The question is pretty short. I have been told that unlike for flat spacetime, in curved spacetime we cannot write
$g^{\mu \nu}\partial_{\nu}f=\partial^{\mu}f$
With $f$ an scalar function, but I don't really get why.
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1$\begingroup$ What is the definition of $\partial^\mu$? $\endgroup$– MBNOct 6, 2020 at 11:05
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$\begingroup$ I thought it was $g^{\mu \nu} \partial_{\nu}$ but apparently it isn't. $\endgroup$– alexsanchezcastOct 6, 2020 at 11:09
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1$\begingroup$ What else would it be? $\endgroup$– Qmechanic ♦Oct 6, 2020 at 15:29
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3 Answers
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Well I'll answer but I would be interested to know if my answer is wrong. I thought we defined $\partial^a$ such that $$ \partial^a f \equiv g^{a\mu} \partial_\mu f . $$ The thing to be careful about is then the fact that $$ \partial_\mu (g^{a\mu} f) = (\partial_\mu g^{a\mu}) f + g^{a\mu} \partial_\mu f $$ so usually $$ \partial_\mu (g^{a\mu} f) \ne \partial^a f . $$ It also follows that you cannot assume that $\partial^a$ operators commute, nor even that $\partial^\mu \partial_\mu$ is equal to $\partial_\mu \partial^\mu$ unless you make some other definition of $\partial^a$.
answered Oct 6, 2020 at 11:17
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$\begingroup$ So, if I didn't misunderstand, we might actually define $\partial^{\alpha} f \equiv g^{\alpha \mu} \partial_{\mu} f$ but by doing so we would lose some interesting properties of our usual derivative $\partial_{\alpha}$ and therefore it is not a convenient way of proceeding. $\endgroup$ Oct 6, 2020 at 11:35
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$\begingroup$ Yes that is my understanding. But there is nothing wrong with the definition; it makes reasonable sense so is arguably no more inconvenient than it has to be. If in doubt you can always avoid $\partial^a$ completely in your own work ... but you still need to know what others mean by it. $\endgroup$ Oct 6, 2020 at 12:22
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For a scalar function; $\nabla_{\mu}f=\partial_{\mu}f$, its a definition, we know $\nabla_{\mu}f$ is a $(0,1)$ tensor so $g^{\mu\nu}\nabla_{\mu}f=\nabla^{\nu}f$ but there doesn't exist any similar definition where we identify contravariant version of $\nabla^{\mu}f$ with $\partial^{\mu}f$. So ultimately it boils down to the definition of $\partial^{\mu}$.
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The definition $$ \partial^{\mu} = g^{\mu \nu} \, \partial_{\nu} \tag{1} $$ doesn't have more sense than $$ x_{\mu} = g_{\mu \nu} \, x^{\nu}. \tag{2} $$ Using (1) may lead to many troubles that you want to avoid at all cost. Don't use it!
However, you may use the same for the covariant derivative: $$ \nabla^{\mu} = g^{\mu \nu} \, \nabla_{\nu}, \tag{3} $$ which may be usefull since $\nabla_{\lambda} \, g^{\mu \nu} = 0$.
