Find the velocity of a particle from the Lagrangian

AI Thread Summary
The discussion revolves around finding the velocity of a particle from a given Lagrangian that describes its motion in a D-dimensional space under a central potential. The Lagrangian is expressed as L = -mc^2 √(1 - v^2/c^2) - (α/r) exp(-βr). Participants are attempting to apply Lagrange's equations of motion but encounter difficulties in integrating the resulting expressions. There's a suggestion to simplify the problem by using a consistent set of generalized coordinates, preferably Cartesian, to clarify the calculations. The conversation emphasizes the importance of sticking to one coordinate system and considering the Hamiltonian approach if necessary, although the consensus leans towards using the Lagrangian for this problem.
Lightf
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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?
 
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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}},<br />
or you write everything in spherical coordinates r, \vartheta,\varphi!
 
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.
 
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.
 
Now I am confused. Should I use Hamiltion's equations then if I use the Hamiltonian? \dot{q}=\frac{dH}{dp}?
 
Lightf said:
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.
 
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