Solving a Relativistic Momentum Problem with Differential Equations

[la6ki]

I've been trying to work on differential equations using several books and one of the first exercise questions I encountered already has me stuck.

The momentum p of an electron at speed v near the speed c of light increases
according to the formula p=mv/√(1-v^2/c^2), where m is a constant. If an electron is subject to a constant force F, Newton's second law describing its motion is

$\frac{dp}{dt}$=$\frac{d}{dt}$$\frac{mv}{√(1-v^2/c^2)}$=F

Find v(t) and show that v → c as t→∞. Find the distance traveled by the electron
in time t if it starts from rest.

I started by taking the derivative of the momentum with respect to time and obtained:

F=ma/(1-v^2/c^2)^3/2

I'm not sure how to continue from here. This differential equation doesn't look simple to solve but it's actually from the very first set of exercises of an introductory ODE chapter so I don't think I'm actually supposed to be solving it. I think I am probably missing something that makes this really simple to solve and am confusing myself.

Any help would be appreciated.

 

[Curious3141]

I've been trying to work on differential equations using several books and one of the first exercise questions I encountered already has me stuck.

The momentum p of an electron at speed v near the speed c of light increases
according to the formula p=mv/√(1-v^2/c^2), where m is a constant. If an electron is subject to a constant force F, Newton's second law describing its motion is

$\frac{dp}{dt}$=$\frac{d}{dt}$$\frac{mv}{√(1-v^2/c^2)}$=F

Find v(t) and show that v → c as t→∞. Find the distance traveled by the electron
in time t if it starts from rest.

I started by taking the derivative of the momentum with respect to time and obtained:

F=ma/(1-v^2/c^2)^3/2

I'm not sure how to continue from here. This differential equation doesn't look simple to solve but it's actually from the very first set of exercises of an introductory ODE chapter so I don't think I'm actually supposed to be solving it. I think I am probably missing something that makes this really simple to solve and am confusing myself.

Any help would be appreciated.

The derivative looks correct.

To avoid confusion, let's state it more clearly: $\displaystyle \gamma^3 m \frac{dv}{dt} = F$, where $\displaystyle \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$ using the usual definitions.

That's a simple separable ODE, and you have to solve it (no way around it). But it's very easy. You can integrate wrt $v$ by a simple trigonometric substitution. Then find the limit as $t \to \infty$.

(By the way, in one of your intermediate steps in finding $v(t)$, you will derive an expression relating $m,v,F$ and $t$ that has a very simple form that is actually immediately obvious from the relationship between force and impulse).

To find the distance, what would you do with $v(t)$?

 

[HallsofIvy]

I don't understand why you took "the derivative of the momentum". You are given that
$$\frac{dp}{dt}= \frac{d}{dt}\frac{mv}{\sqrt{1- \frac{v^2}{c^2}}}= F$$

Integrating that is trivial:
$$\frac{mv}{\sqrt{1- \frac{v^2}{c^2}}}= Ft+ C$$

Now, solve that for v. Of course, then, v= dx/dt.

 

[Curious3141]

I don't understand why you took "the derivative of the momentum". You are given that
$$\frac{dp}{dt}= \frac{d}{dt}\frac{mv}{\sqrt{1- \frac{v^2}{c^2}}}= F$$

Integrating that is trivial:
$$\frac{mv}{\sqrt{1- \frac{v^2}{c^2}}}= Ft+ C$$

Good point.

 

[la6ki]

As I suspected, I was definitely making it more complicated for myself :) Thanks for the help guys, I managed to solve it. I'm still not too experienced and sometimes obvious things don't immediately hit me in the face.