhtSiP-E/WMW26H4-tNI/AAAAAAAACOk/QdRLe7XYl1g7fo-60Fj4Xik5ljeX7IqzQCLcB/s1600/10.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-TpgkhtSiP-E/WMW26H4-tNI/AAAAAAAACOk/QdRLe7XYl1g7fo-60Fj4Xik5ljeX7IqzQCLcB/s1600/10.png" /></a></p><p>Time reverses! Your watch is now running backwards. In the diagram above, the particles fly apart, then comes the photon, and finally, the particles collide. </p><p>Before we conclude that time is (or is proportionate to) the speed of photons relative to the other particles in a system, let's look at how gravity impacts time, and take a closer look at light, i.e., electromagnetic waves. Once again we define some variables: </p><p><a href="https://2.bp.blogspot.com/-97pq9-Z_HWk/WMW7MN0M-RI/AAAAAAAACO0/hQt-exprbbIb85r_nag8SkUSlZ7l4CUqwCLcB/s1600/11.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-97pq9-Z_HWk/WMW7MN0M-RI/AAAAAAAACO0/hQt-exprbbIb85r_nag8SkUSlZ7l4CUqwCLcB/s1600/11.png" /></a></p><p>We include the variables for permittivity and permeability. Taken together, they determine the speed of an electromagnetic wave. A photon's relative speed (u) has its own corresponding permittivity and permeability. Free space is a vacuum. That is where light has a velocity of c. If the permittivity of free space, for example, had a lower value, light would go faster. Increasing velocity v causes permittivity and permeability to increase (so does additional mass/energy). As a result, EM waves (light) slow down or are relatively slower. The mathematical proof below provides further insight into the essence of time: </p><p><a href="https://1.bp.blogspot.com/-AWkyL_R5_y8/WMXB5xvcasI/AAAAAAAACPI/n6NBHKgdjckf671ycXbI4oOsMV0_tddMACLcB/s1600/12.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-AWkyL_R5_y8/WMXB5xvcasI/AAAAAAAACPI/n6NBHKgdjckf671ycXbI4oOsMV0_tddMACLcB/s1600/12.png" /></a></p><p><a href="https://4.bp.blogspot.com/-3cQASx87k6Q/WMXB-Xe053I/AAAAAAAACPM/nTtom7s3tF0mPlfl_NR9ftRWHtdOyJ-5wCLcB/s1600/13.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-3cQASx87k6Q/WMXB-Xe053I/AAAAAAAACPM/nTtom7s3tF0mPlfl_NR9ftRWHtdOyJ-5wCLcB/s1600/13.png" /></a></p><p>Equations 23) and 24) show that time is the reciprocal of frequency. Equations 24) and 26) define the rate of time (with variable t[sub o] set to 1) in one of two ways: The ratio of the relative speed (u) of a photon to light speed (c); or, the ratio of permittivities and permeabilities. Both ways are equivalent. To put it more simply, time (at the quantum level) is the measure of the rate bosons can carry force between particles. If that process is disrupted by high speeds or increased mass/energy, that process will slow down. Anything that is a function of that process will also slow down--including your watch. </p><p>Update: Quantum particle-waves are transverse waves: their oscillations are perpendicular to their propagation direction. The following is a mathematical proof that shows that the Lorentz equations above work for transverse waves. </p><p><a href="https://2.bp.blogspot.com/-NrE1YCgUh8o/WMiiG5kAIQI/AAAAAAAACPs/rKUuJDd3O4ozHUdrFC6vh9wOLtO2iQqUgCLcB/s1600/13b.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-NrE1YCgUh8o/WMiiG5kAIQI/AAAAAAAACPs/rKUuJDd3O4ozHUdrFC6vh9wOLtO2iQqUgCLcB/s1600/13b.png" /></a></p><p><a href="https://4.bp.blogspot.com/-A4AfOcTOM9s/WMiiK-DLzRI/AAAAAAAACPw/ZlPD3zlS9F09PewvJgek11dSY17gDaiXwCLcB/s1600/14.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-A4AfOcTOM9s/WMiiK-DLzRI/AAAAAAAACPw/ZlPD3zlS9F09PewvJgek11dSY17gDaiXwCLcB/s1600/14.png" /></a></p><p><a href="https://3.bp.blogspot.com/-RgVEzXfvZG0/WMiiOUH5T4I/AAAAAAAACP0/IrvfgEg-jPM_LoKPQ3hFFEeXwaFUeUWDACLcB/s1600/15.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-RgVEzXfvZG0/WMiiOUH5T4I/AAAAAAAACP0/IrvfgEg-jPM_LoKPQ3hFFEeXwaFUeUWDACLcB/s1600/15.png" /></a></p><p><a href="https://4.bp.blogspot.com/-I9AHpZZbPkw/WMiiSTN4p_I/AAAAAAAACP4/M3IoAL3HxhIsGWxfW4T09Il3I100kwOGACLcB/s1600/16.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-I9AHpZZbPkw/WMiiSTN4p_I/AAAAAAAACP4/M3IoAL3HxhIsGWxfW4T09Il3I100kwOGACLcB/s1600/16.png" /></a></p><p>Equation 37) is the formula used to calculate the velocity of a transverse wave. We were able to derive it from the Lorentz equation for relative mass. This shows that the two are intrinsically connected. Equation 39) predicts what we expect: increased mass (m') reduces the time rate (t'). </p><p>Update: Here are a couple of equations that enable you to calculate time rate (t') when there is light moving between any two stars in a galaxy: </p><p> <a href="https://3.bp.blogspot.com/-x0T-pWZH2Sk/WMs8_YZ9uKI/AAAAAAAACRI/mykPQzlRPGk7euxrqaJp135NfnNGtzYowCLcB/s1600/16b.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-x0T-pWZH2Sk/WMs8_YZ9uKI/AAAAAAAACRI/mykPQzlRPGk7euxrqaJp135NfnNGtzYowCLcB/s1600/16b.png" /></a></p><p><a href="https://1.bp.blogspot.com/-XphvfpotvP4/WMs9N9uKdoI/AAAAAAAACRM/ujtBmmMC21YS5h9o4ngThCxv6wCO1LpFQCLcB/s1600/17.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-XphvfpotvP4/WMs9N9uKdoI/AAAAAAAACRM/ujtBmmMC21YS5h9o4ngThCxv6wCO1LpFQCLcB/s1600/17.png" /></a></p><p><a href="https://3.bp.blogspot.com/-XMRjTN_kqXs/WMmd50lBg7I/AAAAAAAACQc/2RaIdrApNygFFCsDgcH_O17Nl9iZEy3ygCLcB/s1600/18.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-XMRjTN_kqXs/WMmd50lBg7I/AAAAAAAACQc/2RaIdrApNygFFCsDgcH_O17Nl9iZEy3ygCLcB/s1600/18.png" /></a></p><p>Diagram A above shows light moving between two points at a 90 degree angle to velocity v. The first term of equation 40) drops out and the equation reduces to the Pythagorean theorem used for the Lorentz transform. If the light and the galaxy are going in the same direction or opposite directions (zero degrees), then equation 40's second term drops out. The equation reduces to a linear relation between c and v which always equals c according to the math below: </p><p><a href="https://4.bp.blogspot.com/-R5FaaINCloY/WMs-39_cDNI/AAAAAAAACRY/niDiHia90qcWMnXMJh70_XJl34mlfSebQCLcB/s1600/17b.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-R5FaaINCloY/WMs-39_cDNI/AAAAAAAACRY/niDiHia90qcWMnXMJh70_XJl34mlfSebQCLcB/s1600/17b.png" /></a></p><p><a href="https://1.bp.blogspot.com/-7S0YELOjTN8/WMtANvVweqI/AAAAAAAACRk/Y1cgwe6SnrouofC3dHrMxVxNGxG8mvCYgCLcB/s1600/17c.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-7S0YELOjTN8/WMtANvVweqI/AAAAAAAACRk/Y1cgwe6SnrouofC3dHrMxVxNGxG8mvCYgCLcB/s1600/17c.png" /></a></p><p>Angles between zero and 90 degrees yield some combination of both equation-40) terms. </p><p>Diagram B above assumes the galaxy is rotating and moving to the right at varied angles to the direction of the light. We take the sum of all those angles to get the average angle. We calculate u and plug it into equation 41) to get time rate t'. </p><p> </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-2068813013258741672017-02-27T18:26:00.001-08:002017-06-24T17:08:53.736-07:00Using Fluid Mechanics to Explain Expanding Spacetime and Gravity<p><iframe width="560" height="315" src="https://www.youtube.com/embed/IAWsEr4CyjY" frameborder="0" allowfullscreen></iframe></p><p>The above video is an excellent demonstration of Bernoulli's principle. Notice how the current flows to the right and how the leaves get caught in the eddies. The current is slowed by the obstacles and this causes a back flow. Now imagine a current of expanding spacetime flowing in all directions, and matter swirling in eddies we call galaxies. The outward flow is slowed by the presence of matter--there is a back flow we call gravity. </p><p>Below is Bernoulli's equation along with a couple of diagrams. The top diagram shows a boat going merrily down an unimpeded stream. Pressure (P) is zero. This is analogous to an expanding universe devoid of matter. The second diagram shows a stream with a big black boulder in the way. It slows the stream down. The current pushes and brushes the boulder, so pressure (P) is greater than zero. There is a back flow indicated by the red arrow. The boat is pushed upstream to the boulder. This is analogous to matter in an expanding universe causing a gravitational effect. </p><p><a href="https://1.bp.blogspot.com/-b77bXyPFjtk/WLTPx2IRraI/AAAAAAAACI8/h0oYKL7QgQgB6uDH2l1_AFhXQH31s4NTwCLcB/s1600/1.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-b77bXyPFjtk/WLTPx2IRraI/AAAAAAAACI8/h0oYKL7QgQgB6uDH2l1_AFhXQH31s4NTwCLcB/s1600/1.png" /></a></p><p>Bernoulli's equation isn't very practical if applied to spacetime and gravity. Bernoulli didn't take into account relativity, for instance. However, we learn a very useful concept from Bernoulli: Mass causes pressure to rise and the current velocity (v) to slow. If pressure falls, velocity increases, but in this case, pressure can only fall if mass is reduced. Kind of sounds like momentum conservation, doesn't it? </p><p>In the diagram below we feature a momentum equation. If spacetime were devoid of all matter, the outflow velocity (v) would equal the back flow velocity (c) (see arrows). The mass pressure density (pm) would equal the vacuum mass density (ps). There would be no gravity--just expansion. </p><p><a href="https://3.bp.blogspot.com/-87MO-9yhbn0/WLTVBOXzOMI/AAAAAAAACJM/hxLHuy7LbNEVCF8Ms0GkKyq4wcim0YSNQCLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-87MO-9yhbn0/WLTVBOXzOMI/AAAAAAAACJM/hxLHuy7LbNEVCF8Ms0GkKyq4wcim0YSNQCLcB/s1600/2.png" /></a></p><p>The next diagram shows a universe with a black hole. Expanding spacetime flows around and puts pressure on this black hole. To conserve momentum, the outflow velocity is reduced. The result is a back flow indicated by the arrows in red. We have gravity. </p><p><a href="https://2.bp.blogspot.com/-nN11xQPMLA0/WLTXdomvuhI/AAAAAAAACJY/qE5pYSypUW4Wy9CSuhy4U0iLSJRjNHKzwCLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-nN11xQPMLA0/WLTXdomvuhI/AAAAAAAACJY/qE5pYSypUW4Wy9CSuhy4U0iLSJRjNHKzwCLcB/s1600/3.png" /></a></p><p>The momentum equation seems like a pretty good model for spacetime and gravity, but is it valid? Let's see if we can prove its validity. First let's define some variables: </p><p><a href="https://2.bp.blogspot.com/-CDLlWMvlc-g/WLTafgnWGSI/AAAAAAAACJk/PFIpP2LBac43kiifYIKD3pI-VsA76MkkACLcB/s1600/4.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-CDLlWMvlc-g/WLTafgnWGSI/AAAAAAAACJk/PFIpP2LBac43kiifYIKD3pI-VsA76MkkACLcB/s1600/4.png" /></a></p><p>And here's the proof: </p><p><a href="https://1.bp.blogspot.com/-Pzt6RogYm2c/WLTbdvSbIPI/AAAAAAAACJw/98MTK7gcUxcZDUVQYTAmcSFQRmev72fJgCLcB/s1600/5.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-Pzt6RogYm2c/WLTbdvSbIPI/AAAAAAAACJw/98MTK7gcUxcZDUVQYTAmcSFQRmev72fJgCLcB/s1600/5.png" /></a></p><p><a href="https://2.bp.blogspot.com/-rSXgNC9vzQY/WLTbhwlshiI/AAAAAAAACJ0/exzX0eSfEqAWFWobcVsDOPpgk1soAx2NgCLcB/s1600/6.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-rSXgNC9vzQY/WLTbhwlshiI/AAAAAAAACJ0/exzX0eSfEqAWFWobcVsDOPpgk1soAx2NgCLcB/s1600/6.png" /></a></p><p><a href="https://2.bp.blogspot.com/-mNtFU3svWCs/WLTbl-ggyXI/AAAAAAAACJ4/DXZuoqthozwafGB4RbBU7mZeFy-oNwi1wCLcB/s1600/7.