1. Jul 9, 2006

### RabidSmurf

This may have been asked before, but nonetheless I am unable to find it if so.

We know Einsteins theory of relativity is confirmed experimentally, and so any postulations made in theory must be true. I guess what I am saying is, I am having difficulty understanding why nothing can travel at the speed of light and yet photons can travel at the speed of light.

from E=mc^2 we know that Energy and Mass are basically equivalent, and that the faster something travels the more mass it gains, thus at 99.999%
the speed of light, it's mass is increased by a factor of 224, and at 99.9999999999% it's mass is increased by a factor more then 70,000, and at the speed of light it's mass is increased by a factor of infinity, and therefore would require an infinite amount of energy to push anything to the speed of light. Thus making it impossible for anything to travel at the speed of light.

Now I know a photon is massless and this is why photons and gravitons (if they exist) are the exception to this rule, but it still carries energy, my question is, if energy is equivalent to mass then how is it that a photon can travel at the speed of light? Does this not violate the very laws set forth in Special Relativity, or more likely, have I missed something very fundamental in the theory itself?

Last edited: Jul 9, 2006
2. Jul 9, 2006

### Mute

An object's total energy is given by E^2 = (pc)^2 + (mc^2)^2, where m is the object's rest mass. E = mc^2 is the rest energy of an object (i.e., its total energy in the reference frame where it is at rest); photons are never at rest in any frame, so they have no rest energy. Thus a photon's total energy is E = pc, which is consistent with the photon's mass being zero (although I guess it's a bit more subtle than that given that p = mu/(1-u^2/c^2)^(1/2), which goes to 0/infinity for m=0, v = c. Of course, that's an indeterminate form so can give way to a definite value.)

Last edited: Jul 9, 2006
3. Jul 9, 2006

### masudr

I guess the problem arises from the understanding of the equation $E=mc^2.$ It's correct form is

$$E_{\mbox{rest}}=m_0c^2,\ \ \ \ \mbox{(1)}$$

where E is the energy of a particle in it's rest frame, and $m_0$ is it's rest mass.

Now, of course, the photon is moving at speed c in any frame, and has no rest mass, so the above relation clearly cannot apply.

The more correct relation is (as Mute has pointed out)

$$E^2 = m_0^2c^4 + p^2c^2\ \ \ \ \mbox{(2)}$$

which is valid in any frame. Here, p is given by

$$p=\gamma m_0 v \ \mbox{ where } \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

Alternatively, you can write

$$E=\gamma m_0 c^2$$

but it has less physical content than eqn. (2)

4. Jul 10, 2006

### pmb_phy

You're confusing the meaning of m in E = mc2. This m is the (relativistic) mass of the particle. It is the rest mass of a photon which is zero, not the mass. You can think of mass, m, as being defined by the relationship p = mv. When the particle is a tardyon (a particle which can travel at v < c) then the mass as measured by an observer is a function of velocity of the particle as measured by that observer. If the particle is luxon (a particle which travels at v = c) then m = p/c. When, as for a photon, E = pc then m = E/c2.

Pete

Last edited by a moderator: Jul 10, 2006
5. Jul 10, 2006

Staff Emeritus
This is your standard answer of course, and I think you owe it to the newbies to frankly state it's a minority view. The majority view is that $$e^2 = p^2c^2 + c^2m^4$$, the right side of which is just the squared magnitude of the momentum energy vector, and hence an invariant, is true of all inertial particles, and massless ones too. Just plug in the invariant mass 0 to retrieve the equation e = pc, or plug in p = 0 to retrive e= mc^2 in the massive particle's rest frame.

6. Jul 10, 2006

### pmb_phy

Please clarify. What is the "it" in "it's a minority view"? I made no statement or assertion of a particular view. It appears to me that your complaint is merely a disagreement on my use of notation, i.e. I use m to mean "relativistic mass" whereas you use it to mean "proper mass." In my humble opinion one should first clarify the meaning of a term and then use that term consistently. Using a qualifier as I did made my post easier to read and write. Its akin to defining the term "momentum" in a quantum mechanics text. A good QM text will let the reader know that the term "momentum" refers to canonical momentum rather than linear momentum. Once the author has clarified the issue he need not do so again.

Beyond that I choose not to participate in discussions on the meaning of "mass." If I need to state my viewpoint then I simply post a link to an article I wrote. In that way I waste no space and people can choose to read it if they're interested or ignore it if they are not.

In any case I no longer post often enough to know who is a newbie and who isn't.

Pete

Last edited: Jul 10, 2006
7. Jul 10, 2006

### actionintegral

Please bear with me. I am one of the slower-witted participants in this newsgroup, but my heart is in the right place.

Your first sentence is quite meaningless!

My starting point for the theory of special relativity is the following:
The speed of light is the same for all observers.

8. Jul 10, 2006

### pmb_phy

To be exact: The speed of light is the same for all inertial observers. There is a significant difference.

