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Photon mass destroyed?

  1. Apr 6, 2008 #1
    I'm still struggling with the concept of a mass of a photon. I understand that photons have momentum because of their inertial mass...m = p/v. This makes sense to me but what does it mean to say they have zero 'rest mass'. If they have momentum at some time t, and thus inertial mass at that time, how can they have zero 'rest mass' at some later time? Has mass not been destroyed? Where did this inertial mass go? Thanks.
  2. jcsd
  3. Apr 6, 2008 #2
    The term rest mass is a misnomer when it refers to a photon, since the photon can never be at rest. A better term is proper mass. The reason its call rest mass is due to the relation which holds for luxons and tardyons namely

    E2 - (pc)2 = (m0c2)2

    where m0 = proper mass. For a photon E = pc and the above relation yields m0 = 0.

  4. Apr 6, 2008 #3


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    Your confusion comes from your use of this nonrelativistic equation.
    The only mass that is useful is the m_0 in pmb's post.
  5. Apr 6, 2008 #4


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    the way i think about it is:

    that inertial mass of the photon is

    [tex] m = \frac{E}{c^2} = \frac{h \nu}{c^2} [/tex]

    the momentum is

    [tex] p = m v = m c = \frac{h \nu}{c}[/tex]

    this "inertial mass" or "relativistic mass" of photons is a term that is out of vogue by people who would say "photons are massless" without qualification. the mass they mean is the rest mass or the proper mass (sometimes referred to as m0). the relationship between rest mass and inertial mass or relativistic mass for any other particle than a photon (those particles not at a speed of c for any observer's frame of reference) is:

    [tex] m_0 = m \sqrt{1 - \frac{v^2}{c^2}} [/tex]

    if the velocity of the photon is c, then the rest mass is zero as long as the inertial mass is finite.
  6. Apr 6, 2008 #5


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    if m is the relativistic mass (for particles slower than photons, [itex]m=m_0/\sqrt{1-v^2/c^2}[/itex]) so deprecated of late, the equation p=mv is valid in the relativistic context.
  7. Apr 7, 2008 #6
    m = p/v is the definition of inertial mass regardless of what theory one is focusing on just as F = dp/dt is the definition of force regardless of whether one is considering Newtonian mechanics or relativistic mechanics.

    As far as m0 being the only useful mass, that is incorrect. In fact there are cases when m0 can't even be defined. Only in certain special cases, such as isolated systems/objects or for structureless point particles does m0 take on a well defined meaning.

  8. Apr 7, 2008 #7
    And that has led to errors. For example; in some relativity texts authors will attempt to define the mass of an object m as p = [itex]\gamma[/itex]mv then go on and suggest that mass density is given by energy density over c2. Such an assumption is wrong and originates from the misuse of E0 = mc2. In general inertial mass is a function of stress and in an EM field there is stress, even when it is merely a gas of photons which has a rest frame and thus a rest energy definable. In the case of a photon gas there is pressure which adds to the inertial mass of the photon gas. This is covered in Schut's GR text.

    There was a very interesting article in the American Journal of Physics which gets into this topic. The article is

    The inertia of stress, Rodrigo Medina, Am. J. Phys. 74,1031 (2006)

    This article is available online at http://arxiv.org/abs/physics/0609144

    To have a good understanding of inertia this is a must read article!

    Best wishes

    Last edited: Apr 7, 2008
  9. Apr 7, 2008 #8


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    but this is correct:

    E0 = m0c2


    i dunno, but anything can be misused. i think the blanket and unqualified statement that "photons are massless particles" leads more (lay) people to error and contradiction than having a concept of relativistic mass vs. "proper" mass (or "invariant mass" or whatever they be calling it). but if the only mass in our semantic is the rest mass, i think that leads most lay physikers and beginners into saying some pretty dumb things than if such concepts and differentiation were made.
  10. Apr 7, 2008 #9
    Yes. That is quite correct and is identical to what I wrote as you can see by noting that the m I used in the above expression, i.e. E0 = mc2, is identical to the one you used here, i.e. m0.
    While you're correct it is not what I had in mind. Nearly every single relativity text and article I've read on mass has made the error of implying that E0 = m0c2 holds in every concievable circumstance, which it does not. I'm not absolutely certain how this came to be but I'm pretty sure its a result of students beleiving that newer texts are better than older ones in certain ways. I agree that in a lot of cases that is true. But what is missing is certain generalities that have not made it into newer texts. The general case of mass is a perfect example. E.g. for continuous media the mass is completely defined by the stress-energymomentum tensor and, in the most general case, the mass density is not proportional to the energy density. Since this seems to have been missed by most relativity authors then readers of those texts never learn about it. Thus when one studies general relativity one can easily become confused, or at least come to the wrong conclusions, regarding the fact that the active gravitational mass of matter is a function of both energy and pressure. I.e. inertial mass depends on pressure and pressure is a source of gravity.
    I agree. If someone simply says either "photons are massless particles" or "photons have mass" then they are not learning the full story. Learning only part of a theory and its comnsequences is bound to lead one to make errors. I've seen too many people on physics forums/newsgroups take the notion that "photons have no mass" and conclude from that that "light does not generate a gravitational field" which of course it does.

    Regarding invariant mass vs proper mass; Caution is warranted here. These two terms should not be confused with each other. Typically the term invariant mass is used to refer to the value of the total 4-momentum of a system where proper mass refers to the value of the 4-momentum of a single particle. As such the invariant mass of a particle equals the proper mass. I learned this from Hans C. Ohanian, one of the authors of Gravitation and Spacetime - 2nd Ed., i.e. that invariant mass is a term used for a system of particles. Its unclear to me if he extended that to systems containing both particles and fields but I assume this to be the case.
    I think people will make mistakes when they only learn part of the ramifications of a theory, no doubt. Ideally one will make very few errors if they have studied it in general and completely, and that includes studying the stress-energy-momentum tensor. For the beginner this tensor can be avoided by studying certain derivations which were first done by Einstein. Unfortunately those derivations are hard to find. I can show you such a derivation if you'd like?

    Best wishes

    Last edited: Apr 7, 2008
  11. Apr 7, 2008 #10


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    I agree with this. The concept itself has uses, although I generally prefer the invariant mass. I think generally that people should always specify "invariant mass" or "relativistic mass" to avoid confusion. Although I believe that it is acceptable to use the unqualified term "mass" to refer to the "invariant mass" it never hurts to be explicit.
  12. Apr 7, 2008 #11
    The reason I like the term proper mass is because it goes along nicely with the name of the magnitude of other 4-vectors such as the spacetime displacement whose magnitude is proportional to either proper distance or proper time.

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