Relativistic Momentum: Force Accelerating a proton

[Dr. Science]

Homework Statement

Find the force necessary to give a proton an acceleration of 10^19m/s^2 when the proton has a velocity ( along the same direction as the force) of 0.9c

Homework Equations

p=gamma * m * U

F = d/dt [gamma * M * U]

F=gamma^3 *m*a

The Attempt at a Solution

initial momentum = 1/(sqrt(1-0.9c^2/c^2)) * 1.67*10^-27kg * 0.9c = 1.03 *10^-18

 

[Dr. Science]

guys..

 

[rude man]

hang on, relief is on its way, I will do it by tonight MST (but I stay up till at least 3 a.m.!).

 

[rude man]

OK, here we go, 2 glasses of Shiraz or not:

We have F = dp/dt = (d/dt){m0v/(√(1 - v2/c2)}
= m0(d/dt){v/√(1 - v2/c2)}
= m0*du/dt

where u = v/√(1 - v2/c2)

So now du/dt = du/dv*dv/dt where dv/dt = 1e19m/s2

Compute du/dv by standard way, then
du/dt is known as a function of v.
Let v = 0.9c and you have du/dt as a definite number.
Finally F = m0*du/dt
where m0 is the rest mass of a proton.

Make sense?

 

[Dr. Science]

I think so, the answer I got was 2 * 10^-7 N

Is that correct?

 

[rude man]

OK, I got
du/dv = v(-1/2)(-2v/c2)[(1 - v2/c2)^(-3/2)] + (1 - v2/c2)^(-1/2)

= 1/(1 - v2/c2)^(1/2) + 0.81/(1 - v2/c2)^(3/2)
= 2.294 + 9.780 = 12.074
Then, F = 12.074*1e19*1.67e-27 = 2.016e-7N
Yay team!

 

[Dr. Science]

Haha, Thank you so much :)