Diffeomorphism invariance and contracted Bianchi identity

  • #1
"Don't panic!"
601
7
I've been reading Straumann's book "General Relativity & Relativistic Astrophysics". In it, he claims that the twice contracted Bianchi identity: $$\nabla_{\mu}G^{\mu\nu}=0$$ (where ##G^{\mu\nu}=R^{\mu\nu}-\frac{1}{2}g^{\mu\nu}R##) is a consequence of the diffeomorphism (diff) invariance of the Einstein-Hilbert (EH) action. Now, I can show (I think) that the EH action is diff invariant, by considering a infinitesimal diff, generated by a vector field ##X##: $$\delta_{X}S_{EH}=\phi^{\ast}S_{EH}-S_{EH}=\int_{M}\mathcal{L}_{X}\left(d^{4}x\sqrt{-g}R\right)=\int_{M}\left[\mathcal{L}_{X}\left(d^{4}x\sqrt{-g}\right)R +d^{4}x\sqrt{-g}\mathcal{L}_{X}\left(R\right)\right]\\ \qquad\qquad\qquad\quad\;\;=\int_{M}d^{4}x\sqrt{-g}\left[\nabla_{\mu}X^{\mu}R+X^{\mu}\nabla_{\mu}R\right]=\int_{M}d^{4}x\sqrt{-g}\,\nabla_{\mu}\left(X^{\mu}R\right)\\ =\int_{\partial M}d^{3}x\sqrt{h}\,n_{\mu}X^{\mu}R=0\;\;\qquad\qquad\qquad\quad$$ where ##h_{ij}## is the induced metric on the boundary ##\partial M## of the manifold ##M##, with ##n^{\mu}## the normal vector to the boundary. The last equality follows upon the assumption that ##X^{\mu}## has compact support in ##M##.

However, I'm unsure how one uses this fact to derive the (twice-contracted) Bianchi identity?

Straumann simply writes: $$\delta S=\int_{M}d^{4}x\sqrt{-g}\left(\frac{1}{\sqrt{-g}}\frac{\delta S_{EH}}{\delta g^{\mu\nu}}\right)\delta g^{\mu\nu}=\int_{M}d^{4}x\sqrt{-g}\,G_{\mu\nu}\delta g^{\mu\nu}=-\int_{M}d^{4}x\sqrt{-g}\,G^{\mu\nu}\delta g_{\mu\nu}$$ and notes that for an infinitesimal diff (generated by some vector field ##X##), ##\delta g_{\mu\nu}=2\nabla_{(\mu}X_{\nu)}##, such that $$\delta_{X} S=-2\int_{M}d^{4}x\sqrt{-g}\,G^{\mu\nu}\nabla_{\mu}X_{\nu}=2\int_{M}d^{4}x\sqrt{-g}\,X_{\nu}\nabla_{\mu}G^{\mu\nu}=0$$ and so, since ##X^{\nu}## is arbitrary, it must be that ##\nabla_{\mu}G^{\mu\nu}=0##.

What confuses me about this, is that one neglects the effect of the Lie derivative on ##d^{4}x## in this case (in the proof that the EH action is diff invariant, it was taken into account). Is the point that an infinitesimal diff is carried out in the *same* coordinate chart, and so ##d^{4}x## doesn't change in this case?
 

Answers and Replies

  • #3
"Don't panic!"
601
7
Yes. See e.g.

http://web.mit.edu/edbert/GR/gr5.pdf

page 9 bottom, where this is explained in detail

Thanks. I've actually read these notes before, but I was left with some doubts. For example, the author states that one can always carry out a coordinate transformation, such that the coordinate values of the new point are the same as those of the old point. Wouldn't this also introduce a Jacobian factor in the action (when one shows that the action is diff invariant)?

Also, what further confuses me about this, is that the Lie derivative of the volume form is ##\mathcal{L}_{X}\left(d^{4}x\sqrt{-g}\right)=\nabla_{\mu}X^{\mu}d^{4}x\sqrt{-g}##. This suggests that ##d^{4}x## does transform under a diffeomorphism.
 
  • #4
haushofer
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To be honest, it has been a while since I've read those notes, and I mainly remember the huge (HUGE) confusion I had (have?) concerning diffemorphisms, coordinate transformations and Lie-derivatives. I guess it all goes back to the passive v.s. active interpretation.

However, ##d^4 x \sqrt{-g}## is a scalar, and under a coordinate transformation ##x \rightarrow x + \xi(x)## a general scalar ##\phi(x)## transforms as ##\delta \phi (x) = \xi^{\rho}\partial_{\rho}\phi(x) = \xi^{\rho}\nabla_{\rho}\phi(x)## (could be a minus sign mistake). From that you should be able to deduce the transformation law of the covariant volume element.
 
  • #5
"Don't panic!"
601
7
To be honest, it has been a while since I've read those notes, and I mainly remember the huge (HUGE) confusion I had (have?) concerning diffemorphisms, coordinate transformations and Lie-derivatives. I guess it all goes back to the passive v.s. active interpretation.

