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A Photons relativistic mass

  1. Oct 13, 2011 #1
    I read somewhere else on this forum that photons have zero invariant mass, only relativistic mass. Here is my problem, Einstein's equation for relativistic mass says that is invariant mass is zero then so will relativistic mass at all velocities. If a photon does have relativistic mass, then invariant mass would be greater than zero and before a photon could reach light speed it's mass would diverge to infinity. What am I missing here?

    Thanks in advance.
  2. jcsd
  3. Oct 14, 2011 #2


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    A photon has kinetic energy, or simply energy. I've never heard anyone try to treat this a relativistic mass. The formula for relativistic mass you refer to cannot be applied to photons. You must use the more general relation: E^2 = p^2 c^2 + m^2 c^4 . Here, since m is zero, you just get the familiar E = p c = h * frequency. (p is momentum).
  4. Oct 14, 2011 #3
    You are probably missing an up-to-date book. We do not talk in terms of "relativistic vs rest mass" anymore. The mass is a constant number which defines what a particle is, as a characteristic of the representation of the symmetry under the Lorentz group. If you like to have fun with historical terms, there are also "transverse" and "longitudinal" masses :
    I think this formalism emerges from an attempt to recover the Galilean/Newtonian equations and are mostly a curiosity, although it is still possible to find senior scientists using them :smile:
  5. Oct 14, 2011 #4


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    Einstein wrote in a letter (1948)

    "It is not good to introduce the concept of the mass M = m/(1-v²/c²)1/2 of a body for which no clear definition can be given. It is better to introduce no other mass than 'the rest mass' m. Instead of introducing M, it is better to mention the expression for the momentum and energy of a body in motion."

    My observation is that whenever this M is introduced, people are happy b/c they are able to write down 'well-known' expressions like E=Mc² and p=Mv. But later they get confused b/c these are the only two equations in SR where this M really 'helps'. One should instead write down E=Mv²/2, look at it for a moment and then never talk about this M anymore.
  6. Oct 14, 2011 #5


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    I also agree that use of relativistic mass should be discuraged. But if you urge me to do so,
    I shall comment on your question: As a photon has m=0 and v=c the expression for M is undefined as the denominator is zero, too. Hence any M is compartible with m=0.
  7. Oct 14, 2011 #6
    Thank you for the responses everyone. I think I understand what you are trying to say. If I could ask one follow up question to ensure that my understanding is sound. My understanding is that photons have energy and momentum, but never mass. Is this correct?
  8. Oct 14, 2011 #7


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    Yes, they have energy E and momentum p but no rest mass m.

    In SR the following equation always holds:

    E² = p²c² + (mc²)²

    For photons you have

    E = hf
    p = hf/c

    and therefore vanishing rest mass m = 0.
  9. Oct 14, 2011 #8


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    Please start by reading the https://www.physicsforums.com/forumdisplay.php?f=210" [Broken] in the Relativity forum.

    Last edited by a moderator: May 5, 2017
  10. Oct 14, 2011 #9
    ok sorry :/
  11. Oct 14, 2011 #10
    Although I believe the concept of relativistic mass should be avoided, I'll try to answer your question. Relativistic mass is given by:

    [itex]m_r= \gamma m[/itex]

    where [itex]m[/itex] is the rest mass and [itex]\gamma = (1-v^2/c^2)^{-1/2}[/itex]. You'll notice that when [itex]m=0[/itex] for any particle traveling at less than [itex]c[/itex], [itex]m_r[/itex] is obviously zero. This, however, isn't valid for particles traveling at [itex]c[/itex] (as photons do) because when [itex]v=c[/itex] is used in the Lorentz factor then [itex]m_r= \frac{0}{0}[/itex] which is indeterminate. We need some other way of expressing [itex]m_r[/itex] if we wish to calculate the relativistic mass of a photon.

    Similar to the expression for [itex]m_r[/itex], the energy is given by:

    [itex]E= \gamma m c^2[/itex]

    Solving for Eliminating [itex]\gamma[/itex] we obtain:

    [tex]\gamma = \frac{m_r}{m} = \frac{E}{mc^2}[/tex]

    Solving for [itex]m_r[/itex], we get the expression:

    [tex]m_r = \frac{E}{c^2} = \frac{h}{c \lambda}[/tex]
  12. Oct 14, 2011 #11
    Heh. I'm not quite a "senior", though the AARP keeps sending me invitations. I did grow up in the era when "relativistic mass" was a fixture of the lore of relativity, just like length contraction and time dilation.

