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Insights What is relativistic mass and why it is not used much? - Comments

  1. Aug 9, 2015 #1

    Orodruin

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  3. Aug 9, 2015 #2

    PAllen

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    Just a comment that if you use 4-force instead of 3-force, you do recover a Newtonian like equation:

    F = dp/dτ = d(m0U)/dτ = m0 dU/dτ = m0A

    where U is 4-velocity and A is 4-acceleration.
     
  4. Aug 9, 2015 #3

    Orodruin

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    Indeed, this was previously an FAQ and the target audience is mainly people who are not familiar with 4-vectors. It was written due to the endless stream of people we get asking questions on the subject of changing mass in relativity.

    It should be noted that if the system is not closed (e.g., an object absorbing external radiation), as in classical mechanics, you would obtain
    $$
    \frac{d(mU)}{d\tau} = m A + V \frac{dm}{d\tau},
    $$
    also analogous to the Newtonian case.
     
  5. Aug 9, 2015 #4
    Could you clarify whether 'a' in your final equation represents proper or coordinate acceleration? I always have to rack my brains about this when reading about acceleration in SR ;)
     
  6. Aug 9, 2015 #5

    Orodruin

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    a is always the acceleration in a particular inertial frame for the purposes of this post. No knowledge about proper acceleration is assumed.
     
  7. Aug 9, 2015 #6

    Dale

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    Yes. But usually by the time someone learns four-vectors they are not confused about relativistic mass vs. invariant mass.

    In fact, I sometimes wonder if it isn't easier to just teach four-vectors than have such discussions at all.
     
  8. Aug 10, 2015 #7
    Well done Orodruin! :smile:
     
  9. Aug 10, 2015 #8

    Orodruin

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    I think it should be pointed out that while I wrote the original FAQ, several PF science advisors contributed opinions and suggestions on how to refine it.
     
  10. Aug 10, 2015 #9
    Great as always Orodruin!
     
  11. Aug 11, 2015 #10
    Another way of expressing SR kinematics in Newtonian terms (using 3-vectors in some one frame) is by phrasing the second law as F=dp/dt, and defining momentum by p=Ev/c^2, with E the total energy. The gamma factor can even be derived from this using the work/energy relation. Since these equations are simple (no square roots!), this has led me to speculate that such a definition of momentum should be seen as more "causative" than mv*gamma, although of course they are both frame dependent. This would give E/c^2, the relativistic mass, a status that would justify its onetime popularity. ;-)
     
  12. Aug 12, 2015 #11
    That doesn't need to be defined. It can be derived from the classical definition of momentum.
     
  13. Aug 12, 2015 #12
    my point is that you can describe the dynamic laws of SR, in terms of three-vectors, without mentioning the gamma factor.
     
  14. Aug 12, 2015 #13

    PAllen

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    A peculiarity of this approach (which, of course, does not make it wrong) is that the simple non-radiating case of forced motion is not simple at all. You have:

    F = dp/dt = E' v + E v'

    with all terms important for the simple non-radiating impulsive force (I use c=1). Meanwhile, with 4-vectors, the non-radiating case becomes the very simple:

    F = m A

    which has to do with the fact that the 4-vector approach sees gamma as an intrinsic feature velocity, by virtue of proper time. It is metric in nature rather than part of dynamics. In the language of 4-vectors, if we talk about velocity or acceleration divorced from any particular mass or energy, the gamma factors are built in.
     
  15. Sep 26, 2015 #14
    Can you write more about discreancy between special relativity and Newtonian gravity?
    Maybe, if a very fast comet is flying close to earth, its gravitational force is proportional to earth to m_0 and not to \gamma m_0
    ##m_0## and not to ##\gamma m_0##?
    Is something else?
     
  16. Sep 26, 2015 #15
    [itex]\left( {1 + \beta ^2 } \right) \cdot \gamma \cdot m_0[/itex]
     
  17. Sep 26, 2015 #16

    Orodruin

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    In general relativity gravitation is not coupled to mass only. Instead, the source of space-time curvature is energy, momentum, and stress. Only for weak gravitation and small velocities does it reduce to the Newtonian case where mass is the source of gravity. You simply cannot describe gravity in special relativity by changing the mass for the relativistic mass. There should be some threads here on that subject if you use the search function.
     
  18. Sep 26, 2015 #17
    I searched these threads, and partly I understand better, but I do not find everything. As Dr. Stupid wrote above for the above example of Schwarchild geometry, factor ##1+\beta^2## means the main discrepancy between Newtonian gravity + SR vs. GR+"SR". Can you have still any concrete examples for this discrepancy. What is at big velocities and small gravity. (One example gave Dr. Stupid.)
     
  19. Sep 26, 2015 #18

    Orodruin

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    The perhaps most famous differences between the predictions of Newtonian gravity and GR are:
    - The perihelion shift of Mercury's orbit
    - The amount of gravitational deflection of light passing by the Sun
    In the first case, both the velocity and gravitational fields are relatively small so the effect is very small - it requires very good precision to make the measurement. The second case was one of the classic tests of GR and involves light, so the speeds involved are large (large means comparable to light speed).
     
  20. Sep 26, 2015 #19

    pervect

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    I have a suspicion that a significant number of the readers here on PF are rather rusty on their calculus. (Some readers may not have had calculus at all - but trying to totally eliminate calculus from the exposition seems to me to require a totally different approach, if it's possible at all).

    As a consequence of this (presumed) rustiness, simply mentioning that F = dp/dt may not entirely get through to such readers, who are still mentally attached to the idea of F = ma, both because of both familiarity and because it deosn't require them to try and remember their calculus.

    A little more hand-holding, reviewing the classical perspective and explaining that the origin of F=ma was in fact F=dp/dt, might be helpful. As a consequence, it might be more clear when one explains that F=dp/dt no longer leads directly to F=ma in relativistic physics. If nothing else, it may inspire some readers to brush up on the necessary mathematical underpinings (this is probably rather optimistic). In other cases it may at least get across the idea of what's missing and preventing a fuller understanding of the issues.

    I'm not sure of the details of how to actually go about writing all of this, but I thought I'd mention the idea in case someone was motivated.
     
  21. Sep 28, 2015 #20
    Some textbooks (such Alonso&Finn which I used as a student) present the SR derivations of acceleration from F=dp/dt; that's a useful stating point.
     
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