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Relativistic Acceleration Contradiction

  1. Nov 8, 2009 #1
    Hello everyone! I apologize if my first post is of a problem that may seem like a really silly one. But it was something we were stuck with today in Modern Physics class. It is NOT a homework assignment!

    We have proved that relativistic force, F = m*(du/dt)*(gamma)3 where gamma is the Lorentz factor. This definition of relativistic force was derived by differentiating the definition of relativistic linear momentum with respect to time.

    Now the silly problem that must have a simple solution somehow:
    Theoretically, we say that as we increase the velocity of an object up to the speed of light then the acceleration will tend to get to zero. Meaning we can never accelerate beyond the speed of light.

    This is the theory.
    By looking at the mathematics involved:

    We know from the definition of F above that a = (du/dt)*(gamma)3, the relativistic acceleration.
    Also, gamma = 1/(1 - (v2/c2))0.5.
    If we let v approach c (let's say v=c at the extreme) then this would result in the denominator approaching 0 and thus gamma approaching infinity. Which is in contradiction to the previous statement we made regarding acceleration.

    My question is simple: The theoretical statement is definitely right so what went wrong in the mathematics and messed it all up?
  2. jcsd
  3. Nov 8, 2009 #2
    If force F = m*(du/dt)*(gamma)3

    then acceleration a = (du/dt) = F/(m(gamma)3),

    which tends to zero as v tends to c.
  4. Nov 8, 2009 #3
    Well this is where I'm puzzled!
    Isn't the relativistic acceleration supposed to be gamma^3*(du/dt)?
    So why exclude gamma^3?

    Sorry if I'm being rather annoying :(
  5. Nov 8, 2009 #4
    [tex]a_x = \gamma^{-3} a_x^\prime[/tex]

    [tex] \lim_{v \to c} \frac{1}{\gamma^3} = \lim_{v \to c} \sqrt{1- \frac{v^2}{c^2}}^3 [/tex]

    You are neglecting the reciprocal, which can be easy to do since gamma is already in a reciprocal form. The reciprocal of gamma approaches zero as v approaches c.
    Last edited: Nov 8, 2009
  6. Nov 8, 2009 #5
    Thank you blkqi...That cleared it up a bit!
    If I understood this correctly:
    -Relativistic acceleration = du/dt and NOT (gamma^3)*(du/dt)
    -Therefore du/dt = (F/m)(1/(gamma)^3)
    From which we can easily get the answer that agrees with the theory.

    Please correct me if I'm wrong but that's about it, right?
  7. Nov 8, 2009 #6
    Sorry, I added the acceleration transformation above.
  8. Nov 8, 2009 #7
  9. Nov 8, 2009 #8
    EDIT: Checking the link!
    We didn't really take the acceleration transformation equation but I will try and digest it fully and come back to this thread. But, just by the looks of it, I think I pretty much understand intuitively what you're saying!
  10. Nov 8, 2009 #9
    The acceleration transformation is just a direct consequence of the velocity transformations.

    You probably know the tranform for [tex]v_x^\prime[/tex] and [tex]t^\prime[/tex]

    You also know [tex]a_x^\prime = \frac{dv_x^\prime}{dt^\prime}[/tex].

    So you just differentiate [tex]v_x^\prime[/tex] via quotient rule. Then substitute in the transform [tex]dt^\prime[/tex]
  11. Nov 8, 2009 #10
    Perhaps by "relativistic acceleration" you are thinking of "four acceleration" which is change in the "four velocity" of an object with respect to its own proper time, where in turn the "four velocity" is the change in distance relative to the change in proper time of the object. In other words it is the acceleration felt by observers on the accelerating object. The distance is the non length contracted distance measured by the observers in its initial frame and the proper time of the object is continually slowing down relative to clocks in the initial frame.

    So if the acceleration measured by observers in the initial frame is (a) the acceleration measured by observers undergoing the acceleration is [itex]a^\prime = a * \gamma^{3}[/tex]. The cubed gamma factor come about like this. In the momentarily comoving frame of the accelerating object a' = (du')/(dt') = [itex](dx')/(dt ')^2[/itex]. In the initial frame [itex]a = (dx '*\gamma^{-1})/(dt '*\gamma)^2 = (dx ')/(dt ')^2 * \gamma^{-3}[/itex] [itex] = (du '/dt ')* \gamma^{-3}[/itex].

    The confusion and apparent contradiction comes about because you have not specified who measures the acceleration and how it is measured.
    Last edited: Nov 8, 2009
  12. Nov 9, 2009 #11
    blkqi: Yup! That's clear enough now, thank you!
    kev: I think that last post of yours clears everything I had doubts about! Thank you!
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