However, ##d^4 x \sqrt{-g}## is a scalar, and under a coordinate transformation ##x \rightarrow x + \xi(x)## a general scalar ##\phi(x)## transforms as ##\delta \phi (x) = \xi^{\rho}\partial_{\rho}\phi(x) = \xi^{\rho}\nabla_{\rho}\phi(x)## (could be a minus sign mistake). From that you should be able to deduce the transformation law of the covariant volume element.

I similarly am very confused over diffeomorphisms, and how to interpret passive and active coordinate transformations correctly.

I'm pretty sure, that under an infinitesimal diffeomorphism, tensors (of all rank) transform by a Lie derivative, i.e. ##\delta T^{\mu_{1}\ldots\mu_{m}}_{\;\;\nu_{1}\ldots\nu_{n}}=\mathcal{L}_{\xi}T^{\mu_{1}\ldots\mu_{m}}_{\;\;\nu_{1}\ldots\nu_{n}}##. And I know that the Lie derivative of ##d^{4}x\sqrt{-g}## is ##\mathcal{L}_{\xi}\left(d^{4}x\sqrt{-g}\right)=\nabla_{\mu}\xi^{\mu}d^{4}x\sqrt{-g}##. However, if one goes through the derivation of this, it implies that the Lie derivative of ##d^{4}x## is non-zero, implying that it does change under a diffeomorphism. Hence my confusion :frown:
 
  • #6
haushofer
Science Advisor
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Yes, I'm confused also now. I think it's a matter of definition. Somehow, the author performs an active transformation and after that performs a passive one, such that the new coordinates in the new system equal the values of the old coordinates in old system. But it was a long time for me since I did this kind of stuff, so probably I'm sloppy, but maybe it helps you in some way. If not, I apologize.

Forget about those notes. I'd say that under ##x^{\mu '}=x^{\mu}+\xi^{\mu}##, one has

[tex]
\delta \sqrt{-g} = \frac{1}{2} \sqrt{-g} g^{\mu\nu} \delta g_{\mu\nu} = \sqrt{-g} g^{\mu\nu} \nabla_{\mu} \xi_{\nu} = \sqrt{-g} \nabla_{\mu} \xi^{\mu}
[/tex]

and, since

[tex]
d^{4}x' = det|\frac{\partial x^{\mu'}}{\partial x^{\nu}}| d^4 x \ \ \rightarrow \delta (d^{4}x) = \partial_{\mu} \xi^{\mu} d^{4}x
[/tex]
(so yes: the volume element ##d^4 x## does change!)

we get in total

[tex]
\delta \int (\sqrt{-g} d^4 x ) = \int \delta (\sqrt{-g} d^4 x ) = \int \Bigl( \sqrt{-g} \nabla_{\mu} \xi^{\mu} d^4 x + \sqrt{-g} \partial_{\mu} \xi^{\mu} d^{4}x \Bigr)
[/tex]

which becomes

[tex]
2 \int \Big \nabla_{\mu} ( \sqrt{-g} \xi^{\mu} \Bigr) d^4 x
[/tex]

But to be honest, I don't understand the factor 2. Maybe someone can give more insight.
 
  • #7
"Don't panic!"
601
7
Yes, I'm confused also now. I think it's a matter of definition. Somehow, the author performs an active transformation and after that performs a passive one, such that the new coordinates in the new system equal the values of the old coordinates in old system. But it was a long time for me since I did this kind of stuff, so probably I'm sloppy, but maybe it helps you in some way. If not, I apologize.

Forget about those notes. I'd say that under ##x^{\mu '}=x^{\mu}+\xi^{\mu}##, one has

[tex]
\delta \sqrt{-g} = \frac{1}{2} \sqrt{-g} g^{\mu\nu} \delta g_{\mu\nu} = \sqrt{-g} g^{\mu\nu} \nabla_{\mu} \xi_{\nu} = \sqrt{-g} \nabla_{\mu} \xi^{\mu}
[/tex]

and, since

[tex]
d^{4}x' = det|\frac{\partial x^{\mu'}}{\partial x^{\nu}}| d^4 x \ \ \rightarrow \delta (d^{4}x) = \partial_{\mu} \xi^{\mu} d^{4}x
[/tex]
(so yes: the volume element ##d^4 x## does change!)

we get in total

[tex]
\delta \int (\sqrt{-g} d^4 x ) = \int \delta (\sqrt{-g} d^4 x ) = \int \Bigl( \sqrt{-g} \nabla_{\mu} \xi^{\mu} d^4 x + \sqrt{-g} \partial_{\mu} \xi^{\mu} d^{4}x \Bigr)
[/tex]

which becomes

[tex]
2 \int \Big \nabla_{\mu} ( \sqrt{-g} \xi^{\mu} \Bigr) d^4 x
[/tex]

But to be honest, I don't understand the factor 2. Maybe someone can give more insight.

Thanks for your response. Yeah, I'm still confused over this whole thing. I'm yet to find any satisfactory explanation of it all from any source.
 
  • #8
Alvarballon
1
0
Did anyone find a good answer for this? I'm stuck in the same thing...
 

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