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### Why the Ground State Ain't Zero

If you are familiar with Max Planck's work or Albert Einstein's photo-electric effect, then you know energy comes in discrete pac...

## Monday, August 28, 2017

### Deriving Newton's Constant G From Quantum Physics

Previously, we did a classical-relativistic derivation of Newton's constant G. Today we will derive it from Schrodinger's equation and Dirac's quantum field equation. Below are the variables we will use:

To derive G from quantum mechanics (which covers low-energy individual particles) we begin with Schrodinger's time-independent equation:

Consider two particles with the same total energy (H), but their potential (V,V') and kinetic (h-bar terms) energies differ. We set their respective Hamiltonians (H,H') equal to each other:

Below we define their respective wave functions:

At equation 6 above we calculated the double derivative of each kinetic-energy term. Below we define the complex conjugate of each wave function. We use these complex-conjugates to find the particles' expectation values (see equation 10):

At equation 11 we equate the particles' kinetic-energy difference with their potential-energy difference. Equation 12 is a quantum-mechanical version of Einstein's field equations: The wave numbers (k,k') have the same units as spacetime curvature.

Equation 15 above is the square of the gravitational velocity (v^2) between the two particles. At equation 16 below it is perfectly legal to multiply the right side by the particles' masses (2m) and the distance (r) between said particles if we also divide by those figures. From there we can derive equation 20.

At equation 20, we can see why Newton's constant (G) is constant. Any change in the mass (m) or distance (r) between the particles causes an offsetting change in the wave-number (k) ratio; i.e., spacetime curvature changes when the distance-mass ratio changes. However, you may have wondered what would happen to G if the two particles had the same wave numbers (and the same potential and kinetic energies). The wave-number difference would be zero (or at some ground state) and so would G!

Quantum field theory to the rescue. From here on, mass (m) shall be the mass (or mass equivalent) of a field within a given volume, not an individual particle. This field could be a vacuum (not necessarily a perfect vacuum) where half-spin particles are created and annihilated. We use Dirac's equation below to assist us:

Below we derive equation 26--the square of the field's energies.

Let's suppose that this field interacts with another field, but this other field contains more mass (m') and a lower average wave number (k'). At 29 below we compare the squared energy differences of these two fields. They may or may not be equal.

Using a similar procedure we once again derive G (see equation 36):

If the two sides of the equation are not equal, we throw in a factor of n to make them so. (If the sides are equal, n = 1.)

Now Newton's constant can be constant even if the two fields have identical masses and wave numbers. A zero difference in wave numbers is offset by a zero difference in mass. (A minimum non-zero ground state also helps G to be constant.) Note that a difference in mass is the mass we measure above a given field mass. For example, when calculating Earth's gravity, the Earth's mass does not include the vacuum field mass (m). However, the total mass (m') includes the Earth and the vacuum field mass (m). Thus the Earth mass is the mass difference or net mass.

## Thursday, August 17, 2017

### Deriving the Gravitational Constant G

Today we will derive the gravitational constant G, also known as Newton's constant. Here are the variables we will be working with:

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.

Our starting point shall be the Lorentz equation, courtesy of Einstein's theory of special relativity:

By doing some algebra we can derive equation 7 below:

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.

We manipulate Einstein's energy equation to get equation 8:

We make a substitution, then do some more algebra until we derive G at equation 15:

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.

## Saturday, August 12, 2017

### How to Derive a Black Hole From Einstein's Field Equations

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:

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.

Let's begin the derivation with equation 1:

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.

Now let's do the subtraction at equation 4 to get equation 5:

At equation 6 we set g equal to 8pi, so we can divide both sides of equation 5 by 8pi to get equation 7:

At 8 we set R equal to 1/r^2 and make a substitution to get equation 9:

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).

Taking equation 12 and applying limits gives the black-hole equation 14:

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.

Below are bonus equations that show the maximum potential velocity is light speed and the maximum rest-mass energy is still mc^2.

## Monday, August 7, 2017

### How to Conserve Dark Energy and the Rest

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.

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:

Equation 1 below shows how photon energy (Ep) is a function of its wavelength (lambda). The bigger lambda gets, the smaller the photon energy.

Equation 2 is dark energy (Ed)--a function of energy density (pd) times volume (V). As volume gets bigger, so does dark energy.

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:

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:

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:

You might recognize the momentum-energy term on equation 7's left side. It appears in this famous equation:

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:

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.

## Tuesday, August 1, 2017

### Adding Gravity to the Standard Model Lagrangian

Here is one version of the Standard Model Lagrangian. Click on the image below to learn more details. Lagrangian L =

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:

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:

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:

We can now see the relationship between gravity and the other forces. If we do a little more algebra, we get an interesting result:

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.