# An equivalent lagrangian 1

1. Jan 19, 2010

Hi everyone!
again problem with d'inverno's equation!!

ok let me see, in chapter 11 section 11.6, it is said that
(∂L ̅)/(∂g^ab )=Γ_ac^d Γ_bd^c-Γ_ab^c Γ_cd^d

as we see at first if you derive (∂L ̅)/(∂g^ab ) from equation 11.37 you can have the above result.
but my professor said it's not true as you should consider that every Γ_ab^c has a g^ab term in it and you should consider that too.

as this is simple derivative I think what I did was correct!!
can anyone help me with it?

2. Jan 19, 2010

### Altabeh

Your Professor's claim is simply wrong and he's confused which has a gaudy reason. As you can see, the equation (11,42) represents the derivative of the quantity $$\bar{\mbox{L}}_G$$ (this is not a true scalar density, it only transforms like a scalar density for linear transformations) with respect to the metric density $$\mbox{g}^{ab}$$; so here one must distinguish all the quantities involved in $$\bar{\mbox{L}}_G$$, including $$g^{ab}$$, $$\mbox{g}^{ab}$$ and $$g^{ab}_{,c}$$. But these three just boil down to two quantities, one always being the metric tensor or the metric density and the second is the first derivative of the other with respect to coordinates. So I guess here you are a little bit flabbergasted why the the derivatives of $$\mbox{g}^{ab}$$ would also be allowed to take a role as a variable of $$\mbox{g}^{ab}$$ as D'inverno himself does so and this is the reason why probably your Professor is mistaken; first you got to know the fact that

$$\mbox{g}^{ab}_{,c}=(\sqrt{-g}g^{ab})_{,c}=(\sqrt{-g})_{,c}g^{ab}+\sqrt{-g}g^{ab}_{,c}=$$
$$1/2\sqrt{-g}g^{fd}g^{ab}g_{fd,c}+\sqrt{-g}g^{ab}_{,c}= 1/2\sqrt{-g}{g^{fd}g^{ab}g_{fd,c}+g^{ab}_{,c}}=$$
$$\sqrt{-g}(-1/2g^{fd}_{,c}g^{ab}g_{fd}+g^{ab}_{,c})$$.

Thus we recovered a factor $$\sqrt{-g}$$, which if was multiplied by that free metric tensor in another Christoffel symbol in the equation (11.38) would leave a metric density, and the metric derivatives $$g^{ab}_{,c}$$.

Second, your Professor seems to be missing this fact and that actually D'inverno makes use of the above setup in which the only two variables of $$\bar{\mbox{L}}_G$$ are supposed to be the metric density and its derivatives.

So you are right and don't be worried about anything. Btw may I ask where you are taking courses of GR and where you hail from? Just curious.

AB

Last edited: Jan 19, 2010
3. Jan 19, 2010

Thank you so much. this really helped me.

But could you tell me what is exactly the difference between \mbox{g}^{ab} and g^{ab}?
I thought it's just a \sqrt{-g} Coefficient.
and that means I can work with one of them!

By the way I am a master student and am taking the course in Tehran university. I'm iranian.
Where are you from?

4. Jan 19, 2010

oh! you're right! $$\sqrt{-g}$$ is not just a Coefficient!
but can you tell me what does it mean physically?
I mean as I know the metric is actually the potential.
what does $$\sqrt{-g}$$ change?

5. Jan 19, 2010

### Altabeh

Hi my friend

\mbox{g}^{ab} is actually a contravariant density of weight w=2, so it is different from the dual metric g^ab which contains information as to how a spacetime is distorted and warped due to the presence of gravitational masses. The first one does not give away such vital information but it rather helps the Einstein's equations be able to get extracted from the variational principle and, more importantly, it provides a platform for the deduction of the Landau-Lifschitz identities through taking advantage of the transformation of it into another coordinates system and putting the idea of variational principle into use when doing so.

$$\sqrt{-g}$$ is the square root of the determinant of metric tensor, -g. It is also said to be the determinant of the Jaccobi matrix of coordinates transformation of determinant of metric tensor. (To get it, simply take the determinant of the coordinate transformation of metric tensor.) And metric is not actually a potential and it ain't gonna be ever. (Just roll your eyes up this last quote I made to see the most primitive explanation of metric tensor in terms of GR.)

I'm a compatriot of yours!

AB

6. Jan 20, 2010

hey compatriot !!

thanks for explaining it so clear!
but you know, I've heard from someone that the metric somehow gives us the potential which is in the space-time, gamma is the force and the riemannian tensor R is the change of the gravitational field.
Is this completely wrong?!

7. Jan 20, 2010

### Altabeh

1- If you by potential mean the gravitational potential, then a perturbed metric tensor can lead us to the potential as in a perturbed spacetime due to a weak gravitational field, that fairly is metric determining the gravitational potential -2MG/c^2r.

2- If the Christoffel symbols were representative of gravitational forces, then they wouldn't enter the geodesic equation with being coefficients of [velocity]^2 so as to obtain components of acceleration. Their dimension is (length)^(-1) so let alone force and stuff. They just expose the nature of curvature of spacetime in the presence of matter and gravitation and help us gain the equations of motion in the spacetime they belong to.

3- Riemann tensor is somehow related to the change of gravitational field in the fabric of spacetime; it exactly shows how moving on a trajectory of spacetime would be affected by the gravitational matter and in a mathematical point of view, it comes in handy in describing the geometry of spacetime when it comes to shape and curvature even if gravitation is not involved, e.g. the curvature of an isotropic spacetime without a scale factor is intrinsic and does not have anything to do with gravitation iff k=1 or -1.

AB

8. Jan 20, 2010

Thanks alot!

you explained this very clear and it really helped me to understand these things better!

are you an expert in GR?