    I think that in part the notion was introduced to facilitate applications of Newtonian mechanics in the relativistic regime. A typical conundrum was this: consider a box floating in deep space. A thermal photon is emitted inside the box, from the left wall, headed to the right, straight for the opposite wall. Since we know photons have momentum, we know the box will start moving to the left because of the reaction from the photon emission. As soon as the photon hits the right wall, the impulse stops the box. Question: if you look at the box before and after, it has moved to the left, so how is it that the center of mass of the box moves to the left? There are no outside forces acting on the box. This gedanken cries out for us to ascribe a mass to the photon.

    How would you talk about this without using relativistic mass?
  13. Oct 14, 2011 #12


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    Just use energy instead of mass. Call it center of energy. Normally, the more important thing is the center of momentum frame, where where the total momentum is zero, rather than where the center of energy is.
  14. Oct 14, 2011 #13
    Answering from an also not-so-ancient textbook:

    E = c √(m02c2 + p2)
    => for m0=0:
    E = c p

    Thus the mass equivalent is E/c2 = p/c.
    That also follows directly from p = mr v, for v=c.

    Of course, elfmotat already presented roughly the same, as does bcrowell:

    Compare also the usenet physics faq:
  15. Oct 14, 2011 #14
    One important aspect of talking about relativistic physics is that mass is not conserved. We have to use the conservation of mass and energy.

    The resolution of the thought experiment is this. In order to produce the photon, the left side of the box has to lose a small amount of mass which, in the end, is delivered to the right side of the box. Therefore, the center of mass of the box shifts to the right thus causing the box to be shifted to the left to keep the center of mass constant.

    However, during each process (the production of the photon, say), rest mass is not conserved. Momentum is. Same for the absorption of the photon. Only after all of the relativistic processes does the rest mass once again emerges as a conserved quantity.
  16. Oct 14, 2011 #15
    Relativistic mass and rest mass are conserved but rest mass is not additive.
  17. Oct 15, 2011 #16


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    Rest mass (invariant mass is) conserved.

    Before the box has emitted the photon you have

    [itex]p^\mu = (m_B c^2,0)[/itex]

    After the box has emitted the photon you have

    [itex]p^\mu = (E_B + E_\gamma,0)[/itex]

    The momentum of box and photon cancel.

    For the energy that means

    [itex]m_B c^2 = E_B + E_\gamma[/itex]

    [itex](m_B c^2 - E_\gamma)^2 = E_B^2[/itex]

    [itex](m_B c^2)^2 + E_\gamma^2 - 2 m_B c^2 E_\gamma = (m^\prime_B c^2)^2 + c^2p_B^2[/itex]

    Momentum of photon and box (squared) are equal, p = E/c for photons, therefore

    [itex](m_B c^2)^2 + E_\gamma^2 - 2 m_B c^2 E_\gamma = (m^\prime_B c^2)^2 + c^2p_\gamma^2 = (m^\prime_B c^2)^2 + E_\gamma^2 [/itex]

    I introduced m' for the box after the photon has been emitted.

    [itex](m_B c^2)^2 - 2 m_B c^2 E_\gamma = (m^\prime_B c^2)^2 [/itex]

    [itex]2 m_B E_\gamma = c^2[m_B^2 - {m^\prime_B}^2] [/itex]

    Now we could introduce the 'relativistic mass of the photon' using M = E/c²

    [itex]2 m_B M_\gamma = m_B^2 - {m^\prime_B}^2 [/itex]

    It is clear that masses do no longer add up as usual (thanks to DrStupid for that remark). In addtion it does not help to introduce the 'relativistic mass of the photon'. I think it is confusing in this context.

    What is more interesting is to express the rest mass of the box after the emission of the photon using m and E

    [itex]{m^\prime_B}^2 = m_B^2\left(1- \frac{2 E_\gamma}{m_bc^2}\right) [/itex]

    Of course the rest mass of the box itself is an upper bound for the photon energy :-)
  18. Oct 15, 2011 #17


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    In case anyone is still confused, "conservation of (rest) mass" could be interpreted in two different ways for a composite system that consists of a number parts (and does not interact with any external objects).

    If you mean "the sum of all the masses of the parts remains constant", that is incorrect in relativity theory.

    If you mean the total "system mass" remains constant, that is correct in special relativity (and also in static spacetimes in general relativity). The system mass need not equal the sum of the parts.
  19. Oct 15, 2011 #18


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    In terms of the example with the box and the photon:

    Before the photon has been emitted the rest mass of the box and system mass are identical.

    After the photon has been emitted the rest mass of the box has decreased m' < m as a function of the energy E of the photon; this should be clear from my calculation. So the sum of the two rest masses m' + 0 (for the photon) is not identical to the rest mass of the box m before the photon has been emitted.

    But the system mass (= invariant mass of the system box' + photon) after the photon has beem emitted is identical to the mass of of the box before the emission. I used this in the beginning when writing down my four-vectors.
    Last edited: Oct 15, 2011
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