Trajectory of a particle in a field (special relativity)

Kynio
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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.

Homework Statement


I have to find a trajectory of a particle in field of force:

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

Homework 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.


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)?
 
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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?
 
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
 
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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 think you'll want to use the work-energy theorem and conservation of energy instead...
I don't really have an idea how to use it. Can you at least tell me how to start?
 
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Kynio said:
After separating variables and integrating, I've got:

\dot{x}=c\sqrt{1-\frac{c^4m^4x^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.

You've now got a separable 1st order ODE...surely you know how to solve that:wink:EDIT: Shouldn't you have m^2 in the numerator, and not m^4?

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

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.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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