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Relativistic mass in Six Not-so-easy Pieces (Feynman)

  1. Apr 14, 2015 #1
    I'm hoping that someone can help me understand a topic on page 86 of Feynman's "Six Not-so-easy-pieces", in the "4-4 Relativistic mass" section. He says, "It is an interesting exercise to now check whether or not Eq.(4.9) is indeed true for arbitrary values of w, assuming that Eq.(4.10) is the right forumula for the mass." http://www.feynmanlectures.caltech.edu/I_16.html#Ch16-S4

    I can follow the derivations of (4.9,10), but I don't understand how to perform the exercise. My best result was to use
    $$ v^2 = u^2 + w^2(1 - u^2/c^2)$$
    to expand (4.9) to
    $$\frac{m_w}{m_v} = \sqrt{1 - \frac{w^2 + v^2}{c^2 - w^2}}$$
    which becomes (4.10) in the limit of the vertical components approaching zero. But I don't think that's the exercise Feynman was proposing. Any ideas?
  2. jcsd
  3. Apr 15, 2015 #2

    Simon Bridge

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    eq4.9 is in terms of w and has a "relativistic mass" term in it.
    He seems to be proposing that you input 4.10 for the relativistic mass and check the range of w for which the equation is true.

    FYI: In practice we no longer use the "relativistic mass" concept: mass is mass and the word refers to the rest mass.
  4. Apr 22, 2015 #3
    Thanks for the help - I took a little break and was able to find a solution. I set up equations for each "relativistic mass" and plugged them in to 4.9 as you said, and it looks like it applies for any value of w < c. With $$m_w = \frac{m_0}{\sqrt{1 - w^2/c^2}}$$ and $$m_v = \frac{m_0}{\sqrt{1 - v^2/c^2}} = \frac{m_0}{\sqrt{1 - \frac{u^2 + w^2(1 - u^2/c^2)}{c^2}}}$$ you get $$\frac{m_w}{m_v} = \sqrt{1 - u^2/c^2}$$ automatically, as Feynman said.

    I found https://www.physicsforums.com/threads/what-is-relativistic-mass-and-why-it-is-not-used-much.796527/ [Broken] and if I understood it correctly, the "relativistic mass" would depend on both the direction of the velocity of the object as well as the direction of the force acting on it, so it becomes ambiguous as a general attribute of the object. Would there be a similar argument for "relativistic length" or "time"? It seems a little different in that time dilation and length contraction are only dependent on the velocity of the object (in the observer's frame), so you could still say that something is shorter or takes longer as long as you specify the frame of reference.
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  5. Apr 23, 2015 #4

    Simon Bridge

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    Mass is not usually thought of as a geometric construct while it is easy to show that time is.
    The association of gamma with m is not like its association with t or L though.
    We dont talk about relativistic lengths and times either though, more like how relativistic mass is now thought of as a manifestation of momentum and kinetic energy. Its about whats useful. Thinking of the gamma as somehow increasing an objects mass causes more confusion than it saves.
  6. Apr 24, 2015 #5
    Perhaps that post is not clear enough: γm0 depends as little on direction as m0.
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  7. Apr 24, 2015 #6


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    ...however the relativistic mass used in the force equation varies between ##\gamma m## and ##\gamma^3m## depending on the angle between the force and the mass' current velocity. These are the "transverse" and "longitudinal" relativistic masses (if memory serves), which are two more reasons not to use relativistic mass, IMO.
  8. Apr 24, 2015 #7
    That is wrong: "relativistic mass" replaced those pre-SR mass concepts for that reason. By chance I know that as I learned SR with relativistic mass. :wink:
  9. Apr 24, 2015 #8


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    transverse and longitudinal relativistic mass are pure SR concepts - they can't possibly have meaning pre-relativity. They were being taught as late as 1967 in SR courses - I can post the textbook pages if desired.

    They are not inherently required in using relativistic mass . What they come from is trying rescue: F=ma, with 'a' being coordinate acceleration in standard inertial coordinates. Then what you put for 'm' varies with the orientation of the force in relation to the object's velocity. The alternative (while still using relativistic mass, which I don't), is to simply abandon this equation and use only F= dp/dt. [edit: but in using this equation with relativistic mass (in which p=mv, m being relativistic mass), one must realize dm/dt can almost never be ignored, because m is really m(v). In contrast, the modern formulation only needs to worry about dm/dt for bodies that emit or absorb radiation or particles. You have, in modern formulation:

    p = m U, where m is rest mass or invariant mass, and U is 4 velocity = dxi/dτ
    F = dp/dτ = m A, where A is 4-acceleration = dU/dτ (as long as the body is not emitting or absorbing something).
    Last edited: Apr 24, 2015
  10. Apr 25, 2015 #9

    Simon Bridge

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    The question being responded to (in posts 5-8) starts "if I have understood [the cited thread] correctly..."
    The short answer, therefore, seems to be "no, you have not understood the thread."