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-mNtFU3svWCs/WLTbl-ggyXI/AAAAAAAACJ4/DXZuoqthozwafGB4RbBU7mZeFy-oNwi1wCLcB/s1600/7.png" /></a></p><p><a href="https://1.bp.blogspot.com/-d6ebQVq6ArQ/WLTbpgCreNI/AAAAAAAACJ8/uabcip8U_AkTyicSZv3EmSVbCIjSW_JegCLcB/s1600/8.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-d6ebQVq6ArQ/WLTbpgCreNI/AAAAAAAACJ8/uabcip8U_AkTyicSZv3EmSVbCIjSW_JegCLcB/s1600/8.png" /></a></p><p>Within the proof we derived another useful equation: </p><p><a href="https://4.bp.blogspot.com/-ik9YLNgg1Qc/WLTc5Y3iw6I/AAAAAAAACKI/dDekKbyLFjIcf6AJ0YouLRGu2JBlnFtgQCLcB/s1600/9.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-ik9YLNgg1Qc/WLTc5Y3iw6I/AAAAAAAACKI/dDekKbyLFjIcf6AJ0YouLRGu2JBlnFtgQCLcB/s1600/9.png" /></a></p><p>The above equation tells us the instant gravitational velocity squared (gx) at a given location (x)--or, fluid-mechanically speaking, the net back flow rate squared. </p><p>Update: The following proof shows that spacetime momentum is conserved but gravitational momentum is not: </p><p><a href="https://2.bp.blogspot.com/-Ju3roXV5lZU/WLmr023I6OI/AAAAAAAACKc/ILEYpTjp9T4K0X8ma_ymk1sSBpC6duixwCLcB/s1600/10.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-Ju3roXV5lZU/WLmr023I6OI/AAAAAAAACKc/ILEYpTjp9T4K0X8ma_ymk1sSBpC6duixwCLcB/s1600/10.png" /></a></p><p><a href="https://2.bp.blogspot.com/-DIMgewKMnAA/WLmr6cWgB3I/AAAAAAAACKg/Qa-P2B2z4b4t5s2r8P-1zg5TM6jCJBIygCLcB/s1600/11.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-DIMgewKMnAA/WLmr6cWgB3I/AAAAAAAACKg/Qa-P2B2z4b4t5s2r8P-1zg5TM6jCJBIygCLcB/s1600/11.png" /></a></p><p>The first equation above clearly shows that any change to pressure mass density (pm) causes outflow velocity (vo) to change, so momentum is conserved. The last equation above shows that any change in a falling mass (m) has no effect on velocity (v), so different masses fall at the same rate in a gravitational field. </p><p>Update: The above math suggests that spacetime expands at a steady rate up to light speed (c) and is slowed by the presence of matter. What about Hubble's constant and expansion velocity v = Hr? </p><p><a href="https://3.bp.blogspot.com/-dD2ZhzyWvzE/WLm_X1LXQeI/AAAAAAAACK8/oDbhlUhyQmEkpL9-9KLFT2daFEhKoesSQCLcB/s1600/12.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-dD2ZhzyWvzE/WLm_X1LXQeI/AAAAAAAACK8/oDbhlUhyQmEkpL9-9KLFT2daFEhKoesSQCLcB/s1600/12.png" /></a></p><p>How fast spacetime expands depends on whether you use Hubble's constant or time rate (t'). Time rate (t') grows as spacetime grows. Hubble's constant does not. The diagram below shows that spacetime can expand no more than light speed but appear to be expanding at an accelerated rate. </p><p><a href="https://1.bp.blogspot.com/-SfaRQlVq5Xs/WLnDdCMK9SI/AAAAAAAACLI/CXUGarH_XGYRuFZbukoeKSWZwDYEFhCWQCLcB/s1600/13.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-SfaRQlVq5Xs/WLnDdCMK9SI/AAAAAAAACLI/CXUGarH_XGYRuFZbukoeKSWZwDYEFhCWQCLcB/s1600/13.png" /></a></p><p>If we assume all the dots above are in motion, the expansion is a steady rate. If we assume that the red dot is at rest, then the blue dot appears to be moving away at light speed, and the green dot appears to be moving twice light speed! To get the right numbers for gravity, we need to go with the first assumption: spacetime expands at a steady maximum rate of light speed when it is not slowed by matter. </p><p>Update: The following equation was derived above. </p><p><a href="https://3.bp.blogspot.com/-YjtGzfO-5tE/WLxpdNaSyXI/AAAAAAAACLw/lBD5h9YWrL4ecKLGQ4YL9iOczaGZ6kScQCLcB/s1600/14.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-YjtGzfO-5tE/WLxpdNaSyXI/AAAAAAAACLw/lBD5h9YWrL4ecKLGQ4YL9iOczaGZ6kScQCLcB/s1600/14.png" /></a></p><p>The diagram below shows spacetime expanding in a gravitational field (see arrows). The net backflow or gx is c^2-v^2. As mentioned earlier, the presence of matter (a planet, star, etc.) slows the outflow rate of v. </p><p><a href="https://3.bp.blogspot.com/-vjMItEQy9X8/WLxrOpj3rgI/AAAAAAAACL4/Wr3LmxBdRxEZSkO5tVKpRgptJ9mjwLbNACLcB/s1600/15.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-vjMItEQy9X8/WLxrOpj3rgI/AAAAAAAACL4/Wr3LmxBdRxEZSkO5tVKpRgptJ9mjwLbNACLcB/s1600/15.png" /></a></p><p>If any mass is falling in this gravitational field, we could use these equations to describe it: </p><p><a href="https://2.bp.blogspot.com/-ObDviXS4oi0/WLxs480YQbI/AAAAAAAACME/d3uzLoekB_0PtEsF40fedH450QfqOeOGwCLcB/s1600/16.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-ObDviXS4oi0/WLxs480YQbI/AAAAAAAACME/d3uzLoekB_0PtEsF40fedH450QfqOeOGwCLcB/s1600/16.png" /></a></p><p>We could also use the diagrams below: </p><p><a href="https://1.bp.blogspot.com/-HjcHNa0E76E/WLxuS9y_xiI/AAAAAAAACMM/5_lR3Mf1hU8ea8QHGyPAb2Z1jyI6gFhIwCLcB/s1600/17.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-HjcHNa0E76E/WLxuS9y_xiI/AAAAAAAACMM/5_lR3Mf1hU8ea8QHGyPAb2Z1jyI6gFhIwCLcB/s1600/17.png" /></a></p><p>Diagram 1) above represents a small mass falling in a gravitational field. The variable u is the spacetime outflow velocity where the falling mass is located at a given moment of time. Diagram 2) represents a larger mass falling in the same field. Its corresponding outflow velocity is slower (indicated by a shorter double arrow). If we do the math for each mass, we find that the total falling velocity-squared (gx) is the same for both. This once again confirms that different masses fall at the same rate. </p><p>Update: The equations below include the dark matter effects of spacetime. As volume (V) grows, the significance of spacetime mass (psV) becomes more significant and pressure mass (increased spacetime mass caused by matter) becomes less significant. At galactic scales and beyond, most of the gravity is caused by spacetime rather than matter. </p><p><a href="https://2.bp.blogspot.com/-l0596BcYd64/WMBqBNtqXdI/AAAAAAAACMo/PXqAgl6DL-wNq1ETXa8-jveN9myBUZrUQCLcB/s1600/M3.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-l0596BcYd64/WMBqBNtqXdI/AAAAAAAACMo/PXqAgl6DL-wNq1ETXa8-jveN9myBUZrUQCLcB/s1600/M3.png" /></a></p><p> </p><p> GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-78231361652251340032017-02-12T17:48:00.001-08:002017-06-24T17:12:28.603-07:00The Dark Matter and Dark Energy Effects of Spacetime<p><iframe width="560" height="315" src="https://www.youtube.com/embed/QAa2O_8wBUQ" frameborder="0" allowfullscreen></iframe></p><p>Alice and Bob don't always agree. Alice is outside our universe, looking at the big picture. She sees the universe expanding at a steady rate, like a balloon attached to a helium tank. From her point of view, what we classically think of as energy is conserved. Here's what she sees at time one (t1): </p><p><a href="https://2.bp.blogspot.com/-N7obz5cAS6E/WKDf3hVF42I/AAAAAAAACC4/XfKdebtsyqcJr5WSg6iY44jdSGk7wK4XQCLcB/s1600/1.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-N7obz5cAS6E/WKDf3hVF42I/AAAAAAAACC4/XfKdebtsyqcJr5WSg6iY44jdSGk7wK4XQCLcB/s1600/1.png" /></a></p><p>Here's what she sees at time two (t2): </p><p><a href="https://2.bp.blogspot.com/-XGOfT-B3_o0/WKDgUxIdX-I/AAAAAAAACC8/l8nNGPpYLMEcI4iBt4n81mVOZFL_dmalwCLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-XGOfT-B3_o0/WKDgUxIdX-I/AAAAAAAACC8/l8nNGPpYLMEcI4iBt4n81mVOZFL_dmalwCLcB/s1600/2.png" /></a></p><p>Bob has a different take. He's on earth looking outward. He sees the universe expanding at an accelerated rate--the red shift of distant galaxies is greater than that of closer galaxies. Energy is not conserved--it is increasing! Here's what he sees at t1 and t2: </p><p><a href="https://4.bp.blogspot.com/-NfARUx00cgU/WKDiFB60v-I/AAAAAAAACDM/8iMDSFv7Ujc-LtseIlhBLUq0sxZWqqJsQCLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-NfARUx00cgU/WKDiFB60v-I/AAAAAAAACDM/8iMDSFv7Ujc-LtseIlhBLUq0sxZWqqJsQCLcB/s1600/3.png" /></a></p><p>In the diagram above, the blue dot represents Bob's galaxy. The red dot is a galaxy far far away. As far as Bob is concerned, that red dot is moving the fastest. Bob uses equation 1) below to model what he sees; whereas, Alice uses equation 2): </p><p><a href="https://1.bp.blogspot.com/-j6qTXglIv9g/WKDlKIRhVBI/AAAAAAAACDY/ysqnR9DgZag8Kx266Qo4-hAHMcpJ5ACXgCLcB/s1600/4.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-j6qTXglIv9g/WKDlKIRhVBI/AAAAAAAACDY/ysqnR9DgZag8Kx266Qo4-hAHMcpJ5ACXgCLcB/s1600/4.png" /></a></p><p>Equation 1) shows velocity increasing as the radius (r) increases. However, radius (r) fails to tell us the effect gravity has on time. Where gravity is strong, time runs more slowly. Where gravity is weak, time runs faster. Gravity is weaker where the radius is larger, and vice versa. So we can substitute ct' for r in equation 1). </p><p>Alice sees the universe expanding at a steady rate. However, we can change this to ct'/t' in equation 2) to show that time (t') cancels itself. By contrast, Bob, uses Hubble's constant, a fixed time--it doesn't cancel time (t'). As a result, Alice sees a short expansion over a short time where gravity is strong, and a long expansion over a long time where gravity is weak. She sees a steady expansion velocity. Bob sees a slower velocity where gravity is strong and a faster velocity where gravity is weak. </p><p>Both Alice and Bob notice that spacetime has energy density or pressure. They both use the following equations: </p><p><a href="https://2.bp.blogspot.com/-0ApMHCV4Q3o/WKDsf8BVNHI/AAAAAAAACDo/BynPk_p1itsqIkKV2KqPyeRuDn0KVLE0gCLcB/s1600/5.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-0ApMHCV4Q3o/WKDsf8BVNHI/AAAAAAAACDo/BynPk_p1itsqIkKV2KqPyeRuDn0KVLE0gCLcB/s1600/5.png" /></a></p><p>Normally when there is pressure, the volume increases and that relieves the pressure. In the case of space, increasing volume just adds more energy. This keeps the pressure constant, so expansion is continuous. (Equation 5) above shows the volume (V) cancelling itself.) </p><p>Notwithstanding the added energy, Alice sees conserved energy. Using Einstein's field equations, we can derive something that shows why. </p><p><a href="https://1.bp.blogspot.com/--EIdzDnNaCo/WKDv4dUjiHI/AAAAAAAACD0/-2znrNC-GWcaVG5rtl2oUXiL0BJZP6fCwCLcB/s1600/6.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/--EIdzDnNaCo/WKDv4dUjiHI/AAAAAAAACD0/-2znrNC-GWcaVG5rtl2oUXiL0BJZP6fCwCLcB/s1600/6.png" /></a></p><p><a href="https://4.bp.blogspot.com/-vZjKUAZmNMA/WKDweIiT29I/AAAAAAAACD8/UT3XAIbdDbo-HUDY3NJHycyPxl7sVC7XgCLcB/s1600/7.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-vZjKUAZmNMA/WKDweIiT29I/AAAAAAAACD8/UT3XAIbdDbo-HUDY3NJHycyPxl7sVC7XgCLcB/s1600/7.png" /></a></p><p><a href="https://1.bp.blogspot.com/-oUE1x7UGZfg/WKDxRu59xJI/AAAAAAAACEI/AMfK8rMsvJoZFrQKBxaPjLigHF4BHJtygCLcB/s1600/8.