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Time derivative of relativistic momentum

  1. Sep 16, 2008 #1
    How does one take the time derivative of ϒmv ?

    I tried treating gamma and mv as separate functions but it just gets messy and ultimately wrong.
     
  2. jcsd
  3. Sep 17, 2008 #2

    gabbagabbahey

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    Well, the product rule simply gives:
    [tex] \dot {\vec{p}} = \dot {\gamma} m \vec{v} + \gamma \dot {m} \vec{v} + \gamma m \dot {\vec{v}} [/tex]

    For systems that conserve mass, [tex] \dot m =0 [/tex]. While,
    [tex] \dot {\gamma} = \frac{d}{dt} \left( 1- \frac{v^2}{c^2} \right) ^{-1/2} = \left( \frac{-1}{2} \right) \left( \frac{2v \dot {v} }{c^2} \right) \left( 1- \frac{v^2}{c^2} \right) ^{-3/2} = - \gamma ^3 \left( \frac{v \dot {v}}{c^2} \right) [/tex]

    And so,

    [tex]\dot {\vec{p}} = \gamma m \dot {\vec{v}} -\gamma ^3 m \left( \frac{v \dot {v}}{c^2} \right) \vec{v} =\gamma m {\vec{a}} -\gamma ^3 m \left( \frac{v a}{c^2} \right) \vec{v} [/tex]
     
  4. Sep 18, 2008 #3
    Why is [tex] \left( \frac{2v \dot {v} }{c^2} \right) [/tex] not [tex]\left( \frac{-2v \dot {v} }{c^2} \right)[/tex]? What happened to the minus sign?
     
  5. Sep 19, 2008 #4

    gabbagabbahey

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    OOps,. my bad:redface:. Yes I accidentally dropped a negative sign. The correct answer should be
    [tex]\dot {\vec{p}} = \gamma m \dot {\vec{v}} +\gamma ^3 m \left( \frac{v \dot {v}}{c^2} \right) \vec{v} =\gamma m {\vec{a}} +\gamma ^3 m \left( \frac{v a}{c^2} \right) \vec{v}[/tex]
     
  6. Sep 20, 2008 #5
    Ah yes, thank you
     
  7. Oct 31, 2011 #6
    i can't get (−2*v*v/c^2). what rule do we use on this? can anybody show a more detail step on this part?
     
    Last edited: Oct 31, 2011
  8. Oct 31, 2011 #7
    Chain rule. Since v is a function of t, the time derivative of 1-v^2/c^2 is -2v(dv/dt)/c^2.

    https://www.physicsforums.com/showthread.php?t=343032
    I posted a similar question a year after this one if you want a second angle.
     
    Last edited: Oct 31, 2011
  9. Nov 1, 2011 #8
    thanks master newbie, i was confused on the c^2 actually. so we just leave it as it is bcoz we want the derivative of v not c. correct me if i'm wrong.
     
  10. Nov 1, 2011 #9
    It is left there because c (and c^2) is a constant.
     
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