How does Newton's Second Law Relate Velocity and Force?

[beecher]

Homework Statement

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

Homework Equations

p = gamma*mv
gamma = 1/(1-v²/c²)^1/2

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

 

[gabbagabbahey]

Homework Statement

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

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?

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

$\gamma$ 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 $\frac{dp}{dt}$

 

[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 a lot closer to the final solution, however I still can't 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

 

[gabbagabbahey]

which is another product rule
d/dt (1-v²c^-2) = -2vc^-2+2c^-3v²

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

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

 

[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(v[gamma³v/c²dv/dt] + gammadv/dt)
=m(gamma³v²/c²dv/dt + gamma dv/dt)
=mgammadv/dt(gamma²v²/c²+1)
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 I am looking for.

 

[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...

 

[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)

 

[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?

 

[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!