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Relativistic mass, how did Einstein thought of it?

  1. Jun 5, 2005 #1
    I've a question here, is there a way to derive [tex] m'= \gamma m [/tex]? So far, i have not seen any derivation to this. I have heard that relativistic mass is a definition, but how did Einstein thought of this?
  2. jcsd
  3. Jun 5, 2005 #2

    Meir Achuz

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    In order to LT momentum, it has to be made part of a four-vector.
    The fourth (usually the 0th component) is energy (as identified by the NR reduction).
    The invariant length of the momentum four-vector is given by
    m^2=E^2-p^2 (letting c=1), with m being the (Invariant) mass of the object.
    Using v=p/e, this can be rewritten as E=m\gamma.
    Einstein thought up "relativisitic mass" by identifying the increase in E with v as an increase in "relativisitic mass". But the energy increase is no longer described this way. (I am pretty sure even Einstein stopped using "relativisitic mass".)
    Mass is considered a relativistic invariant, with E increasing because of the \gamma factor.
  4. Jun 5, 2005 #3
    When it was discovered after Einstein's mass-energy relation that force could no longer be expressed as F = ma (m = resst mass) Max Planck showed that it could be expressed as F = dp/dt where p = mv. This relation actually defines mass in that m = p/v. It was then that Richard C. Tolman, in an article in Nature, concluded that his should be considered THAT definition of mass. Herman Weyl had a very similar idea which I believe is a better one but still equivalent. All of the comments to now are with regards to objects which may be considered point objects. For the afore mentioned relationship to hold true m must be relativistic mass. Later on when 4-vectors came along the temporal component became relativistic mass*c. Then in 1907 Einstein showed that even light has mass. When all was said and done Einstein concluded that the correct expression for mass was fully described by a second rank tensor, the stress-energy-momentum tensor.

    Note: Word of caution. E = mc^2 holds only in special cases.

    Last edited: Jun 5, 2005
  5. Jun 6, 2005 #4

    Meir Achuz

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    The history is interesting, but m=p/v or light having mass are not part of current SR theory. There is an energy-momentum tensor, but mass is a scalar invariant.
  6. Jun 6, 2005 #5
    That is true of proper mass. Not (inertial/relativistic) mass itself. Its easier for some to drop the subscript and drop the adjective "proper" for ease when it can be assumed with relative ease. Makes for easy reading. Don't confuse ease with physics. This can only be done in a limited number of situations.

    Consider this example: Take a rod with equal and opposite charges on each end. Let the proper mass of this "dumbell" be m0. Let there be a uniform electric field in S where the rod is at rest and laying on the x-axis. The the E-field be parallel to the x-axis. Now transform to S' which is moving in the +x direction. The relation

    [tex]E^2 - (pc)^2 = m_0^2 c^4[/tex]

    Does not hold for this dumbell.

    There is also an interesting problem in Ohanian's new SR text. He states the energy density of a magnetic field and then asks you to find the mass density. The answer in the back of the text is incorrect. He uses energy density = "rest mass density" *c^2 which is an invalid relationship in this situation.

    Last edited: Jun 6, 2005
  7. Jun 6, 2005 #6
    A beautiful demonstration due to Paul LANGEVIN exists (in French language) in:" Mecanique des particules; champs; LENNUIER - GAL - PERRIN; collection U; ed. 1970; pages 256-260". It starts very classically with considerations about the kinetic energy of a particle. Lorentz transformations of the coordinates are sufficient and supposed to be known to be able to "follow" the demonstration.
  8. Jun 6, 2005 #7
    I created a derivation on my web site. This one is pretty straight forward.


    You'll notice that it is Weyl's definition of mass that is used when one derives [itex]m = \gamma m_0[/itex]. Notice also that it is assumed from the start that the particles are tardyons (particles which move at v < c) and not luxons (particles which move at v = c).

  9. Jun 7, 2005 #8

    Meir Achuz

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    pmbphy: Of course m^2=E^2-p^2 has to be changed if there are potentials present.
    Look at any good text to see how.
    Ohanion is a good popularizer, but should not be quoted as an authority.
  10. Jun 7, 2005 #9
    This has nothing to do with potential. It has to do with stress. I never said that the particles were so close, or the rod so short, that the potential energy had to be accounted for. In fact Einstein assumed that the charges were such that the interaction potential could be ignored.

    However if you're speaking about a single charged 'point' particle in an electric field for which the potential energy of position was non-zero then "m^2=E^2-p^2" is still valid.

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