- #1

wildemar

- 20

- 0

## 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:

[tex]

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

\tag{a}

[/tex]

## Homework Equations

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

[tex]

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

[/tex]

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

Elsewhere he proves that

[tex]

\partial_c g = g g^{ab} \partial_c g_{ab}

\tag{b}

[/tex]

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

[tex]

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

\ ,

[/tex]

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

[tex]

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

\ .

[/tex]

Then, by seeing

*g*as [itex]g(g_{ab}(x^c))[/itex] 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

[tex]

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

= - \frac{1}{2 \sqrt{-g}} \delta g

\ .

[/tex]

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

[tex]

- \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}

\ ,

[/tex]

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