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Proving F=y^3ma in relativity

  1. Feb 12, 2006 #1

    Pengwuino

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    The question I'm given is:

    Newton's second law is given by [tex] \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
    \over F} = \frac{{d\vec p}}{{dt}}[/tex]. If the force is always parallel to the velocity, show that [tex] \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
    \over F} = \gamma ^3 m\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
    \over a} [/tex].

    Now, how do I get started on this thing?

    Also, what I'm really wondering is how this is actually applied. When they say [tex] \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
    \over F} = \gamma ^3 m\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
    \over a} [/tex]… what gamma is being used? Also, are we looking at the change in momentum from wrt to the K frame?
     
  2. jcsd
  3. Feb 12, 2006 #2

    George Jones

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    Since [itex]\vec{F}[/itex] and [itex]\vec{p}[/itex] are always parallel, you can remove the arrows, and just work with magnitudes. Write [itex]p[/itex] in terms of [itex]v[/itex], and differentiate using the product rule and the chain rule. After differentiating, find a common denominator

    Regards,
    George
     
  4. Feb 12, 2006 #3

    Pengwuino

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    What is the 'v'. Is v the speed of the moving frame?
     
  5. Feb 13, 2006 #4

    George Jones

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    [itex]v[/itex] is the speed of the particle with respect to the lab frame, so that [itex]p = \gamma m v[/itex].

    Regards,
    George
     
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