#### SimonsCat

**Registered Member**

I've been able to become confident enough in this work now to actually post it officially here as I have had to get it checked out; and because of James' very excellent set of questions, I have been able to insert ''critical question time'' excerpts explaining anything that may be unclear to a reader.

We need to first derive a new definition of binding energy density for gravity, which is what the first part consists of. The second part consists of interpreting the results in terms of Wien's displacement law, which would translate to how redshift (wavelengths of radiation) is effected by the temperature of the system (blackbody) characterized by Weins constant.

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.

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 for a blackbody 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 blackbody 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 Wien's displacement constant $$b$$

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

**abstract**We need to first derive a new definition of binding energy density for gravity, which is what the first part consists of. The second part consists of interpreting the results in terms of Wien's displacement law, which would translate to how redshift (wavelengths of radiation) is effected by the temperature of the system (blackbody) characterized by Weins constant.

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

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

**critical question time**

''why do we need this.''

it helps us understand that the redshift can be described in terms of energy, instead of the usual description of $$\phi$$. It just means that we can expand to find the energy of the system in the same way we find the gravitational corrections from an expansion in $$\phi$$.''why do we need this.''

it helps us understand that the redshift can be described in terms of energy, instead of the usual description of $$\phi$$. It just means that we can expand to find the energy of the system in the same way we find the gravitational corrections from an expansion in $$\phi$$.

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),

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

**critical question time**

''what kind of star are we talking about''

We are speaking generally about most types of stars that can exhibit a redshift. Note that not all stars have the time dilation property, like a quasar.''what kind of star are we talking about''

We are speaking generally about most types of stars that can exhibit a redshift. Note that not all stars have the time dilation property, like a quasar.

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

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

**critical question time**

''What difference of two masses have we taken here?''

This is the difference between $$M_{total}$$ and $$\mathbf{M}$$.''''What difference of two masses have we taken here?''

This is the difference between $$M_{total}$$ and $$\mathbf{M}$$.''

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

**critical question time**

''why are we using a dimensionless gravitational potential.''

It should be interesting to note that only dimensionless parameters in physics are truly physical systems. There is another reason, we have identified that the Schwarzschild metric is dimensionless and the gravitational potential plays this role as well.''''why are we using a dimensionless gravitational potential.''

It should be interesting to note that only dimensionless parameters in physics are truly physical systems. There is another reason, we have identified that the Schwarzschild metric is dimensionless and the gravitational potential plays this role as well.''

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

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:

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

**critical question time**

''whose previous work?''

My own previous work, I worked out this formula a year back.''whose previous work?''

My own previous work, I worked out this formula a year back.

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

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

**critical question time**

''what is the significance of a_g?''

This is known as the gravitational acceleration, which has the same dimensions as a surface charge M/R^2.''''what is the significance of a_g?''

This is known as the gravitational acceleration, which has the same dimensions as a surface charge M/R^2.''

Due to equivalence principle, the temperature is

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

**critical question time**

''How does this follow from the eqivalence principle''

When a detector is uniformly accelerated through spacetime, with an acceleration a_g, it registers a thermal blackbody radiation at temperature characterized by the equation above. It has been come to be named as the Unruh effect.''How does this follow from the eqivalence principle''

When a detector is uniformly accelerated through spacetime, with an acceleration a_g, it registers a thermal blackbody radiation at temperature characterized by the equation above. It has been come to be named as the Unruh effect.

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

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

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

**critical question time**

''which second radiation law?''

https://en.wikipedia.org/wiki/Planck's_law#First_and_second_radiation_constants

''which second radiation law?''

https://en.wikipedia.org/wiki/Planck's_law#First_and_second_radiation_constants

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 blackbody 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 Wien's displacement constant $$b$$

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

**critical question time**

''what is Wiens law?''

https://en.wikipedia.org/wiki/Wien's_displacement_law

''what is Wiens law?''

https://en.wikipedia.org/wiki/Wien's_displacement_law

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