# Variation of the metric's determinant [General Relativity, Variational Calculus]

Hello all :)

## Homework Statement

I'm trying to understand the fundamentals of General Relativity, but alas, I seem to be unable to grasp the fundamentals of variational calculus.

Specifically, I'd like to prove the following relation for the square root of the negated determinant of the metric:

$$\delta \sqrt{-g} = \frac{1}{2} \sqrt{-g} g^{ab} \delta g_{ab} \tag{a}$$

## Homework Equations

I'm using d'Inverno's book, and he defines the variation symbol as

$$\delta f(x) = \bar{f}(x) - f(x)$$

that is, the difference between a function and a slightly altered function.

Elsewhere he proves that

$$\partial_c g = g g^{ab} \partial_c g_{ab} \tag{b}$$

by using Laplace's formula for the determinant (no sum convention)

$$g = \sum_b g_{ab} \tilde{g}^{ab} \ ,$$

taking the derivative for a specific element and using $g^{ab} = \frac{1}{g} \tilde{g}^{ab}$ (the formula for the inverse matrix by means of the determinant and the cofactor matrix) to get

$$\frac{ \partial g }{ \partial g_{ab} } = \tilde{g}^{ab} = g g^{ab} \ .$$

Then, by seeing g as $g(g_{ab}(x^c))$ he differentiates g with regard to x (using of course the chain rule) and gets the above equation (b) for derivative of the the metric determinant.

## The Attempt at a Solution

So I thought that a similar route must be taken for the variation of the metric. The problem is that I don't really know how to treat the variation. D'Inverno says that variation "behaves like a derivative". From the above definition I have been able to prove the sum rule (trivial) and the product rule (almost trivial). The chain rule however, I did not get. I don't even have an idea.

Assuming I have a chain rule, I can then easily do

$$\delta\sqrt{-g} = \frac{1}{2 \sqrt{-g}} \delta(-g) = - \frac{1}{2 \sqrt{-g}} \delta g \ .$$

And if I had the variational eqivalent of eq. (a) I could say

$$- \frac{1}{2 \sqrt{-g}} \delta g = - \frac{1}{2 \sqrt{-g}} g g^{ab} \delta g_{ab} = \frac{1}{2} \sqrt{-g} g^{ab} \delta g_{ab} \ ,$$

which is the expected result.

So if I could prove the chain rule for this definition of variation I'd be set. Or maybe I need to take a wholly different route?

thanks in advance for any pointers
/W

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AEM
Hello all :)

## Homework Statement

I'm trying to understand the fundamentals of General Relativity, but alas, I seem to be unable to grasp the fundamentals of variational calculus.

Specifically, I'd like to prove the following relation for the square root of the negated determinant of the metric:

$$\delta \sqrt{-g} = \frac{1}{2} \sqrt{-g} g^{ab} \delta g_{ab} \tag{a}$$

## Homework Equations

I'm using d'Inverno's book, and he defines the variation symbol as

$$\delta f(x) = \bar{f}(x) - f(x)$$

that is, the difference between a function and a slightly altered function.

Elsewhere he proves that

$$\partial_c g = g g^{ab} \partial_c g_{ab} \tag{b}$$

by using Laplace's formula for the determinant (no sum convention)

$$g = \sum_b g_{ab} \tilde{g}^{ab} \ ,$$

taking the derivative for a specific element and using $g^{ab} = \frac{1}{g} \tilde{g}^{ab}$ (the formula for the inverse matrix by means of the determinant and the cofactor matrix) to get

$$\frac{ \partial g }{ \partial g_{ab} } = \tilde{g}^{ab} = g g^{ab} \ .$$

Then, by seeing g as $g(g_{ab}(x^c))$ he differentiates g with regard to x (using of course the chain rule) and gets the above equation (b) for derivative of the the metric determinant.

## The Attempt at a Solution

So I thought that a similar route must be taken for the variation of the metric. The problem is that I don't really know how to treat the variation. D'Inverno says that variation "behaves like a derivative". From the above definition I have been able to prove the sum rule (trivial) and the product rule (almost trivial). The chain rule however, I did not get. I don't even have an idea.

Assuming I have a chain rule, I can then easily do

$$\delta\sqrt{-g} = \frac{1}{2 \sqrt{-g}} \delta(-g) = - \frac{1}{2 \sqrt{-g}} \delta g \ .$$

And if I had the variational equivalent of eq. (a) I could say

$$- \frac{1}{2 \sqrt{-g}} \delta g = - \frac{1}{2 \sqrt{-g}} g g^{ab} \delta g_{ab} = \frac{1}{2} \sqrt{-g} g^{ab} \delta g_{ab} \ ,$$

which is the expected result.

So if I could prove the chain rule for this definition of variation I'd be set. Or maybe I need to take a wholly different route?

thanks in advance for any pointers
/W
Well, it looks to me like you're on the right track. Did you do something like the following to expand

$$\delta f(x) = \bar{f}(x) - f(x)$$,

assume that $$\bar{f}(x) = f(x + \delta x)$$

and then expand $$f(x+ \delta x)$$ in a Taylor series so that

$$\delta f(x) = f(x) - \frac{\partial f(x)}{\partial x} \delta x -f(x)$$

Then,

$$\delta f(x) = \frac{\partial f(x)}{\partial x} \delta x$$.

This might be the chain rule you desire.

There's a rather old and somewhat quirky text on relativity, Principles of Relativity Physics by James Anderson that has a couple identities that might be useful for comparison with what you're doing:

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

Hope this all helps.

Last edited:
Hello,

I'm also struggling in getting through with the calculation of variations in GR for deriving, from the action principle, the dynamic equations and energy-momentum tensor of a given field, scalar or tensorial.

Do you recommend any good book, that really goes into the details of these calculations.

Hello,

I'm also struggling in getting through with the calculation of variations in GR for deriving, from the action principle, the dynamic equations and energy-momentum tensor of a given field, scalar or tensorial.

Do you recommend any good book, that really goes into the details of these calculations.

Putting aside the dynamic equations, see for instance "Lectures on General Relativity" by A. Papapetrou, Chapter 8 that is a beautiful treatment of the problem you are struggling with.

AB

As for books, salparadise, I really haven't read a lot. I only really know D'Inverno's book, which isn't detailed, but easy to follow. So no luck there. He certainly doesn't do it for a general Lagrangian (which I doubt is even possible). But then the variation principle boils down to a couple of derivatives of the Lagrangian, which you should find in just about every book on the topic. From then on it's just brute force calculation, and this can easily be done with Maple (or Mathematica, I guess).

Thanks AEM,

While I can't get my hand on the book you mentioned, I using among others Weinberg, and I just notices the following. In page 364 there's this equation that confused me, maybe it's obvious but I can't see it...

$$\delta g^{\mu \nu}= - g^{\mu \rho} g^{\nu \sigma} \delta g_{\rho \sigma} \ .$$

Why is there a minus sign there. Shouldn't the metric just simply raise and lower index without additional signs. Is it because of the variations?

Thanks,

haushofer
A late reply :) But the minus sign there is simply because $\delta (g_{\mu\nu}g^{\nu\rho})=0$, so you have to be carefull whenever you switch from the variation with upper indices to the one with lower indices and vice versa!