Variation with respect to the metric

[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.

 

[Bill_K]

δ(√-g)/δg_μν = ½(√-g) g^μν

δg^αβ/δg_μν = - ½ g^αμg^βν - ½ g^ανg^βμ

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

 

[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 ?

 

[Bill_K]

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^a}_c {δ^b}_d ?

Sure. (Except they need to be symmetrized.)

What will you use for δ(√-g)/δg^cd ?

 

[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.

 

[Bill_K]

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.

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

If you're going to hold u^a fixed, write the expression as g_ab u^a u^b and use δg_ab/δg^cd.

Otherwise, if you're going to hold u_a fixed, write it as g^ab u_a u_b and use δg^ab/δg^cd. The two results will differ by a sign.

 

[ryan albery]

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

 

[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?