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E = mc2 is not really true in all cases?

  1. Oct 26, 2006 #1
    E = mc2 is not really true in all cases??

    I am new to the field of SR and GR so my knowledge is not much but in the last thread I red following .

    "That is an incorrect statement. It assumes that E = mc2 in all cases, which it does not. It therefore "relativistic mass" cannot be replaced by "mass-energy". This assumption can lead a person to make statements which are very serious errors. E.g. people will almost always believe that mass density is identical to energy density. It is not. Rindler explains this in his 1982 intro to SR book."

    So please explain, E = mc2 is not really true in all cases, in which cases this doesnt hold true.
  2. jcsd
  3. Oct 26, 2006 #2


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    Two answers, which are consistent with one another:

    1. If I look at an object which is moving with a velocity v with respect to me, and somehow I measure its energy, in my frame, then its energy will be [tex]\gamma E[/tex], where E is its energy in its own frame, and [tex]\gamma = \frac {1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]. So if I equate this to [tex]mc^2[/tex] then [tex]\gamma E = mc^2[/tex] or [tex]E = \frac{mc^2}{\gamma}[/tex].

    2. The general law of energy for particles in SR is
    [tex]E^2 = p^2c^2 + m^2c^4 [/tex] (p is the magnitude of the three momentum), so for a massless particle (m=0 in its own frame), the second term drops out and we would have [tex]E = pc[/tex]. On the other hand, for a massive particle in its own frame, the first term drops out (in its own frame its momentum is zero), and we have [tex]E = mc^2[/tex].

    Both answers give you the same info: the famous equation refers to a massive particle in its own rest frame. In all other cases the law is subject to relativistic corrections. This is not at all academic, either. Particle physicsts have to use these relatiionships in calcualting the effects of the fast moving paricles in their experiments.
  4. Oct 26, 2006 #3


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    the notation we used when i was in school 3 decades ago was

    [tex] E = T + E_0 = \frac{m_0 c^2}{\sqrt{1 - \frac{v^2}{c^2}}} = m c^2 [/tex]

    where [itex]T = E - E_0 [/itex] is the kinetic energy (which depends on whose frame of reference you are in), [itex]E_0 = m_0 c^2[/itex] is the rest energy (the "energy in its own frame") and [itex]m_0[/itex] is the rest mass or the apparent mass in the frame of the body (also called "invariant mass"). we also were presented with the out of vogue concept of "relativistic mass"

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

    so [tex]E = m c^2[/tex]

    can hold in all cases with [itex]E[/itex] and [itex]m[/itex] both having corresponding meaning in the same frame. if [itex]m[/itex] is rest mass, than [itex]E[/itex] is rest energy. we were able to equate this [itex]E[/itex] (total energy, the sum of kinetic and rest energies) to the same [itex]E[/itex] that relates particle energy to wave function frequency:

    [tex]E = h \nu = \hbar \omega[/tex]

    resulting in an apparent inertial mass of

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

    and given that relativistic mass we always defined momentum to be

    [tex] p = mv [/tex]

    and assuming that photons always move at a speed of [itex]c[/itex] for any reference frame, then the momentum was simply

    [tex] p = mv = \frac{h \nu}{c^2} v = \frac{h \nu}{c} [/tex]

    and we got the same expression (keeping notational consistency):

    [tex]E^2 = p^2 c^2 + m_0^2 c^4 [/tex]

    but this notation has evidently been frowned on in current times. now [itex]m[/itex] always seems to mean only "rest mass" (what we used to call "[itex]m_0[/itex]") and there is no "relativistic mass" to talk about.

    we didn't call photons "massless", we said only that their rest mass was 0.
  5. Oct 27, 2006 #4
    Explaining it is not quite that easy. For this reason I created a two web pages to give a proper explanation of why E does not always equal mc2. The case when it does hold is for a isolated system. If you have a rod and you squeeze it from the ends then this is not a closed system and the energy is not related to mass by E = mc2. Please see the following for why this is true


    Example #1

    Example #2

    Example #3 - (Published by Wolfgang Rindler et al in the American Journal of Physics)

    Also, the proper mass of an object which is not a closed system (e.g. a box that radiates two photons) is not related to the energy by m2c4 = E2 - (pc)2 which is the magnitude of the 4-momentum. For expaination please see the bottom of the following web site, the section labeled An Incorrect Application of Invariant Mass


    Best wishes


    ps - I can scan that part of Rindler's text in which explains this and e-mail it too you if you'd like.
  6. Oct 28, 2006 #5
    I also forgot to mention that "relativistic mass" is the time-component of the 4-momentum. To see its general form (in time orthogonal fields) please see Eq. 14 at


    When the system of coordinates corresponds to that of a non-inertial frame of reference (i.e. a gravitational field) then it the time-component of the 4-momentum is no longer proportional to the energy of the particle and once again we do not have an equality. The energy is always proportional the time-component of the 1-form corresponding to the particle's 4-momentum.

    Best wishes

  7. Nov 9, 2006 #6
    Last edited: Nov 9, 2006
  8. Nov 9, 2006 #7
    jainabhs - I recently recieved a paper from an acquaintance of mind which makes me question my assertion that E does not always equal mc^2. There seems to be two definitions floating around in the physics literature by some very respectable relativists on both sides of the claim. I walked through the derived one author gave with a fine toothed comb. I found it a solid arguement and one that appeared to be agreement with a paper by Einstein himself. It will take me sometime to review all this literature so I'm placing my opinion on this on hold until I walk through this new definition very carefully. It will take a while since this is not the only project I'm working on at this time.

    Best wishes

  9. Nov 9, 2006 #8
    I wish I could but for now I'm swamped with work. Perhaps we could discuss it in PM? I will probably leave this forum soon due to a certain problem which arose yesterday but will check in every now and then to check my PM. However that seems like a long paper so I may not dig into it deeply for several weeks. But after that I would be able to give it more attention. I'm writing a new version of a paper I wrote on this subject so it is in my best interest to read your work. :smile:

    Best wishes

  10. Nov 9, 2006 #9
    information request

    how could I see the paper you mention? you do not answer messages you receive?
    sine ira et studio
  11. Nov 10, 2006 #10
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