# Relativistic Moment F = dp/dt

1. Sep 13, 2007

### lylos

1. The problem statement, all variables and given/known data
Newton's second law is given by $$\vec{F}$$=$$\frac{d\vec{p}}{dt}$$. If the force is always parallel to the velocity, show that $$\vec{F}$$=$$\gamma^{3}m\vec{a}$$.

2. Relevant equations
$$p=\gamma m\vec{v}$$
$$\frac{d\vec{v}}{dt}=\vec{a}$$
$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

3. The attempt at a solution
I figure all you need to do would be to find d/dt of p. After a couple tries, I'm still unable to get the result that it provides.

2. Sep 13, 2007

### Dick

Yes, that's what you do, and it's straightforward enough. No real tricks. Maybe you could show us an attempt?

3. Sep 13, 2007

### lylos

Here's a scan of what I did... at the very bottom I was then lost.

Last edited by a moderator: May 3, 2017
4. Sep 13, 2007

### Dick

You missed a factor of dv/dt in the first term when you did the chain rule. So factoring out all of the common terms (gamma*m*dv/dt), you only need to show that (v^2/c^2)/(1-v^2/c^2)+1=gamma^2. You are almost there.

5. Sep 13, 2007

### lylos

Thanks for your help... I didn't realize that when I would take the derivative of -v^2/c^2 that it would be (-2v/c^2)(dv/dt). I think that's what you were getting at. I was able to get the answer that way :)

6. Sep 13, 2007

### Dick

That's what I was getting at.