Pete

9. Jul 10, 2006

### George Jones

Staff Emeritus
The (local physical) speed of light is the same for all (inertial or non-inretial) observers.

10. Jul 10, 2006

### pmb_phy

...whereas the coordinate speed of light is not.

Pete

11. Jul 10, 2006

### clj4

I am sorry , Pete but you cannot use m=E/c2 for the photon.
The reason is very simple, the above relationship would give a "photon mass" of about 3eV. In reality, there are experiments that put an upper limmit on "photon mass" of 6*10^-17 eV.
The above should close the door on the issue of "photon mass". The photon is a massless particle.

A limit on the photon mass can be obtained through satellite
measurements of planetary magnetic fields. The Charge Composition
Explorer spacecraft was used to derive a limit of 6x10^-16 eV with high
certainty. This was slightly improved in 1998 by Roderic Lakes in a
laborartory experiment which looked for anomalous forces on a Cavendish
balance. The new limit is 6x10^-17 eV. See here:

http://pdg.lbl.gov/2005/tables/gxxx.pdf

Studies of galactic magnetic fields suggest a much better limit of less
than 3x10^-27 eV but there is some doubt about the validity of this
method.
References:

[1]E. Fischbach et al., Physical Review Letters, 73,
514-517 25 July 1994.

[2]http://pdg.lbl.gov/2005/tables/gxxx.pdf

12. Jul 14, 2006

### pmb_phy

Sorry clj4, but whether m=E/c2 can be used for a photon depends on a precise definition of m. And as I mentioned above this is possible when m is relativistic mass. It appears to me from this comment that you totally ignored the explanation regarding this point, i.e. that this m when used in this way, is relativistic mass. Or do you claim that the relativistic mass of a photons is zero?
As I mentioned above, this depends on what one means by "mass." 3eV/kg for a photon is the relativistic mass of a photon defined as m = p/v. A zero photon mass refers to the proper mass of the photon which is indeed zero.
Nope. All you've effectively done, in essenance, is to choose a definition of Mass = m for which m = 0 for luxons (e.g. photon). Look at any text which uses the term "mass" to mean "relativistic mass" and you'll find that the author states that a photon has "mass". E.g. See

Relativity; Special, General and Cosmological, Wolfgang Rindler, Oxford University Press (2001), page 120
Introducing Einstein's Relativity, Ray D'Inverno, Oxford University Press (2002), page 50.
Special Relativity, A.P. French, Norton Press (1968), page 20.
Special Relativity, Albert Shadowitz, Dover Pub. (1968), page 91.
Basic Relativity], Richard A. Mould, Springer Verlag (1994), page 120.

For further information please see the article I wrote on this topic at

http://www.geocities.com/physics_world/mass_paper.pdf

That article addresses all aspects of this topic that I could think of and have seen discussed. Since this topic is discussed too often I will no longer respond to comments in this thread on this point. For further discussion/questions/comments please PM me as much as you'd like.

Pete

Last edited: Jul 14, 2006
13. Jul 14, 2006

### clj4

There is no point in talking about "the relativistic mass of a photons"

Do you see that anywhere in today's mainstream physics? What you see is this:

http://pdg.lbl.gov/2005/tables/gxxx.pdf

...and only this.

We all know that. The point is that very few, if any (maybe you being the exception) talk about the "photon relativistic mass"

I am very familiar with your paper (did AmJPhys publish it?) and with your excellent website. Let's agree to disagree on this one, ok? :-)

Last edited: Jul 14, 2006
14. Jul 14, 2006

### pmb_phy

No. They didn't think that it was a good enough topic for their journal. Also, AJP has a kind of hidden policy lately. When they send the peer-reviewers a paper whose topic is relativity it comes with the instruction "Try to reject this paper." One of the refs wasn't that familiar with the concept of mass so he made some serious mistakes in his review.

I had already stated that I wouldn't discuss that topic outside of PMs. You may interpret that on agreeing to disagree if you wish. If you want a response to the comments above then see you in PM.

Pete

15. Jul 14, 2006

### clj4

Interesting, do you know this for a fact? Does it come from the chief editor?

16. Jul 14, 2006

### pmb_phy

Read your PM messages. That is where I responded to this question.

Pete

17. Jul 15, 2006

### higheriration

Here's one for you. Let's say that you are traveling at a fraction of the speed of light, for the sake of argument we'll put it at 51% (approx. 92,820 mps) in an easterly direction, and you happen to pass some object (a 1957 chevrolet bel-air) travelling in a westerly direction moving at a speed of -for the sake of simplicity- 92,820 miles per second, relatively speaking would you percieve the vehichle to be moving faster than the speed of light? I digress.

Last edited: Jul 15, 2006
18. Jul 15, 2006

### pmb_phy

In special relativity one never measures anything moving at speeds greater than the speed of light. It seems that you're using the Galilean velocity transformation (pre-relativity) when you should be using the Lorentz transformation of velocities, which is relativistically correct.

Pete