- #1
wildemar
- 20
- 0
Hello all :)
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]
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.
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
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