# Show that F = gamma^3*ma

1. Sep 19, 2009

### beecher

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

Newton's second law is given by F = dp/dt. If the force is always parallel to the velocity, show that F = gamma3ma

2. Relevant equations

p = gamma*mv
gamma = 1/(1-v2/c2)1/2

3. The attempt at a solution

I really have no clue where to begin. This is what I've done so far, but I don't think I'm even on the right track.

F = dp/dt = gamma3ma = [1/(1-v2/c2)1/2]3ma
=1/(1-v2/c2)3/2ma
=1/(1 - 3v2/c2 + 3v4/c4 - v6/c6)1/2ma

Thats as far as i get on the right side.

For the left, all i can manage is dp/dt = d/dx p = d/dx gamma*mv
This is where I get confused. gamma and m are constant, so does dp/dt = gamma*m*dv/dt?

Some clarification would be great.

Thanks

2. Sep 19, 2009

### gabbagabbahey

Actually, Newton's second law says that $$\textbf{F}=\frac{d\textbf{p}}{dt}$$, this is why the condition that "force is always parallel to velocity" is important. ....Do you see why this condition means $F=\frac{dp}{dt}$?

Hint: Think of a particle undergoing uniform circular motion, its speed is constant, yet its momentum is still changing...why?

$\gamma$ is dependent on the speed of the particle, right? Does the speed of the particle change with time when a force is applied parallel to its motion? If so, then $\gamma$ will have implicit time dependence, and you will need to use the product rule to take the derivative $\frac{dp}{dt}$

Last edited: Sep 19, 2009
3. Sep 19, 2009

### beecher

That makes sense, I wasnt thinking when I said gamma was constant.
So now I have dp/dt = d/dt(gamma*mv) = m d/dt (gamma*v) and I use product rule
= m (v*d/dt gamma + gamma dv/dt) and now I need to find d/dt gamma
gamma = (1-v2/c2)-1/2 and I use chain rule
d/dt gamma = -1/2 (1-v2/c2)-3/2d/dt(1-v2c-2) which is another product rule
d/dt (1-v2c-2) = -2vc-2+2c-3v2
so now plugging back in I get:
dp/dt = m (v[-1/2 (1-v2/c2)-3/2(-2vc-2+2c-3v2)] + gamma dv/dt)
This = m [v([(-1)(-2v/c2+2v2/c3)]/2gamma3) + gamma*dv/dt]
=m[v([v/c2-v2/c3]/gamma3)]+gamma*dv/dt
=m[(v2/c2-v3/c3)(gamma-3)+gamma*dv/dt]
=(v2/c2-v3/c3)gamma-3m + gamma*dv/dt*m
I am now alot closer to the final solution, however I still cant get the last step.
I have gamma*dv/dt*m which = gamma*ma on the right side.
I need to make it gamma3ma.
One more tip should do it.
Thanks

4. Sep 19, 2009

### gabbagabbahey

Ermm...isn't the speed of light a constant? You don't need the product rule here, but you do need the chain rule:

$$\frac{d}{dt}\left(1-\frac{v^2}{c^2}\right)=\frac{-2v}{c^2}\frac{dv}{dt}$$

5. Sep 19, 2009

### beecher

Your right, I've been doing this for so long my brain is sluggish.
So dp/dt = m(v[-1/2(1-v2/c2)-3/2(-2v/c2dv/dt)] + gammadv/dt)
=m(gamma3v2/c2dv/dt + gamma dv/dt)
So thanks to your help I can now see the end in sight but still one last hump to get over to finally get the gamma3ma that im looking for.

6. Sep 19, 2009

### gabbagabbahey

Concentrate on the term in brackets:

$$\gamma^2\frac{v^2}{c^2}+1=\frac{v^2}{c^2\left(1-\frac{v^2}{c^2}\right)}+1=\frac{v^2}{c^2-v^2}+1$$

put everything over a common denominator and simplify...

7. Sep 19, 2009

### beecher

[v2/(c2-v2)]+[(c2-v2)/(c2-v2)]
=c2/c2-v2
Sub back into the rest
m*gammac2dv/dt/(c2-v2)
=[(mc2dv/dt)/(1-v2/c2)1/2] / c2-v2
=[(mc2dv/dt)/(1-v2/c2)1/2] * 1/(c2-v2)

8. Sep 19, 2009

### gabbagabbahey

Sure, but if you divide both the numerator and denominator of $\frac{c^2}{c^2-v^2}$ by $c^2$, what do you get?

9. Sep 19, 2009

### beecher

Aha!

Then it becomes gamma2 giving me mgammadv/dt(gamma2)
=mgamma3dv/dt
=gamma3ma
=Solution! :)

Thanks you very much, your help has been greatly appreciated!