# Gravitational redshift derivation

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 $\widetilde{m}$?

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Is there a particular reason you are trying to avoid treating these problems correctly? I mean, from time to time it may be pedogogically useful, but... –  dmckee Jan 18 '13 at 19:48
Also, "Photon has mass m=0 period! And should be unaffected by gravitational field." implies that you are assuming that gravitation couples to (rest) mass (a perfectly reasonable place to start if you've learned Newtonian gravity), but in general relativity gravitation couples to the stress-energy tensor which is a more general object and has non-zero components for the photon. –  dmckee Jan 18 '13 at 20:39
I think I touched "flag" by mistake in my tiny, stupid phone touchscreen. Please some administrator undo that flag. I am very sorry. –  Eduardo Guerras Valera Jan 18 '13 at 21:36
Agreed with dmckee. Honestly, you should question whatever the source is coming up with all these mass-based manipulations claiming to get at general relativity. Properly doing anything of this sort involving gravity involves carefully setting up coordinates and applying the geodesic equation. –  Chris White Jan 19 '13 at 10:14
You have "the right to use relativistic mass" within the framework of Special Relativity (Electrodynamics) for instance to calculate the radius of a particle trajectory inside a mass spectrometer, and similar problems where inertial mass takes part but gravity simply doesn't exist. You cannot mix newtonian gravity or the result is nonsense. Finding the solution to that nonsense took Einstein a decade, and the result was the General Theory of Relativity. –  Eduardo Guerras Valera Jan 20 '13 at 3:25

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.