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Force to Lorentz Force

  1. Jan 31, 2014 #1
    Greetings,

    I am trying to to show that the force in this form:

    [itex]\dot{P}=e(\vec{∇}\vec{A})\vec{\dot{x}}-e\vec{∇}\phi[/itex]

    Is equal to the lorentz force. I have been trying the approach of some sort of vector identity but have not gotten anywhere.

    The equation and where I am trying to get with it are posted on wikipedia towards the bottom:
    http://en.wikipedia.org/wiki/Hamiltonian_mechanics

    I appreciate any help!
     
  2. jcsd
  3. Feb 1, 2014 #2

    vanhees71

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    The equation of motion is given by (in Heaviside-Lorentz units)
    [tex]m \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{\dot{\vec{x}}}{\sqrt{1-\dot{\vec{x}}/c^2}} \right)=q \left (\vec{E}+ \frac{\vec{v}}{c} \times \vec{B} \right )=q \left [-\vec{\nabla} \phi + \frac{\vec{v}}{c} \times (\vec{\nabla} \times \vec{A}) \right ].[/tex]
    To derive it from the principle of least action, I prefer the Lagrangian form, because there the action is manifestly covariant:
    [tex]L=-mc^2 \sqrt{1-\vec{v}^2/c^2}-q \left (\phi-\frac{\dot{\vec{x}}}{c} \cdot \vec{A} \right ).[/tex]
    The action is manifestly covariant, because
    [tex]\sqrt{1-v^2/c^2} \mathrm{d} t = \mathrm{d} \tau = \sqrt{\mathrm{d} x_{\mu} \mathrm{d} x^{\mu}}[/tex]
    and
    [tex](\phi-\frac{\dot{\vec{x}}}{c} \cdot \vec{A}) \mathrm{d} t = \mathrm{d} t \frac{1}{c} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} t} A_{\mu}=\frac{1}{c} \mathrm{d} x^{\mu} A_{\mu}.[/tex]
    The four potential [itex](A^{\mu})=(\phi,\vec{A})[/itex] is a Mikowski four-vector.

    To get the equations of motion, you just write down the Euler-Lagrange equations
    [tex]\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{\vec{x}}}=\frac{\partial L}{\partial \vec{x}}.[/tex]
    Just taking the derivatives, gives
    [tex]\vec{P}=\frac{\partial L}{\partial \dot{\vec{x}}}=\frac{m \dot{\vec{x}}}{\sqrt{1-\dot{\vec{x}}^2/c^2}}+\frac{q}{c} \vec{A}.[/tex]
    Here it is important to note that [itex]\vec{P}[/itex] is the canonical momentum, not the mechanical momentum, which is given by the first term only.

    Further we have
    [tex]\frac{\partial L}{\partial \vec{x}}=-q \vec{\nabla} \phi + \frac{q}{c} \vec{\nabla} (\dot{\vec{x}} \cdot \vec{A}).[/tex]

    Written in the three-dimensional Ricci calculus (with all indices as subscripts) the equation of motion thus reads
    [tex]\dot{P}_j=\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m \dot{x}_j}{\sqrt{1-\dot{\vec{x}}^2/c^2}} + q A_j \right ) = \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m \dot{x}_j}{\sqrt{1-\dot{\vec{x}}^2/c^2}} \right ) + \frac{q}{c}\dot{x}_k \partial_k A_j = -q \partial_j \phi + \frac{q}{c} \dot{x}_k \partial_j A_k.[/tex]
    Combining the two terms with the vector potential on the right-hand side gives
    [tex]\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m \dot{x}_j}{\sqrt{1-\dot{\vec{x}}^2/c^2}} \right )=-q \partial_j \phi + \frac{q}{c} \dot{x}_k (\partial_j A_k-\partial_k A_j).[/tex]
    The last term can be rewritten as
    [tex]\dot{x}_k (\partial_j A_k-\partial_k A_j) =\dot{x}_k \epsilon_{jkl} (\vec{\nabla} \times \vec{A})_l=[\dot{\vec{x}} \times (\vec{\nabla} \times \vec{A})]_j.[/tex]
    Rewriting everything in vector notation yields the correct equation of motion.

    Note that the notation in Wikipedia, from where you copied your equation, is not very clear. That's why I used the Ricci calculus in the intermediate steps, where it is clear what are free indices and which are repeated indices to be summed.
     
  4. Feb 1, 2014 #3

    Meir Achuz

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    [tex]{\bf F}=-e\nabla\phi-e\partial_t{\bf A}+e{\bf v\times(\nabla\times A)}[/tex]
    [tex]=-e\nabla\phi-e\partial_t{\bf A}+e\nabla{\bf(v\cdot A)}-e{\bf(v\cdot\nabla)A}[/tex]
    [tex]=-e\nabla\phi+e\nabla{\bf(v\cdot A)}-e\frac{\bf dA}{dt}.[/tex]
     
    Last edited: Feb 1, 2014
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