If $A_\mu$ is a one-form, then is there a widely accepted convention among physicists about whether the notation $$\partial_\mu A^\mu \tag{1}$$ means "the partial-derivative four-divergence of the four-vector $A^\mu$ corresponding to $A_\mu$", i.e. $$\partial_\mu (g^{\mu \nu} A_\nu),\tag{2}$$ or just $$g^{\mu \nu} \partial_\mu A_\nu~?\tag{3}$$

The former definition corresponds more naturally to our usual definition of the partial derivative, but has the unfortunate property that $\partial_\mu A^\mu \neq \partial^\mu A_\mu$. For higher partial derivatives, do we adopt the convention that all partial derivatives are taken before raising or lowering any indices, so that the the contractions are invariant under the interchange of which index is raised and which is lowered? Or are partial (as opposed to covariant) derivatives used rarely enough in GR that there's no need to adopt a general convention for how they work (after all, the quantity $\partial_\mu A^\mu$ is non-tensorial under either convention)?

(Please don't close this question as being about math rather than physics. This question is asking whether there is a notational convention accepted among physicists, and has nothing to do with math.)

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    $\begingroup$ I don't know of any conventions and I don't think there are any, precisely for the reason you state. But if I had to guess, I'd say that most people would agree that the metric goes inside the derivative. $\endgroup$ – Javier Jul 21 '17 at 21:51
  • $\begingroup$ Perhaps the question could be better-posed/more interesting from the math pov if you replaced those partials by covariant derivatives, but with a connection that need not be metric compatible. Thoughts? $\endgroup$ – AccidentalFourierTransform Jul 21 '17 at 22:10
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    $\begingroup$ @AccidentalFourierTransform Partial derivatives are covariant derivatives with a connection that need not be metric compatible. Technically, any system of coordinates defines a connection in which parallel-transport is defined by simply keeping the partials with respect to each coordinate direction constant. That's just not usually a very physically useful connection (except in the case of Cartesian coordinates on flat spacetime). $\endgroup$ – tparker Jul 21 '17 at 23:05
  • $\begingroup$ @tparker good point. $\endgroup$ – AccidentalFourierTransform Jul 22 '17 at 9:37

There isn't one, because partial derivatives are not meaningful in GR.

Partial derivatives can appear in two places:

  • Exterior derivatives
  • Lie derivatives.

Obviously they can also appear if you expand a covariant derivative but you really shouldn't raise or lower individual incides then.

For covariant derivatives, it doesn't matter, because $\nabla g=0$, so you can freely move $g$ in or out of the derivative and then we have $\nabla_\mu A^\mu=\nabla^\mu A_\mu$.

For Lie-derivatives, you can express them with covariant derivatives. However it does matter, because Lie-derivatives do not commute with $g$, unless your vector field is a Killing-field, so we have $\mathcal L_X A^\mu\neq(\mathcal L_X A_\nu)g^{\mu\nu}$, in this case, you need to specify whether you raise/lower before or after the Lie-derivative. However I need to say that the index notation meshes really badly with the Lie-derivative notation anyways.

For exterior derivatives, you can express that with covariant derivatives, and also, the exterior derivative is meaningful if and only if, you calculate it on a differential form, which are, by definition, lower-indexed.

As AccidentialFourierTransform said in the comments, the issue is more interesting if you have multiple connections and/or multiple metrics and/or a non-compatible connection. Every time I have seen such situations in the physics literature, the raisings/lowerings were written out explicitly, or a convention was declared beforehand, but because these occurrences are rather specific, one cannot really make a definitive convention en general.

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    $\begingroup$ @AccidentalFourierTransform I have noted non-metric connections in the last paragraph. $\endgroup$ – Bence Racskó Jul 22 '17 at 14:01
  • $\begingroup$ Note that, as I mentioned in a comment to the OP, partial derivatives technically are covariant derivatives with respect to a connection that is not necessarily metric compatible. $\endgroup$ – tparker Jul 22 '17 at 15:34
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    $\begingroup$ @tparker And that is not necessarily globally defined, and whose existence entirely depends on the whim of choosing a chart. While technically $\partial_\mu$ is indeed a local connection, in terms of function it has no internal meaning. It is only used as a "reference device" because we know how to calculate it. $\endgroup$ – Bence Racskó Jul 22 '17 at 15:39

Most of the doubts in your questions can be solved if you avoid calling a vector (or a form) their coordinates.

