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Lorentz force law from action

  1. Jun 19, 2014 #1
    Hi, I am trying to derive the Lorentz force law in the following form:
    [tex]
    q \frac{dw^\mu}{d\tau} = q w^\mu \partial_\nu A_\sigma \epsilon_\mu^{\nu \sigma}
    [/tex]
    by varying the following Lagrangian for a classical particle:
    [tex]
    S = \int d^3 x \left( -m \int d\tau \delta(x-w(\tau) ) + q \int d\tau \frac{dw^\mu}{d\tau} A_\mu \delta(x - w(\tau) ) \right)
    [/tex]
    where w tracks the position of the particle as a function of proper time. Note that there may be a couple terms/indices missing from the above expressions (still trying to figure that out).

    I've read on a different thread that Feynman has that derivation in "QM and Path Integrals", which I have handy right now, however I couldn't find the derivation I am looking for.

    Any push in the right direction would be more than appreciated.
     
  2. jcsd
  3. Jun 20, 2014 #2

    vanhees71

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    The covariant formalism is a bit tricky. The problem is that you cannot use the proper time directly in the Lagrangian, because it implicitly contains the constraint that [itex]\mathrm{d} \tau=\mathrm{d} t \sqrt{1-v^2}[/itex] (I'm setting the speed of light to 1 here).

    You can, however use an arbitrary "world parameter" and write down a parameter-independent action functional. For the free particle it's
    [tex]S_0[x]=-m \int \mathrm{d} \lambda \sqrt{\dot{x}_{\mu} \dot{x}^{\mu}}.[/tex]
    I'm using the "west-coast convention" for the pseudometric, i.e., [itex]\eta_{\mu \nu}=\text{diag}(1,-1,-1,-1)[/itex].
    The interaction with an external electromagnetic field is determined by
    [tex]S_i[x]=-q \int \mathrm{d} \lambda \dot{x}^{\mu} A_{\mu}(x).[/tex]
    Note that the total Lagrangian,
    [tex]L=-m \sqrt{\dot{x}_{\mu} \dot{x}^{\mu}} - q \dot{x}^{\mu} A_{\mu}(x)[/tex]
    is a homogeneous function of degree 1 wrt. [itex]\dot{x}[/itex]. This implies the parameter independence. Now you can derive the equations of motion from the Hamilton principle of least action as usual, using the Euler-Lagrange equations. The canonical momenta are given by
    [tex]p_{\mu}=\frac{\partial L}{\partial \dot{x}^{\mu}}=-m \frac{\dot{x}_{\mu}}{\sqrt{\dot{x}_{\nu} \dot{x}^{\nu}}}-q A_{\mu}[/tex]
    and thus the equation of motion
    [tex]\dot{p}_{\mu}=\frac{\partial L}{\partial x^{\mu}} \; \Rightarrow \; -m \frac{\mathrm{d}}{\mathrm{d} \lambda} \left (\frac{\dot{x}_{\mu}}{\sqrt{\dot{x}_{\nu} \dot{x}^{\nu}}} \right ) - q \dot{x}^{\nu} \partial_{\nu} A_{\mu}(x)=-q \dot{x}^{\nu} \partial_{\mu} A_{\nu}.[/tex]
    Now you can bring this into a more familiar form by rearranging the terms to
    [tex]m \frac{\mathrm{d}}{\mathrm{d} \lambda} \left (\frac{\dot{x}_{\mu}}{\sqrt{\dot{x}_{\nu} \dot{x}^{\nu}}} \right ) = q \dot{x}^{\nu} (\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu})=q F_{\mu \nu} \dot{x}^{\nu}.[/tex]
    Now you can choose for [itex]\lambda[/itex] whatever parameter you like. It is crucial to note that the equation of motion, if derived from a Lagrangian that is homogeneous in [itex]\dot{x}^{\mu}[/itex] of degree one, automatically fulfills the constraint equation
    [tex]\dot{x}^{\mu} \frac{\mathrm{d}}{\mathrm{d} \lambda} \left (\frac{\dot{x}_{\mu}}{\sqrt{\dot{x}_{\nu} \dot{x}^{\nu}}} \right )=0,[/tex]
    which implies that the 4 equations of motion are not independent from each other and thus you can choose the parameter [itex]\lambda[/itex] as you like after the variation is done, i.e., on the level of the equations of motion.

    If you choose the proper time, [itex]\lambda=\tau[/itex], then you get [itex]\dot{x}_{\mu} \dot{x}^{\mu}=1[/itex] and thus
    [tex]m \frac{\mathrm{d} u_{\mu}}{\mathrm{d} \tau}=q F_{\mu \nu} u^{\nu} \quad \text{with} \quad u^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.[/tex]
    You can as well chose a coordinate time, [itex]t[/itex] of an inertial frame. Then the equation of motion appears in a form that is not longer manifestly covariant, but you get, because of [tex]\dot{x}^{\mu} \dot{x}_{\nu}=1-\vec{v}^2, \quad \vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t},[/tex]
    for the spatial part of the equations of motion (writing the Faraday tensor in terms of the three-dimensional notation with [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex])
    [tex]m \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{\vec{v}}{\sqrt{1-\vec{v}^2}} \right ) = q (\vec{E}+\vec{v} \times \vec{B}),[/tex]
    and the time component is just the energy-work relation, following from this:
    [tex]m \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{1}{\sqrt{1-\vec{v}^2}} \right ) = \vec{v} \cdot \vec{E}.[/tex]
     
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