Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Origin of the Maxwell energy-momentum tensor?

  1. Aug 11, 2010 #1
    Electrodynamics force is [tex]f_i=F_{ik}j^k=F_{ik}\partial_j F^{jk}[/tex]. I claim that the only way to obtain the Maxwell energy-momentum tensor [tex]T_i^j=-F_{ik}F^{jk}+\delta_i^jF_{kl}F^{kl}/4[/tex] is to write the force as a divergence: [tex]f_i=-\partial_jT_i^j[/tex].
  2. jcsd
  3. Aug 11, 2010 #2
    The energy-momentum tensor of any field must have zero divergence.
  4. Aug 11, 2010 #3
    It is a widely-spread delusion.
  5. Aug 12, 2010 #4
    It can also be derived from Noether's theorem, it is the conserved current of translations [itex]x \rightarrow x+a[/itex].
  6. Aug 12, 2010 #5
    It is a widely-spread delusion as well. The Noether's theorem gives various energy momentum tensors, but not the Maxwell tensor.
    Sorry, tex works badly.
    Canonical Lagrangian L_1[tex]L_1=-F_{ij}F^{ij}/4[/tex] gives canonical tensor
    Dirac’s Lagrangian L_2 [tex]L_2=-F_{ij}F^{ij}/4-(\partial_iA^i)^2/2[/tex] gives [tex]T_2{}_i^j=-\partial_iA_kF^{jk}+\delta_i^jF_{kl}F^{kl}/4-\partial_iA^j\partial_kA^k+\delta_i^j(\partial_kA^k)^2/2[/tex],
    Vector Lagrangian [tex]L_3=-\partial_iA^j\partial^iA_j/2[/tex] gives [tex]T_3{}_i^j=-\partial_iA_k\partial^jA^k+\delta_i^j\partial_kA_l\partial^kA^l[/tex],
    Soper’s Lagrangian [1] [tex]L_4=-F_{ij}F^{ij}/4-A_ij^i [/tex] gives [tex]T_4{}_i^j=-\partial_iA_kF^{jk}+\delta_i^jF_{kl}F^{kl}/4+\delta_i^j A_kj^k[/tex]. But Soper was mistaken: he obtain a false tensor [tex]T_f{}_i^j=-\partial_iA_kF^{jk}+\delta_i^jF_{kl}F^{kl}/4+A_ij^j[/tex] [2].
    [1] D. E. Soper, Classical Field Theory (N.Y.: John Wiley, 1976).
    [2] R.I. Khrapko, Professor Soper's mistake http://khrapkori.wmsite.ru/ftpgetfile.php?id=43&module=files
    Last edited: Aug 12, 2010
  7. Aug 12, 2010 #6
    The fact that the energy momentum tensor you wrote can be obtained by Noether's theorem is NOT a "delusion": take the formalism of general relativity, take the usual electromagnetic field lagrangian, vary not only the A fields but also the metric, and finally impose the flat space-time metric... what will you get?
  8. Aug 12, 2010 #7
    Emmy Noether did not intend to use curvilinear coordinates. And why we must prefer [tex]T_i^j=-F_{ik}F^{jk}+\delta_i^jF_{kl}F^{kl}/4[/tex] to [tex]T_1{}_i^j=-\partial_iA_kF^{jk}+\delta_i^jF_{kl}F^{kl}/4[/tex], or others?
    By the way, variation of coordinates does not give spin tensor!
  9. Aug 12, 2010 #8
    It's just a conservation law. The symmetric tensor is preferred because it is...symmetric, I guess. Variation of coordinates (and fields) by Lorentz transformations DOES give you the spin tensor. I don't know what Noether intended or not, I havent't read her biography, but using her theorem you get a lot of conservation laws (all, as far as I know) of a lagrangian field theory, regardless if the lagrangian describes a drum, a water wave, a Higgs boson or a general relativistic matter system.
  10. Aug 12, 2010 #9
    You're right that there's a problem with the normal derivation from Noether's theorem, the tensor


    is not gauge invariant. But this can be solved with a couple of tricks. One is to perform a gauge transformation when you vary A, instead of

    [tex]A_\mu (x) \rightarrow A_\mu (x+a) = A_\mu (x) + a^\nu \partial_\nu A_\mu[/tex]

    you can subtract a gauge term like this

    [tex]A_\mu (x) \rightarrow A_\mu (x) + a^\nu \partial_\nu A_\mu - \partial_\mu (a^\nu A_\nu) = A_\mu (x) + a^\nu F_{\nu\mu}[/tex]

    because a is constant, and this gives you the standard gauge invariant stress energy tensor.
  11. Aug 29, 2010 #10
    Sorry, I do not understand your tricks. They gives [tex]T_5{}_i^j=-\partial_iA_kF^{jk}-a^l\partial_iF_{lk} F^{jk}+\delta_i^jF_{kl}F^{kl}/4[/tex] (Sorry, tex works badly)
    My thought is the Lagrange formalism with the Noether's theorem cannot give the electromagnetic energy-momentum tensor. And the deriving of this tensor, which is described in all textbooks, is a mistake. See [1]
    [1] R.I.Khrapko, “Mechanical stresses produced by a light beam,” J. Modern Optics, 55, 1487-1500 (2008) http://khrapkori.wmsite.ru/ftpgetfile.php?module=files&id=9
    Last edited: Aug 29, 2010
  12. Aug 30, 2010 #11
    Sorry but I don't understand how you got that expression. I may have messed up, I don't know. I got the standard Maxwell stress energy tensor from the general equation for a Noether current.

    If you have some fields [itex]\phi_a (x)[/itex] and a lagrangian [itex]\mathcal{L}(\phi_a (x),\partial_\mu \phi_a (x))[/itex], and if the action is invariant under some symmetry, you can transform the fields like this:
    [tex]\phi_a (x) \rightarrow \phi_a (x) + \epsilon^\alpha \Phi_{a\alpha}(x)[/tex]
    where epsilon is an infinitesimal parameter and Phi encodes the transformation in terms of phi. The lagrangian transforms like
    [tex]\mathcal{L} \rightarrow \mathcal{L} + \epsilon^\alpha \partial_\mu \Lambda^\mu_\alpha[/tex]
    So that the action is invariant. Then the Noether current is
    [tex]j^\mu_\alpha = \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi_a)}\Phi_{a\alpha} - \Lambda^\mu_\alpha[/tex]
    To get the Maxwell stress energy tensor, set [itex]\phi_a = A_\mu[/itex] so that a is a spacetime index, and set [itex]\epsilon^\alpha = a^\nu[/itex] where a is a constant infinitesimal vector so that alpha is also a spacetime index. Then to get the standard Maxwell stress energy tensor we require [itex]\Phi_{a\alpha} = F_{\mu\nu}[/itex] which I got by Taylor expanding A as normal, [itex]A_\mu (x) \rightarrow A_\mu (x) + a^\nu \partial_\nu A_\mu[/itex] then subtracting a gauge term, which doesn't affect the lagrangian, so it won't affect Lambda either. That gives [itex]a^\nu \Phi_{\mu\nu} = a^\nu \partial_\nu A_\mu - a^\nu \partial_\mu A_\nu = a^\nu F_{\mu\nu}[/itex] I think that works but I'm not 100%, if you found a mistake let me know...
  13. Aug 31, 2010 #12
    Usually, when deriving the Noether current (energy-momentum or angular momentum tensor), they use an infinitesimal coordinate transformation rather than transformation of field (the field satisfies Euler-Lagrange equations, i.e. field equations). Please present your calculation in details as an attachment.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook