# Need help with Special Relativity: Force and Energy

1. Jun 20, 2010

### unam1292

1. The problem statement, all variables and given/known data

I'm trying to show that

$$\frac{dE}{dT}$$= F x V where F is force and V is velocity

2. Relevant equations

F=$$\frac{dP}{dT}$$=M x V x $$\gamma$$

All of the other equations from special relativity (contraction, dilation, energy) are probably important as well.

3. The attempt at a solution

Honestly, I have no idea where to start. Can someone please point me in the right direction. I'm a high school junior and I have very little experience with relativity. I understand that I should show some work, but I truly don't have any but I have been trying for a while.

2. Jun 20, 2010

### vela

Staff Emeritus
You should know the formula for the relativistic energy E of an object. Try differentiating that with respect to time and see what you get.

3. Jun 20, 2010

### unam1292

The formulas for relativistic energy don't involve time (t) as far as I can tell.
I'm not sure how taking the derivative of that would help unless I'm
overlooking something.

4. Jun 20, 2010

### vela

Staff Emeritus
It doesn't depend explicitly on time, but the energy depends on $\gamma$, which, in turn, depends on velocity, which does change with time.

5. Jun 20, 2010

### unam1292

gosh, now I'm even more stuck, thanks anyway.

so you meant that I should use $$\frac{x}{t}$$ instead of v?

well after doing that, and taking the derivative, I can see no way for it to manipulate in F x V.

6. Jun 20, 2010

### vela

Staff Emeritus
No, I think that'll make it more complicated. Keep it in terms of just the velocity v.

$$\gamma = \frac{1}{\sqrt{1-(v/c)^2}}$$

What do you get if you differentiate it with respect to time, remembering that v is a function of time?

Last edited: Jun 20, 2010
7. Jun 20, 2010

### unam1292

ok, this is what I have

$$\frac{dE}{dt}$$=$$mc^{2}\frac{-\gamma^{3}v}{c^{2}}\frac{dv}{dt}$$= F x V

am I doing this correctly? when equate the equation above with F x V (where F = gamma*mass*acceleration*v)
i get that -$$\gamma^{3}=\gamma$$ so I get the feeling I messed up somewhere...

Last edited: Jun 20, 2010
8. Jun 20, 2010

### vela

Staff Emeritus
Hmm, not sure how you came up with that. Can you show more steps?

9. Jun 20, 2010

### unam1292

Sure,

So here's the derivative of gamma
$$\dot {\gamma} = \frac{d}{dt} \left( 1- \frac{v^2}{c^2} \right) ^{-1/2} = \left( \frac{-1}{2} \right) \left( \frac{-2v \dot {v} }{c^2} \right) \left( 1- \frac{v^2}{c^2} \right) ^{-3/2} = \gamma ^3 \left( \frac{v \dot {v}}{c^2} \right)$$

hence
$$\frac{dE}{dt}=mc^{2}\frac{-\gamma^{3}v}{c^{2}}\frac{dv}{dt}$$

because E = mc$$^{2}\gamma$$.

$$F\bulletV=m\gamma\frac{dv}{dt}v = \frac{dE}{dt}=mc^{2}\frac{\gamma^{3}v}{c^{2}}\frac{dv}{dt}$$

and then I simplified this

Last edited: Jun 20, 2010
10. Jun 20, 2010

### vela

Staff Emeritus
You're getting there. You need to be a bit more careful when calculating F. You know that the momentum is given by $p=\gamma mv$. When you differentiate it, you need to use the product rule.