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

• wildemar
In summary, the conversation discusses the difficulty in understanding variational calculus in relation to General Relativity. The individual is trying to prove a relationship involving the negated determinant of the metric and is seeking pointers on how to approach the variation symbol. They mention using d'Inverno's book and possibly needing to prove the chain rule. Another person suggests expanding the function using a Taylor series and recommends a book with useful identities for comparison. The conversation concludes with a request for recommendations on a book that goes into detail on these calculations.
wildemar
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

wildemar said:
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,

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!

The whole thing is explained in Carroll's book on relavtivity in a very simple and elegant manner..I believe its the chapter on gravitational waves

## 1. What is the metric determinant in general relativity?

The metric determinant is an important quantity in general relativity that represents the volume scaling factor in a given coordinate system. It is used to calculate the proper distance between points in spacetime and is a crucial component in the Einstein field equations.

## 2. How is the metric determinant defined mathematically?

In general relativity, the metric determinant is defined as the determinant of the metric tensor, which is a mathematical object that describes the curvature of spacetime. It is denoted by g and is calculated by taking the product of the diagonal elements of the metric tensor.

## 3. What is the significance of variation of the metric's determinant in general relativity?

The variation of the metric's determinant is a key concept in variational calculus, which is used to find the equations of motion in general relativity. It allows us to determine how the metric determinant changes as we vary the coordinates, which is important in understanding the behavior of particles and light in curved spacetime.

## 4. How is the variation of the metric's determinant calculated?

The variation of the metric's determinant is calculated by taking the derivative of the metric determinant with respect to the coordinates and then integrating over the entire spacetime. This allows us to determine the changes in the metric determinant due to variations in the coordinates.

## 5. What are some real-world applications of understanding the variation of the metric's determinant in general relativity?

Understanding the variation of the metric's determinant is crucial in many real-world applications, such as predicting the behavior of objects in strong gravitational fields, such as around black holes. It is also important in cosmology, where it helps us understand the expansion of the universe and the evolution of large-scale structures. Additionally, it is used in the development of GPS systems, which rely on precise measurements of spacetime to function correctly.

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