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Time derivative of relativistic momentum

  1. Jan 11, 2009 #1
    Did I understand correctly that relativistic momentum is

    [tex]p(t) = m\cdot\gamma(t)\cdot v(t)[/tex],​

    where [itex]\gamma = c/\sqrt{c^2-v^2}[/itex] and [itex]c[/itex] is the speed of light? For the fun of it I wrote down the time derivative and got

    [tex]{d\over dt}p(t) = \gamma^3(t)\cdot a(t)[/tex]​

    with [itex]a(t) = d v(t)/dt[/itex]. Yet I cannot find the funny exponent of 3 of a [itex]\gamma[/itex] anywhere in a book. Am I missing a trivial transformation of [itex]\gamma^3 a[/itex] into a better known form? Or is it that the time derivative of relativistic momentum is not an important concept, which is why I did not come across this formula before?

    Additional question: is [itex]p[/itex] as define above is a conserved quantity?

    Last edited: Jan 11, 2009
  2. jcsd
  3. Jan 11, 2009 #2


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  4. Jan 11, 2009 #3


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    No, the mass is constant. It's the relativistic mass (=[itex]\gamma m[/itex]) that isn't constant.

    Birulami, your calculation is fine, and yes, the p you have defined is conserved. I don't know why you can't find it in a book. One possible reason is that it's often convenient to talk about rapidity instead of velocity when you're describing the motion of an accelerating particle. E.g. in post #15 in this thread, the ^3 appears on a hyperbolic function (eq. 12).

    You may also find the calculation I did in #15 in this thread interesting.
    Last edited: Jan 11, 2009
  5. Jan 11, 2009 #4


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    You're right, I plead temporary brain lapse. The mass is constant. [tex] \gamma [/tex] ensures that m is the rest mass.
  6. Jan 11, 2009 #5


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    [tex]\frac{d\bf p}{dt}= m\frac{d}{dt}\left[\frac{\bf v}
    {\sqrt{1-{\bf v}^2}}\right]
    = m[\gamma{\bf a}+\gamma^3{\bf v(v\cdot a)}]
    =m\gamma^3[{\bf a}+{\bf v\times(v\times a)}][/tex].
    with c=1. Either final form is useful.
    dp/dt is important, and is discussed in advanced texts.
    Last edited: Jan 11, 2009
  7. Jan 12, 2009 #6
    Indeed. I'll take a closer look. Thanks,
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