Dark Energy In - Dark Energy Out = Gravity

In the diagram below there is a body falling to earth. After a distance of dr, what is its instant velocity?

Let's derive a formula that can give us the instant velocity:

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.

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:

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:

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?

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.)

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.

The mathematical proof below confirms the squared velocities we derived above are unaffected by different masses. We start with equation 27:

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:

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:

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.

Now, take a close look at equation 22:

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.

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:

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.

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.

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.

The diagram below is a one-dimensional representation of how gravity emerges from the expanding vacuum:

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'.

The next diagram demonstrates how mass m' causes a tidal effect:

In the above diagram, gravity is weaker above mass m and stronger below. The next diagram demonstrates the gravity of mass m'. (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.)

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:

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.

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).

On a cosmological scale, galaxies are moving apart but they also have gravity. The diagram below shows how this possible:

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).

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.

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.

Update: Here's a mathematical proof showing the correlation between wave number and time rate:

Update: Below are dark energy-gravity equations that take into account Hubble's observations. First we define the variables:

The first equation below is power per area at any location in spacetime. Check out what can be derived from this equation.

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.)

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.

Update: When in a gravitational or dark-energy field, if objects move with space instead of through space, terms like pressure, energy and force only give us units. Read the new post entitled, "Does Gravity Really Exist?--Dark Energy's Equivalence Principle."

Deriving the Schwarzschild Solution to Einstein's Field Equations

Step one: Beginning with Einstein's field equations, derive the Scharzschild radius (equation 13 below):

Next, we call on Pythagoras and a right triangle to derive a basic metric equation (equation 15 below):

Using the same right triangle we derive the Lorentz factor (equation 19 below):

Now check out equation 20:

Because of equation 20, we can make a substitution and derive equations 22 and 23:

Equations 22 and 23 allow us to make more substitutions. The result is something that resembles the Schwarzschild metric (equation 24):

Here's the actual Scharzschild metric:

We can replace equation 24's cdt' with dr (differential radius) to get the following:

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:

So far we've used a triangle with only two space dimensions. We are one dimension short, but we can fix that:

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).

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:

If we take the limit of these variables we get a point in spacetime indicated by the red dot in the diagram below:

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.

We can find the value of each g component within the Scharzschild metric:

Thus the metric tensor is as follows:

With the information we have, we can derive equation 42 below.

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).