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Deriving the Fourier Transform

In the field of quantum mechanics the Fourier transform shows the relationship between momentum/wave-number space, phi(k), and position sp...

Monday, September 18, 2017

Deriving the Fourier Transform

In the field of quantum mechanics the Fourier transform shows the relationship between momentum/wave-number space, phi(k), and position space psi(x). Equations 1 and 2 below are a typical example of this relationship:

Let's see if we can derive equations 1 and 2 from scratch. We start with perhaps the simplest transforms:

At equation 3 above, we make the right side a function of x by multiplying by e^ikx. If we multiply both sides by e^-ikx, we get equation 4 and equation 4's right side transforms from a function of x to a function of k.

Next, we take equation 3 and find the integral of both sides with respect to k:

We solve the integral on the left side first.

It looks like we are going to get infinity. Darn! We want something finite. Here's what we are dealing with: Imagine a line segment with point zero at the center. The furthest point to the right is an infinite number of points from point zero. Going to the left, there are a minus infinite number of points.

Suppose we bend the line segment into a half circle like this:

We deduce that pi/2 radians is equivalent to an infinite number of points.

We take the square root of both sides of 7 to get 8:

It is most convenient that the square root of infinity is still infinity and the square root of pi/2 radians is still equivalent to an infinite number of points from zero to (pi/2)^.5. To make the math less cluttered we do the following:

Using some high-school algebra we derive equation 14 below:

By repeating the steps above we derive equation 2 (aka: equation 24):

Saturday, September 16, 2017

Gravity is Weak Because of Light Speed

According to the above video, gravity lives outside our universe in a higher dimension or parallel universe. It then leaks into our universe. By the time it gets to us, it has a weak signal. However, using extra dimensions and universes to explain what we don't understand only increases the overall mystery. It also increases the burden of proof of the scientists who postulate such a hypothesis. Surely higher dimensions and parallel universes are more poorly understood than the gravity we can actually observe and measure. For these reasons, they make a poor hypothesis.

Today we are going to explain the weakness of gravity using what we already know. First we define the variables we will be using:

Time for a thought experiment. Imagine a cubic section of our universe. Divide this cube into two halves: A and B. A and B contain a vacuum with a few photons whizzing around with momentum p. A and B each have fairly equal numbers of these photons.

From another part of our universe we take mass m and add it to section B:

At equation 1 above we add section B's momentum and mass energies. It's obvious that A's total energy is less than section B's (see inequality 2 below):

Now, look what happens when we divide both sides by p^2:

Number 3 above is an absurdity. How can light speed (c^2) be less than another speed? According to Einstein, light speed is the maximum speed limit in our universe. Maxwell showed mathematically that light is electromagnetism. The speed of light is a function of the permittivity and permeability of free space. Since these values are constant, light speed is also constant in a vacuum.

To date, no particles have been discovered that can exceed the speed of light. But is light speed really the top speed? The inequality at 3 suggests otherwise. Here's why we currently believe nothing goes faster than light in a vacuum. Below is a Minkowski-ish spacetime diagram.

Velocity vector A is at rest; it is moving through time (t'), so its velocity is zero (see equation 3b). B is moving through both time and space, so its velocity is greater than zero (see 3c). C is moving through space only. It's proper time (t') is zero and its velocity ... are you ready for this? ... is infinite (see 3d).

Of course this infinite speed is from the photon's point of view. We calculate what an observer sees at 3e above. At 3f, we assert that light speed is infinite speed if proper time is used. Can't beat infinity, therefore light speed is the top speed.

Since light speed is the top speed, the inequality at 3 is wrong. Equation 4 below makes better sense:

Notice we simply added a reduction factor of epsilon^2. What can we learn from this? We learn that the full energy of section B does not come into play. It is weakened by the fact that nothing can go faster than light. To restore equilibrium completely, let's multiply the left side by a time-dilation ratio. (To see how this ratio was derived, click here.)

From equation 5 we can derive equation 7 below (Newtonian gravity). Equation 9 reveals the nature of the reduction factor.

Equation 9 tells us that gravity is weak due to time dilation and the ratio of section A momentum to section B momentum. In particular, gravity is weak because momentum p is weak. Why is p so weak? It's the momentum you find in the vacuum of space (with a few photons thrown in). Where there is low vacuum energy there is weak gravity.

Update: Here is a mathematical proof showing why Newton's constant is such a small number:

Friday, September 8, 2017

A New Mathematical Dark Matter Discovery

While fooling around with some gravity equations, I discovered some intriguing results. First, let's define the variables we will use:

Dark matter and dark energy are currently treated as two different entities, but the diagram below shows they could be two sides of the same coin:

The above diagram shows galaxies (circles) being pushed apart by dark energy (arrows). But notice how the dark-energy arrows are pushing in on the center galaxy. Any galaxy in the universe could be deemed the center and experience inward pressure from dark energy. Of course there is also bound to be outward pressure. If the two are equal and opposite, nothing interesting happens. But if the two are unequal, things get mighty interesting!

