Differential equation on relativistic momentum (ML Boas)

[agnimusayoti]

Homework Statement

The momentum $p$ of an electron at speed $v$ near the speed $c$ of light increase s according to the formula $p=\gamma mv$, where $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$; m is a constant (mass of an electron). If an electron is subject to a constant force F, Newton's second law describing its motion is: $$\frac{dp}{dt}=\frac{d}{dt} \gamma mv=F$$. Find (a) v(t) and show that v approaches c as t approaches infinite. (b) Find the distance traveled by the electron in time t if it starts from rest.

Relevant Equations

For $dy/dx = c$, the solution is $$y=\int {c dx}$$

$$p=\gamma m v$$
$$F = \frac {md (\gamma v}{dt}$$
$$\int{F dt} = \int{md (\gamma v}$$
$$F t= \gamma mv$$

At this step, I don't know how to make v as explicit function of t, since gamma is a function of v too. Thankss

 

[agnimusayoti]

Gee, I think I just have to take quadrat on both sides. Pfft..

 

[agnimusayoti]

Now, I will update what I get and still can't get the distance function.
I get:
$$v^2=\frac{Ft}{m^2+\frac{Ft}{c^2}}$$, then
$$v=\sqrt{\frac{Ft}{m^2+\frac{Ft}{c^2}}}$$

 

[agnimusayoti]

Writing $v=\frac{dx}{dt}$ gives:
$$\frac{dx}{dt}=\sqrt{\frac{Ft}{m^2+\frac{Ft}{c^2}}}$$

 

[agnimusayoti]

Any suggestion how to determine this integral?

 

[fresh_42]

You can integrate it:
https://www.wolframalpha.com/input/?i=int+(root(x/(x+a)))dx=
but I think the questions could be answered without integration.

 

[pasmith]

Any suggestion how to determine this integral?

It's easier if you start with the correct expression for v.

Now, I will update what I get and still can't get the distance function.
I get:
$$v^2=\frac{Ft}{m^2+\frac{Ft}{c^2}}$$, then
$$v=\sqrt{\frac{Ft}{m^2+\frac{Ft}{c^2}}}$$

Try again. What does squaring both sides of mv = Ft/\gamma(v) give you?

 

[agnimusayoti]

Shoot. I forget to square Ft. Sorry,

 

[agnimusayoti]

Youre right.
I get this:
$$dx=\frac{cF t}{\sqrt{m^2 c^2 + F^2 t^2}} dt$$
Then, integrating this equation, I get:
$$x=\frac{c}{F}\sqrt{m^2c^2 + F^2t^2} + C_1$$
Because electron starts from rest, therefore at t=0 x = 0. So, I get $C_1=-\frac{mc^2}{F}$.
Using this constant, I get the same answer with the solution.
Thanks. Sorry for not being meticulous.

 

[neilparker62]

https://www.wolframalpha.com/input/?i=F=m/sqrt(1-v^2/c^2)*dv/dt

Well we do get c as the maximum but we also get an oscillation which is interesting (?).

Power = F.v(t) and we can integrate power to obtain energy. Another interesting maximum emerges:

https://www.wolframalpha.com/input/?i=Integrate+++F*c*sin(F*t/(m*c))+dt