Question About Special Relativity

In summary: If you want to get technical, you could say that photons have zero rest mass, but they do have a finite energy (due to the Heisenberg Uncertainty Principle).
  • #1
RabidSmurf
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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?
 
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  • #2
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.)
 
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  • #3
I guess the problem arises from the understanding of the equation [itex]E=mc^2.[/itex] It's correct form is

[tex]E_{\mbox{rest}}=m_0c^2,\ \ \ \ \mbox{(1)}[/tex]

where E is the energy of a particle in it's rest frame, and [itex]m_0[/itex] 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)

[tex]E^2 = m_0^2c^4 + p^2c^2\ \ \ \ \mbox{(2)}[/tex]

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

[tex]p=\gamma m_0 v \ \mbox{ where } \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

Alternatively, you can write

[tex]E=\gamma m_0 c^2[/tex]

but it has less physical content than eqn. (2)
 
  • #4
RabidSmurf said:
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?
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
 
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  • #5
pmb_phy said:
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

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 [tex]e^2 = p^2c^2 + c^2m^4[/tex], 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
selfAdjoint said:
This is your standard answer of course, and I think you owe it to the newbies to frankly state it's a minority view.
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
 
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  • #7
RabidSmurf said:
why nothing can travel at the speed of light and yet photons can travel at the speed of light.

have I missed something very fundamental in the theory itself?

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
actionintegral said:
My starting point for the theory of special relativity is the following:
The speed of light is the same for all observers.
To be exact: The speed of light is the same for all inertial observers. There is a significant difference.

Pete
 
  • #9
pmb_phy said:
To be exact: The speed of light is the same for all inertial observers. There is a significant difference.

Pete

The (local physical) speed of light is the same for all (inertial or non-inretial) observers.
 
  • #10
George Jones said:
The (local physical) speed of light is the same for all (inertial or non-inretial) observers.
...whereas the coordinate speed of light is not.

Pete
 
  • #11
pmb_phy said:
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

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
clj4 said:
I am sorry , Pete but you cannot use m=E/c2 for the photon.
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?
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.
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.
The above should close the door on the issue of "photon mass". The photon is a massless particle.
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
 
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  • #13
pmb_phy said:
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?
There is no point in talking about "the relativistic mass of a photons"
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.

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.

A zero photon mass refers to the proper mass of the photon which is indeed zero.

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? :-)
 
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  • #14
clj4 said:
I am very familiar with your paper (did AmJPhys publish it?) ..
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.

Let's agree to disagree on this one, ok? :-)
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
pmb_phy said:
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."

Interesting, do you know this for a fact? Does it come from the chief editor?
 
  • #16
clj4 said:
Interesting, do you know this for a fact? Does it come from the chief editor?
Read your PM messages. That is where I responded to this question.

Pete
 
  • #17
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) traveling 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.
 
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  • #18
higheriration said:
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) traveling 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.
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
 

1. What is special relativity?

Special relativity is a theory proposed by Albert Einstein in 1905 that explains the relationship between space and time. It states that the laws of physics are the same for all observers moving at a constant speed, and that the speed of light in a vacuum is the same for all observers regardless of their relative motion.

2. How does special relativity differ from Newtonian mechanics?

Special relativity differs from Newtonian mechanics in several ways. It takes into account the effects of time dilation and length contraction at high speeds, while Newtonian mechanics assumes that time and space are absolute. Additionally, special relativity introduces the concept of spacetime, where space and time are interconnected.

3. How does special relativity affect our understanding of the universe?

Special relativity has revolutionized our understanding of the universe by challenging our previous notions of time and space. It has led to the development of new theories, such as the theory of general relativity, which explains gravity and the structure of the universe on a larger scale.

4. What are some practical applications of special relativity?

Special relativity has practical applications in various fields, such as GPS technology, particle accelerators, and nuclear energy. It also plays a crucial role in understanding the behavior of objects at high speeds, such as spacecraft and satellites.

5. Is special relativity proven?

Special relativity has been extensively tested and has consistently been found to accurately describe the behavior of objects at high speeds. However, like all scientific theories, it is open to further testing and refinement as our technology and understanding of the universe advances.

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