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I Can someone explain to me relativistic mass?

  1. Jul 11, 2016 #1
    I was told in special relavity that f=ma does not always hold true so using the f=delta p/delta t is better. So why is that? They said it had something to do with relativistic mass, at least when traveling at 99.9% of the speed of light.
     
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  3. Jul 11, 2016 #2

    PeterDonis

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  4. Jul 11, 2016 #3
  5. Jul 11, 2016 #4

    PeterDonis

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    F = ma isn't used in SR because it isn't correct in SR. It's only correct in the approximation where all velocities are much less than the speed of light. F = dp/dt is correct for all possible velocities.
     
  6. Jul 11, 2016 #5
    Why is f=ma not correct for speeds much less than c? And what qualifies as "much less"? Is it because the mass is dynamic?
     
  7. Jul 11, 2016 #6

    PeterDonis

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    I said f = ma is correct for speeds much less than c. It is not correct (more precisely, it becomes a worse and worse approximation) for speeds that get closer and closer to c.

    For a brief discussion of why, see here:

    https://en.wikipedia.org/wiki/Relativistic_mechanics#Force
     
  8. Jul 12, 2016 #7

    Ibix

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    @ReptileBaird - Pretty much all of Newton's formulae are wrong, in short. They are approximately correct at low speeds and low energies and weak gravitational fields. In the latter half of the nineteenth century we began to do experiments and work out theories that could reach the regimes where it was obvious that Newton was wrong, and Einstein put it all together in 1905-1916.

    We mostly continue to use Newton's equations because they're good enough for very nearly everything. For example, a rocket accelerating at 15g for 100s would, according to Mission Control, be accelerating slightly less than Newton would predict. But the difference is in the 7th or 8th significant figure, if my mental arithmetic is correct. So, even NASA uses Newton because the maths is so much simpler and the error is negligible even at the speeds they sling around.

    The GPS is a famous exception - it is a sufficiently high precision system that it includes a general relativistic calculation in its design. But the rule is that Newton is close enough unless you're doing something ridiculously high energy or high precision.
     
  9. Jul 12, 2016 #8

    vanhees71

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    Well, your question is very easily answered. The answer has two parts:

    (a) Relativistic mass is an idea that is outdated since 1907 and should not be used and taught anymore. It leads to confusion. The mass in relativity is alwasy understood as the invariant mass of an object and as the name tells you it's a scalar, i.e., independent of the frame of reference.

    (b) ##\vec{F}=m \vec{a}## doesn't hold in general, even not in Newtonian physics. The correct law, already stated by Newton, is ##\vec{F}=\mathrm{d} \vec{p}/\mathrm{d} t##. Now in the theory of relativity this is a very complicated quantity, and it is easier to use proper time (assuming we have a massive particle; the massless case is more complicated and of no practical relevance since the only massless particle-like objects are photons and shouldn't be treated as classical particles to begin with). The proper time of a particle is defined by
    $$\mathrm{d} \tau=\mathrm{d} t \sqrt{1-\vec{v}^2/c^2},$$
    where ##t## is the coordinate time wrt. an inertial reference frame. Now you define the four-vector
    $$p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.$$
    This is the four-momentum vector. It's time component is the energy of the particle (up to a factor ##c## which is due to the unfortunate joice of SI units), including its rest energy
    $$p^0=m c \gamma=\frac{m c}{\sqrt{1-\vec{v}^2/c^2}}.$$
    The three spatial components are relativistic momentum
    $$\vec{p}=\frac{m \vec{v}}{\sqrt{1-\vec{v}^2/c^2}}.$$
    The mass is a scalar and the covariant energy-momentum relation of a classical particle is given by the "on-shell condition"
    $$p_{\mu} p^{\mu}=m^2 c^2,$$
    which leads to the energy-momentum relation
    $$p^0=\frac{E}{c}=\sqrt{m^2 c^2 +\vec{p}^2}.$$
    This shows that the relativistic energy includes the rest energy of the particle
    $$E_0=c p^0|_{\vec{p}=0}=m c^2.$$
     
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