# Variation with respect to the metric!

1. Apr 26, 2014

### gravity90

Hi everyone!

There is something that I would like to ask you. Suppose you have

$$\frac{1}{\sqrt{-g}} \frac{\delta (\sqrt{-g} (g^{ab} u_a u_b + 1))}{\delta g^{cd}}$$

The outcome of this would be $u_{c}u_{d}$ or $-u_{c} u_{d}$ ?

I am really confused.

Last edited: Apr 26, 2014
2. Apr 26, 2014

### Bill_K

δ(√-g)/δgμν = ½(√-g) gμν

δgαβ/δgμν = - ½ gαμgβν - ½ gανgβμ

From these you can derive the δ/δgμν's. (Do you really want to do that??)

3. Apr 26, 2014

### gravity90

Thank you very much for your response. But why don't we use:

$\frac{δg^{ab}}{δg^{cd}}$ = {δ^a}_c {δ^b}_d

or

$\frac{δg_{ab}}{δg_{cd}}$ = {δ^c}_a{δ^d}_b ?

Last edited: Apr 27, 2014
4. Apr 26, 2014

### Bill_K

Sure. (Except they need to be symmetrized.)

What will you use for δ(√-g)/δgcd ?

5. Apr 26, 2014

### gravity90

I will use:

$\frac{1}{√-g}$ $\frac{δ(√-g)}{δg^{μν}}$ =$- 1/2 g_{μν}$

This is what I found in the Carroll's textbook and it seems right.

But the thing is that the indices a,b are mute $(g^{ab} u_a u_b)$ so upon variation with respect to $δg^{cd}$ or $δg_{cd}$ we should get the same result.

Last edited: Apr 26, 2014
6. Apr 26, 2014

### Bill_K

Well, you have to decide which thing you want to hold fixed - ua or ua. This will be determined by the physics of the problem you're trying to solve.

If you're going to hold ua fixed, write the expression as gab ua ub and use δgab/δgcd.

Otherwise, if you're going to hold ua fixed, write it as gab ua ub and use δgab/δgcd. The two results will differ by a sign.

7. Apr 26, 2014

### ryan albery

I respect your math skills... but what the hell are are you guys talking about? Little g? What is little g?

8. Apr 27, 2014

### gravity90

Thank you again for your response. But what is the difference between keeping $u^a$ or $u_a$ from a physical point of view?