A Landau Lifshitz Gravitational field equation

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1. Aug 9, 2016

Mr-R

Book: Landau Lifshitz, The Classical Theorey of Fields, chapter 11, section 95.

I have gone through the derivation of Einstein field equations but not without holes to fill and fix in my understanding. Lets start with the action for the grtavitational field $S_g$ which after some explanation they defined it as: $$\int R\sqrt{-g}d\Omega= \int G\sqrt{-g}d\Omega+\int \frac{\partial (\sqrt{-g}w^i)}{\partial x^i}d\Omega$$
Where as it is explained in the book, $R$ is the Ricci scalar. On the RHS, $G$ contains the metric tensor and its first derivative. The second term on the RHS has the form of the divergence of some function $w^i$ and by Gauss' theorem and the variational principle, it vanishes at the limits (Over all space-time coordinates) of the region of integration.
Amid varying we get :
$$\delta \int \sqrt{-g}Rd\Omega=\delta \int \sqrt{-g}Gd\Omega$$
Now they note that the quantity on the LHS is a scalar and therefore the RHS is also a scalar. But "G itslef is not a scalar". This confuses me because the last equality has the same form, which to me makes $G$ looks like a scalar when compared to the LHS ($R$). Does the very first (unvaried) equality show that $G$ is not a scalar? And $G$ does not have any indices. Doesn't that make it a scalar?
That's it for now. Thanks in advance.

2. Aug 10, 2016

haushofer

I'm not familiar with L&L's derivation and notation, but as it stands I'd say your G is indeed a scalar. But which one is not clear to me, as I cannot think of a scalar quantity contain only up to first derivatives of the metric. So I guess there's your confusion.

3. Aug 10, 2016

samalkhaiat

Your first equation tells you that G is not a scalar. And your second equation says that $\delta \left(\int d^{4}x \ \sqrt{-g} \ G \right)$ is a scalar. In special relativity you have something "similar":
$$\int dS_{\mu} \ J^{\mu}(x) = \int d^{3}x \ J^{0}(x) .$$
The LHS and RHS are both Lorentz scalars, while $J^{0}(x)$ is obviously not a scalar.

4. Aug 10, 2016

haushofer

Isn't the divergence term also a scalar? Why would G not be a scalar then?

5. Aug 10, 2016

Mr-R

$G$ was calculated to and found to be $G=g^{ik}( \Gamma^m_{il}\Gamma^l_{km}-\Gamma^l_{ik}\Gamma^m_{lm})$. Isn't this a scalar??

Your explanation makes sense. But I can't relate it to my question. Please refer to the form of $G$ I provided above.

6. Aug 10, 2016

Mr-R

haushoer, that's what I am saying. Maybe the author means it's not a Lorentz invariant scalar? I doubt it though.

7. Aug 10, 2016

samalkhaiat

No. It is not a scalar because one can always choose a coordinate system in which $g^{\mu\nu} = \eta^{\mu\nu}$ and $\Gamma^{\rho}_{\sigma \tau} = 0$ at any point. So, $G$ can be made to vanish by simple choice of coordinates. But scalar, vector and higher rank tensor that vanish in one particular coordinate system, they vanish in any coordinate system. So, $G$ can not be a scalar. You also have the choice to prove that fact by brute force.
A scalar can be obtained by integrating a non-scalar quantity.

8. Aug 10, 2016

Mr-R

Thanks to you, I think that I understand now,. The Ricci scalar $R=g_{ik}R^{ik}$ is a scalar because even if we chose an inertial frame it might not vanish due to the derivatives of Christoffel symbols contained in $R$, $\partial \Gamma$. Which means $R\propto \partial^2g$. So even a flat coordinate system could have a non vanishing second derivative of the metric tensor. Is this correct?
I think my problem was naively thinking that a zero rank tensor is automatically regarded as a scalar.
Another question: Can one tell if $G$(as defined above) is a tensor just from "looking" at it? Or does one has to perform a coordinate transformation and see if it transforms like a tensor?

