# Diffeomorphism invariance and contracted Bianchi identity

• I
• "Don't panic!"
In summary, the conversation discusses Straumann's claim that the twice contracted Bianchi identity can be derived from the diffeomorphism invariance of the Einstein-Hilbert action. The conversation also considers the derivation of the EH action's diff invariance, and the confusion surrounding the effect of the Lie derivative on the volume form.
"Don't panic!"
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?

haushofer said:
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.

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.

haushofer said:
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

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

$$\delta \sqrt{-g} = \frac{1}{2} \sqrt{-g} g^{\mu\nu} \delta g_{\mu\nu} = \sqrt{-g} g^{\mu\nu} \nabla_{\mu} \xi_{\nu} = \sqrt{-g} \nabla_{\mu} \xi^{\mu}$$

and, since

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

we get in total

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

which becomes

$$2 \int \Big \nabla_{\mu} ( \sqrt{-g} \xi^{\mu} \Bigr) d^4 x$$

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

haushofer said:
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

$$\delta \sqrt{-g} = \frac{1}{2} \sqrt{-g} g^{\mu\nu} \delta g_{\mu\nu} = \sqrt{-g} g^{\mu\nu} \nabla_{\mu} \xi_{\nu} = \sqrt{-g} \nabla_{\mu} \xi^{\mu}$$

and, since

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

we get in total

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

which becomes

$$2 \int \Big \nabla_{\mu} ( \sqrt{-g} \xi^{\mu} \Bigr) d^4 x$$

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.

Did anyone find a good answer for this? I'm stuck in the same thing...

## 1. What is diffeomorphism invariance?

Diffeomorphism invariance is a property of a physical theory or mathematical model that remains unchanged under transformations of the coordinate system used to describe it.

## 2. Why is diffeomorphism invariance important in physics?

Diffeomorphism invariance is important in physics because it allows us to describe the same physical phenomenon using different coordinate systems, making it easier to understand and analyze the underlying principles and laws governing the system.

## 3. What is the contracted Bianchi identity?

The contracted Bianchi identity is a mathematical equation that describes the relationship between the curvature of a space-time manifold and its energy-momentum content. It is used in general relativity to ensure that the laws of physics are invariant under coordinate transformations.

## 4. How does diffeomorphism invariance relate to the contracted Bianchi identity?

Diffeomorphism invariance is a necessary condition for the contracted Bianchi identity to hold. This means that if a physical theory is diffeomorphism invariant, then the contracted Bianchi identity will also hold, ensuring that the theory is consistent and valid.

## 5. Can the contracted Bianchi identity be derived from diffeomorphism invariance?

Yes, the contracted Bianchi identity can be derived from diffeomorphism invariance using mathematical techniques. This is an important step in verifying the consistency and validity of a physical theory, as it shows that the theory is able to describe the same physical phenomenon in different coordinate systems.

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