# Relativistic mass of a photon

1. Mar 3, 2005

### eNathan

what is the relativistic mass of a single photon

2. Mar 3, 2005

### kirovman

Try using mass-energy equivilence.

$$E = h f = m c^2$$

3. Mar 3, 2005

### eNathan

How do I plug in those variables?

4. Mar 3, 2005

### dextercioby

What do you mean...?"c" and "h" are fundamental constants.They're known.U need either the photon's energy (E),or its frequency (f),or its wavelength ($\lambda$).Since the photon is a particle,the last 2 #-s refer to the classical em.wave...So the photon is described properly only by E (energy).(actually momentum & helicity in quantum treatment).

Daniel.

5. Mar 3, 2005

### eNathan

I said relavistic mass, not energy. Can someone give me an example? Or is that too much to ask?

6. Mar 3, 2005

### dextercioby

Relativistic mass (for a photon,at least) IS energy.In Heaviside-Lorentz units...

Daniel.

7. Mar 3, 2005

### Staff: Mentor

Visible light is on the order of 500nm, the speed of light is 300,000km/sec, and Planck's constant is 6.626068 × 10-34 m2 kg / s. Plug, chug.

8. Mar 3, 2005

### rbj

nothing is too much to ask. but you might want to accept it when people answer.

the only mass property of a photon comes from

$$E = m c^2 = \hbar \omega$$

resulting in

$$m = \frac{E}{c^2} = \frac{\hbar \omega}{c^2}$$

that's it. the mass of a photon. because the speed of a photon in vacua is $c$ for all observers and because the relationship of relativistic mass to rest mass for any particle is

$$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

where $m_0$ is the rest mass and $v$ is the particle's velocity relative to some observer, if there was any non-zero rest mass, the particle's relativistic mass would be $\infty$ not $\frac{\hbar \omega}{c^2}$, so the rest mass of a photon has to be zero.

we can't spell it out any further for you, as best as i can tell.

r b-j

Last edited: Mar 3, 2005
9. Mar 3, 2005

### eNathan

Russ_Watters, you said
As far as I know, units of speed are always in Meters/Second, so why are you using KM?

10. Mar 4, 2005

### dextercioby

There are exceptions.So-called tolerated units.In the case of velocity:km/h,km/s...For pressure mm col.Hg/Torrs,etc.

Daniel.

11. Mar 4, 2005

### pmb_phy

That's incorrect. relmass is not idetical to energy. This is equality under some specfc cases

The relativistic mass ofa photo can be expressed in two ways one of which is m = p/c.

pete

12. Mar 4, 2005

### dextercioby

I'm not gonna argue with you,Pete,but don't u think it's weird that u're the only person who has this opinion...?

Daniel.

13. Mar 4, 2005

### marlon

How is m = p/c for a photon ??? isn't m divergent here ??? How do you renormalize this ???

What is the second expression ???

marlon

14. Mar 4, 2005

### K.J.Healey

thats a pointless equation

<p> = d/dt(m*x) = mv

m = mv/c, v=c, m=m

15. Mar 4, 2005

### marlon

http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

this should explain a lot
the (invariant or rest-) mass of a photon is ZERO, c'est tout, point final
the relativistic mass depends on the frame of reference, but since all frames of reference are equivalent, only the invariant mass matters...

marlon

16. Mar 5, 2005

### pmb_phy

No. I don't find it weird at all. If I see Rindler state this in no uncertain terms in his text then who am I to argue with Rindler?

Besides, this is something that one can demonstrate rather easily by a simple example which I've stated. The only think weird is that nobody chose to actually look at the problem I stated and then find the energy and mass.

You'd agree that a photon has momentum, 'p', would you not? You'd also agree that the speed of a photon is c, do you not? Then the relation m = p/c is well defined.

The energy of a photon is related to the photons momentum, 'p', by E = pc --> p = E/c. Plug this into "m = p/c" and you'll get m = E/c^2 which is fairly obvious. The relation m = p/c is found in many texts on relativity. E.g. see French's text "Special Relativity" on page 16 where in a foot note French states

*By inertial mass we mean the ratio of linear momentum to velocity.

Last edited: Mar 5, 2005
17. Mar 5, 2005

### pervect

Staff Emeritus
I was reviewing this whole argument, when I ran across Steve Carlip's paper

http://lanl.arxiv.org/abs/gr-qc/9909014

A general argument is presented that for a closed physical system to which the virial theorem applies, the total "gravitational mass" of the system is going to be the total energy of the system, E - the sum of mc^2 for the constituent particles plus the total energy E

Carlip's "gravitational mass" appears to be identical to pmb's "relativistic mass", it's the mass that appears in the low velocity limit of p=mv or f=ma.

