# Differential equation on relativistic momentum (ML Boas)

## 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

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

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
$$v=\sqrt{\frac{Ft}{m^2+\frac{Ft}{c^2}}}$$

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

Any suggestion how to determine this integral?

pasmith
Homework Helper
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?

Shoot. I forget to square Ft. Sorry,

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.