Gravitational Redshift, Temperature and Weins Law

Discussion in 'Alternative Theories' started by SimonsCat, Jan 20, 2017.

  1. SimonsCat Registered Member

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

    Part One

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

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

    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

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


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

    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 \(b\)

    \(b = T \lambda^{\max}_g\)
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  3. SimonsCat Registered Member

    Just a small correction, just looked through,

    \(W = \frac{Gm^2}{\lambda_g} = 2 \pi S T \)
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  5. SimonsCat Registered Member

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  7. SimonsCat Registered Member

  8. paddoboy Valued Senior Member


    The abstract:

    On the Interpretation of the Redshift in a Static Gravitational Field

    The classical phenomenon of the redshift of light in a static gravitational potential, usually called the gravitational redshift, is described in the literature essentially in two ways: on the one hand the phenomenon is explained through the behaviour of clocks which run the faster the higher they are located in the potential, whereas the energy and frequency of the propagating photon do not change with height. The light thus appears to be redshifted relative to the frequency of the clock. On the other hand the phenomenon is alternatively discussed (even in some authoritative texts) in terms of an energy loss of a photon as it overcomes the gravitational attraction of the massive body. This second approach operates with notions such as the "gravitational mass" or the "potential energy" of a photon and we assert that it is misleading. We do not claim to present any original ideas or to give a comprehensive review of the subject, our goal being essentially a pedagogical one.

    Hmmm, I see it actually as six of one and half a dozen of the other.......
  9. SimonsCat Registered Member

    Let's just revisit an equation, we take a look back at a formula derived which was equivalent to the second radiation law, or can be seen as the same thing as a product of the radius and the Hawking temperature \(R_H T_H\) and takes the form of:

    Just to add some physics to this, the current density is related to the RHS via

    \(\nabla \phi = \frac{4 \pi G}{c^2}\ j\)

    Just need to make a quick correction to this, because we use dimensionless form of \(\phi\). This is simply averted by stating the previous equation, in a new way

    \(\nabla \phi = \frac{4 \pi G}{c^4}\ j\)

    and actually it seems very appealing to express the equation like this, as we have retrieved the Einstein factor

    \(k = \frac{4 \pi G}{c^4}\)


    \(\phi = \frac{Gm}{c^2R}\)

    so we gain an extra factor of \(c^2\) on the RHS and we have tweaked this.

    but what we really have is

    \((\nabla \cdot \nabla) \phi =\nabla \cdot \frac{4 \pi G}{c^4}\ j\)

    which is just

    \(\Delta \phi = \nabla \cdot \frac{4 \pi G}{c^4}\ j\)
  10. SimonsCat Registered Member

    There is quite a bit to digest in this next post so you might want to make a cuppa! The energy we defined was:

    \(E = \frac{c^4}{2G} \int dV\ \phi \Delta \phi\)

    where \(\phi\) is dimensionless. Think of it this way

    \(E = \frac{c^4}{2G} \int dV\ \frac{\phi}{c^2} \Delta \frac{\phi}{c^2}\)

    keep one of the potentials dimensionless:

    \(Ec^2 = \frac{c^4}{2G} \int dV\ \phi \Delta \frac{\phi}{c^2}\)

    where the last \(c^2\) factor is taken into account for this rendition of the formula

    \(\Delta \phi = \nabla \cdot \frac{4 \pi G}{c^4}\ j_m\)

    The current is defined here as:

    \(j_m = \frac{mv}{t} = \frac{d}{dt} \int dV\ \rho \mathbf{v}\)

    \(E = \frac{c^4}{2G} \int dV\ \frac{\phi}{c^2} \Delta \frac{\phi}{c^2}\)

    distribute one speed of light squared (just for now)

    \(Ec^2 = \frac{c^4}{2G} \int dV\ \phi \Delta \frac{\phi}{c^2}\)


    \(\Delta \phi = \nabla \cdot \frac{4 \pi G}{c^2}\ j_m\)

    then divide \(c^2\) on both sides to give

    \(\Delta \frac{\phi}{c^2} = \nabla \cdot \frac{4 \pi G}{c^4}\ j_m\)


    \(Ec^2 = \frac{c^4}{2G} \int dV\ \phi \Delta \frac{\phi}{c^2}\)

    we replace directly now into our equation we get:

    \(Ec^2 = \frac{c^4}{2G} \int dV\ \phi \nabla \cdot \frac{4 \pi G}{c^4}\ j_m\)

    cancel like terms:

