Tensor Derivatives in Index Notation in Special Relativity

Question

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?

Answer 1

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.