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-oUE1x7UGZfg/WKDxRu59xJI/AAAAAAAACEI/AMfK8rMsvJoZFrQKBxaPjLigHF4BHJtygCLcB/s1600/8.png" /></a></p><p>At equation 14), when the spacetime mass (ms)in the numerator increases, so does V * bar-lambda in the denominator. The increase in spacetime curvature caused by spacetime mass(energy) cancels or is always proportionate to the spacetime mass. Also, when volume (V) increases, spacetime curvature (K) decreases, so V * K is constant. The remaining variables are also constant. Thus, energy (E) is conserved. </p><p>Bob, of course, disagrees. He says the energy is growing. If we perform an operation on the conserved-energy equation, we can see why: </p><p><a href="https://1.bp.blogspot.com/-BWyS1rz2AVo/WKD2NJwlz7I/AAAAAAAACEc/NNAhgq3OuJ0rsYWnUjhw086k1f78ntrRQCLcB/s1600/9.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-BWyS1rz2AVo/WKD2NJwlz7I/AAAAAAAACEc/NNAhgq3OuJ0rsYWnUjhw086k1f78ntrRQCLcB/s1600/9.png" /></a></p><p>According to equation 17), as the universe's radius (r) increases, so does energy (E'). So who's right? Alice or Bob? Answer: they both are. What we observe depends on our frame of reference and whether we use Hubble's constant or time (t'). What Alice and Bob observed can be conveniently labeled the dark energy effects of spacetime. </p><p>Equation 15) reveals the dark matter effect. We mentioned earlier, when volume (V) grows, so does spacetime mass (ms). Variable m, however, remains constant. This means spacetime mass becomes more significant at greater volumes and matter mass becomes less significant. When we crunch the numbers, spacetime makes up most of the mass and energy within the volume of the known universe. It follows that it would cause most of the universe's gravity. </p><p>One possible mechanism for the extra gravity we observe is spacetime's expansion pressure. Imagine a weightless environment. Imagine a transparent balloon being filled with gas. Floating in the middle of that balloon is a marble. The gas pressure presses outward against the balloon's inner surface. It expands the balloon, but the pressure goes inward against the marble's surface as well. </p><p><a href="https://1.bp.blogspot.com/-VkOmAoZz9_E/WKD932BYpiI/AAAAAAAACEw/nEvJzaujezIxewAKEpCg6uDcJdHlspnewCLcB/s1600/10.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-VkOmAoZz9_E/WKD932BYpiI/AAAAAAAACEw/nEvJzaujezIxewAKEpCg6uDcJdHlspnewCLcB/s1600/10.png" /></a></p><p>The arrows in the above diagram represent the pressure going outward and inward. If the marble is a metaphor for a galaxy, then, in addition to gravity caused by matter, the galaxy is receiving pressure from the outside. There is also counter-pressure from within. This could give the impression of additional gravity. </p><p>Another cause of additional gravity is time (t'). Since spacetime adds more mass, the rate of time must be slower, and slower time should produce some gravitational effects. </p><p>One has to wonder, though: if the galaxies had enough gravity to attract each other, would the universe still expand? Back to the balloon. Imagine the balloon has a bunch of marbles floating inside it. They are held together by strings. Hot gas is pumped into the balloon. The balloon expands anyway. The hot gas flows around the marbles' surfaces and creates pressure on them and between them. If the strings are strong enough, the marbles won't separate. If the strings are weak, the marbles may separate and go with the flow of the hot gas in the expanding balloon. Our universe may work the same way. </p><p><a href="https://3.bp.blogspot.com/-7d2O7oET-Cs/WKELkjXW6lI/AAAAAAAACFE/KdfAfKYadPMVOnSikU4Bg2VC7lm9mVESwCLcB/s1600/11.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-7d2O7oET-Cs/WKELkjXW6lI/AAAAAAAACFE/KdfAfKYadPMVOnSikU4Bg2VC7lm9mVESwCLcB/s1600/11.png" /></a></p><p>Update: Here are the complete equations that model the dark matter and dark energy effects of spacetime. First we introduce some new variables: </p><p><a href="https://4.bp.blogspot.com/-mxdEbtisxW8/WKSaDGkzt4I/AAAAAAAACF0/Qx_9XPYYRaYQ9r2G2C5otPGuL43voHSMwCLcB/s1600/17.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-mxdEbtisxW8/WKSaDGkzt4I/AAAAAAAACF0/Qx_9XPYYRaYQ9r2G2C5otPGuL43voHSMwCLcB/s1600/17.png" /></a></p><p>To conserve energy we assume expanding spacetime goes in equal and opposite directions. These equal and opposite directions cancel each other. We use +/- signs to indicate that. </p><p><a href="https://3.bp.blogspot.com/-zZR6NALpLc4/WKSaucG493I/AAAAAAAACF4/3PMFUvx8_hcJXCyLICyggPB-uhkMmjp5ACLcB/s1600/13.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-zZR6NALpLc4/WKSaucG493I/AAAAAAAACF4/3PMFUvx8_hcJXCyLICyggPB-uhkMmjp5ACLcB/s1600/13.png" /></a></p><p>In the diagram above we arbitrarily label one half the spacetime volume (V) on the left as "-" and the right half as "+." Now here are the equations: </p><p><a href="https://3.bp.blogspot.com/-JvT8lVszdlg/WKSblKp4K9I/AAAAAAAACGI/t_c-s-cghy0Eq_XJV0JW5Bo4BLIyCihVgCLcB/s1600/14.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-JvT8lVszdlg/WKSblKp4K9I/AAAAAAAACGI/t_c-s-cghy0Eq_XJV0JW5Bo4BLIyCihVgCLcB/s1600/14.png" /></a></p><p>The following equation was designed to show that energy is truly conserved. No matter how big or small spacetime energy (Es) gets, overall energy is conserved. Es appears in both the numerator and denominater; it is the energy of both space and time, so it cancels itself. </p><p><a href="https://1.bp.blogspot.com/-ZY1LeNMLRc8/WKSgXnw1CaI/AAAAAAAACGs/OGsb9ilIgVoXdSFees_H2l5_Q8qWBx06QCLcB/s1600/15.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-ZY1LeNMLRc8/WKSgXnw1CaI/AAAAAAAACGs/OGsb9ilIgVoXdSFees_H2l5_Q8qWBx06QCLcB/s1600/15.png" /></a></p><p>Although, Bob insists that energy overall is increasing, so his equation is as follows: </p><p><a href="https://3.bp.blogspot.com/-eMr0o_AtUbg/WKSdw2XISAI/AAAAAAAACGc/WmHS8XN7GF8T4TQ20Y2GDb1Wytvp2s9iQCLcB/s1600/16.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-eMr0o_AtUbg/WKSdw2XISAI/AAAAAAAACGc/WmHS8XN7GF8T4TQ20Y2GDb1Wytvp2s9iQCLcB/s1600/16.png" /></a></p><p></p><p> </p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-40566178727464926192017-02-03T18:25:00.005-08:002017-05-07T09:29:41.832-07:00Testing My Dark Matter Hypothesis<p><iframe width="560" height="315" src="https://www.youtube.com/embed/lC3J8xXjHYk" frameborder="0" allowfullscreen></iframe></p><p></p><p>According to my hypothesis, dark matter isn't matter at all. This is why I believe Isaac Newton didn't discover it and work it into his famous equation: </p><p><a href="https://2.bp.blogspot.com/-7XmA_TZxudo/WJUlNIlsAkI/AAAAAAAAB-s/cMKN4Z65XIgt10IbbxdaYvsM8S1yDibngCLcB/s1600/1.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-7XmA_TZxudo/WJUlNIlsAkI/AAAAAAAAB-s/cMKN4Z65XIgt10IbbxdaYvsM8S1yDibngCLcB/s1600/1.png" /></a></p><p>Had he known that time, space or spacetime has ground-state (zero-point) energy he might have included a tiny correction term in his equation. Then again, the correction is so small it is insignificant at a local scale--hence the reason his equation works so well at such a scale. A major clue about the nature of so-called dark matter can be found in the following quote: </p><p> <a href="http://assets.zombal.com/03a5cab4/GalaxyDensity.pdf" target="blank">"When you go out to much larger distances, you encapsulate the galactic halo and a lot more dark matter."--Darin Ragozzine.</a> </p><p>Yes, if you go out further into space you find more (cough) dark matter. This makes perfect sense if dark matter isn't matter, but spacetime. According to the theory, dark matter makes up around 85% of all matter. If this is true why the heck aren't we swimming in it? Don't you find it strange that something so plentiful should be so elusive? Here's a crude illustration of how "dark matter" comes into play: </p><p><a href="https://2.bp.blogspot.com/-OptBLFz1780/WJUqOAx-oMI/AAAAAAAAB-8/TGFU1DJlYZQdbVdWxW884feTVqDHB2XZACLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-OptBLFz1780/WJUqOAx-oMI/AAAAAAAAB-8/TGFU1DJlYZQdbVdWxW884feTVqDHB2XZACLcB/s1600/2.png" /></a></p><p>Note the big black dot at the center, and take note of the little dots surrounding it. Let's pretend the big black dot is what we call ordinary, visible matter, and the little dots are dark matter. Within the smaller circle, the big dot accounts for most of the density. The little dots are insignificant within that region. There, Newton's equation works just fine. But when we go out further along radius (r), the little dots make up most of the density--the big dot becomes more and more insignificant. </p><p>So clearly this dark matter stuff doesn't make itself known to us at the local scale. It takes a greater region of "space" before we begin to notice its effects. To test my hypothesis, I took Einstein's field equations and did some algebra to put the variables in scalar form and to express them in terms of energy density. Why? Because energy density causes gravity (curved spacetime) and it is easier to find and plug in actual data (which is in scalar form). </p><p><a href="https://4.bp.blogspot.com/-4u5sQIokkNQ/WJUvPqXteaI/AAAAAAAAB_M/j_ZYEgKcN3E44A9ReMoBo6HuglEl2fp3gCLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-4u5sQIokkNQ/WJUvPqXteaI/AAAAAAAAB_M/j_ZYEgKcN3E44A9ReMoBo6HuglEl2fp3gCLcB/s1600/3.png" /></a></p><p>At equation 4), we have the sum of two energy densities: spacetime (Ts) and visible matter (Tm). Together they make the total energy density (T). If dark matter is in fact spacetime, then we expect its energy density to be insignificant locally and more significant at large scales. Equation 5) reveals that spacetime's energy density is a very small number. Let's add it to and compare it with earth's energy density: </p><p><a href="https://3.bp.blogspot.com/-lb5G35qpXa8/WJU1BxVqlKI/AAAAAAAAB_c/go9j0Bw4NXsOroD_ftqz_YwS8obCktXKwCLcB/s1600/4.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-lb5G35qpXa8/WJU1BxVqlKI/AAAAAAAAB_c/go9j0Bw4NXsOroD_ftqz_YwS8obCktXKwCLcB/s1600/4.png" /></a></p><p>The above equations would make Newton very happy. Earth's energy density (ED) is huge compared to spacetime's. In fact, spacetime's ED is so insignificant, we can ignore it when determining earth's gravity. But what about the galaxy's gravity? </p><p><a href="https://1.bp.blogspot.com/-ryG9V_i96uw/WJU23OSO1hI/AAAAAAAAB_o/N3czPG6azBYvZOBniiRPXmuPlPa1NHLLACLcB/s1600/5.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-ryG9V_i96uw/WJU23OSO1hI/AAAAAAAAB_o/N3czPG6azBYvZOBniiRPXmuPlPa1NHLLACLcB/s1600/5.png" /></a></p><p>Ah! The tables have turned! Spacetime's energy density now causes a significant portion of the gravity. There is more than one measurement for our galaxy's energy density, depending on the method and the source. The range I found is what you see at equation 8). Taking this range into account, equation 9) shows that spacetime accounts for approximately ten percent to nearly 100% of the total ED. By contrast, "ordinary matter" constitutes approximately zero to 90% of the total ED. This confirms that spacetime has a huge impact on gravity at galactic scales. Dark matter has been under our noses all along. You can find it in any empty space. </p><p>Update: <a href="http://sci-technews99.blogspot.com/2017/05/astronomers-have-discovered-invisible.html" target="blank">Scientists believe they have discovered a dark matter galaxy. Click here for details.