The next diagram takes a rectangular volume of a spiral galaxy and some surrounding space. The volume is divided into section A, a volume of space, and section B which includes space and baryonic matter (stars, gases, etc.). Each section has a length of distance r. The arrows represent the directions of expanding dark energy.

Since the galaxy is rotating, it has angular momentum q. The angle of this momentum is approximately 90 degrees relative to r. In the next slide we zoom in closer and see more details:

Section A has dark energy, mass m and little or no baryonic matter m' to slow down the expansion momentum (p) along r. Section B has baryonic matter m' and mass m. To conserve energy, outgoing momentum p' is less than incoming momentum p. Mass m normally is attributed to dark matter, but below we redefine it and define the other variables in more detail:

Equation 1 describes the total momentum at the cosmological horizon(light speed * vacuum mass or mass equivalent). Equation 2 shows that dark energy momentum is a function of distance r and Hubble's constant. As r increases, the speed of expansion increases; i.e., dark energy is more kinetic relative to the observer. So momentum p increases. Equation three shows that the converse is true: when r is relatively short, expansion velocity is slower. When r is zero, expansion velocity is zero. In that case dark energy is less kinetic and behaves more like mass or potential energy. So variable m becomes dark-energy rest mass (or something equivalent if it is mass-less). Equation 4 shows variable p' is equal to itself plus an angular momentum q term.

At equation 5 (Einstein's energy equation) below, the left side represents section A, the right--section B. Both sections have distance r which is short compared to the distance to the cosmological horizon. Applying equations 2 and 3 we get a dark energy that has nearly zero momentum (p) and nearly 100% mass (m). We start with the assumption that sections A and B are filled with nothing but dark energy.

At equation 6 we add some baryonic matter (m') and angular momentum (q) to the right side. Doing so reduces dark-energy momentum from p to p'. (Caveat: If momentum is too small, then dark-energy mass is reduced.) Equations 7 and 8 show that, at 90 degrees and on average, the momentum q term is zero. That leaves us with equation 9.

And from equation 9 we can derive equation 22:

You will note equation 22 has an extra mass term which is necessary if you are measuring the gravity of a galaxy. If you measure gravity on solar and planetary scales, the extra mass term becomes insignificant--the equation reduces to Newton's law of gravitation. Also, at the cosmological horizon, the extra term vanishes. Equations 23 and 24 demonstrate these contingencies.

Now let's have some fun! Below are some typical numbers you will find on the web. However, we are going to assume there is no dark matter and substitute dark energy.

We want to calculate percentages and proportions, so we divide the above numbers by 4. That way baryonic matter is equal to 1. We divide dark-energy rest mass again by 2, so sections A and B can have equal portions. We plug and chug below:

As you can see, the result is fairly consistent with dark matter estimates. Now, let's do our galaxy. Here's some more numbers pulled off the web:

We plug and chug again:

According to our calculations, our galaxy consists of only 12% baryonic matter. The extra mass is called dark matter, but at least some of it could be dark-energy rest mass (or mass equivalent).

Sunday, September 3, 2017

A Quantum Gravity Lagrangian without the Graviton

According to Einstein, mass, momentum and energy cause spacetime to curve and curved spacetime tells matter how to move. Since gravity is considered one of the five fundamental forces (interactions), it ought to have its own boson--the graviton.

Unfortunately, particle physicists have had no luck finding this very elusive particle. Is it possible the graviton does not exist? If so, how does gravity work? That's what we will explore today. First, we define some variables:

Imagine a universe devoid of gravitons. Such a universe has mostly photons (radiation), and a little bit of matter here and there. The vacuum of space isn't much of a vacuum. We use Einstein's energy equation (equation 2) to describe the mass, energy and momentum in this universe.

Equation 2's first term represents the photon radiation; the second term is matter. Let's assume this universe is not entirely homogeneous and isotropic. There are places where there is more or less mass, more or less radiation. We compare two such places at equation 3:

Using a bit of algebra, we derive equation 7 below:

Equation 7 reveals an increase in net mass causes the second term's wave-number ratio to shrink. This correlates nicely with the slowing of time and with spacetime curvature. From 7 we can derive Newton's gravitational constant:

Wait! We derived Newton's constant? How is that possible without gravitons? Note the wave numbers we used to derive G are from photons, not gravitons. Doing some more algebra leads to the Lagrangian (L) below:

Equation 17 is the gravity Lagrangian for our universe filled with photons and a little matter thrown in. Still no graviton in sight. When we take partial derivatives with respect to momenta, here's what we get:

Equation 18 shows the photon radiation velocity in a gravitational field. Equation 19 shows the velocity of fermions (mass particles) and satellites in a gravitational field. These results are consistent with our current understanding of gravity. However, we arrived at these results without using gravitons, gravitinos, strings, d-branes, extra dimensions, sparticles and all the fairy dust modern-theoretical physics has to offer.

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.