Time derivative of relativistic momentum

[birulami]

Did I understand correctly that relativistic momentum is

p(t) = m\cdot\gamma(t)\cdot v(t),​

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

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

with a(t) = d v(t)/dt. Yet I cannot find the funny exponent of 3 of a \gamma anywhere in a book. Am I missing a trivial transformation of \gamma^3 a 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 p as define above is a conserved quantity?


Harald.

 

[AEM]

Two things, (1) the mass is not a constant, so you need to differentiate that as well, and (2) go to this previous thread: https://www.physicsforums.com/showthread.php?t=256933

 

[Fredrik]

No, the mass is constant. It's the relativistic mass (=\gamma m) 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.

 

[AEM]

You're right, I plead temporary brain lapse. The mass is constant. \gamma ensures that m is the rest mass.

 

[Meir Achuz]

\frac{d\bf p}{dt}= m\frac{d}{dt}\left[\frac{\bf v}<br /> {\sqrt{1-{\bf v}^2}}\right]<br /> = m[\gamma{\bf a}+\gamma^3{\bf v(v\cdot a)}]<br /> =m\gamma^3[{\bf a}+{\bf v\times(v\times a)}].
with c=1. Either final form is useful.
dp/dt is important, and is discussed in advanced texts.

 

[birulami]

You may also find the calculation I did in #15 in this thread interesting.

Indeed. I'll take a closer look. Thanks,
Harald.