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Variational principle & lorentz force law

  1. Sep 27, 2010 #1
    1. The problem statement, all variables and given/known data
    Show that the Lorentz force law follows from the following variational principle:
    [tex]S=\frac{m}{2}\int\eta_{\mu\nu}u^\mu u^\nu ds-q\int A_\mu u^\mu ds[/tex]

    2. Relevant equations
    Definition of Field Strength Tensor
    Integration by Parts
    Chain Rule & Product Rule for Derivatives

    3. The attempt at a solution
    In order to find the force equation, we need to vary the action and find when [itex]\delta S=0[/itex].

    So, we begin by varying the action integral,
    [tex]\delta S=\frac{m}{2}\int\eta_{\mu\nu}\delta(u^\mu u^\nu)ds-q\int\delta(A_\mu u^\mu)ds[/tex]

    Then apply product rule to expand all the terms,
    [tex]\delta S=m\int\eta_{\mu\nu}u^\nu\frac{d(\delta x^\mu)}{ds}ds-q\int\left(\partial_\nu A_\mu u^\mu\delta x^\nu+A_\mu\frac{d(\delta x^\mu)}{ds}\right)ds[/tex]

    Simplified and renamed indices in the second term,
    [tex]\delta S=m\int u_\mu\frac{d(\delta x^\mu)}{ds}ds-q\int\left(\partial_\mu A_\nu u^\nu\delta x^\mu+A_\mu\frac{d(\delta x^\mu)}{ds}\right)ds[/tex]

    Reverse integration by parts to get the variation out of the derivatives,
    [tex]\delta S=m\int\left(\frac{d}{ds}\left(u_\mu\delta x^\mu\right)-\frac{du_\mu}{ds}\delta x^\mu\right)ds-q\int\left(\partial_\mu A_\nu u^\nu\delta x^\mu+\frac{d}{ds}\left( A_\mu\delta x^\mu\right)-\frac{dA_\mu}{ds}\delta x^\mu\right)ds[/tex]

    Grouping like terms we have,
    [tex]\delta S=\int\left(\frac{d}{ds}\left(mu_\mu\delta x^\mu-qA_\mu\delta x^\mu\right)\right)ds+\int \left( q\left(\frac{dA_\mu}{ds}\delta x^\mu-\partial_\mu A_\nu u^\nu\delta x^\mu\right)-m\frac{du_\mu}{ds}\delta x^\mu\right)ds[/tex]

    After integration, the first term should be zero since the variations must vanish at the endpoints. Also, using the chain rule on the dA/ds derivative gives me
    [tex]\delta S=\int \left( q\left(\partial_\nu A_\mu u^\nu-\partial_\mu A_\nu u^\nu\right)-m\frac{du_\mu}{ds}\right)\delta x^\mu ds[/tex]

    Rearranged terms to simplify further,
    [tex]\delta S=\int \left( -q\left(\partial_\mu A_\nu-\partial_\nu A_\mu\right)u^\nu -m\frac{du_\mu}{ds}\right)\delta x^\mu ds[/tex]

    Then I use the definition of the field strength tensor to get,
    [tex]\delta S=\int \left( -qF_{\mu\nu}u^\nu -m\frac{du_\mu}{ds}\right)\delta x^\mu ds[/tex]

    Then [itex]\delta S=0[/itex] for any arbitrary variation [itex]\delta x^\mu[/itex] only if the integrand itself is zero, so

    Rearranging the expression gives

    This is the correct form for the Lorentz force law, however I have an additional minus sign. It's probably a trivial problem, but I've double checked all the work several times and I can't figure out where it's coming from.
    Last edited: Sep 27, 2010
  2. jcsd
  3. Sep 28, 2010 #2
    Perhaps the problem is this:

    On the right hand side you have [tex]u^\nu[/tex]
    On the left hand side [tex]u_\mu[/tex]

    The relation between the two is given by [tex]\eta_{\mu\nu}[/tex] and that depends on your signature convention. This signature is expressed in the sign of the first term of your action principle.
  4. Sep 28, 2010 #3
    I checked my notes and it's supposed to come out to

    We've been using the convention that the spatial terms have the minus signs in the Minkowski metric (+---). In an earlier step I used [itex]u_\mu=\eta_{\mu\nu}u^\nu[/itex] to get the lower mu index in the first term with the mass. I can't find anything that says how the signature affects the process of raising/lowering indices though.
  5. Sep 29, 2010 #4
    Question is why do you have minus sign in front of the second (or plus in front of the first) term in:

    S=\frac{m}{2}\int\eta_{\mu\nu}u^\mu u^\nu ds-q\int A_\mu u^\mu ds

    Landau and Lifgarbagez has this action


    The first term is [tex]\sqrt{\eta_{\mu\nu}u^{\mu}u^{\nu}}\,ds[/tex] but it should not matter.
    Last edited: Sep 29, 2010
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