Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\widetilde{m}$?

share|cite|improve this question

closed as unclear what you're asking by ACuriousMind, MAFIA36790, DilithiumMatrix, CuriousOne, Asher May 25 at 17:24

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

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
up vote 4 down vote accepted

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.

share|cite|improve this answer
Thank you. It helped a lot. I haven't yet learned GR, so i think this will be my next book on Amazon. It is sad our professor used above derivation (which is wrong but accidentaly gives the right anwser) but never warned us about this. I really don't know why did i deserve negative points... – 71GA Jan 20 '13 at 14:01
No problem. I personally didn't downvote you, so I don't know about that.. Actually, to add a few more comments on your question: so one can derive that redshift formula you wrote, which is a first order approximation to the actual formula in a Schwarzschild metric, by assuming that 1) the Einstein equivalence principle, which states that the laws of physics reduce to those of special relativity in small regions of spacetime, 2) the gravitational fields are weak. the upshot of 1) is that you can't 'tell' gravity from acceleration,, and of 2) is that you can describe the gravitational – nervxxx Jan 20 '13 at 22:48
field by a Newtonian potential, like you have done. but nowhere in that derivation will one require a mass of the photon. This is detailed in Sean Carroll's book: Spacetime and Geometry, an introduction to General Relativity, in chapter 2, which I recommend. – nervxxx Jan 20 '13 at 22:51

Not the answer you're looking for? Browse other questions tagged or ask your own question.