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When we derive gravitational potential it is dependant only on (rest) mass $m$, but i have seen a derivation of gravitational redshift equation placing relativistic mass $\widetilde{m}$ instead of rest mass $m$ like this:

$$ \begin{split} W_1 &= W_2\\ h\nu_1 + 0 &= h \nu_2 + \left(-\frac{GM \widetilde{m}}{r}\right)\\ h\nu_1 &= h \nu_2 -\frac{GM \widetilde{m}}{r}\\ h\nu_1 &= h \nu_2 -\frac{GM h \nu_{2}}{c^2r}\\ \nu_1 &= \nu_2 \left(1 -\frac{GM}{c^2r} \right)\\ \end{split} $$

I think this is wrong! What led me to this conclusion? In my other topic people warned me that i can't use $\frac{1}{2}mv^2$ for a photon. So i think that i have no right to swap $m$ with $\widetilde{m}$ than state that photon has mass and use $\widetilde{m}$ in equation:

$$ W_p = -\frac{GMm}{r} $$

Photon has mass $m=0$ period! And should be unaffected by gravitational field. I know experiments show us different story but still...

QUESTION 1: I would need some explaination on this fragile topic.

QUESTION 2: Could anyone tell me allso when i have the right to use relativistic mass $\widetilde{m}$?

Community
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71GA
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1 Answers1

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Like the Schwarzschild radius, the gravitational redshift formula can only be gotten properly from GR. All the Newtonian analysis, in which you use the Newtonian potential, Newtonian kinetic energy etc., while giving the right answers, must be regarded as just a mnemonic to 'get' the answers.

It's a matter of philosophy really. Once we have identified that a theory is not adequate in more general purposes (Newtonian, here) and is superseded by a theory (GR, here) which is consistent, we must learn not to use the old theory unless we can show that the laws are valid in that regime (for example, Newtonian is valid for $v \ll 1$).

In this case, you have identified that the photon has no mass, so any 'derivation' in which a photon's 'mass' is used is not a derivation. It just so happens that identifying $E = h\nu$ and $E = m$ (not a valid step) and using Newtonian analysis gets you the right answer, but since we have agreed not to use the Newtonian analysis, we must learn to accept that the steps are not valid.

I hope that helped.

nervxxx
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