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I’m working through Special Relativity by V. Faraoni, and am puzzled by something in his chapters on tensors. He tells us that the partial derivative of a tensor field, e.g. $T_{\alpha, \gamma}$, is not a tensor, because

\begin{align*}\frac{\partial T_{\alpha'}}{\partial x^{\gamma'}} &= \frac{\partial x^{\delta}}{\partial x^{\gamma'}}\frac{\partial}{\partial x^{\delta}}\left[\frac{\partial x^{\nu}}{\partial x^{\alpha'}}T_{\nu}\right]\\ &= \frac{\partial x^\delta}{\partial x^{\gamma'}}\left[\frac{\partial}{\partial x^{\delta}}\left(\frac{\partial x^{\nu}}{\partial x^{\alpha'}}\right)\right]T_{\nu} + \frac{\partial x^{\delta}}{\partial x^{\gamma'}}\frac{\partial x^{\nu}}{\partial x^{\alpha'}}\frac{\partial T_{\nu}}{\partial x^{\delta}}, \end{align*}

and generally $\frac{\partial^2 x^{\nu}}{\partial x^{\delta}\partial x^{\alpha'}}\ne 0$ (I’ve simplified his more general mixed-tensor version here). That is perfectly fine by me.

But on the next page he sets an example of the preceding few sections, using fluid mechanics’ velocity vector field, $v(\mathbf{x},t)$, and from this, states that the components

$$v_{ij} = \frac{\partial v_i}{\partial x^j}$$

form a 2-tensor. He gives no justification for this in terms of the above warning.

Am I misunderstanding the definition of $v_{ij}$ here—is it not equivalent to $v_{i,j}$? Or is it the case for the fluid velocity field that the mixed derivatives vanish? My background in fluid mechanics is sketchy at best.

I’m also unsure why he’s using covariant indices for the velocity vector field; might that just be typographical error?

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It seems to me that the $T_{\alpha,\gamma}$-expression is in arbitrary coordinates. However, the $v_{ij}$-expression is in Cartesian coordinates... and has likely implicitly used the metric to raise and lower indices.

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  • $\begingroup$ Yes, thank you. I had misinterpreted the text as implying that the given expression (from a cartesian basis) was tensor invariant Re-reading, what he actually referred to was “the tensor whose components are <...> in one given cartesian basis”. So the expression is not invariant, but the tensor is. Groovy. $\endgroup$
    – Bob Arthur
    Jun 8, 2020 at 3:17

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