# Homework Help: Show that m varies with v in theory of relativity

1. Mar 16, 2013

### Feodalherren

Am I supposed to take dm/dv? That means m0 is constant, correct?

If I do that I end up getting a m' on the Left hand side... What the heck does that mean?! Also where the crap is the "a" coming from?!

2. Mar 16, 2013

### rude man

Consider: F = d/dt (p) = m dv/dt + v dm/dt = m a + v dm/dt
Now introduce your idea of taking dm/dv in coming up with dm/dt as a function of a and dm/dv.

Warning: I have not succeeded myself in deriving this result. And I know it's correct, dadblame it!
Somebdy - help!

3. Mar 16, 2013

### rude man

a = dv/dt.

4. Mar 16, 2013

### tms

Correct.
Since there is no $m'$ in the problem, the $m'$ means whatever you said it meant when you wrote it down.

You'll also get a better response if you control your frustration and avoid even mild profanities here.

5. Mar 16, 2013

### tms

It does work.

6. Mar 16, 2013

### rude man

Yeah, I know it has to, I just could not get the result.

7. Mar 17, 2013

### TSny

I think the problem should have stated that the force is assumed to act parallel to the velocity. Perhaps that was meant to be evident by the lack of vector notation. But, anyway, it's worth noting that if the force does not act parallel to the velocity, then you will get a different result for the relationship between the force and acceleration. (In fact, the acceleration is generally not even in the same direction as the force).

8. Mar 17, 2013

### Feodalherren

I'm still not getting it right. There must be something I'm missing.

The constants are m0 and c, correct?

when I do d/dv I end up with [ -2m0vc^(-2) ] / (1-(v^2/c^2))^3/2

9. Mar 17, 2013

### ehild

F=d(mv)/dt.

dm/dt=dm/dv dv/dt = a dm/dv .

Your formula for dm/dv is not correct. The first "2" should be 1/2.

ehild