Find the velocity of a particle from the Lagrangian

[Lightf]

Homework Statement

Consider the following Lagrangian of a relativistic particle moving in a D-dim space and interacting with a central potential field.

$$L=-mc^2 \sqrt{1-\frac{v^2}{c^2}} - \frac{\alpha}{r}\exp^{-\beta r}$$

...

Find the velocity v of the particle as a function of p and r.

Homework Equations

Lagrange's Equations of motion

$$\frac{d}{dt}(\frac{dL}{dv})= \frac{dL}{dr}$$

The Attempt at a Solution

$$\frac{dL}{dr} = \frac{\alpha \exp^{-\beta r}(r+1)}{r^2}$$
$$\frac{dL}{dv} \equiv p = \frac{mc^2 v}{\sqrt{1-\frac{v^2}{c^2}}}$$
$$\frac{d}{dt}(\frac{mc^2 v}{\sqrt{1-\frac{v^2}{c^2}}}) = \frac{\alpha \exp^{-\beta r}(r+1)}{r^2}$$

I am not sure what to do next. If I try to differentiate the left side I get

$$mc^2 \dot{v}(v^2(1-\frac{v^2}{c^2})^{-\frac{3}{2}} + (1-\frac{v^2}{c^2})^{-\frac{1}{2}} ) = \frac{\alpha \exp^{-\beta r}(r+1)}{r^2}$$

Which seems very hard to integrate.. Any ideas to find v easier?

 

[vanhees71]

Think about the first integrals! There is one very obvious from the fact that $L$ is not explicitly dependent on time.

Further it is important to write out everything in one set of generalized coordinates and their time derivatives. Either you use Cartesian coordinates and
$$\vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t}, \quad r=\sqrt{\vec{x} \cdot \vec{x}},$$
or you write everything in spherical coordinates $r, \vartheta,\varphi$!

 

[Lightf]

Since the $L$ is not explicitly dependent on time $\frac{dL}{dt}=0$. I cannot see the obvious :(

I will try to redo my work with generalised coordinates and see if I makes it clearer.

 

[vanhees71]

No! The Hamiltonian,
$$H=\vec{x} \cdot \vec{p}-L,$$
where
$$\vec{p}=\frac{\partial L}{\partial \dot{\vec{x}}}$$
is the canonical momentum of the particle.

 

[Lightf]

Now I am confused. Should I use Hamiltion's equations then if I use the Hamiltonian? $\dot{q}=\frac{dH}{dp}$?

 

[tannerbk]

Now I am confused. Should I use Hamiltion's equations then if I use the Hamiltonian? $\dot{q}=\frac{dH}{dp}$?

No, use the Lagrangian, just make sure you are using only one set of generalized coordinates. I would use Cartesian.