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Electrodynamics Lagrangian. differences in sign in online references.

  1. Dec 29, 2008 #1


    It is written:

    [tex] \mathcal{L} \, = \, \mathcal{L}_{\mathrm{field}} + \mathcal{L}_{\mathrm{int}} = - \frac{1}{4 \mu_0} F^{\alpha \beta} F_{\alpha \beta} + A_{\alpha} J^{\alpha} \,.

    a personal calculation with this Lagrangian, I get an off by -1 sign error, so I initially came to the conclusion that this should be:

    [tex] \mathcal{L} \, = \, \mathcal{L}_{\mathrm{field}} + \mathcal{L}_{\mathrm{int}} = \frac{1}{4 \mu_0} F^{\alpha \beta} F_{\alpha \beta} + A_{\alpha} J^{\alpha} \,.

    Searching online I find disagreement as well. Two examples are:

    http://www.wooster.edu/physics/lindner/Ph377Spring2003HW/HW2.pdf [Broken]

    Am I in error, or are there differences in conventions/definitions that can account for this sign variation?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 29, 2008 #2


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    I've always found getting these signs right very trying and it is very easy for sign errors to occur. Be wary of online sources.

    There is a choice of convention in the metric used to contract indices. Some prefer the time component of the metric to be positive and some the spatial component. This however will not affect the sign of the doubly contracted [itex]F_{\mu\nu}F^{\mu\nu}[/itex]. But it will affect the sign of the singly contracted interaction term, [itex] A_\mu J^\mu[/itex].

    This plus the fact that the over all sign of the Lagrangian density is immaterial in so far as its resulting Euler-Lagrange equations are concerned would allow both to be correct. (However one needs to be careful with the sign of the Lagrangian when one is going to add it to another Lagrangian term.)

    I'm not inclined to re-derive your three references right now so I can't tell you if they or you are in error. I will suggest that you carefully compare definitions of F, A, J, and if used the metric g in each reference and see where they may disagree. Then if so try to rework each case.

    The ultimate convention to follow is that you want the Hamiltonian density to come out positive. In my weak recollection this has always required a minus sign in front of the [itex]F_\mu\nu F^\mu\nu[/itex] term. But also recall that this tensor is anti-symmetric so simply reversing one of the pairs of indices can absorb the sign. Pay close attention to this in your references.
    [Edit: Actually the first thing to make certain of is that you get the right E-L equations, then check the over-all sign of the Lagrangian for the convention which makes H positive.]

    (Actually I might take a deeper look but I won't make any promises.)
    Last edited: Dec 29, 2008
  4. Dec 29, 2008 #3
    I think I know what this is about. While I seem to recall that

    F^{\alpha \beta} F_{\alpha \beta}

    isn't metric dependent (I'll have to confirm), the dot product part [itex]A_\mu J^\mu[/itex] is. If that's the case, then this also likely explains the sign variation that I've seen in resulting tensor equation too.
  5. Dec 29, 2008 #4
    thanks James (I think we posted at the same time;)

    wikipedia is particularily bad seeming for properly defining all the related quantities, but the process of deciphering what's posted there can be educational.

  6. Dec 30, 2008 #5
    I didn't get too far trying to calculate the electrodynamic Hamiltonian density for the general case, so I tried it for a very
    simple special case, with just an electric field component in one direction:

    &= \frac{1}{2}(E_x)^2 \\
    &= \frac{1}{2}(F_{01})^2 \\
    &= \frac{1}{2}(\partial_0 A_1 - \partial_1 A_0)^2 \\

    Goldstein gives the Hamiltonian density as

    \pi &= \frac{\partial \mathcal{L}}{\partial \dot{n}} \\
    \mathcal{H} &= \dot{n} \pi - \mathcal{L}

    If I try calculating this I get

    &= \frac{\partial}{\partial (\partial_0 A_1)} \left( \frac{1}{2}(\partial_0 A_1 - \partial_1 A_0)^2 \right) \\
    &= \partial_0 A_1 - \partial_1 A_0 \\
    &= F_{01} \\

    So this gives a Hamiltonian of
    &= \partial_0 A_1 F_{01} - \frac{1}{2}(\partial_0 A_1 - \partial_1 A_0)F_{01} \\
    &= \frac{1}{2} (\partial_0 A_1 + \partial_1 A_0 )F_{01} \\
    &= \frac{1}{2} ((\partial_0 A_1)^2 - (\partial_1 A_0)^2 ) \\

    For a Lagrangian density of [itex]E^2 - B^2[/itex] we have an energy density of [itex]E^2 + B^2[/itex], so I'd have expected the Hamiltonian density here to stay equal to [itex]E_x^2/2[/itex], but it
    doesn't look like that's what I get (what I calculated isn't at all familiar seeming).

    If I haven't made a mistake here, perhaps I'm incorrect in assuming that the Hamiltonian density of the electrodynamic Lagrangian should be the energy density?
  7. Jan 1, 2009 #6


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    A follow-up note. I did find a sign error (mixing of conventions) in the wikipedia article and left a comment in the discussion page. I have been playing with an edit of the article and may fix it if I can figure the best combination of conventions to fit with other articles.

    I also found a foot-note in my old SR text: Rindler's Introduction to Special Relativity.
    In the text:
    (38.1) ----> [tex]F_\mu = \frac{q}{c} E_{\mu\nu}U^\nu[/tex]

    (38.3)-----> [tex]{E^{\mu\nu}}_{,\mu} = k\rho_0 U^\nu = kJ^\nu[/tex]

    (38.5)-----> [tex] E_{\mu\nu} = \Phi_{\nu,\mu} - \Phi_{\mu,\nu}[/tex]

    He uses [itex]E_{\mu\nu}[/itex] instead of [itex]F_{\mu\nu}[/itex] for the electromagnetic field tensor, and [itex]\Phi^\mu \sim (\phi,c\mathbf{A})[/itex] as the 4-potential. F is the force on a test particle. (He uses lower-case bold e and b for the 3-vector electric and magnetic fields and uppercase [itex]B_{\mu\nu}[/itex] for the dual electro-magnetic field [itex]B_{mu\nu}=\frac{1}{2}\varepsilon_{\mu\nu\alpha\beta}E^{\alpha\beta}[/itex] ).

    He also uses the [itex] g \sim diag(1,-1,-1,-1)[/itex] metric convention.

    One problem in this text is that Rindler uses the test-particle Lagrangian:

    [tex] \Lambda = -cm_0\sqrt{g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu} - \frac{q}{c}\Phi_\mu \dot{x}^\mu[/tex]
    whereas I think the negative of this would have been better. His choice gives positive canonical spatial momentum when you take [itex]\tilde{p}_k=\frac{\partial}{\partial \dot{x}^k} \Lambda[/itex] but since this is a lowered index component of the full canonical 4-momentum it should be the negative under the choice of metric convention.
    Last edited: Jan 1, 2009
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