Variation of Einstein-Hilbert action

1. Dec 10, 2008

jdstokes

The Einstein field equations $\mathsf{G} = \kappa \mathsf{T}$ can be derived by considering stationary metric variations of the Einstein Hilbert action,

$S = \int \mathrm{d}^4x \sqrt{-g} (R/2\kappa + \mathcal{L}_\mathrm{M})$.

$0 = \delta S = \int\mathrm{d}^4 x\left(\frac{1}{2\kappa}\frac{\partial (\sqrt{-g} R)}{\partial g^{\alpha\beta}}+ \frac{\partial (\sqrt{-g}\mathcal{L}_\mathrm{M})}{\partial g^{\alpha\beta}}\right)\delta g^{\alpha\beta}$

etc.

In conventional field theory, however, we consider variations of the action integrand with respect to both the field $\varphi$ as well as its first derivative $\partial_\mu \varphi$.

Why can we avoid doing this when $\varphi = g^{\alpha\beta}$?.

2. Dec 10, 2008

Fredrik

Staff Emeritus
The action is always a functional that takes the field(s) to a number. The Lagrangian on the other hand is just a (multi-variable) polynomial (it takes a bunch of numbers to a number). The variables that we plug into it are the field and its derivatives at some spacetime point x. Isn't that the case here too? Aren't there derivatives of g in the definition of R?

3. Dec 10, 2008

jdstokes

Yes, there are derivative terms in e.g., the Christoffel symbols.

My question is why don't we treat the field derivatives as independent variables as we do in, for example scalar field theory (or even classical mechanics for that matter).

Let me give you an example. In electromagnetism the equation of motion is

$\frac{\partial \mathcal{L}}{\partial A_\mu} - \partial_\nu \frac{\partial \mathcal{L}}{\partial \partial_\nu A_\mu} = 0$.

In general relativity the equation of motion is

$\frac{\partial \mathcal{L}}{\partial g_{\mu\nu}} = 0$. I don't understand why is it not

$\frac{\partial \mathcal{L}}{\partial g_{\mu \nu}} - \partial_\rho \frac{\partial \mathcal{L}}{\partial \partial_\rho g_{\mu\nu}}= 0$?

4. Dec 10, 2008

Fredrik

Staff Emeritus
OK, I see what you mean. Unfortunately I don't know the answer.

5. Dec 10, 2008

haushofer

I'm not sure if I understand your question correctly. If you want to use explicitly the EL equations instead of doing the variation directly, you consider the metric, it's first derivative and second derivative as independent fields. So t he EL equations become

$\frac{\delta\mathcal{L}}{\delta g_{\mu\nu}} = \frac{\partial \mathcal{L}}{\partial g_{\mu \nu}} - \partial_\rho \frac{\partial \mathcal{L}}{\partial (\partial_\rho g_{\mu\nu})} + \partial_{\rho}\partial_{\lambda}\frac{\partial\mathcal{L}}{\partial(\partial_{\rho}\partial_{\lambda}g_{\mu\nu})}= 0$

in which I've set the boundary conditions to zero. Using now the Lagrangian

$\mathcal{L} = \sqrt{g}R$

gives you the right equations of motion after a very tedious calculation. In d'Inverno, chapter 11 you can find some more information on this.

Last edited: Dec 10, 2008
6. Dec 10, 2008

haushofer

Maybe it's the confusing notation between partial derivatives and functional derivatives that disturbs you. In your

$0 = \delta S = \int\mathrm{d}^4 x\left(\frac{1}{2\kappa}\frac{\partial (\sqrt{-g} R)}{\partial g^{\alpha\beta}}+ \frac{\partial (\sqrt{-g}\mathcal{L}_\mathrm{M})}{\partial g^{\alpha\beta}}\right)\delta g^{\alpha\beta}$

I would have used functional derivatives $\delta$ instead of partial derivatives $\partial$.

7. Dec 10, 2008

jdstokes

Hang on, the Einstein field equations are

$\frac{\partial \mathcal{L}}{\partial g_{\mu\nu}} = 0$,

where $\mathcal{L} = \sqrt{-g}(R/2\kappa + \mathcal{L}_{\rm matter})$

this is a partial derivative, not a functional derivative. Compare with with electromagnetism:

$\frac{\partial \mathcal{L}}{\partial A_\mu} - \partial_\nu \frac{\partial \mathcal{L}}{\partial \partial_\nu A_\mu} = 0$

these are also partial derivatives.

