Use of covariant derivative in general relativity.

  • #1
When we derive equation of motion by variation of the action, we use rules of ordinary differentiation and integration. So only ordinary derivatives can appear in the equation. Now in general relativity we are supposed to replace all those ordinary derivatives by covariant derivatives. Is that how one rigorously get equations valid in general relativity?

Second, suppose there is a term like \del_m(1/ \sqrt{1-g^{ab}T,aT,b}). Will I have have to replace the ordinary derivatives in the denominator also in this case?

I have these doubts. Can anybody clear these? Also I am not sure if this is the best place to ask such questions, just trying since I am trying to learn things alone. Are there other websites where one can ask such questions and get help? Thanks.

Answers and Replies

  • #2
It's my impression that the use of covariant derivatives is necessary in GR to compensate for the non-linearity of the coordinate systems that are often used (like spherical coordinates). If you want to see how a vector or tensor changes as it changes position in a curved space -- or even in a flat space with curved coordinates -- the use of ordinary derivatives (or gradients) can give misleading results, since the components will change as the basis vectors change, even if the direction and magnitude of the actual vector (or tensor) remains fixed -- that is, even if it is parallel transported.

I don't know about the T,a T,b matter; I will have to think about it. Perhaps someone else can jump in now...
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  • #3
\del_m(1/ \sqrt{1-g^{ab}T,aT,b

This is unreadable - can you tex it, please ?
  • #4
I think I can answer your question, but I don't know how familiar you are with varying actions in curved spaces. In any case, the Lagrangian density gets multiplied by [tex]\sqrt{|g|}[/tex] if you recall. Now, go ahead and perform your variation, and you'll end up with ordinary derivatives. Let's look at a scalar field for simplicity. There your variation, after integration by parts, yields
[tex]\partial_\alpha ( \sqrt{|g|} \partial^\alpha \phi ) \delta \phi[/tex]

Now, making use of the identity
[tex]\nabla_\alpha v^\alpha = \frac{1}{\sqrt{|g|}} \partial_\alpha \left ( \sqrt{|g|}\,v^\alpha \right)[/tex]
we are in a good position to turn that partial derivative into a covariant derivative.

Incidently, this is why covariantly conserved vector currents lead to conserved charges (Gauss' theorem), whereas covariantly conserved tensor currents (e.g. stress energy tensor) do not.
  • #5
For Mentz114: The expression is:
\\partial_m(1/\\sqrt{1-g^{ab}\\partial_a T \\partial_b T})
In fact I need to learn how to use latex properly in this forum. Can anybody help? I didnot get the proper link by searching.
  • #6
just use [ tex ] \partial_\mu ... [ / tex ] without the added spaces.

You can also get the complete markup for the latex used by othes by clicking on the equation and copying/pasting (including the tex tags)
  • #7
Thanks. Let me try again. The expresion was:
\partial_m(\frac{1}{\sqrt{1-g^{ab}\partial_a T(x) \partial_b T(x)}})
One more question: If I make a mistake in writing in latex here, can I correct it later? If yes, then how do I do that?
  • #8
There should be an edit button next to your post, provided too much time hasn't elapsed since you hit submit, and no one has replied.

The equation you wrote down is pretty scary. Care to identify?
  • #9
Hi lbrits,
Thanks a lot for clarifications. Actually I was doing some calculations with scalar Tachyon field whose action is
S_{T}=-\int d^4x \sqrt{-g} \, \, V(T)\sqrt{1-g^{ab}\partial_a T(x) \partial_b T(x)}
The field had been used to model dark matter and energy though doesnot seem popular now a days. For one example you may see eqn(1) in : astro-ph/0212198

Now regarding your first post, yes I'm familiar with variation of the metric tensor etc though there are some gaps in my understanding. Earlier I thought I could use the equation of motion obtained by varying the action and I did not care to replace the ordinary derivatives by covariant derivatives. But after I saw cov derivatives were used to express the eqn of motion in some paper, I had this doubt that may be I didn't understand properly. But even after your clarification I have a new doubt. I could follow what you wrote. But the fact that I am using some identity to replace ord. der. by cov. der. probably means that I am just writing in a different way but the content of equations remain same. Or is that wrong? (I may be wrong in complicated cases or more general cases.) Moreover in case it is compulsory to replace all ord derv by cov. derv. after variation of action, then a generalization of the identity you wrote is necessary, for other kinds of fields. Can you give the general formula?

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