</a></p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com4tag:blogger.com,1999:blog-3628542335168231161.post-42653929556193887322017-02-01T18:33:00.000-08:002017-06-18T10:26:13.868-07:00What Dark Matter Really Is<p><iframe width="560" height="315" src="https://www.youtube.com/embed/e4nnpg4N35o" frameborder="0" allowfullscreen></iframe></p><p>What exactly is dark matter? If you watch the above video, you will be introduced to some strange and bizarre theories involving parallel universes and higher vibrating string octaves. Ordinary matter doesn't explain the amount of gravity observed. It is assumed that galaxy clusters, and the whole universe contain far more matter than can be observed using electromagnetic signals. Such an assumption is based on the theory of General Relativity which propounds that gravity is caused by the presence of matter. </p><p>Matter and energy density curve spacetime and spacetime tells matter how to move. Gravitational acceleration (g) is currently understood and defined by the following equations (a=acceleration; t=time; x=distance; r=radius; G=Newton's constant; m=mass; gij=metric tensor; T=stress-energy tensor; v=velocity; gamma=Christoffel symbol): </p><p><a href="https://1.bp.blogspot.com/-Iy5Ov-D9EfY/WUamTE6E3uI/AAAAAAAAC0U/QTIOlbCLL1MPCeTWS8KlyCe3OYatzov3wCLcBGAs/s1600/1.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-Iy5Ov-D9EfY/WUamTE6E3uI/AAAAAAAAC0U/QTIOlbCLL1MPCeTWS8KlyCe3OYatzov3wCLcBGAs/s1600/1.png" data-original-width="576" data-original-height="793" /></a></p><p>It is believed that if no matter and/or energy is present, spacetime is flat and there is no gravity. Given this belief, it is not surprising that a theory of dark matter would emerge. There must be something out there responsible for all that extra gravity. But whatever it is, it does not behave like ordinary matter. In fact, it is undetectable--hence the name: dark matter. </p><p>As of this writing, no dark matter particle has been conclusively identified. I am going to go out on a limb here and make a prediction: no dark matter particle will ever be discovered. "Dark matter" is not matter at all. Something else is causing the unexplained gravity. Let me illustrate. Let's start with Einstein's field equations: </p><p><a href="https://2.bp.blogspot.com/-u5VluGYtQu8/WJKKrid9NWI/AAAAAAAAB9M/7qIBGHLqaaIa0zcuDQLSrCeQFOitRcWJACLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-u5VluGYtQu8/WJKKrid9NWI/AAAAAAAAB9M/7qIBGHLqaaIa0zcuDQLSrCeQFOitRcWJACLcB/s1600/2.png" /></a></p><p>The left side contains Einstein's tensor (Gii) which is the measure of the spacetime curve. The right side contains the stress-energy tensor (Tii) which measures energy density. When energy density is zero, spacetime curvature is also zero--no gravity. Now let's turn the equation into a constant (c^4: light speed): </p><p><a href="https://4.bp.blogspot.com/-SWDKBAQd5Fw/WJKNAyJW2ZI/AAAAAAAAB9Y/hErgvOyFmNwBdItYxmEeIUcVaZUjbzEqwCLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-SWDKBAQd5Fw/WJKNAyJW2ZI/AAAAAAAAB9Y/hErgvOyFmNwBdItYxmEeIUcVaZUjbzEqwCLcB/s1600/3.png" /></a></p><p>In the equation above, any change to the right side has no effect on the left side. Why is this important? You will see. Let's continue: </p><p><a href="https://4.bp.blogspot.com/-DTLmSqIRGmU/WJKOUF33fkI/AAAAAAAAB9k/5IL-rdsSg28ksH-sGQ9av_nC2yDhzmQBwCLcB/s1600/4.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-DTLmSqIRGmU/WJKOUF33fkI/AAAAAAAAB9k/5IL-rdsSg28ksH-sGQ9av_nC2yDhzmQBwCLcB/s1600/4.png" /></a></p><p>Take a close look at equation 4) above. The left side is an acceleration term. If time (t') is held constant, then no changes we make on the right side will change the acceleration term. What we have is acceleration that is independent of matter and energy. This is significant; however, we know that gravitational acceleration is not independent of matter and energy--or is it? Let's keep going. We need to define what we mean by time (t'): </p><p><a href="https://2.bp.blogspot.com/-7HzOpF9CcxQ/WJKQjjWKXvI/AAAAAAAAB9w/9CxuggvvX7Mba5bs4A7mFyhXi3t8fG_uwCLcB/s1600/5.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-7HzOpF9CcxQ/WJKQjjWKXvI/AAAAAAAAB9w/9CxuggvvX7Mba5bs4A7mFyhXi3t8fG_uwCLcB/s1600/5.png" /></a></p><p>Time (t') is time (t) multiplied by a Lorentz factor. We make the substitution in equation 8) above. Since both sides are acceleration terms, we can set the left side equal to gravitational acceleration (g): </p><p><a href="https://2.bp.blogspot.com/-yLvfAiwBPlA/WJKSHPr4mCI/AAAAAAAAB98/g5eCToJFpY809__OP2cH8mWo0BQTo4GMgCLcB/s1600/6.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-yLvfAiwBPlA/WJKSHPr4mCI/AAAAAAAAB98/g5eCToJFpY809__OP2cH8mWo0BQTo4GMgCLcB/s1600/6.png" /></a></p><p>According to equation 9), gravitational acceleration is not independent of changes in energy/matter (E). When E increases, so does g and when E decreases, so does g. But what happens if E is zero? </p><p><a href="https://2.bp.blogspot.com/-Z8GJPUaMXPY/WJKUL1AbIHI/AAAAAAAAB-I/4LHIJ3Yj5NEBGVyy6G0tmqdd9XhJmmZgQCLcB/s1600/7.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-Z8GJPUaMXPY/WJKUL1AbIHI/AAAAAAAAB-I/4LHIJ3Yj5NEBGVyy6G0tmqdd9XhJmmZgQCLcB/s1600/7.png" /></a></p><p>When E is zero, we still have time (t). Equation 11) above is equivalent to dark matter; albeit, it is not matter at all. We set matter to zero, so what is it? It is pure acceleration as a function of time. Gravity is not only caused by the presence of matter, it is also caused by time alone. Matter will contract time and make gravity stronger, but gravity exists even in the absence of matter. To have zero gravity requires an infinite change in time (according to equation 11). We can say with confidence that time intervals, so far, have been finite--so gravity, even in the absence of matter, is greater than zero. </p><p>Equation 11) shows that most of the gravity in our universe is caused directly by time. The remainder is indirectly caused by the relatively tiny amount of matter that make up the galaxies. </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com2tag:blogger.com,1999:blog-3628542335168231161.post-60321663202345406602017-01-27T16:39:00.001-08:002017-01-28T12:07:38.851-08:00Something vs. Nothing vs. Mixed States<p><iframe width="560" height="315" src="https://www.youtube.com/embed/xdZMXWmlp9g" frameborder="0" allowfullscreen></iframe></p><p>Yes, we could ask, "Why is there something rather then nothing?" However, if we take quantum physics into account, we shouldn't limit ourselves to pondering something or nothing--we also need to ponder mixed states. Before we do, let's work out the probability of there being something or nothing using a kind of information theory. We can let '1' represent something, and '0' represent nothing. </p><p>Imagine a coin with a '1' on one side and a '0' on the other. We assume the probability of either side being face up is 0.5. However, the coin might favor one side over the other. How do we fix this? We get a trillion coins where one side might be more or less probable than the other. The probabilities could range from 0 to 1, but the average probability will be at or near 0.5 for each side. So it's safe to assume, if we perform the coin toss many many times (and switch coins), on average, we will get each side 50 percent of the time. </p><p>To make our analysis easier, let's assume we are using a coin that has the expected (average) value of 0.5 for each side. Half the time we are going to get a universe with something, and half the time we will get nothing. If we define "nothing" as absolutely nothing, we can improve our odds of getting "something" by adding additional coins. Let "n" in the last two equations below be the number of coins: </p><p><a href="https://4.bp.blogspot.com/-ZfoWMM5A20Y/WIu1YKM-IQI/AAAAAAAAB5o/Zh3x2YaDTmgOl0mncnZPeDXg6mwcYOEeQCLcB/s1600/1.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-ZfoWMM5A20Y/WIu1YKM-IQI/AAAAAAAAB5o/Zh3x2YaDTmgOl0mncnZPeDXg6mwcYOEeQCLcB/s1600/1.png" /></a></p><p>Using two coins yields the following results: </p><p><a href="https://3.bp.blogspot.com/-eursbBXF-Pc/WIu6XDYyYYI/AAAAAAAAB54/t_J9UgvSwbAvHBwaEF3S-22oasIBd7bMQCLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-eursbBXF-Pc/WIu6XDYyYYI/AAAAAAAAB54/t_J9UgvSwbAvHBwaEF3S-22oasIBd7bMQCLcB/s1600/2.png" /></a></p><p>The [0 0] above is nothing. Those pairs above it have a '1'--so they are something. By only using a pair of coins we have improved the chance of getting something to 0.75. What happens if we use three coins? </p><p><a href="https://1.bp.blogspot.com/-u6gHAN-cxD4/WIu6dTlsi9I/AAAAAAAAB58/XnnhjSdlIF0OXpK_PxA9qM2yz_RciulqACLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-u6gHAN-cxD4/WIu6dTlsi9I/AAAAAAAAB58/XnnhjSdlIF0OXpK_PxA9qM2yz_RciulqACLcB/s1600/3.png" /></a></p><p>The probability of getting something jumps to 0.875. If we use an infinite number of coins, the probability of getting something is one--a sure thing. The probability of getting nothing is zero. </p><p><a href="https://4.bp.blogspot.com/-BukOBTpE-jc/WIu6if_fNFI/AAAAAAAAB6A/XaTdVAyC0icFcFI1AfVymU6hrwWvptJngCLcB/s1600/4.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-BukOBTpE-jc/WIu6if_fNFI/AAAAAAAAB6A/XaTdVAyC0icFcFI1AfVymU6hrwWvptJngCLcB/s1600/4.png" /></a></p><p>Thus we have a reason why there is something rather than nothing. But what if what we call "something" is really something else: a mixed state? What exactly is a mixed state? You could think of it as being both something and nothing--an undecided state, a state with both ones and zeros. By contrast, a pure state has all ones or all zeros: </p><p><a href="https://1.bp.blogspot.com/-Puh1AEQXjnw/WIu6mo_0E-I/AAAAAAAAB6E/6Z58SLXk6908afRacPiFFTV5a0JQLiFZACLcB/s1600/5.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-Puh1AEQXjnw/WIu6mo_0E-I/AAAAAAAAB6E/6Z58SLXk6908afRacPiFFTV5a0JQLiFZACLcB/s1600/5.png" /></a></p><p>If we dare to ask where something or nothing came from, one answer is a mixed state. The mixed state evolves into a pure state. When we flip a coin, we don't know if it's heads or tails when it is in the air. We can say it's both heads and tails--a mixed sate. When it hits the ground and comes to rest, it's in a pure state. </p><p>Another example analogous to a mixed state is a ball sitting on a hill. It can roll down one side or the other. It stays in a mixed state until something disturbs it: </p><p><a href="https://4.bp.blogspot.com/-WPJrpZ0KuHY/WIu6r2sX1oI/AAAAAAAAB6I/G3Mz2wrXYfoiySllWLV-sNZaKCGlrh-IwCLcB/s1600/6.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-WPJrpZ0KuHY/WIu6r2sX1oI/AAAAAAAAB6I/G3Mz2wrXYfoiySllWLV-sNZaKCGlrh-IwCLcB/s1600/6.png" /></a></p><p>In the above diagram, the ball's state can evolve into the pure state of '1' or '0.' </p><p>So instead of imagining a universe with just something or nothing, imagine a universe filled with ones and zeros, and, to make things more interesting, let's add an additional state of '-1.' </p><p><a href="https://4.bp.blogspot.com/-sgjfp-rkjk4/WIu6xakrIbI/AAAAAAAAB6M/qLKJzyLprco-6JGTl0rNb66WdTnRzyFGwCLcB/s1600/7.