Therfore I think it is quite reasonable to say that relativistic mass is just the energy of a system. The only caveat is that the system has to be closed.

18. Mar 6, 2005

### pmb_phy

By the way, besides being in such texts by, say, Rindler and perhaps Mould and it can also be found in Einstein's paper on this from 1906 as well as currently in journals such as the American Journal of Physics. I recall one article called Mass renormalization in classical mechanics by Griffith. He also noted the difference between E/c2 and p/v. What I'm surprised at is that more people aren't familiar with this? Have you ever read Rindler's intro SR text?

I created a page a while back as an example. See

http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm

Pete

19. Mar 7, 2005

### pervect

Staff Emeritus
In what manner does your webpage conflict with Carlips paper, or are you not disagreeing with Carlip?

20. Mar 7, 2005

### Staff: Mentor

I use km because its a prettier looking number. Besides, "kilo" is just a prefix, meaning "thousand" - so km are still meters (dropping the "k" and adding 3 zeroes is not a conversion) - its a very common simplification. I would hope you could reconcile the units when plugging into the equations...

And its not polite to be argumentative when asking a question and people are trying to help you.

Last edited: Mar 7, 2005
21. Mar 8, 2005

### pmb_phy

It appears to me that Carlip is not speaking of any cases in which the energy defined mass is different than the momentum defined mass so no problem arises in that paper that I can see from a quick read. I read it closely last year and have forgotten the details.

Pete

22. Mar 9, 2005

### pervect

Staff Emeritus
Well, as I see it, what comes out of Carlips equations is an isotropic gravitational / inertial mass (which are the same because of the principle of equivalence) which is the same in all directions.

I'm not so sure your example yields an isotropic relationship between momentum and velocity, however your calculation wasn't very detailed so it's unclear what you think.

But I also think that the correct calculation of inertial/gravitational mass will be the same in all directions - that the relationship between system momentum and system velocity will be a scalar and not a tensor.

23. Mar 9, 2005

### pmb_phy

I gave you an example of a case when E/c^2 does not equal p/v. Then you referred to Carlips paper as if it was the same topic. Carlip was speaking of the mass of closed systems. I was not. I was speaking of mass in general and in the case I gave the object is not isolated. Pertaining to the mass of a closed system Rindler states

What part do you not understand. I'll elaborate for you. But the example is quite clear and is not lacking in details.

As for Carlip? ... I'm not 100% sure of the connection but I'll take an educated crack at it: It can be shown that in the week field limit

$$\nabla^2 \Phi = \frac{4\pi G}{3}(\rho + 3p^/c^2) = \frac{4\pi G}{3}\rho_{grav }$$

Notice the inertial mass from the result I obtained here, http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm, for one dimensional motion is identical to $\frac{4\pi G}{3}(\rho + 3p^/c^2)$

So add one for each dimension, i.e. 3, then you'll end up with the expression above for the week field limit.

But that's just a guess. Seems right though.

Pete

Last edited: Mar 9, 2005
24. Mar 9, 2005

### pervect

Staff Emeritus
Well, an example of where your exposition could use some clarification is

"It can be shown that the momentum density as measured in S is given by [1].

"It can be shown that" is not a particularly strong derivation of a result, sorry :-).

But it does look like you are both getting the same results. Carlip gets

m_grav = (integral of) T_00 + T_11 + T_22 + T_33 (using geometric units where c=1).

You get m_grav = (integral of) rho + 3P (again, getting rid of the pesky factors of 'c' by using geometric units.). Assuming that the stress energy tensor being used is that of a perfect fulid, then T_00 = rho, and T_11 = T_22 = T_33 = P.

Looks the same to me.

Carlip goes on further to say some more interesting things about how the virial theorem sorts the whole mess out.

25. Mar 10, 2005

### pmb_phy

I was addressing that one someone who knew the topic and is familiar with that stress-energy momentum tensor.

m_grav = (integral of) T_00 + T_11 + T_22 + T_33 (using geometric units where c=1).

You get m_grav = (integral of) rho + 3P (again, getting rid of the pesky factors of 'c' by using geometric units.). Assuming that the stress energy tensor being used is that of a perfect fulid, then T_00 = rho, and T_11 = T_22 = T_33 = P.

Looks the same to me.

Carlip goes on further to say some more interesting things about how the virial theorem sorts the whole mess out.[/QUOTE]