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Find the velocity of a particle from the Lagrangian

  1. Aug 19, 2013 #1
    1. The problem statement, all variables and given/known data

    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.


    2. Relevant equations

    Lagrange's Equations of motion

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


    3. 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?
     
  2. jcsd
  3. Aug 19, 2013 #2

    vanhees71

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    2016 Award

    Think about the first integrals! There is one very obvious from the fact that [itex]L[/itex] 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
    [tex]\vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t}, \quad r=\sqrt{\vec{x} \cdot \vec{x}},
    [/tex]
    or you write everything in spherical coordinates [itex]r, \vartheta,\varphi[/itex]!
     
  4. Aug 19, 2013 #3
    Since the [itex]L[/itex] is not explicitly dependent on time [itex]\frac{dL}{dt}=0[/itex]. I cannot see the obvious :(

    I will try to redo my work with generalised coordinates and see if I makes it clearer.
     
  5. Aug 19, 2013 #4

    vanhees71

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    Science Advisor
    2016 Award

    No! The Hamiltonian,
    [tex]H=\vec{x} \cdot \vec{p}-L,[/tex]
    where
    [tex]\vec{p}=\frac{\partial L}{\partial \dot{\vec{x}}}[/tex]
    is the canonical momentum of the particle.
     
  6. Aug 19, 2013 #5
    Now I am confused. Should I use Hamiltion's equations then if I use the Hamiltonian? [itex]\dot{q}=\frac{dH}{dp}[/itex]?
     
  7. Aug 21, 2013 #6
    No, use the Lagrangian, just make sure you are using only one set of generalized coordinates. I would use Cartesian.
     
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