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The energy-momentum tensor $T^{\mu\nu}$ is not uniquely defined because we can add a term $\partial_{\lambda}X^{\lambda\mu\nu}$ to it, where $X^{\lambda\mu\nu} = - X^{\mu\lambda\nu}$, and show that it is still divergenceless.

I know $\partial{\mu}T^{\mu\nu} = 0$ because $T^{\mu\nu}$ is a conserved current. I also know that I should use the antisymmetry condition above along with the fact that the repeated indices need to be summed over.

I also know that $$\partial_{\nu}x^{\mu} \equiv \delta_{\nu}^{\mu} \tag{1}$$ and $$\partial_{\mu}x^{\mu} \equiv \delta_{\mu}^{\mu}=4\tag{2}$$ due to the number of spacetime dimensions in the problem.

Question: How would I explicitly be able to take the derivative of a tensor using index notation? I want to show the above statement, but I would also like to explicitly be able to calculate the term $\partial_{\mu}\partial_{\lambda}X^{\lambda\mu\nu}$. Is there a way to extend the definitions (1) and (2) above for taking derivatives / multiple derivatives of tensors?

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  • $\begingroup$ I am not sure what you want to do. $\partial_\mu\partial_\lambda X^{\lambda\mu\nu}$ vanishes identically because $\partial_\mu\partial_\lambda=\partial_\lambda\partial_\mu$ (i.e., it is symmetric in the indices $\mu,\lambda$) while $X^{\lambda\mu\nu}$ is anti-symmetric in the indices $\mu,\lambda$. $\endgroup$ – Dvij Mankad Mar 6 at 11:17
  • $\begingroup$ FWIW, (1) & (2) only depend on the number of spacetime dimensions, not the Minkowski metric per se. $\endgroup$ – Qmechanic Mar 6 at 11:28
  • $\begingroup$ Moreover, if you want to add something to $T^{\mu\nu}$, it should have two free-indices. $\partial_\mu\partial_\lambda X^{\lambda\mu\nu}$ has only one free-index so adding it to $T^{\mu\nu}$ is simply not allowed. $\endgroup$ – Dvij Mankad Mar 6 at 11:29
  • $\begingroup$ I used the above example from a book as context. My question might be better stated as asking if there is a way I can extend rule (1) above to calculate a term like $\partial_{\rho}X^{\mu\nu}$. $\endgroup$ – user570877 Mar 6 at 11:31
  • $\begingroup$ Well, if you want to calculate such a term explicitly, you need to know what $X^{\mu\nu}$ is explicitly as a function of $\{x^{\kappa}\}$. $\endgroup$ – Dvij Mankad Mar 6 at 11:34
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I think you mean to add a term $\partial_\lambda X^{\lambda\mu\nu}=Y^{\mu\nu}$ to the energy-momentum tensor: Terms to be added need to have the same index structure. You also require the antisymmetry condition $X^{\lambda\mu\nu}=-X^{\mu\lambda\nu}$

The the divergence of the sum is $$\partial_\mu\left(T^{\mu\nu}+Y^{\mu\nu}\right)=\partial_\mu T^{\mu\nu}+\partial_\mu Y^{\mu\nu}$$ The first term vanished by assumption ($T^{\mu\nu}$ is a conserved current). The second term vanished due to the antisymmetry of $X^{\lambda\mu\nu}$: We can write it out as $$\partial_\mu Y^{\mu\nu}=\partial_\mu \partial_\lambda X^{\lambda\mu\nu}$$ Now keep in mind that in this expression, only $\nu$ is a free index, while $\lambda$ and $\mu$ are "dummy" indices that are summed over and can be renamed.

Then we expand the term as $$\partial_\mu Y^{\mu\nu}=\partial_\mu \partial_\lambda X^{\lambda\mu\nu}\\=\frac{1}{2} \left(\partial_\mu \partial_\lambda X^{\lambda\mu\nu}+\partial_\mu \partial_\lambda X^{\lambda\mu\nu}\right)\\=\frac{1}{2} \left(\partial_\mu \partial_\lambda X^{\lambda\mu\nu}+\partial_\lambda \partial_\mu X^{\mu\lambda\nu}\right) \\ =\frac{1}{2} \left(\partial_\mu \partial_\lambda X^{\lambda\mu\nu} -\partial_\lambda \partial_\mu X^{\lambda\mu\nu}\right)\\ =\frac{1}{2} \left(\partial_\mu \partial_\lambda X^{\lambda\mu\nu} -\partial_\mu \partial_\lambda X^{\lambda\mu\nu}\right)\\=0$$ From teh first to the second line, we have used $x=\frac12(x+x)$, from the second to the third we have renamed the indices on the second term, from the third to the fourth we have used the antisymmetry of $X^{\lambda\mu\nu}$ to exchange the first two indices, and from the fourth to the fifth we have used that the partial derivatives can be interchanged. Hence, the sum is zero.

In summary, wht we have shown is that the contraction of a symmetric pair of indices (on $\partial_\mu\partial\lambda$ and an antisymmetric one (on $X^{\lambda\mu\nu}$) results in zero.

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  • $\begingroup$ Okay, this is much satisfying. I was endlessly confused with OP's original notation! Anyway, I don't seem to recall this ambiguity in the definition of $T_{\mu\nu}$. Can you suggest what does it correspond to either physically or in the framework of Noether's theorem? $\endgroup$ – Dvij Mankad Mar 6 at 13:11
  • $\begingroup$ Sorry about that Dvij, that was my fault when I was scribbling down the work on paper. It then confused me. If I understand your question, $T^{\mu\nu}$ is an antisymmetric tensor. If we add a term $\partial_{\lambda}X^{\lambda\mu\nu}$ to $T^{\mu\nu}$, so that we redefine $T^{\mu\nu}$ as $T^{\mu\nu} + \partial_{\lambda}X^{\lambda\mu\nu}$ we can make our new energy-momentum tensor symmetric. $\endgroup$ – user570877 Mar 6 at 13:25
  • $\begingroup$ @user570877 It's fine, of course! :) No, $T^{\mu\nu}$ is not an anti-symmetric tensor--it is already a symmetric tensor. And adding $\partial_\lambda X^{\lambda\mu\nu}$ does not keep it symmetric unless $ X^{\lambda\mu\nu}$ is symmetric in $\mu,\nu$ indices. $\endgroup$ – Dvij Mankad Mar 6 at 13:48
  • $\begingroup$ @DvijMankad: The standard derivation of $T^{\mu\nu}$ as Noether current is not, in general, symmetric, but you can add "improvement terms" along the lines above to make it symmetric and keep it conserved (see e.g. en.wikipedia.org/wiki/…). ... $\endgroup$ – Toffomat Mar 6 at 13:54
  • $\begingroup$ ... (continued) On the other hand, all this is in flat spacetime, and does not always generalise well to GR (curved space, covariant derivatives etc.). There, it is more convenient to use the variation of the action w.r.t. the metric, see e.g. Carroll's lecture notes (arxiv.org/abs/gr-qc/9712019), discussion around Eqns. (4.70) and (5.38). $\endgroup$ – Toffomat Mar 6 at 13:54

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