# Homework Help: Proving F=y^3ma in relativity

1. Feb 12, 2006

### Pengwuino

The question I'm given is:

Newton's second law is given by $$\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over F} = \frac{{d\vec p}}{{dt}}$$. If the force is always parallel to the velocity, show that $$\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over F} = \gamma ^3 m\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over a}$$.

Now, how do I get started on this thing?

Also, what I'm really wondering is how this is actually applied. When they say $$\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over F} = \gamma ^3 m\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over a}$$… what gamma is being used? Also, are we looking at the change in momentum from wrt to the K frame?

2. Feb 12, 2006

### George Jones

Staff Emeritus
Since $\vec{F}$ and $\vec{p}$ are always parallel, you can remove the arrows, and just work with magnitudes. Write $p$ in terms of $v$, and differentiate using the product rule and the chain rule. After differentiating, find a common denominator

Regards,
George

3. Feb 12, 2006

### Pengwuino

What is the 'v'. Is v the speed of the moving frame?

4. Feb 13, 2006

### George Jones

Staff Emeritus
$v$ is the speed of the particle with respect to the lab frame, so that $p = \gamma m v$.

Regards,
George