    \(\frac{c^4}{2G} \frac{4 \pi G}{c^4}\ = 2 \pi\)

    which leaves:

    \(Ec^2 = 2 \pi \int dV\ \phi \nabla \cdot j_m\)

    \(c^2\) can be absorbed back into the remaining potential making it dimensionless again and we have;

    \(E = 2 \pi \int dV\ \Phi \nabla \cdot j_m\)

    You can check the dimensions of this equation, I calculate it as:

    \(E = R^3 \frac{1}{R^2}\ \frac{mv}{t}\)

    which appears consistent, by noticing that the RHS simplifies to a simple:

    \(E t = R^3 \frac{1}{R^2}\ mv = mvR = \hbar\)

    If you have kept up so far, good. Now, how does this translate into the gravitational binding equation in the OP?

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

    by defining the energy density from the previous work:

    \(\rho_{energy} = 2 \pi \phi \nabla \cdot j_m\)

    the entire RHS is completely changed where the binding energy is now related to momentum rate of change (or also called the momentum current),

    \(\delta E_{binding} = 2 \pi \phi( \nabla \cdot j_m - \frac{\nabla\ \cdot\ j_m}{(1 - \frac{2Gm}{E}\frac{M}{R})})\)

    just to finish, an important equation to understand that I haven't covered is the continuity equation.
    Last edited: Jan 22, 2017
  11. Xelasnave.1947 Valued Senior Member

    I don't pretend to follow but may I ask now that you arrive at the last equation how would you apply it and or could you provide an example of it in an application to a specific problem in astrophysics.

  12. SimonsCat Registered Member


    ''I don't pretend to follow but may I ask now that you arrive at the last equation''

    Well, it came be noticing the curl of the potential \(\nabla \phi\) is just

    \(\nabla \phi = \frac{4 \pi G}{c^4}\ j\)

    from standard theory, but in our energy binding equation, we have the presence of an extra \(\nabla \phi\) which is the same as \(\phi \Delta \phi \) so I drew on these relationships to derive what I think would be the right form of the equation.

    ''how would you apply it and or could you provide an example of it in an application to a specific problem in astrophysics.

    I am actually still looking into this.
  13. SimonsCat Registered Member

  14. SimonsCat Registered Member

    My friend Matti Pitkanen (creator of geometrodynamics) took a look as well and made a valuable contribution of knowledge:

    \(Q= \in j^0 \sqrt{g_4} dV_3, j^0\)

    is time component of conserved current associated with time translation

    \(x^{\alpha} \rightarrow x^{\alpha} +\epsilon \delta^{\alpha}_0 (Kronecker\ delta)\) .

    This gives

    \(j^0= T^{00}g_{00}\).

    And this in turn gives

    \(E= \int T_{00}g^{00} \sqrt{g_4} dV_3\)
  15. SimonsCat Registered Member


    Just thinking about your question again... and I don't have an exact application yet but there is something to note about momentum flux, and that is, it is defined by the equation of continuity

    \(\nabla \cdot j_m = -\frac{\partial \rho}{\partial t}\)

    which does indeed have applications in cosmology. The equation above is straight from hydrodynamics.
  16. Xelasnave.1947 Valued Senior Member

    Thanks but "now" was correct I did not mean "how".
    But my main question was really how or where you would use it.
    Thank you.
  17. SimonsCat Registered Member

    I felt like a cheated you a bit, so I have given you a little something to work with in post 12.
  18. Xelasnave.1947 Valued Senior Member

    I don't understand simple maths I was hoping you could show how all your work solved some astronomy problem.
  19. Xelasnave.1947 Valued Senior Member

    Thank you
  20. SimonsCat Registered Member

    My other threads attempt to solve problems. This thread was never really about solving a cosmological problem, but instead, it was supposed to draw upon relationships between temperature, entropy and Wein's Law. My thread had somewhat evolved since then, and I was just looking into how that gravitational binding energy translated into the momentum flux.
  21. Q-reeus Banned Valued Senior Member

    Thanks for letting me know right there you do not know what you are talking about. Thought it best to skim through till noticing a particularly obvious giveaway.
  22. exchemist Valued Senior Member

    When I look up geometrodynamics, I find references to well-known people such as John Wheeler and of course Einstein. Unaccountably, there seems to be no mention of any Matti Pitkanen:

    If I look up Matti Pittkanen, he appears to be a cross-country skiier:änen

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  23. SimonsCat Registered Member


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