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-sgjfp-rkjk4/WIu6xakrIbI/AAAAAAAAB6M/qLKJzyLprco-6JGTl0rNb66WdTnRzyFGwCLcB/s1600/7.png" /></a></p><p>Instead of coin tosses we have binary number sequences that continuously change over time (t). The above diagram is an example of a two-digit binary sequence. The diagram below models the evolution of these binary numbers: </p><p><a href="https://3.bp.blogspot.com/-Ibn57j16-Ow/WIu63KclYYI/AAAAAAAAB6Q/mAUWd5oDhQYbO6nD3tFmgieBTrK7raTMgCLcB/s1600/8.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-Ibn57j16-Ow/WIu63KclYYI/AAAAAAAAB6Q/mAUWd5oDhQYbO6nD3tFmgieBTrK7raTMgCLcB/s1600/8.png" /></a></p><p>The diagram above starts with the pure state of [1 1]. It then moves through mixed states to get to [0 0], then moves through more mixed states to get to [-1 -1], and so on. Basically we have something that resembles a sine wave. There is a problem, however: it is too predictable for our probabilistic universe. The evolution of states should be random: </p><p><a href="https://3.bp.blogspot.com/-MebbgICskiM/WIu660bCT3I/AAAAAAAAB6U/SqNVPBJIZ3ktbSyU1Teeeev2EAHDY1WfwCLcB/s1600/9.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-MebbgICskiM/WIu660bCT3I/AAAAAAAAB6U/SqNVPBJIZ3ktbSyU1Teeeev2EAHDY1WfwCLcB/s1600/9.png" /></a></p><p>The diagrams above are more realistic. The different states should have varied time duration(t00,t11,etc.). If, say, state [00] jumps to [11] without passing through [01] and [10], We say that these mixed states each have a zero duration. It is also possible for any state to have an exceptionally long duration. </p><p>So how do we get the predictable sine wave we started with? We take a trillion random waves and calculate the average time duration for each state. We can then add those average times together to get the total time and average wavelength: </p><p><a href="https://2.bp.blogspot.com/-i41aF9LixPM/WIu6-0BmJQI/AAAAAAAAB6Y/kFOMoMjM41QrTvtVqlev9iHPTPswNbtcwCLcB/s1600/10.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-i41aF9LixPM/WIu6-0BmJQI/AAAAAAAAB6Y/kFOMoMjM41QrTvtVqlev9iHPTPswNbtcwCLcB/s1600/10.png" /></a></p><p>Our predictable binary sine wave is really the expectation value of a bunch of random binary waves: </p><p><a href="https://2.bp.blogspot.com/-2zE0cmIBLuk/WIu7FJtF1YI/AAAAAAAAB6g/YhsPUq8bntcdgyFbjKneLtGa1FcL4BcdQCLcB/s1600/11.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-2zE0cmIBLuk/WIu7FJtF1YI/AAAAAAAAB6g/YhsPUq8bntcdgyFbjKneLtGa1FcL4BcdQCLcB/s1600/11.png" /></a></p><p>Now, our binary sine wave isn't very smooth nor continuous. It has a clunky, stair-step quality. We can make it smoother if we add more digits to the states: </p><p><a href="https://2.bp.blogspot.com/-IdMNntjbejI/WIu7KiN8TAI/AAAAAAAAB6k/OD78eqQNjnwIlnVf_IsahNCmTtVr1ZSIwCLcB/s1600/12.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-IdMNntjbejI/WIu7KiN8TAI/AAAAAAAAB6k/OD78eqQNjnwIlnVf_IsahNCmTtVr1ZSIwCLcB/s1600/12.png" /></a></p><p>In the diagram above, notice the wave with only one digit is the clunkiest. The smooth wave at the bottom has an infinite number of digits or at least a really big number of digits. So let's take the bottom wave and make it our official sine wave: </p><p><a href="https://1.bp.blogspot.com/-s_reazhrTTk/WIu7P_Z8TbI/AAAAAAAAB6o/XfoLNGxOjwQi3eiKbsrGTZGUckf1SzEJgCLcB/s1600/13.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-s_reazhrTTk/WIu7P_Z8TbI/AAAAAAAAB6o/XfoLNGxOjwQi3eiKbsrGTZGUckf1SzEJgCLcB/s1600/13.png" /></a></p><p>Sooner or later, our sine wave will interact with another wave (superposition). At one extreme the phase difference between the waves could be 180 degrees, resulting in destructive interference: </p><p><a href="https://3.bp.blogspot.com/-z3IJcV2LfJU/WIu7g7fVJoI/AAAAAAAAB64/EbQtgbmpDd48WYB_JD8hUNpOzYcyO8JCwCLcB/s1600/14.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-z3IJcV2LfJU/WIu7g7fVJoI/AAAAAAAAB64/EbQtgbmpDd48WYB_JD8hUNpOzYcyO8JCwCLcB/s1600/14.png" /></a></p><p>At the other extreme the phase difference could be zero degrees, resulting in constructive interference: </p><p><a href="https://3.bp.blogspot.com/-cgYZMzbEEyk/WIu7UxoEgII/AAAAAAAAB6w/Uh6eQuZkqus4wtxDu72EAVGS2Rpsu4gpwCLcB/s1600/15.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-cgYZMzbEEyk/WIu7UxoEgII/AAAAAAAAB6w/Uh6eQuZkqus4wtxDu72EAVGS2Rpsu4gpwCLcB/s1600/15.png" /></a></p><p>If we have a bunch of waves, the average or median phase between each pair will be 90 degrees. Thus, we end up with both a sine and cosine wave. </p><p><a href="https://1.bp.blogspot.com/-10VHblv46QI/WIu7mXdoWGI/AAAAAAAAB68/dYSVFg8ldS4nnUvAZD0-_9brepnar0hKwCLcB/s1600/16.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-10VHblv46QI/WIu7mXdoWGI/AAAAAAAAB68/dYSVFg8ldS4nnUvAZD0-_9brepnar0hKwCLcB/s1600/16.png" /></a></p><p>Having sine and cosine waves lead to wave functions and matter--the stuff we call "something." </p><p><a href="https://3.bp.blogspot.com/-QqP-1jktJvY/WIu7q8CdTrI/AAAAAAAAB7A/tm2R0G18RRIqUD0U66o9L5Lv_Lxp8emkQCLcB/s1600/17.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-QqP-1jktJvY/WIu7q8CdTrI/AAAAAAAAB7A/tm2R0G18RRIqUD0U66o9L5Lv_Lxp8emkQCLcB/s1600/17.png" /></a></p><p>However, as we have witnessed, the stuff we call "something" is really a mixture of both something and nothing: zeros and ones. The stuff we call "something" also oscillates between pure and mixed states. </p><p><a href="http://gmjacksonphysics.blogspot.com/2016/11/why-there-is-something-rather-than.html" target="blank">You can find more on this topic if you click here.</a></p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-13859350694218962092017-01-22T18:44:00.000-08:002017-01-27T16:42:38.786-08:00Quantizing Photons and Gravitational Lensing<p><iframe width="560" height="315" src="https://www.youtube.com/embed/QByYDIQezg8" frameborder="0" allowfullscreen></iframe></p><p></p><p>According to General Relativity theory, curved spacetime causes the mass-less photon to follow a curved geodesic path when it is near a planet, star or black hole. Newton's equation, F = GMm/r^2 seems limited to masses (M and m). The way Einstein figured it, if particles follow a geodesic curve due to warped spacetime, then mass (or the lack thereof) doesn't matter. Gravity attracts photons as well as massive objects. But what exactly is curved spacetime at the quantum scale and how does it move photons around? </p><p>The diagram below depicts a photon we will call "a." It's entering the gravitational field of a planet surrounded by circular field lines. The red circles have less energy than the blue circles. You could say the field lines are blue-shifted towards the planet's surface and red-shifted away from it. </p><p><a href="https://2.bp.blogspot.com/-EH8rqQ4xpz8/WIVQz_LasjI/AAAAAAAAB30/VDTAOBUz1YE5WH8qP3FJItQntOwNyO-WwCLcB/s1600/1.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-EH8rqQ4xpz8/WIVQz_LasjI/AAAAAAAAB30/VDTAOBUz1YE5WH8qP3FJItQntOwNyO-WwCLcB/s1600/1.png" /></a></p><p>Photon "a" will pick up energy, increase its wave frequency if it moves toward the surface, and will lose energy if it goes up and away. Since we are covering quantum physics, let's assume photon "a" can go in any direction it wants within the gravitational field: </p><p><a href="https://3.bp.blogspot.com/-CmsK93z7ssM/WIVQ9VkG12I/AAAAAAAAB34/Stivn5EBm8IvVkCOhxpr_JKO8-7mG6sGQCLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-CmsK93z7ssM/WIVQ9VkG12I/AAAAAAAAB34/Stivn5EBm8IvVkCOhxpr_JKO8-7mG6sGQCLcB/s1600/2.png" /></a></p><p>Photon a's position (y) is very uncertain. Uncertainty Principle to the rescue! Equation 1) below shows that if energy (E) increases, the position of "a" becomes more certain. Thus, a's position will most likely be where there is the most energy, and more energy is found if the photon heads down. </p><p><a href="https://2.bp.blogspot.com/-jHaMhjTereg/WIVRCIT_4kI/AAAAAAAAB38/VVvcy19KM10J5aIvQ3O--jnDCxKm5-wWACLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-jHaMhjTereg/WIVRCIT_4kI/AAAAAAAAB38/VVvcy19KM10J5aIvQ3O--jnDCxKm5-wWACLcB/s1600/3.png" /></a></p><p>Equations 2) and 3) show the relationship between energy (E), the wave number (k) and Einstein's field equations. At the quantum scale, curved spacetime is wave numbers that become larger when a photon moves down and smaller when a photon moves up. Equation 4) enables us to determine the probability that photon "a" will be at position "y." As we hinted at above, the probability is large where energy (E) is large. That seems reasonable when you consider it takes energy to make a photon. </p><p>Even though photon a's position is uncertain, let's imagine it is somewhere just outside the planet's gravitational field. We plot its probability distribution below. It makes a nice bell curve. We could even think of the photon as a distribution of energy. </p><p><a href="https://2.bp.blogspot.com/-xI2eHBxkwKI/WIVRGh6EC9I/AAAAAAAAB4A/D0f1gTnnv_EM7fF28PDGWYqfL54ECHzKgCLcB/s1600/4.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-xI2eHBxkwKI/WIVRGh6EC9I/AAAAAAAAB4A/D0f1gTnnv_EM7fF28PDGWYqfL54ECHzKgCLcB/s1600/4.png" /></a></p><p>The expectation value is at the bell curve's center. This is where there is the most energy; this is where we will most likely find "a." Let's add some energy to the right side of the bell curve: </p><p><a href="https://2.bp.blogspot.com/-UpCzWNlIZQs/WIVRKvZytbI/AAAAAAAAB4E/iLUm7CKYGLgCLmYtBDxGo9m7Siy2J-CAgCLcB/s1600/5.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-UpCzWNlIZQs/WIVRKvZytbI/AAAAAAAAB4E/iLUm7CKYGLgCLmYtBDxGo9m7Siy2J-CAgCLcB/s1600/5.png" /></a></p><p>The energy distribution is now changed. This creates a new expectation value for "a.". </p><p><a href="https://4.bp.blogspot.com/-aiNnfVsMUKg/WIVRRdNU8uI/AAAAAAAAB4I/c86rUxbiZXUcMgluxdK3lh3VoR3if9pHgCLcB/s1600/6.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-aiNnfVsMUKg/WIVRRdNU8uI/AAAAAAAAB4I/c86rUxbiZXUcMgluxdK3lh3VoR3if9pHgCLcB/s1600/6.png" /></a></p><p>Photon "a" has shifted to the right, or, more precisely, photon "a" can more likely be found to the right of its previous most-likely position. If we turn the graph on its side, we get a picture of "a" falling towards the energy field. </p><p><a href="https://1.bp.blogspot.com/-quEGU7Kl0rQ/WIVRVGCLCZI/AAAAAAAAB4M/GxLzB5qO_WAOAIGM5MDgAyVU9gCIaV_igCLcB/s1600/7.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-quEGU7Kl0rQ/WIVRVGCLCZI/AAAAAAAAB4M/GxLzB5qO_WAOAIGM5MDgAyVU9gCIaV_igCLcB/s1600/7.png" /></a></p><p>But it's not just the single, classical-looking photon "a"--the entire energy and probability distribution of "a" is falling towards the energy field. </p><p><a href="https://2.bp.blogspot.com/-DM2Bm9zoFDI/WIVRYryOt3I/AAAAAAAAB4Q/fRF1QO2fUiQ4vXNyemFtjTw6tMtRrZSeACLcB/s1600/8.