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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?

Qmechanic
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symanzik138
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  • More on symmetric stress-energy-momentum tensor. Related: http://physics.stackexchange.com/q/27048/2451 , http://physics.stackexchange.com/q/119838/2451 and links therein. – Qmechanic Feb 03 '15 at 23:02
  • Note that the Noether energy-momentum tensor is not the same as the Belinfante-Rosenfeld energy-momentum tensor. – Ryan Unger Feb 03 '15 at 23:15
  • 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 ? – symanzik138 Feb 03 '15 at 23:29
  • @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. – Ryan Unger Feb 03 '15 at 23:47
  • 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 http://physics.stackexchange.com/q/119838/39942 – symanzik138 Feb 03 '15 at 23:56
  • By "coordinate independent," we mean that $\int \mathcal{L}(x),dv=\int\mathcal{L}(x'),dv'$ where $dv$ is the canonical measure. The choice $dv:=d^nx,\sqrt{|g|}$ reduces to the ordinary Jacobian transformation rule. – Ryan Unger Feb 04 '15 at 00:06
  • Thanks Ocelo. Let's say I have the chern simons term ada, then I don't insert a sqrt(g). When I have Maxwell term F^\mu\nu F_\mu\nu then I do add the sqrt(g) term that you mentioned. I wanted to know what the general idea is. Let's say I have a \partial_\mu phi \partial^\mu phi term. Do I add a Sqrt (g) term? Is that all that I add? What about if I have a coupling to a current j^\mu a_\mu ? – symanzik138 Feb 04 '15 at 00:18
  • Disclaimer: I'm not well-versed in topological field theory. The reason why we don't need a $\sqrt{|g|}$ in the Chern-Simons action is because $a\wedge da$ is a density. It actually transforms under a coordinate change in such a way that it always cancels the $\sqrt{|g|}$ term. You have to add $\sqrt{|g|}$ whenever the Lagrangian density is a scalar. Something like $a\wedge da$ is a 3-form, not a scalar. – Ryan Unger Feb 04 '15 at 00:41
  • If the metric is not written into the Lagrangian, then you don't actually know the Lagrangian. For example, if you think the Maxwell Lagrangian is $\frac{1}{2}(E^2 - B^2)$, you are going to have a bad time when your space-time is curved. The correct expression necessarily involves the metric. – Brian Bi Feb 07 '15 at 05:26

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