# Time derivative of relativistic momentum

1. Jan 11, 2009

### 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?

:surprised
Harald.

Last edited: Jan 11, 2009
2. Jan 11, 2009

### AEM

3. Jan 11, 2009

### Fredrik

Staff Emeritus
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.

Last edited: Jan 11, 2009
4. Jan 11, 2009

### AEM

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

5. Jan 11, 2009

### clem

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

Last edited: Jan 11, 2009
6. Jan 12, 2009

### birulami

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