#### SimonsCat

**Registered Member**

**Part One**

We need to establish the new metric coefficient, which is just a reinterpretation of the Schwarzschild factor:

$$1 - \frac{2GM}{E} \frac{M}{R} = 1 - \frac{E_g}{E}$$

The gravitational field inside a radius $$r = r(0)$$ is given as

$$\frac{dM}{dR} = 4 \pi \rho R^2$$

and the total mass of a star is

$$M_{total} = \int 4 \pi\rho R^2 dR$$

and so can be understood in terms of energy (where $$g_{tt}$$ is the time-time component of the metric),

$$\mathbf{M} = 4 \pi \int \frac{\rho R^2}{g_{tt}} dR = 4 \pi \int \frac{ \rho R^2}{(1 - \frac{2Gm}{E}\frac{M}{R})} dR$$

The difference of those two mass formula is known as the gravitational binding energy:

$$\Delta M = 4 \pi \int \rho R^2(1 - \frac{1}{(1 - \frac{2Gm}{E}\frac{M}{R})}) dR$$

And so, an energy can be obtained by the distribution of the speed of light squared:

$$Mc^2 = 4 \pi \int \frac{\rho c^2}{(1 - \frac{2Gm}{E}\frac{M}{R})} dV = 4 \pi \int \frac{T_{00}}{(1 - \frac{2Gm}{E}\frac{M}{R})} dV$$

The integration is worked out from the following equation:

$$T_{00} dV = \frac{c^4}{8 \pi G} \int dV\ \nabla^2 \phi^2 = \frac{c^4}{8 \pi G} \int dV\ \phi \Delta \phi$$

Plugging this into our mass formula we have a new equation and the final one for this post:

$$E(density) = \frac{c^4}{2G} \int \frac{\phi \Delta \phi}{(1 - \frac{2Gm}{E}\frac{M}{R})}$$

Where direct substitution has simplified the equations presence of the pi-symbol and the remaining quantity $$\frac{c^4}{2G}$$ is exactly 1/2 the classical upper limit of both electromagneism and gravitation.

**Part Two**

It wasn't explained previously, but note that in the integration:

$$T_{00} dV = \frac{c^4}{8 \pi G} \int dV\ \nabla^2 \phi^2 = \frac{c^4}{8 \pi G} \int dV\ \phi \Delta \phi$$

... we use a dimensionless gravitational potential

$$\phi = -\frac{Gm}{c^2R}$$

which means

$$\delta E_{binding} = \frac{c^4}{2G} \int dV\ \phi \Delta \phi - \frac{c^4}{2G} \int dV\ \frac{\phi \Delta \phi}{(1 - \frac{2Gm}{E}\frac{M}{R})}$$

where

$$1 - \frac{2GM}{E} \frac{M}{R} = 1 - \frac{E_g}{E}$$

In previous work, the gravitational binding energy (density) of a distant object, like a star or even a black hole was given in previous work as:

$$\delta E_{binding} = \frac{c^4}{2G} \int \phi \Delta \phi - \frac{c^4}{2G} \int \frac{\phi \Delta \phi}{(1 - \frac{2Gm}{E}\frac{M}{R})}$$

Keep in mind, this full extension involving the Schwarzschild factor (written in terms of energy) is the correct way to describe the physics, but to keep this work nice and simple for the eye, we will work with only the first term (so if you want the full equation, just write it out), the equation we use in this work is a simple energy equation:

$$E = \frac{c^4}{2G} \int \phi \Delta \phi$$

Dividing the mass on both sides, and then using the relationship $$a_g = \frac{c^4}{2Gm}$$

$$\frac{1}{V}c^2 = a_g \int \phi \Delta \phi$$

distribute $$\frac{\hbar}{2 \pi k_B c}$$

$$\frac{\hbar c}{2 \pi k_B} = \frac{\hbar a_g}{2 \pi k_B c} \int dV\ \phi \Delta \phi$$

Due to equivalence principle, the temperature is

$$T = \frac{\hbar a_g}{2 \pi k_B c}$$

$$\frac{\hbar c}{k_B} = 2 \pi T \int dV\ \phi \Delta \phi$$

Which is equivalent to the second radiation law. The Hawking Bekenstein relationship for entropy is:

$$\frac{S}{k_B} = \frac{1}{\hbar c} Gm^2$$

if on the RHS $$\int dV\ \phi \Delta \phi$$ corresponds to a gravitational wavelength then

$$\frac{Gm^2}{S} = 2 \pi T \lambda_g$$

and a work equation

$$W = Gm^2 = 2 \pi S T \lambda_g$$

Which is a form of Weins displacement law (for gravitation) which states that a blackbody radiation curves at different temperatures for a system like a star or a black hole, has been written gravitationally for fittingly, the entropy of the system. Clearly entropy and temperature have already had profound relationships established for them.

$$\lambda^{max}_{g} = \frac{\hbar c}{x} \frac{1}{k_BT}$$

$$\lambda^{max}_{g}T = \frac{\hbar c}{k_B x}$$

$$x$$ is a dimensionless parameter which actually depends on the following:

$$x = \frac{\hbar c}{\lambda_g k_BT}$$

And it must also be mentioned that the product $$T \lambda^{\max}_g$$ is the same thing as Wein's displacement constant $$b$$

$$b = T \lambda^{\max}_g$$