9. Aug 11, 2016

haushofer

Ah, I see now. To answer your last question: no, one needs to perform an explicit coordinate transformation because the expression involves non-tensorial quantities: the connection. Contracting all the indices does not automatically make a quantity a scalar under gct's! Your G turns out NOT to be a scalar, which is not surprising, because whenever connections show up outside the Riemann tensor, you should be alerted. Actually, you should explicitly check for yourself that G is not a scalar by simply performing a gct on it!

Your form of the action is discussed by Zee, chapter VI.5 appendix 7. You take the Einstein-Hilbert action and integrate by parts the derivatives on the connection. These terms on their own are not tensorial, and hence if you throw away boundary terms you end up with a Lagrangian which is not a scalar anymore; it is only modulo boundary terms, and that's why the action is a scalar.

10. Aug 11, 2016

haushofer

Something similar is encountered with the Chern-Simons term which modifies the Einstein-Hilbert term.

11. Aug 11, 2016

Mr-R

This! I was going to ask why the Riemann tensor is a Tensor although it is made up of Christoffel symbols but then remembered that it is obvious from it's derivation. Or from this definition as well $[\nabla_i,\nabla_j] \zeta^k=R^k_{~cij}~\zeta^c$. Thanks for your help and also for the references.

12. Aug 11, 2016

vanhees71

One should note that the RHS is only a scalar if the corresponding charge (i.e., this integral on the right-hand side is conserved), i.e., if the current obeys the continuity equation,
$$\partial_{\mu} J^{\mu}=\partial_t J^0 + \vec{\nabla} \cdot \vec{J}=0.$$
It's very important to keep this in mind to avoid a lot of confusion in such situations. Otherwise only the LHS is a covariant quantity by construction, and you have to define carefully the hypersurface you choose to define this charge. A very careful discussion about these issues can be found in Jackson, Classical Electrodynamics.

13. Aug 12, 2016

samalkhaiat

Yes, that is correct. In curved space-time, you can not make $\partial^{2}g = 0$ by any coordinate transformations. In fact, $\partial^{2}g \neq 0$ means that the space-time is curved.

Zero rank tensor is a scalar, and rank one tensor is a vector. These are called tensorial objects. Tensorial object are frame-independent, i.e., you can not set them to zero by coordinate transformation.

$G$ is not a tensor. It is not a tensorial quantity (i.e., not a scalar) because (as I explained before) you can make $G = 0$ in one frame, and $G \neq 0$ in another frame. Quantities constructed from $g \Gamma \sim g^{2} \partial g$, $\Gamma^{2} \sim (g\partial g)^{2}$ and other similar combinations are not tensorial, i.e., you can not construct genuine scalars, vectors and other tensors out of them. So, yes you can tell that $G = g^{\mu\nu} \left( \Gamma^{\sigma}_{\mu\nu} \Gamma^{\rho}_{\sigma \rho} - \Gamma^{\rho}_{\mu \sigma} \Gamma^{\sigma}_{\nu \rho} \right)$ is not a zero rank tensor (i.e., not a scalar) by just looking at it, because it is of the form $G \sim g \Gamma^{2}$. Similarly, the following object is not a vector $$U^{\nu} = g^{\mu\nu}\Gamma^{\sigma}_{\mu\sigma} - g^{\mu\sigma}\Gamma^{\nu}_{\mu\sigma} .$$
And you can tell that just by looking at it: $U \sim g^{2}\partial g$ is, therefore, a frame dependent object, i.e., $U^{\mu} = 0$ in one frame and non-zero in another. Thus, $U^{\mu}$ is not a vector.
However, the difference between two connections, $\delta \Gamma^{\sigma}_{\mu\nu}$, is a rank-3 mixed tensor. Therefore, $$V^{\nu} = g^{\mu\nu}\ \delta \Gamma^{\sigma}_{\mu\sigma} - g^{\mu\sigma} \ \delta\Gamma^{\nu}_{\mu\sigma} ,$$ is a genuine vector.

14. Aug 12, 2016

samalkhaiat

Yes. And this is the reason for writing "similar" instead of similar.

15. Aug 13, 2016

Mr-R

Got it
Cheers

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