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I have been really staring for a while in a MP-Beiser book and I totally disagree with a statement he does there. On a page 85 he states that photons act as they have a mass $m$. He derives this by stating that:

$$ \begin{split} p &= m v\\ \frac{h\nu}{c} &= m c\\ m&= \frac{h \nu}{c^2} \end{split} $$

But I totally disagree with this. We have learned and derived that momentum of a particle is:

$$ \begin{split} p &= m v \gamma (v)\\ \frac{h \nu}{c} &= m c \gamma{(c)}\\ m &= \frac{h \nu}{c^2 \underbrace{\gamma(c)}_{=0}}\\ m &= 0 \end{split} $$

Something here is totally wrong, but what? How can an author state what he does? I know that on Harvard they did an experiment resulting in different $\nu$ of a photons falling in a gravitational field, but they must have been wrong or something... Please someone explain.

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$\gamma(c)$ is not defined, the momentum of a photon is simply $E/c$ –  Jorge Jan 16 '13 at 16:03
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You are right $\gamma(c) = \frac{1}{\sqrt{1 - \frac{c^2}{c^2}}} = \frac{1}{\sqrt{1-1}} = \frac{1}{\sqrt{0}} = \frac{1}{0}$ still the mass remains zero: $m = \frac{h \nu}{c^2 \frac{1}{0}}=\frac{h \nu 0}{c^2 1} = \frac{0}{c^2}=0$ –  71GA Jan 16 '13 at 16:18
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What book is that? Googling only turns up music albums for me... –  Chris White Jan 16 '13 at 19:39
    
Arthur Beiser - Concepts of modern physics –  71GA Jan 16 '13 at 19:44
    
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2 Answers

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The author is wrong and you're right--photons have no (rest) mass. You can create schemes where it will seem like photons have mass--for instance, if you have two identical photons travelling in opposite directions enclosed in a box, it will appear that the box is holding a mass equal to the two photons' energy.

But, fundamentally, photons have no mass, and while they are affected by gravitational fields (for example, in the famous Arthur Eddington light bending observation), they travel in a fundamentally different way than massive particles (for example, there are no stable orbits of a photon around a black hole).

EDIT:

This actually is much simpler than I thought it was going to be after putting a bit of thought in (note that in everything below, $M$ is code for $\frac{GM}{c^{2}}$ and $t$ is code for $ct$). From the time translation invariance of the schwarzschild metric, we know that the conserved energy of any motion is:

$$E = {\dot t}(1-\frac{2M}{r})$$

Furthermore, the null condition for a radially inward/outward light ray is:

$$0 = -{\dot t}^{2}(1-\frac{2M}{r}) + \frac{{\dot r}^{2}}{1-\frac{2M}{r}}$$

Which we can easily solve (assuming $r \neq 2M$), finding $r = Es + r_{0}$

Where $E$ is still the energy and $r_{0}$ is the radius where we start measuring the null parameter $s$${}^{1}$

Plugging this result back into our equation for $\dot t$, solving for $t$ and integrating, we find:

$$t = Es + r_{0} -2M + 2M\ln(Es +r_{0}-2M) + C$$

Where C is an integration constant. Now, we note that if $s=0$ when $r=2M$, this implies that $r_{0}=2M$, but then, this implies that $t=-\infty$, and we conclude that there is no outgoing null geodesic connecting the horizon to any $r>2M$ (note that, strictly, I should be using coordinates that aren't singular on the horizon like these, but the result would still follow, because if you were using the proper null coordinates, you'd still have $E = {\dot t}(1-\frac{2M}{r})$, which is the source of this problem.

${}^{1}$Note that we are free to rescale $s\rightarrow \lambda s + k$ and none of our results will change except perhaps a rescaling of the integration constants $r_{0}$ and $C$

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There is a problem. Author uses this equation to derive further equations which show a gravitational red shift: $\nu_2 = \nu_1 \left( 1 - \frac{GM}{c^2R} \right)$ Does this mean that these equations are wrong? I mean what follows out of this equation is a famous Schwarzschild radius. I am very surprised i find such contradictions in a modern physics. –  71GA Jan 16 '13 at 15:54
    
The gravitational redshift is real: it was first measured in the 1960's (in an elegant experiment that is worth reading about), and since then has been shown to agree quite well with the prediction. –  dmckee Jan 16 '13 at 16:11
    
I can't believe this if i don't see a derivation. –  71GA Jan 16 '13 at 16:12
    
@71GA: You'll find the derivation of the red shift in any introductory book on GR, or just Google for it. –  John Rennie Jan 16 '13 at 16:49
    
@71GA: what do you want a derivation of? And I should say that there are a lot of ways to derive correct results incorrectly. And that redshift factor is wrong. The frequencies should be multiplied by $\sqrt{1-\frac{2GM}{c^{2}R}}$ –  Jerry Schirmer Jan 16 '13 at 17:49
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There is a lot of confusion here. Photons have no rest mass, but they definitely have a mass from $E=mc^2$. Surprisingly, this is the relevant mass when talking about gravity! However, it's true that, using GR, the problem is more complicated and, as Jerry Schirmer says, you will have to use a completely different family of geodesics.

But I can't see where the author is wrong. People gets confused because GR physicists define mass as relativistic mass, while, for example, particle physicists define it as rest mass.

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GR physicists rarely talk about mass, preferring mass density, and even then energy density is usually what is meant. –  Jerry Schirmer Jan 16 '13 at 17:52
    
On this page i read that even Einstein warned against the concept of mass increasing with velocity. Unfortunately this warning was ignored. physicsandphysicists.blogspot.com/2009/04/… –  71GA Jan 16 '13 at 18:09
    
When you say: "but they definitely have a mass from $E=mc^2$" you should know that $m$ is the rest mass here and as you stated yourself the rest mass is zero so $m=0$. –  71GA Jan 16 '13 at 18:19
    
Sorry, but in that relation $m$ is absolutely NOT the rest mass. In fact, you can see people expanding E in series with respect to velocity. You obtain a first term which is $m_oc^2$ (rest energy), then $\frac{1}{2}m_0v^2$ (kinetic energy) and then higher order terms. See for example en.wikipedia.org/wiki/…. @71GA, if you say that in $E=mc^2$ $m$ is rest mass, then you're saying that photons never have energy. –  Bzazz Jan 16 '13 at 19:15
    
$E$ in the term $E = mc^2$ is a "rest energy" $E_0$ which is an energy which only particles with mass have. Good example is pair production. Photon has no mass and therefore has no rest energy right which is correct i think. –  71GA Jan 16 '13 at 19:54
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