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I can find the Noether current for space time translation symmetry by demanding that the first order correction to the Lagrangian vanishes upon infinitesimal translations of coordinates. But in cases like Maxwell Lagrangian I get a non symmetric energy momentum tensor when I do this. I have seen places that mention that a symmetric energy-momentum tensor can be obtained if one takes the derivative wrt the metric. But if I am given a Lagrangian without the metric explicitly written in, how do I rewrite the Lagrangian with the correct metric dependence? For example say I am thinking of the the Free boson Lagrangian or the Maxwell Lagrangian? Is there a unique way to write the Lagrangian with the space time metric? I presume there has to be metric written in such a way that the action is an integral of a d+1 form. Is that correct?

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  • $\begingroup$ More on symmetric stress-energy-momentum tensor. Related: physics.stackexchange.com/q/27048/2451 , physics.stackexchange.com/q/119838/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Feb 3, 2015 at 23:02
  • $\begingroup$ Note that the Noether energy-momentum tensor is not the same as the Belinfante-Rosenfeld energy-momentum tensor. $\endgroup$
    – Ryan Unger
    Commented Feb 3, 2015 at 23:15
  • $\begingroup$ Thanks for the links. From reading the answers in one the links you posted I am able to rephrase the question : What are the rules to rewrite the action to make the action coordinate independent ? $\endgroup$
    – symanzik138
    Commented Feb 3, 2015 at 23:29
  • $\begingroup$ @symanzik138: Coordinate independent? The Lagrangian density, by definition a scalar, is automatically coordinate independent when integrated. This is the proof of the fact that the energy-momentum tensor has zero divergence. $\endgroup$
    – Ryan Unger
    Commented Feb 3, 2015 at 23:47
  • $\begingroup$ May be the right word I should have used is covariant. I was rephrasing the following comment " .... the action (rendered coordinates independent by the inclusion of the metric in the usual way such ... ) ….." in the answer to physics.stackexchange.com/q/119838/39942 $\endgroup$
    – symanzik138
    Commented Feb 3, 2015 at 23:56

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