# Trajectory of a particle in a field (special relativity)

1. Apr 24, 2010

### Kynio

Hi, I'm new on this forum and I would like to say hello to everybody!
I have a problem with homework from my "Basics of theoretical phisics" class.

1. The problem statement, all variables and given/known data
I have to find a trajectory of a particle in field of force:

$$F = - \frac{\alpha}{x^2}$$

2. Relevant equations

I was said to use:

$$F = \dot{p} = \frac{dp}{dt}$$ and $$p=mv\gamma$$

where $$\gamma$$ is: $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ and m is rest mass.

3. The attempt at a solution

$$\dot{p} = \frac{dp}{dt}=m \frac{d (v\gamma)}{dt} = m (\dot{v}\gamma + v \dot{\gamma})$$

$$\dot{\gamma} = -\frac{1}{2} (1-\frac{v^2}{c^2})^{-3/2} \frac{2v}{c} \dot{v} = \dot{v} \frac{v}{c^2} \gamma^3$$

$$\dot{p} = m(\dot{v} \gamma + v \dot{v} \frac{v}{c^2} \gamma^3) = m \dot{v} \gamma (1+ \frac{v^2}{c^2} \gamma^2)= m \dot{v} \gamma^3$$

$$v=\dot{x} , m \dot{v} \gamma^3 = m \ddot{x} \gamma^3 = -\frac{\alpha}{x^2}$$

$$\ddot{x} x^2 \gamma^3 = -\frac{\alpha}{m}$$

Now I have a problem with that equation. Any ideas how to solve it and get $$x(t)$$?

2. Apr 24, 2010

### Cyosis

This is similar to a classical particle in a gravitational field which cannot be solved for x(t). Adding the gamma factor to it doesn't make things prettier. I don't see how you can solve this for x(t). Are you sure you have to use this method?

3. Apr 24, 2010

### gabbagabbahey

I think you'll want to use the work-energy theorem and conservation of energy instead...

Edit: I suppose since you've come this far with the force method, you may as well continue....just multiply both sides of your equation by $$\frac{\dot{x}}{c^2x^2}$$ and integrate w.r.t. $t$. You won't be able to get an solution in the form $x(t)=\ldots$, but you should be able to get an implicit solution

Last edited: Apr 24, 2010
4. Apr 25, 2010

### Kynio

Since the equation doesn't depend on $$t$$ explicitly, I used a substitution:
$$\dot{x}=u(x)$$
$$\ddot{x}=\frac{du}{dt}=\frac{du}{dx}\frac{dx}{dt}=u'\dot{x}=u'u$$

After separating variables and integrating, I've got:

$$\dot{x}=c\sqrt{1-\frac{c^4m^2x^2}{(\alpha+Dmx)^2}$$ where $$D$$ is a constant.
I think I cannon do much more, tried to solve this in Mathematica, but got a nightmarish outcome.

I don't really have an idea how to use it. Can you at least tell me how to start?

Last edited: Apr 26, 2010
5. Apr 25, 2010

### gabbagabbahey

You've now got a separable 1st order ODE...surely you know how to solve that

EDIT: Shouldn't you have $m^2$ in the numerator, and not $m^4$?

You'll end up with the same DE as above, but a little quicker. Energy is conserved here, so you have $E=\gamma mc^2+U(x)$ where the total energy $E$ is just some constant, and $U(x)$ is the potential energy of the particle due to the conservative force field

$$U(x)=-\int \textbf{F}\cdot d\textbf{r}=\alpha\int \frac{dx}{x^2}=-\frac{\alpha}{x}$$

So,

$$\gamma=\left(1-\frac{\dot{x}^2}{c^2}\right)^{-2}=\frac{E}{mc^2}-\frac{\alpha}{mc^2 x}$$

Solve for $\dot{x}$ and you should end up with the same DE you have above.

Last edited: Apr 25, 2010