Can Newton's Second Law be Applied to Relativistic Motion?

[lylos]

Homework Statement

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 $$\vec{F}$$=$$\gamma^{3}m\vec{a}$$.

Homework Equations

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

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.

 

[Dick]

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

 

[lylos]

http://picasaweb.google.com/barry.farmer/Hw/photo?authkey=byWKUltIRn8#5109819187179884178

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

 

[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.

 

[lylos]

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.

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 :)

 

[Dick]

That's what I was getting at.