    ... for whatever reason.

    The question of why relativistic mass is seldom used these days keeps coming up, and there are well establish answers for students.
    i.e. Baez: Relativity FAQ http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html
    `The question to ask is not whether relativistic mass is fashionable or not, or who likes the idea and who doesn't; rather, as in any area of physics notation and language, we should always ask "Is it useful?". '
    ... usefulness depends on the context.
    In discussions is it important not to mix the contexts.

    The topic is "why it is not used" rather than "why should it be/not be used". (I suspect the latter tends to philosophy anyway.)
    i.e. why do I default to downplaying relativistic mass to students? (short answer: because "geometry" - and this seems to be common, and comes from having to teach physics at a wide range of levels.)

    Specifically: bottom of post #3 is asking about the consistency of the argument found in the cited link.
    In a nutshell: Is it inconsistent to reject a velocity dependent mass when we accept a velocity dependent length and time?

    Generally educators have been concerned that relativistic mass is inconsistent enough with the geometric formulation of general relativity that it becomes a problem for GR students and others attempting to understand statements from GR. i.e. see: Oas G. (2008) On the abuse of relativistic mass [pdf]
    The concept of velocity dependent mass, relativistic mass, is examined and is found to be inconsistent with the geometrical formulation of special relativity. This is not a novel result; however, many continue to use this concept and some have even attempted to establish it as the basis for special relativity. It is argued that the oft-held view that formulations of relativity with and without relativistic mass are equivalent is incorrect. Left as a heuristic device a preliminary study of first time learners suggest that misconceptions can develop when the concept is introduced without basis. In order to gauge the extent and nature of the use of relativistic mass a survey of the literature on relativity has been undertaken. The varied and at times self-contradicting use of this concept points to the lack of clear consensus on the formulation of relativity. As geometry lies at the heart of all modern representations of relativity, it is urged, once again, that the use of the concept at all levels be abandoned.

    ... there are counter arguments of course and Baez (above) recounts a number of them.
  11. Apr 25, 2015 #10
    Apparently the transverse and longitudinal relativistic mass definitions were introduced by Lorentz in 1900 https://de.wikisource.org/wiki/Prinzipien_der_Dynamik_des_Elektrons_(1903)#cite_note-14
    Yes that's the concept used by Feynman (edit: apparently introduced by Tolman in 1912). While it accounts for dm/dt with m=relativistic mass, calculations can still become complicated.
    Last edited: Apr 25, 2015
  12. Apr 25, 2015 #11
    Using relativistic mass in force equation looks like this:

    [itex]F = \frac{{dp}}{{dt}} = \gamma \cdot m \cdot \left( {a + v \cdot \frac{{v \cdot a}}{{c^2 - v^2 }}} \right)[/itex]

    In this equation relativistic mass is still ##\gamma m## and independent from direction. Maybe you refer to the improper use of Eulers's F=M·a under relativistic conditions (I use the symbol M instead of m in order to avoid confusions with rest mass). In this case you get

    [itex]M = \gamma \cdot m \cdot \left( {I + \frac{{v \cdot v^T }}{{c^2 - v^2 }}} \right)[/itex]

    This is not the relativistic mass but an entirely new property with the unit of mass. The eigenvalues of M are called longitudinal and transversal mass.
  13. Apr 25, 2015 #12


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    While that is an interesting approach, I have not seen it taught (or used) except by you on this forum. I have books on SR spanning over 70 years and none uses this approach (and many use longitudinal and transverse relativistic mass). Obviously, Feynman does not use your approach, and it is not mentioned in publishded discussions of this issue (e.g. the papers around the debate introduced by Okun).

    Note, to arrive at it, you have to take dm(v)/dt and notice that the result has a factor of m(v) in it, allowing you to bundle the rest with the acceleration. Also note, that, to me, longitudinal and transverse mass do get at the idea behind relativistic mass: how much coordinate motion is produced by given force. The answer depends on the direction of the applied force in relation to object's motion in a way that does not happen in Newtonian mechanics.
  14. Apr 25, 2015 #13
    That sums it up so well, I "got" it... tyvm ;-)
  15. Apr 25, 2015 #14
    It results from Newtons definitions of mass and force if Galilean transformation is replaced by Lorentz transformation.

    To me, the relativistic mass ##\gamma m## is the ratio of momentum and velocity and it corresponds to Newton's "quantity of matter".

    I know that many sources (e.g Lorentz) refer to the ratio of force and acceleration as the relativistic mass but I do not see a justification for this definition. It does not result from Euler's F=m·a because this is known to be valid for constant m only.
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