$A_{\mu}$ is not a one-form: $A = A_{\mu}dx^{\mu}$ is.

$g^{\mu\nu}$ is not a tensor: $g= g^{\mu\nu}e_{\mu}\otimes e_{\nu}$ is.

As such one thing is just taking partial derivatives of some functions with respect to their variables, namely $$ \sum_{\mu}\frac{\partial}{\partial x^{\mu}} A_{\mu}(x) $$ one other thing is the contraction of a tensor, namely making the tensor act on some dual basis, that is $$ \sum_{\mu\nu\sigma}(g^{\mu\nu}e_{\mu}\otimes e_{\nu})(A_{\sigma}dx^{\sigma}) = \sum_{\mu\nu\sigma}(g^{\mu\nu}A_{\sigma})\, e_{\mu}\, e_{\nu}(dx^{\sigma}) $$ The convention is that you just have to carry the bases things act upon and that is it.

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  • $\begingroup$ So what is your answer to my specific question? $\endgroup$ – tparker Jul 22 '17 at 14:12
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    $\begingroup$ The answer to your question is that there is no "raising or lowering of the indeces" nor there is the need to define what derivatives to take first, because you are doing two different things that you are mistaking by the same: the former is taking a divergence, the latter is contracting a tensor. $\endgroup$ – gented Jul 22 '17 at 14:19
  • $\begingroup$ I'm not doing any things, I'm simply asking about the interpretation of the notation $\partial_\mu A^\mu$. I never actually did any mathematical operations. $\endgroup$ – tparker Jul 22 '17 at 14:34
  • $\begingroup$ $A_\mu$ is a one-form if you use abstract index notation, which is the correct thing to do. The indices are just type annotations. $\endgroup$ – Robin Ekman Jul 28 '17 at 22:37
  • $\begingroup$ "...is a one-form if you use abstract index notation, which is the correct thing to do" No, it isn't and no, it isn't the correct thing to do exactly because you run into the problems that the question is describing. $\endgroup$ – gented Jul 28 '17 at 22:38

Comments to the post (v4):

  1. If $A^{\mu}$ is supposed to be (components of) a vector field, i.e. a (1,0) contravariant tensor field, then the expression (1) is not a divergence. A divergence of a vector field in a pseudo-Riemannian manifold is a scalar field, i.e. a (0,0) tensor field, and has the local form $${\rm div} A~=~ \frac{1}{\sqrt{|g|}}\partial_{\mu} (\sqrt{|g|} A^{\mu}) \tag{A}$$

  2. Similarly, if $A_{\nu}$ is supposed to be (components of) a co-vector field, i.e. a (0,1) covariant tensor field, then the expression (3) is not a (0,0) tensor field.

  3. Apart from the important objection about not working with non-covariant quantities, if OP is merely asking about conventions for a notational short-hand for working with a partial derivative $$\partial^{\mu}\tag{B}$$ with raised index, say, in a general relativistic context, it seems most convenient to let the metric be outside, i.e. $$\partial^{\mu}~:=~g^{\mu\nu}\partial_{\nu}.\tag{C}$$ E.g. the Laplace-Beltrami operator would then become $$\Delta~=~\frac{1}{\sqrt{|g|}}\partial_{\mu}\sqrt{|g|}\partial^{\mu}.\tag{D}$$ But we cannot really recommend the notation (B) outside a special relativistic context in order not to create unnecessary confusion.

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  • $\begingroup$ I was indeed just asking about notational shorthand; as I mentioned in my question, none of the expressions in the OP are tensorial. $\endgroup$ – tparker Jul 22 '17 at 14:41

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