# Difference between $\partial$ and $\nabla$ in general relativity

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I read a lot in Road to Reality, so I think I might use some general relativity terms where I should only special ones.

In our lectures we just had $\partial_\mu$ which would have the plain partial differentials. On a problem set, the Bianchi identity for the Maxwell field tensor is given as: $$\partial^\alpha F^{\beta\gamma} + \partial^\beta F^{\gamma\beta} + \partial^\gamma F^{\alpha\beta} = 0.$$

In Penrose's book, this identity is given as $\nabla_{[a} R_{bc]d}{}^e = 0$ where the square brackets denote the antisymmetrisation like in the previous form. Are those square brackets standard notation in Physics?

Since $\partial_\mu$ is basically the $(\partial_t, \nabla)$ it is a covector or covariant vector. Penrose calls this $\nabla_a$ the covariant derivative (something with a connection and curved manifolds as far as I understood). If I am in a non-curved $\mathbb M$ Minkowski $(1, 3)$ space where I have no curvature (since $\eta_{\mu\nu}$ is $\mathop{diag}(1, -1, -1, -1)$?), I was thingking that $\partial_\mu = \nabla_\mu$. Can I write $\nabla_\mu$ instead of my partial derivatives or do they mean something different?

This post imported from StackExchange Physics at 2014-06-14 13:03 (UCT), posted by SE-user queueoverflow

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Are those square brackets standard notation in Physics?

Yes. See, for example Sean Carroll notes. At least I can tell you from two other classic references using that notation, "General Relativity" by Wald (1984) and "A First Introducion to General Relativity" by Schutz (2009 for the most recent edition)


If I am in a non-curved $\mathbb M$ Minkowski $(1, 3)$ space where I have no curvature (since $\eta_{\mu\nu}$ is $\mathop{diag}(1, -1, -1, -1)$?), I was thingking that $\partial_\mu = \nabla_\mu$. Can I write $\nabla_\mu$ instead of my partial derivatives or do they mean something different?

You can use them indifferently in that case. However, NOT if you switch to non-cartesian coordinates (e.g. Spherical coordinates), because in that case the connection coefficients are not zero in general, even in the absence of curvature, and so the covariant derivatives may differ from the ordinary partial derivatives, even in flat space.

I would simply avoid mixing the symbols or, in the future, it will cost you some extra effort to undo the habit, when you learn about GR and curved spaces.

These are the definitions:

$\partial_\mu V^{\nu} = \frac{\partial}{\partial x^{\mu}}V^{\nu}$

$\nabla_\mu V^{\nu} = \frac{\partial}{\partial x^{\mu}}V^{\nu}+\Gamma^{\nu}_{\mu\alpha}V^{\alpha}$

The so-called connection coefficients are the $\Gamma^{\nu}_{\mu\alpha}$. Their definition consists on a certain combination of partial derivatives of the elements of the metric. In a flat space and cartesian coordinates you can ignore them: they are zero since the elements of the metric are all just constant numbers, $(1,-1,-1,-1)$. It doesn't mean however, that they are zero in Special Relativity in general: the diagonal elements of the metric tensor in spherical coordinates for instance, are functions of the coordinates, namely, $(1, -1, -r^2, -r^2 sin^2 \theta)$ although the space is flat.

If you are interested in getting used to covariant derivatives and tensor calculations in general, without investing too much effort, I suggest you the last chapter (specially the solved problems) of the classic, small book "Vector Calculus" (M.R. Spiegel) from the Schaum series. And, for getting a feeling of the geometrical meaning of the connection coefficients, google for "Parallel Transport". The Schutz book mentioned above has also a very nice explanation.

This post imported from StackExchange Physics at 2014-06-14 13:03 (UCT), posted by SE-user Eduardo Guerras Valera
answered Jan 18, 2013 by (435 points)
I used a contravariant vector in the definitions in order to keep them as simple as possible. Google for the slightly different definition of covariant derivative when applied to covariant vectors, and for the general definitions for tensors of higher ranks. It is not conceptually difficult, but the written expressions are long.

This post imported from StackExchange Physics at 2014-06-14 13:03 (UCT), posted by SE-user Eduardo Guerras Valera
“Parallel Transport” is pictured nicely in “Road to Reality”, so I have an idea what this does. I currently have Arfken & Weber: “Mathematical Methods for Physicists” checkout out, it has a section about that too.

This post imported from StackExchange Physics at 2014-06-14 13:03 (UCT), posted by SE-user queueoverflow
Arfken&Weber has nice explanations and is quite comprehensive. In a more advanced level is Stone&Goldbart (there is a free preprint available in the webpage of one of the authors). The problem with that books is that the authors insist on ignoring the importance of giving solved examples.

This post imported from StackExchange Physics at 2014-06-14 13:03 (UCT), posted by SE-user Eduardo Guerras Valera
...or they give the solved problems only to faculty professors. I hate that.

This post imported from StackExchange Physics at 2014-06-14 13:03 (UCT), posted by SE-user Eduardo Guerras Valera
I am now attending a GR lecture and I understand what you meant :-)

This post imported from StackExchange Physics at 2014-06-14 13:03 (UCT), posted by SE-user queueoverflow

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