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-DM2Bm9zoFDI/WIVRYryOt3I/AAAAAAAAB4Q/fRF1QO2fUiQ4vXNyemFtjTw6tMtRrZSeACLcB/s1600/8.png" /></a></p><p>A new bell curve is formed, but it is only temporary because "a" has moved down into a greater energy field, so the bell curve must shift downward again. </p><p><a href="https://3.bp.blogspot.com/-3iQVM8XRCfM/WIVRcsSYBqI/AAAAAAAAB4U/jmyLGzk7NNwQS3pPrs5I2_jid0dr7dzrQCLcB/s1600/9.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-3iQVM8XRCfM/WIVRcsSYBqI/AAAAAAAAB4U/jmyLGzk7NNwQS3pPrs5I2_jid0dr7dzrQCLcB/s1600/9.png" /></a></p><p>And again. See a pattern yet? </p><p><a href="https://1.bp.blogspot.com/-vs305VIn_iA/WIVRtlyBUhI/AAAAAAAAB4Y/4UOBvc_PJSw7r59mlhI1N8wiwXFSKq6MQCLcB/s1600/10.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-vs305VIn_iA/WIVRtlyBUhI/AAAAAAAAB4Y/4UOBvc_PJSw7r59mlhI1N8wiwXFSKq6MQCLcB/s1600/10.png" /></a></p><p>If photon "a" had enough momentum it could have whizzed right passed the planet along a geodesic curve: </p><p><a href="https://3.bp.blogspot.com/-uunoEMwwPFI/WIVRyCXWF6I/AAAAAAAAB4c/afL5JD8_TuEsq7Rw8DTSSI2Zch-0M3f-QCLcB/s1600/10b.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-uunoEMwwPFI/WIVRyCXWF6I/AAAAAAAAB4c/afL5JD8_TuEsq7Rw8DTSSI2Zch-0M3f-QCLcB/s1600/10b.png" /></a></p><p>We can use a's bell curve to model the geodesic at the quantum scale: </p><p><a href="https://2.bp.blogspot.com/-zgOb-tMdmbA/WIVR3IjqjkI/AAAAAAAAB4g/JRwiA4igUT8NNgaa_pEGu8KCmAuhM4DawCLcB/s1600/11.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-zgOb-tMdmbA/WIVR3IjqjkI/AAAAAAAAB4g/JRwiA4igUT8NNgaa_pEGu8KCmAuhM4DawCLcB/s1600/11.png" /></a></p><p> </p><p> This was part six of the gravity series. To read the other parts, click <a href="http://gmjacksonphysics.blogspot.com/2016/09/how-does-curved-spacetime-cause-linear.html " target="blank1">here</a>, <a href="http://gmjacksonphysics.blogspot.com/2016/09/why-gravity-attracts-quantizing.html" target="blank2">here</a>, <a href="http://gmjacksonphysics.blogspot.com/2017/01/quantizing-gravity-why-different-masses.html" target="blank3">here</a>, <a href="http://gmjacksonphysics.blogspot.com/2017/01/the-quantum-differences-between-gravity.html" target="blank4">here</a>, and <a href="http://gmjacksonphysics.blogspot.com/2016/08/why-spacetime-warps-when-you-add-pinch.html" target="blank5">here</a>. </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-65849023728598947952017-01-19T13:11:00.000-08:002017-07-30T15:33:18.651-07:00The Quantum Differences Between Gravity and Electromagnetism<p><iframe width="560" height="315" src="https://www.youtube.com/embed/4QPRhTfUhPQ" frameborder="0" allowfullscreen></iframe></p><p>Welcome to part five of the gravity series. To read the other parts, <a href="http://gmjacksonphysics.blogspot.com/2017/01/quantizing-gravity-why-different-masses.html" target="blank">click here</a> and <a href="http://gmjacksonphysics.blogspot.com/2016/09/why-gravity-attracts-quantizing.html" target="blank1">here</a>. Are gravity and electromagnetism (EM) the same? There are some who think they are the same force. There may have been a time in the early universe when all the forces were one force, and, as time passed, the one force evolved into the ones we are familiar with. If gravity and EM are the same, they have some distinct differences we will examine at the quantum and cosmic levels. </p><p>In previous posts we discovered that gravity is a field with ever shortening spacetime wavelengths, or, increasing energy as a falling body moves closer to where there is greater mass or energy density. In addition, particle-waves' wave numbers increase. These wave numbers have the same units as curved spacetime. Gravitational acceleration is due to the difference in energy or wave number between an upper surface layer of spacetime and a lower one. </p><p>Imagine a particle falling in a gravitational field. We can model this using Schrodinger's equation with a twist: we take the difference between the Hamiltonian in the upper layer and the Hamiltonian in the lower layer: </p><p><a href="https://1.bp.blogspot.com/-nZhe4NGyxPQ/WID-ru999eI/AAAAAAAAB04/NcBFAjjRlDA3HLYwD-t5ShYxvVsqFVlkwCLcB/s1600/1.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-nZhe4NGyxPQ/WID-ru999eI/AAAAAAAAB04/NcBFAjjRlDA3HLYwD-t5ShYxvVsqFVlkwCLcB/s1600/1.png" /></a></p><p>We want to find the change in wave number (k): </p><p><a href="https://1.bp.blogspot.com/-SMkEb9HCGeU/WID_6t7gdYI/AAAAAAAAB1E/HnxPbOKHOGUi7xB5tmo-lJ9SHp9MFrWCQCLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-SMkEb9HCGeU/WID_6t7gdYI/AAAAAAAAB1E/HnxPbOKHOGUi7xB5tmo-lJ9SHp9MFrWCQCLcB/s1600/2.png" /></a></p><p><a href="https://1.bp.blogspot.com/-XRBa2sOYXic/WIEBOMZybzI/AAAAAAAAB1M/GZHtLFzStYg7A6jWEPIgmAhkjZnICuJ_gCLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-XRBa2sOYXic/WIEBOMZybzI/AAAAAAAAB1M/GZHtLFzStYg7A6jWEPIgmAhkjZnICuJ_gCLcB/s1600/3.png" /></a></p><p>We perform a couple of more steps to get the particle's change in kinetic energy: </p><p><a href="https://4.bp.blogspot.com/-1hJwypHARio/WIECpQKEM4I/AAAAAAAAB1U/tT6KL3UX8dk8ZEySUYhoH-7qD_qRqkNzgCLcB/s1600/4.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-1hJwypHARio/WIECpQKEM4I/AAAAAAAAB1U/tT6KL3UX8dk8ZEySUYhoH-7qD_qRqkNzgCLcB/s1600/4.png" /></a></p><p>If we take the wave number (k) and multiply it by Planck's constant (h-bar) and divide it by the particle's mass, we get the particle's change in velocity (v). </p><p><a href="https://3.bp.blogspot.com/-ZIPxjBu5puc/WIEEdvoBKPI/AAAAAAAAB1c/u2cdqI6TbxMvUOFDSwnYnW3LAvE_EESTQCLcB/s1600/5.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-ZIPxjBu5puc/WIEEdvoBKPI/AAAAAAAAB1c/u2cdqI6TbxMvUOFDSwnYnW3LAvE_EESTQCLcB/s1600/5.png" /></a></p><p>At equation 12) notice that the first term has momentum (p). Here are two kinds of momentum: mv (mass X velocity) and fh/c (frequency X Planck's constant/light speed). Velocity (v) could be a function of one or both of these momenta--or we could have two types of velocity: equations 13 and 14 below are derived from the first term of equation 12). </p><p><a href="https://3.bp.blogspot.com/-aHJfshhFku0/WIENgUqhTYI/AAAAAAAAB14/J-oPwl4XQQk5ZA20m8FGp39e2WWEQi0ZwCLcB/s1600/6.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-aHJfshhFku0/WIENgUqhTYI/AAAAAAAAB14/J-oPwl4XQQk5ZA20m8FGp39e2WWEQi0ZwCLcB/s1600/6.png" /></a></p><p>Equation 13) isn't fully simplified. We want to emphasize that the particle's mass cancels itself. The velocity is the same whether the mass is big or small. We can label this velocity the change in velocity due to gravity (Vg). Take note that momentum is not conserved: a big falling mass has more momentum than a smaller falling mass. </p><p>Equation 14) tells a different tale. A change in mass does change the velocity (Ve). Momentum is conserved. This equation fits the EM force. </p><p>Equations 13) and 14) reveal that when mass (m) is very large, gravity's influence stays the same; whereas, EM's impact diminishes. When mass (m) is small, EM becomes the dominant force--gravity becomes less significant. </p><p>Another key difference between gravity and EM is EM is a function of charge; whereas, gravity is a function of mass or energy density. Gravity attracts but EM obeys Column's law (like charges repel, opposite charges attract). </p><p>The crude diagrams below demonstrate EM interactions. When the field arrows are pointing towards each other, particles A and B push apart. When the arrows point away from each other, A and B separate. Particles A and B attract each other when the field lines (arrows) flow in the same direction, as if B is flowing towards A. </p><p><a href="https://4.bp.blogspot.com/-gk-9QgcsyoI/WIET4GgHkgI/AAAAAAAAB2I/u_x2gXZadWkvmUNxuUWH17ddtj0ViCPQgCLcB/s1600/8.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-gk-9QgcsyoI/WIET4GgHkgI/AAAAAAAAB2I/u_x2gXZadWkvmUNxuUWH17ddtj0ViCPQgCLcB/s1600/8.png" /></a></p><p>The following diagrams show how A and B share gravitational field lines. This sharing causes the energy field between A and B to become more intense than the fields at the far right and left of A and B. A and B want to move away from each other and move closer together. The shared energy between them makes the latter more probable. To see how this works in more detail, <a href="http://gmjacksonphysics.blogspot.com/2017/01/quantizing-gravity-why-different-masses.html" target="blank">click here</a>. As the distance between A and B decreases, the shared energy between them becomes stronger and the gravitational acceleration increases (the inverse square law). </p><p><a href="https://2.bp.blogspot.com/-IKQ16TSbmsw/WIEXSxz5AsI/AAAAAAAAB2U/uOzoiDMleS0ZZrbInX_rqwKcsOzIqm13wCLcB/s1600/9.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-IKQ16TSbmsw/WIEXSxz5AsI/AAAAAAAAB2U/uOzoiDMleS0ZZrbInX_rqwKcsOzIqm13wCLcB/s1600/9.png" /></a></p><p>One thing EM and gravity have in common are mass-less bosons. Since they are mass-less, these bosons have unlimited range and they travel at light speed. This raises a troublesome paradox, for we know EM is much much stronger than gravity. How can atoms and molecules ever get together via gravity when their outer-shell electrons have a repulsive force far greater than gravity's attractive force? </p><p>Consider two hydrogen atoms that are close together. We fully expect the electrons to repel each other, same goes for the protons. But could the proton in one atom be attracted to the electron in the other? If so, that attraction could, to some extent, cancel the repulsive force. We can use the following equations to see if gravity is stronger or weaker than the net EM repulsive force. </p><p><a href="https://3.bp.blogspot.com/-CQx3SFyoUAg/WIEg8T9KzpI/AAAAAAAAB2w/ZFERj6LAfY0IfjkagByzN-CdunQxp1PVACLcB/s1600/10.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-CQx3SFyoUAg/WIEg8T9KzpI/AAAAAAAAB2w/ZFERj6LAfY0IfjkagByzN-CdunQxp1PVACLcB/s1600/10.png" /></a></p><p>When we crunch the numbers we find that as distance (d) between the atoms increases, the EM force (Fe) drops more quickly than gravity (Fg). In the diagram below, where the atoms are close together, the distance between the electrons is very short compared to the distances between opposite charges, so the repulsive force is strong. This is good news! It means the atoms will remain distinct and separate. It means the ground will be solid beneath your feet. </p><p><a href="https://4.bp.blogspot.com/-cJAkGEtuV7c/WIEjcm6BiVI/AAAAAAAAB28/mvWshKOMtKgkJ73ZVPw1r0lOCdYqDEj7wCLcB/s1600/11.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-cJAkGEtuV7c/WIEjcm6BiVI/AAAAAAAAB28/mvWshKOMtKgkJ73ZVPw1r0lOCdYqDEj7wCLcB/s1600/11.png" /></a></p><p>By contrast, when the atoms are far apart, the different distances between opposites and same-charge particles are less dramatic. The charges cancel each other and gravity dominates. </p><p> </p><p> </p><p> </p><p> </p><p> </p><p>GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com1tag:blogger.com,1999:blog-3628542335168231161.post-16464508509166308642017-01-16T17:30:00.001-08:002017-01-30T15:54:51.163-08:00Quantizing Gravity: Why Different Masses Fall at the Same Rate<p><iframe width="560" height="315" src="https://www.youtube.com/embed/CqFdDbLSNaM" frameborder="0" allowfullscreen></iframe></p><p>This is part four of the gravity series. <a href="http://gmjacksonphysics.blogspot.com/2016/09/why-gravity-attracts-quantizing.html" target="blank">Click here to find parts 1 through 3.