Homework Help: 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!