There is a discrepancy.

I don't understand where you got this equation from:

$\frac{\delta\mathcal{L}}{\delta g_{\mu\nu}} = \frac{\partial \mathcal{L}}{\partial g_{\mu \nu}} - \partial_\rho \frac{\partial \mathcal{L}}{\partial (\partial_\rho g_{\mu\nu})} + \partial_{\rho}\partial_{\lambda}\frac{\partial\ma thcal{L}}{\partial(\partial_{\rho}\partial_{\lambd a}g_{\mu\nu})}= 0$

It's not the Einstein field equations.

8. Dec 10, 2008

jdstokes

I retract my last post.

Looking at the variation

$\delta(\sqrt{-g} g^{\alpha\beta}R_{\alpha\beta}) = (\delta \sqrt{-g})g^{\alpha\beta} R_{\alpha\beta} + \sqrt{-g}\delta g^{\alpha\beta}R_{\alpha\beta} + \sqrt{-g}g^{\alpha\beta}\delta R_{\alpha\beta}$

you can see that the first two terms come from partial derivatives with respect to the metric. In order for the last term to vanish, however, you need to consider variations with respect to the Christoffel symbols (or equivalently higher derivatives of the metric).

Thanks.

9. Dec 10, 2008

George Jones

Staff Emeritus
For a careful treatment of variational principles of general relativity, including variation of the Christoffel symbols, I suggest that you look at chapter 4, Lagrangian and Hamiltonian formulations of general relativity, from the book A Relativist's Toolkit: The Mathematics of Black-Hole Mechanics by Eric Poisson.

10. Dec 10, 2008

haushofer

I think you've got your answer, but maybe this helps. Just write the whole calculation explicitly from the start and try to look at how the functional derivative is derived in the first place.

So define

$$\delta\mathcal{L} = \frac{\partial\mathcal{L}}{\partial g_{\mu\nu}}\delta g_{\mu\nu} + \frac{\partial\mathcal{L}}{\partial(\partial_{\lambda}g_{\mu\nu})}\delta(\partial_{\lambda}g_{\mu\nu}) + \frac{\partial\mathcal{L}}{\partial (\partial_{\lambda}\partial_{\rho}g_{\mu\nu})}\delta(\partial_{\lambda}\partial_{\rho}g_{\mu\nu})$$

Here you explicitly use that the field and it's derivative are independent fields. Some partial integrations give you the Euler Lagrange equations plus the boundary terms,

$$\delta\mathcal{L} = \Bigl(\frac{\partial \mathcal{L}}{\partial g_{\mu \nu}} - \partial_\rho \frac{\partial \mathcal{L}}{\partial (\partial_\rho g_{\mu\nu})} + \partial_{\rho}\partial_{\lambda}\frac{\partial\mathcal{L}}{\partial(\partial_{\rho}\partial_{\lambda}g_{\mu\nu})}\Bigr)\delta g_{\mu\nu} + BC$$

Imposing boundary conditions on a hyperplane enables you to define the initial conditions with which you can describe the evolution of the metric to another hyperplane. So if BC=0 this makes people write

$$\frac{\delta\mathcal{L}}{\delta g_{\mu\nu}} \equiv \frac{\partial \mathcal{L}}{\partial g_{\mu \nu}} - \partial_\rho \frac{\partial \mathcal{L}}{\partial (\partial_\rho g_{\mu\nu})} + \partial_{\rho}\partial_{\lambda}\frac{\partial\mathcal{L}}{\partial(\partial_{\rho}\partial_{\lambda}g_{\mu\nu})}$$

I must say that I'm also sometimes confused between the different notations :)

11. Apr 17, 2009

schieghoven

Hello,

The similarity is closer than it first looks... you can integrate (by parts) all of the second derivatives of $g_{\mu\nu}$ in the E-H action, leaving only $g_{\mu\nu}$ and its first partial derivatives. Then you can use Euler-Lagrange equations analogous to the electromagnetic case you've mentioned.

Just beware that the integration by parts breaks general covariance of the integrand (but obviously not the integral as a whole). I presume this is why it is rarely mentioned in introductory courses.

Dave