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Derivation of relativistic acceleration and momentum

  1. May 5, 2008 #1
    Dear all,

    could anyone please show the full derivation of relativistic acceleration and momentum.

    Many thanks n
    happy eqtns
    R
     
  2. jcsd
  3. May 5, 2008 #2

    Hootenanny

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    Welcome to PF,

    Have you tried searching the internet?
     
  4. May 5, 2008 #3
    In relativity, mass is dependent of velocity such that,
    [tex]m=\gamma m_0=\frac{m_0}{\sqrt{1-v^2/c^2}}[/tex].
    [itex]m_0[/tex] is the mass of the object at rest, [itex]c[/itex] is the 299 792 458 m/s.
    Most equations still hold true in relativity, the major exception being F=ma.
    The following are still true:
    [tex]p=mv, F=p', a=v', v=x'.[/tex]
    Using these, we easily find that,
    [tex]p=\frac{m_0v}{\sqrt{1-v^2/c^2}}[/tex]
    and
    [tex]F=p'=(mv)'=m'v+v'm[/tex].
    Now we need to express m' in terms of only v.
    [tex]m'=\left(\frac{m_0}{\sqrt{1-v^2/c^2}}\right)'=\frac{-1/2m_0}{(1-v^2/c^2)^{3/2}}(-2v/c^2)(v')=v\frac{m_0v'}{c^2(1-v^2/c^2)^{3/2}}[/tex].
    Combining this with the above equation for force,
    [tex]F=v^2\frac{m_0v'}{c^2(1-v^2/c^2)^{3/2}}+\frac{m_0v'}{\sqrt{1-v^2/c^2}}[/tex].
    Now you can just factor and solve for [itex]a, v'[/itex].
     
  5. May 5, 2008 #4

    clem

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    You result is for parallel to v. With vectors, there are other terms.
     
  6. Jul 13, 2009 #5
    Newton's second law F/m = a Also recall a = dv/dt Also force is related to relativistic momentum by F = dp/dt
    Relativistic momentum is defined by p = mv(1 - v^2/c^2)^-.5 You need to use implicit derivation to take the derivative of this with respect to t. Thus you should have dp/dt and dv/dt term. Once you are finished getting the derivative and combining terms you should end up with dv/dt = F(1-v^2/c^2)^3/2 /m
     
  7. Jul 13, 2009 #6
    F=ma does work in both SR and GR as long as you are using the 4-vector (tensorial) version.
     
  8. Jul 13, 2009 #7

    clem

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    That result is only valid for a parallel to v.
     
  9. Jul 14, 2009 #8
    Yes, the actual equation is (gamma)ma=(F-F.v/c)v/c as i mentioned in the new thread. What im waiting for is its dervn
     
  10. Jul 14, 2009 #9
    Use the fact that
    [tex]v\frac{dv}{dt}=\vec{v}\cdot \vec{a}[/tex]
     
  11. Jul 14, 2009 #10
    do it
     
  12. Jul 14, 2009 #11
    If you carry out the same calculations I did in the first post I made in this thread, but use vectors, you get this result (you can do it yourself, it's very easy, esp. since I already did it):
    [tex]\vec{F}=m_o \vec{v}\frac{d\gamma}{dt}+m_o\gamma\frac{d\vec{v}}{dt}[/tex]
    and using my above post,
    [tex]\frac{d\gamma}{dt}=\gamma^3\frac{|\vec{a}||\vec{v}|}{c^2}[/tex],
    [tex]\vec{F}=m_o\gamma^3\vec{v}\frac{|\vec{a}||\vec{v}|}{c^2}+m_o\gamma \vec{a}=m_o\gamma^3\vec{v}\frac{\vec{a}\cdot \vec{v}}{c^2}+m_o\gamma \vec{a}[/tex]
    So,
    [tex]m_o\gamma\vec{a}=\vec{F}-[(m_o\gamma^3\vec{a})\cdot\frac{\vec{v}}{c}]\frac{\vec{v}}{c}[/tex].
    According to your author, the following must be true,
    [tex]m_o\gamma^3\vec{a}=\vec{F}[/tex].
    This suggests relativistic mass does not exist and that this author goes against mainstream theory. The only guy I know who goes against this is Levvy. We don't like to trust that guy around here. (Read: http://en.wikipedia.org/wiki/Mass_in_special_relativity#Controversy)

    I wouldn't trust this author if I were you, only if you assume relativistic mass does not exist do you get the result you posted.
     
  13. Jul 14, 2009 #12

    clem

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    As I just posted in the other thread, dotting your m\gamma a equation with v
    shows the result v.F=m\gamma^3(v.a).
    Either you or I are confused about what "mainstream theory" is.
     
  14. Nov 19, 2011 #13
    So I've been told that the speed of light remains constant despite length contraction and dime dilation because they each decrease proportionately. So a velocity of 4 meters/2 seconds in a relativistic scenario might see a halving of values of length and time so it may only traverse 2 meters but only 1 second of time elapses. My question is what happens with acceleration if it is measured in distance per time-squared...Shouldn't it decrease proportionately to square root-t? If time dilates, shouldn't time-squared dilate even more? Or is this too linear of an approach?
     
  15. Nov 19, 2011 #14
    [tex]a \equiv \frac{d^2x}{dt^2}[/tex]

    [tex]p \equiv \frac{\partial L}{\partial v}[/tex]

    with [itex]L[/itex] the relativistic Lagrangian and [itex]v[/itex] velocity
     
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