# Lower vs Upper indices in stress energy tensor

In Goldstein Classical Mechanics, chapter 13 page 56, equations 13.30, the canonical stress energy tensor $$T_\mu^{\,\,\,\nu}$$ is defiend as:

$$T_\mu^{\,\,\,\nu}=\frac{\partial\mathcal{L}}{\partial \eta_{\rho,\nu}}\eta_{\rho,\mu}-\mathcal{L}\delta_{\mu}^{\,\,\,\nu}.\tag{13.30}$$

My question is: how do people know the quantity on the right is $$T_{\mu}^{\,\,\,\nu}$$? Why is it not $$T_{\mu,\nu}$$, or $$T^{\mu,\nu}$$ or $$T^{\mu}_{\,\,\,\nu}$$?

In fact, Peskin and Schroeder, page 19, equation 2.17 gives

$$T^{\mu}_{\,\,\,\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}\partial_\nu\phi-\mathcal{L}\delta^\mu_{\,\,\,\nu}.\tag{2.17}$$

This is different from Goldstein unless the stress energy tensor is symmetric. What difference does it make?

• The canonical stress tensor defined by your equation is symmetry IF $\phi$ is a scalar so at least in that case you don't have to worry about the index positions. Commented May 6, 2022 at 17:56

The functional derivative flips the up/downdedness of the index, which has to be the case, because if you have an expression like [for any two arbitrary tensors $$A$$ and $$B$$]:

$$\frac{\delta}{\delta A_{cd}}\left(A_{ab}B^{ab}\right)$$

you definitely want something like:

$$\frac{\delta}{\delta A_{cd}}\left(A_{ab}B^{ab}\right) = \delta^{c}{}_{a}\delta^{d}{}_{b} B^{ab} = B^{cd}$$

to be the answer for the functional derivative to even make any sense with the tensor indices.

1. In physics the vertical positions (up vs. down, super vs. sub) of tensor indices are usually dictated by the contravariant vs. covariant properties of the tensor.

2. One may prove that in a tensor of the form of differential quotient, the vertical index positions downstairs in the denominator have covariant vs. contravariant properties reversed.

3. The horizontal positions of the tensor indices are more a matter of convention, cf. e.g. the different conventions of Peskin & Schroeder vs. Goldstein for the canonical SEM tensor.