</a> In this part we show how quantum gravity explains why particles that have different masses (or wave frequencies) fall down (not up) at the same acceleration rate when they're in a gravitational field.</p><p>Why do different masses fall at the same rate? It's as if gravity knows the mass of each falling object and adjusts its force to overcome each object's inertia so acceleration (g) is nearly constant. Believe it or not, we can derive the answer to this puzzling phenomenon from Heisenberg's Uncertainty Principle: </p><p><a href="https://3.bp.blogspot.com/-KAzJDTI1IsE/WH1T4avqv2I/AAAAAAAAByI/EEtJalkXsSgi1ic9USgzTg9IN6IITSqpgCLcB/s1600/10.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-KAzJDTI1IsE/WH1T4avqv2I/AAAAAAAAByI/EEtJalkXsSgi1ic9USgzTg9IN6IITSqpgCLcB/s1600/10.png" /></a></p><p>Variable p is momentum, and normally x is position, but for our purposes x shall represent a particle wavelength. Of course h-bar is good old Planck's constant. In equation 4) we end up with k--the wave number. In addition to providing the number of waves in a particle wave, the wave number has the same units as curved spacetime, so do the other terms in equation 4). Below one or more of the terms are set equal to Einstein's tensor (G) and his field equations. </p><p> <a href="https://4.bp.blogspot.com/-98rXQdv8X6w/WH1YIOFfIFI/AAAAAAAAByU/0YWgTJCDouMT5tOsj8q_7iEcgssyIXn-QCLcB/s1600/11.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-98rXQdv8X6w/WH1YIOFfIFI/AAAAAAAAByU/0YWgTJCDouMT5tOsj8q_7iEcgssyIXn-QCLcB/s1600/11.png" /></a> </p><p>If you are wondering how curved spacetime moves matter, the far left side of equation 4 provides a clue. It contains momentum (p). Momentum is all about movement, so any particle that interacts with something with momentum is bound to be set in motion. But why should the motion be necessarily down? We'll cover that later; let's finish what we started: </p><p><a href="https://3.bp.blogspot.com/-suYUA-Tu7Hc/WH1Zgnjy0CI/AAAAAAAAByg/Mn3HvmFkJucSd4ld8FIxJjLftAo8Evi1wCLcB/s1600/12.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-suYUA-Tu7Hc/WH1Zgnjy0CI/AAAAAAAAByg/Mn3HvmFkJucSd4ld8FIxJjLftAo8Evi1wCLcB/s1600/12.png" /></a></p><p>If velocity (v) is held constant, equations 10) and 11) show that when the mass (m) of a particle increases, the wavelength (x) decreases, and vice versa. This anti-correlation between mass and wavelength may explain why different-massed objects fall at the same rate. Equation 12) shows how a photon's velocity (c) is maintained: the higher the frequency (f), the shorter the wavelength (x), and vice versa. So if you take any two particles (A and B), they will have the same velocity from point to point because the A-mass times the A-wavelength equals the B-mass times the B-wavelength. </p><p>They also have the same acceleration (g). This is due to decreasing spacetime and particle wavelengths (increasing wave numbers (k)) as particle A (see diagram 1 below) moves further into the field where there is more energy density. </p><p><a href="https://1.bp.blogspot.com/-Qyu3l4TDTtA/WH1e3YJBLhI/AAAAAAAAByw/-BedhJhWjR0ipGOZJhr6RbaXbxMFk48GwCLcB/s1600/13.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-Qyu3l4TDTtA/WH1e3YJBLhI/AAAAAAAAByw/-BedhJhWjR0ipGOZJhr6RbaXbxMFk48GwCLcB/s1600/13.png" /></a></p><p>Equations 13) and 14) above and 15) through 20) below show how the wave number (k) is derived and how it relates to Schrodinger's and Einstein's equations: </p><p><a href="https://1.bp.blogspot.com/-jP9zMlatDGM/WH_c3se9dMI/AAAAAAAAB0I/l1BeJijDF_sgyUQ5ePRzUBIZf5SspR07ACLcB/s1600/14.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-jP9zMlatDGM/WH_c3se9dMI/AAAAAAAAB0I/l1BeJijDF_sgyUQ5ePRzUBIZf5SspR07ACLcB/s1600/14.png" /></a></p><p>Now let's deal with the problem of motion. Which way does particle A go when it encounters gravity. Why should it go down and not up or sideways? Since we are covering quantum mechanics, it stands to reason that particle A's motion is probabilistic--it can go in an infinite number of directions. </p><p><a href="https://4.bp.blogspot.com/-6tM6iYiWh-A/WH1kVMAL0BI/AAAAAAAABzI/uLv_FFgNfqA-NuzNvmHX5obkpB2RPvJagCLcB/s1600/15.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-6tM6iYiWh-A/WH1kVMAL0BI/AAAAAAAABzI/uLv_FFgNfqA-NuzNvmHX5obkpB2RPvJagCLcB/s1600/15.png" /></a></p><p>The probability of A going in any one direction is 1/infinity--virtually zero. Predicting A's direction seems hopeless, but there is a way to simplify the problem: we divide the possible directions into four sectors--each with a .25 probability density. </p><p><a href="https://2.bp.blogspot.com/-V7v1zUlDDJM/WH1md_z-o3I/AAAAAAAABzU/IKl20LzZbHU1gEzp9qUH9b3Zs7XcPalgwCLcB/s1600/16.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-V7v1zUlDDJM/WH1md_z-o3I/AAAAAAAABzU/IKl20LzZbHU1gEzp9qUH9b3Zs7XcPalgwCLcB/s1600/16.png" /></a></p><p>We can also add up all the vectors in each sector and find the mean or median; i.e., the expectation value: </p><p><a href="https://1.bp.blogspot.com/-F6AfwJcN-S4/WH1nc077iAI/AAAAAAAABzg/-SN1p8P06EM7j-Ivu9o3PL-AjfmyvwV6gCLcB/s1600/17.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-F6AfwJcN-S4/WH1nc077iAI/AAAAAAAABzg/-SN1p8P06EM7j-Ivu9o3PL-AjfmyvwV6gCLcB/s1600/17.png" /></a></p><p>Now we add a gravitational field. Diagram 5 has a blue section (blue shift) and a red section (red shift) to model an increase in energy from top to bottom. We expect particle A to move further and faster in the field's bottom portion than the top portion. </p><p><a href="https://1.bp.blogspot.com/-hWojEdaWw80/WH1pmHPs2MI/AAAAAAAABzs/5pTzSRLGS1sz0LVwa1mRx3_0EygD5m3QwCLcB/s1600/18.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-hWojEdaWw80/WH1pmHPs2MI/AAAAAAAABzs/5pTzSRLGS1sz0LVwa1mRx3_0EygD5m3QwCLcB/s1600/18.png" /></a></p><p>It looks like we've simplified the problem to the point where we can model what particle A is most likely to do when it moves through a gravitational field. </p><p><a href="https://2.bp.blogspot.com/-naWXUidwgpA/WH1sTtKSHoI/AAAAAAAABz4/bmMIBum6IUgFBS1LCANTtX60hRV4kcn8gCLcB/s1600/19.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-naWXUidwgpA/WH1sTtKSHoI/AAAAAAAABz4/bmMIBum6IUgFBS1LCANTtX60hRV4kcn8gCLcB/s1600/19.png" /></a></p><p>In diagram 6 above, particle A starts at the center and goes either up or down at each node. If it heads up, its distance and velocity shrinks due to decreasing field energy. If it goes down its distance and velocity grow due to increasing field energy. The vertical lines in the middle represent the average ups and downs. Note that the average down exceeds the average up, so particle A has a downward bias. Particle A has also added to its net downward velocity--it has accelerated downward or decelerated upwards depending on the direction of its initial velocity. Each time particle A reaches a new field level, on average, it repeats diagram 6 and adds more to its down velocity. Thus, we have gravity. </p><p>Update: The diagrams below show how different masses fall at the same rate in a gravitational field: <p></p><a href="https://1.bp.blogspot.com/-uLJ3BX_Irms/WI-suyXIfwI/AAAAAAAAB74/VC-0-unriSUT6CwPrz3Ag7O9P-y7NF2RgCLcB/s1600/01.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-uLJ3BX_Irms/WI-suyXIfwI/AAAAAAAAB74/VC-0-unriSUT6CwPrz3Ag7O9P-y7NF2RgCLcB/s1600/01.png" /></a><p></p><a href="https://3.bp.blogspot.com/-GwHl51CKtI8/WI-s2VNE8GI/AAAAAAAAB78/mRLUz32nOOEFYRkQG6VEMXm93Ob6RKhSQCLcB/s1600/02.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-GwHl51CKtI8/WI-s2VNE8GI/AAAAAAAAB78/mRLUz32nOOEFYRkQG6VEMXm93Ob6RKhSQCLcB/s1600/02.png" /></a><p></p>As you can see, each particle within each atom interacts with gravity separately (i.e., has its own squiggly field line), so gravity does not distinguish between different masses. The more massive particles have more field lines per time (t) which is equivalent to higher frequency and shorter wavelengths--so big mass times short wavelength falls at the same rate as small mass times long wavelength. <p></p>Update: The following diagram summarizes how mass and mass-less particles behave in a gravitational field: </p><p><a href="https://2.bp.blogspot.com/-rWaE-uU3M8Q/WI-nd-NfD1I/AAAAAAAAB7k/K2Z2sH8Cv0YahTiHvcwBE2CR-nbT2ejTwCLcB/s1600/I0.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-rWaE-uU3M8Q/WI-nd-NfD1I/AAAAAAAAB7k/K2Z2sH8Cv0YahTiHvcwBE2CR-nbT2ejTwCLcB/s1600/I0.png" /></a> </p><p> </p><p> GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-79651280910712627162016-11-21T11:40:00.000-08:002016-11-22T13:53:37.528-08:00Why There is Something Rather Than Nothing<p>Why is there something rather than nothing? Mathematicians use zero to represent nothing. The concept of zero goes way back to ancient Egypt. The hieroglyph known as 'nfr' indicated "emptiness." Aryabhatta, an Indian mathematician, introduced zero in the 5th century AD. </p><p>I was first introduced to zero in the first grade. Our teacher, Mrs. White, taught us that 2 - 2 = 0. Later in life I was introduced to the Lagrange equation, where potential energy is subtracted from kinetic energy. Of course if kinetic energy equals potential energy, you get zero on the right side of the equation. </p><p>It's safe to say that all my life I've been brainwashed to believe you can have something, take it away, and end up with nothing: zero. And I'm not alone; philosophers have asked, "Why is there something rather than nothing?" as if nothing were a viable possibility. </p><p>The closest thing to nothing measured by the Wilkinson Microwave probe is about 6E-10 joules per cubic meter. The energy density of the vacuum of space is close to zero, but not quite. So the question arises, what is the probability of ever having nothing? In the case of the Lagrange equation, what is the probability that a universe could have the same amount of kinetic energy as potential energy? </p><p>Suppose we label kinetic energy "positive," and potential energy "negative." Let's assume all energy is made up of discrete quanta. We take all that quanta (an infinite amount) and put it inside a big cosmic hat. We reach in and pull some out. What is the probability we pulled out the same amount of positive quanta as negative quanta? Or what is the probability we pulled out zero quanta? </p><p>The probability we pulled out zero quanta is easy to figure out. Since there is an infinite number of quanta, the probability is 1/infinity--which is virtually zero. So we are virtually guaranteed to have more than zero quanta outside the hat. To have zero net energy, we need an even number of quanta, and equal amounts of negative and positive quanta. </p><p>If the quanta number is odd, the probability of netting zero is zero. For example, if we pulled out five quanta from our cosmic hat, the closest combination to zero would be three positives and two negatives--or vice versa. If we pulled out four quanta, there's a possibility we could have two positives and two negatives. The equations below enable us to calculate the probability of netting zero: </p><p><a href="https://4.bp.blogspot.com/-R6CTEjWboNM/WDM_yIhv6cI/AAAAAAAABmo/jSFWHLcAh8A7xiThOIFkCtHrEGhRJgrCACLcB/s1600/1.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-R6CTEjWboNM/WDM_yIhv6cI/AAAAAAAABmo/jSFWHLcAh8A7xiThOIFkCtHrEGhRJgrCACLcB/s1600/1.png" /></a></p><p>Equation 2) takes into account that there's a 0.5 probability that the number of quanta (n) could be even, so the probability that is calculated in equation 1) is cut in half. Thus, in equation 2's denominator, there is 2^(n+1) instead of 2^n. </p><p>The above equations show that the larger the number of quanta (n), the less likely we will have zero net energy. But shouldn't the net energy get closer to zero as we add more quanta to the mix? If we flip a coin, we get heads or tails. If we assign a value of minus one to heads and plus one to tails, we get plus one or minus one--never zero. But suppose we toss a trillion coins? We should get an equal amount of heads and tails (or pretty close). </p><p>We know if we increase the number of coins, the variance or deviation from zero is reduced. </p><p><a href="https://1.bp.blogspot.com/-qzvAClkxHO4/WDNMipiyGGI/AAAAAAAABnY/QgKO8N1EvD4BDB5Uox2izZuQy9NJEIJOwCLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-qzvAClkxHO4/WDNMipiyGGI/AAAAAAAABnY/QgKO8N1EvD4BDB5Uox2izZuQy9NJEIJOwCLcB/s1600/2.png" /></a></p><p>Equations 3) and 4) clearly show, that as the number of coins (or quanta) increases, the smaller the variance becomes. The sum of coins or quanta converge to zero. This seems to contradict our earlier finding via equations 1) and 2). Let's crunch some numbers and put the data in a couple of tables to see what we get. </p><p><a href="https://2.bp.blogspot.com/-ZpQOtfsqfL0/WDNHAF4zKTI/AAAAAAAABnA/_GQez6U0BMAr_ype1K2dOFYgRRI0t5buwCLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-ZpQOtfsqfL0/WDNHAF4zKTI/AAAAAAAABnA/_GQez6U0BMAr_ype1K2dOFYgRRI0t5buwCLcB/s1600/3.png" /></a></p><p>The first table above contains data for odd numbers of quanta. We see that the variance (v) decreases as expected. It gets closer to zero as more quanta are added. (This is also true in the second table for even numbers of quanta.) We see in the last two columns the probability of having exactly zero net energy is zero due to an odd number of quanta. </p><p>The second table shows an interesting paradox: As the variance decreases, so does the probability of having exactly zero net energy. This paradox is illustrated in the graphs below: </p><p><a href="https://4.bp.blogspot.com/-2awVDYmFpdc/WDNKYXpwgXI/AAAAAAAABnM/XoXMMExTReEEIU52lF1TSdw4vp1q-IvqACLcB/s1600/P2.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-2awVDYmFpdc/WDNKYXpwgXI/AAAAAAAABnM/XoXMMExTReEEIU52lF1TSdw4vp1q-IvqACLcB/s1600/P2.png" /></a></p><p>So, the closer we are to zero, the less likely we will have precisely zero--this is why there is something rather than nothing. </p><p> </p><p> </p><p> GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0tag:blogger.com,1999:blog-3628542335168231161.post-46020654101768661252016-11-19T13:53:00.000-08:002016-11-19T13:57:26.892-08:00Deriving Maxwell's Equations From Heisenberg and Einstein<p><iframe width="560" height="315" src="https://www.youtube.com/embed/LjY1x5CDvD4" frameborder="0" allowfullscreen></iframe></p><p> </p><p>What is the node that connects Heisenberg's uncertainty principle, Einstein's field equations, and Maxwell's equations? To find out, let's start with Heisenberg's uncertainty principle and see what we can derive: </p><p><a href="https://4.bp.blogspot.com/-7D_KjkOOZ7I/WC9_jsDQw0I/AAAAAAAABjw/A6Fh60q-nHcybiYHwxfm5bT3kVYn3widwCLcB/s1600/1.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-7D_KjkOOZ7I/WC9_jsDQw0I/AAAAAAAABjw/A6Fh60q-nHcybiYHwxfm5bT3kVYn3widwCLcB/s1600/1.png" /></a></p><p>We derive a distance (r). We can do the same if we start with Einstein's field equations: </p><p><a href="https://3.bp.blogspot.com/-8_ZcYkIDz6E/WC9_pMMmtnI/AAAAAAAABj0/mnOa66kooXMJ53f3c0H0T-hGcCRxQ-siQCLcB/s1600/2.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-8_ZcYkIDz6E/WC9_pMMmtnI/AAAAAAAABj0/mnOa66kooXMJ53f3c0H0T-hGcCRxQ-siQCLcB/s1600/2.png" /></a></p><p>We can now equate the right side of equation 6) with equation 11). If we do this, we can discover the above-mentioned node. </p><p><a href="https://4.bp.blogspot.com/-ZDvJiu0eMqc/WC9_tUxsT6I/AAAAAAAABj4/kMIbh8v56kA4u3HqiC00HJBSN0sMI00YACLcB/s1600/3.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-ZDvJiu0eMqc/WC9_tUxsT6I/AAAAAAAABj4/kMIbh8v56kA4u3HqiC00HJBSN0sMI00YACLcB/s1600/3.png" /></a></p><p>Equation 15) reveals the "node" to be energy (E). Equation 16) shows that energy not only connects the uncertainty principle and field equations, it shows the electric field is linked as well. From 16) we can derive the integral form of Gauss's law for electric fields: </p><p><a href="https://3.bp.blogspot.com/-aQsaRh6Cifo/WC9_xD-pFaI/AAAAAAAABj8/mHJm68vvLlI9Z-im0tnLnHkAp6vP_GVvACLcB/s1600/4.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-aQsaRh6Cifo/WC9_xD-pFaI/AAAAAAAABj8/mHJm68vvLlI9Z-im0tnLnHkAp6vP_GVvACLcB/s1600/4.png" /></a></p><p>Equation 20) basically says electric charge produces an electric field. The field flux passing through a closed surface is proportionate to the charge contained within the surface. </p><p> If we take equation 20's integral and divide it by a volume (V), we can derive the differential form of Gauss's law for electric fields: </p><p><a href="https://1.bp.blogspot.com/-UCK-jrGZQiM/WC9_1o6TL6I/AAAAAAAABkA/90O2nqe-F30oFnLmq94fwOK6_1-poYYVgCLcB/s1600/5.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-UCK-jrGZQiM/WC9_1o6TL6I/AAAAAAAABkA/90O2nqe-F30oFnLmq94fwOK6_1-poYYVgCLcB/s1600/5.png" /></a></p><p>Equations 22) and 23) are just different ways of saying the electric field produced by electric charge diverges from a positive charge, or it converges upon a negative charge. </p><p>Going back to energy (E), we can begin again and derive Gauss's law for magnetic fields: </p><p><a href="https://1.bp.blogspot.com/-13EXadnPNDY/WC-GX5baarI/AAAAAAAABkQ/uANsRVx9GAQzVZeXt72FVtevYfCr9pKLACLcB/s1600/6.png" imageanchor="1" ><img border="0" src="https://1.bp.blogspot.com/-13EXadnPNDY/WC-GX5baarI/AAAAAAAABkQ/uANsRVx9GAQzVZeXt72FVtevYfCr9pKLACLcB/s1600/6.png" /></a></p><p> </p><p><a href="https://4.bp.blogspot.com/-h8F7mozoz_E/WC-memg_kqI/AAAAAAAABks/KUAfsSNt0oA2tYq8GyWjj9_O-9BvlIK9gCLcB/s1600/7.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-h8F7mozoz_E/WC-memg_kqI/AAAAAAAABks/KUAfsSNt0oA2tYq8GyWjj9_O-9BvlIK9gCLcB/s1600/7.png" /></a></p><p> </p><p><a href="https://3.bp.blogspot.com/-ceTSg9T2aI8/WC-GgjoLtdI/AAAAAAAABkY/AayRM7ejBGQk-o7sCLikQW36pNg5t1lMwCLcB/s1600/8.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-ceTSg9T2aI8/WC-GgjoLtdI/AAAAAAAABkY/AayRM7ejBGQk-o7sCLikQW36pNg5t1lMwCLcB/s1600/8.png" /></a></p><p>Equation 35) is the integral form. It says the magnetic flux passing through a closed surface is zero. According to equation 36), the magnetic field's divergence at any point is zero. </p><p> Given what we have derived so far, we can also derive the integral form of Faraday's law: </p><p><a href="https://4.bp.blogspot.com/-GJMhsWN9qAI/WC-oOp5GQrI/AAAAAAAABk4/iTYO9D6hT8YbiaeMTJ_t3DnqMXjOnscGwCLcB/s1600/9.png" imageanchor="1" ><img border="0" src="https://4.bp.blogspot.com/-GJMhsWN9qAI/WC-oOp5GQrI/AAAAAAAABk4/iTYO9D6hT8YbiaeMTJ_t3DnqMXjOnscGwCLcB/s1600/9.png" /></a></p><p>Equation 44) describes the electric generator. As the magnetic flux through a surface changes, an electric field is induced. </p><p>The differential form of Faraday's law is derived as follows: </p><p><a href="https://3.bp.blogspot.com/-JkxRxI0f1ZE/WC-rdRG7o5I/AAAAAAAABlE/tER9IMWVctMusMg2ec6-uIfWnnE82bmDgCLcB/s1600/10.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-JkxRxI0f1ZE/WC-rdRG7o5I/AAAAAAAABlE/tER9IMWVctMusMg2ec6-uIfWnnE82bmDgCLcB/s1600/10.png" /></a></p><p>According to equation 51), a magnetic field that changes with time produces a circulating electric field.</p><p> Next, we shall derive the integral form of the Ampere-Maxwell law: </p><p><a href="https://3.bp.blogspot.com/-GGvQ8VQzQTc/WC-vcPzEC3I/AAAAAAAABlM/MSGdHDJdLcUqpM1LhVEyQD7xzbn9mLjlACLcB/s1600/11.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-GGvQ8VQzQTc/WC-vcPzEC3I/AAAAAAAABlM/MSGdHDJdLcUqpM1LhVEyQD7xzbn9mLjlACLcB/s1600/11.png" /></a></p><p> </p><p><a href="https://3.bp.blogspot.com/-t8K7XeBf1nk/WC-vgdJ1gKI/AAAAAAAABlQ/asoiN9hL-MwV4c1xQoIRo4SPfoMGVZm5ACLcB/s1600/12.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-t8K7XeBf1nk/WC-vgdJ1gKI/AAAAAAAABlQ/asoiN9hL-MwV4c1xQoIRo4SPfoMGVZm5ACLcB/s1600/12.png" /></a></p><p>Equation 63) says an electric current (I) produces a circulating magnetic field. The differential form is derived as follows: </p><p><a href="https://3.bp.blogspot.com/-lz3uHuaLGDk/WC-vkuF4vQI/AAAAAAAABlU/C4xOp-7WsSMftFBvr3xloIyteJ1gIY0pQCLcB/s1600/13.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-lz3uHuaLGDk/WC-vkuF4vQI/AAAAAAAABlU/C4xOp-7WsSMftFBvr3xloIyteJ1gIY0pQCLcB/s1600/13.png" /></a></p><p>According to equation 67), an electric field that changes with time produces a circulating magnetic field.</p><p> To put a cherry on top of the work we've done so far, we use Heisenberg's uncertainty principal and Einstein's field equations to derive the electromagnetic tensor (of equation 82) below): </p><p><a href="https://2.bp.blogspot.com/-bEhnU-ApKSE/WC-zi1R58tI/AAAAAAAABlg/qQy7mWt4UNs_kHnf7VfPbt9hLL7lKzffQCLcB/s1600/14.png" imageanchor="1" ><img border="0" src="https://2.bp.blogspot.com/-bEhnU-ApKSE/WC-zi1R58tI/AAAAAAAABlg/qQy7mWt4UNs_kHnf7VfPbt9hLL7lKzffQCLcB/s1600/14.png" /></a></p><p> </p><p><a href="https://3.bp.blogspot.com/-oX-At0wHE9w/WC-zskth5HI/AAAAAAAABlk/DUTV9i3LHyEePQEf1B-2465_yHbE5Y1tgCLcB/s1600/15.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-oX-At0wHE9w/WC-zskth5HI/AAAAAAAABlk/DUTV9i3LHyEePQEf1B-2465_yHbE5Y1tgCLcB/s1600/15.png" /></a></p><p> </p><p><a href="https://3.bp.blogspot.com/-idbkOwvd_9A/WC-zxYQEaTI/AAAAAAAABlo/UmF9hWxOgpIlBB70siMP2uZAsRmI7escACLcB/s1600/16.png" imageanchor="1" ><img border="0" src="https://3.bp.blogspot.com/-idbkOwvd_9A/WC-zxYQEaTI/AAAAAAAABlo/UmF9hWxOgpIlBB70siMP2uZAsRmI7escACLcB/s1600/16.png" /></a></p><p>Notice in equation 82) the Schur product was used to multiply the matrices to get the final result: the electromagnetic tensor. </p><p>To learn more about Maxwell's equations, there is an excellent book entitled "A Student's Guide to Maxwell's Equations" by Daniel Fleisch who explains every detail, symbol, and nuance. There's also this video which is also excellent: </p><p><iframe width="560" height="315" src="https://www.youtube.com/embed/AWI70HXrbG0" frameborder="0" allowfullscreen></iframe> </p> GM Jacksonhttp://www.blogger.com/profile/02363192260461368016noreply@blogger.com0