# Varying determinant of a metric

1. Feb 23, 2013

### pleasehelpmeno

Hi can anyone explain how to find $\delta \sqrt{-g}$ when varying with respect to the metric tensor $g^{\mu\nu}$. i.e why is it equal to $\delta \sqrt{-g} = -\frac{1}{2} \sqrt{-g}g_{\mu\nu} \delta g^{\mu \nu}$

2. Feb 23, 2013

### Bill_K

Because... what else could it be? Seriously, that's a legitimate way to proceed. It has to be a scalar density linear in δgμν, and there is only one such. All that remains is the numerical factor in front, so prove it's -1/2 for a diagonal metric and you're done.

3. Feb 23, 2013

### Morgoth

δ(√g)= (1/2√g) δg

g=εabcdg0ag1bg2cg3d
δg=....

4. Feb 23, 2013

### pleasehelpmeno

Hi i have been lookig at the wiki articles and can anyone explain why $tr(g^{\mu \nu} \delta g_{\mu \nu}) = g^{\mu \nu}\delta g_{\mu \nu}$, is it just a property of metric tensors?

5. Feb 23, 2013

### Bill_K

Do you accept my answer to your original question?

6. Feb 23, 2013

### pleasehelpmeno

well i can prove $\delta \sqrt{-g}=-0.5(\sqrt{-g})^{-1} \delta g$ by creating a chain rule situation for the variation. and i see that $\delta g=g\left( TRACE(g^{\mu \nu} \delta g_{\mu \nu}\right)$, from the jacobi equation. I just don't see how to go from $TRACE(g^{\mu \nu} \delta g_{\mu \nu}) = g^{\mu \nu} \delta g_{\mu \nu})$.

Last edited: Feb 23, 2013
7. Feb 23, 2013

### Bill_K

You don't need Trace here, and in fact Trace can only be applied to a matrix. Whereas gμνδgμν is already summed over all of its indices, and is already a scalar.

(The approach I suggested is much, much easier! )

8. Feb 23, 2013

### Morgoth

the weirdest thing is that if the metric in variation is with indices up it appears with a minus sign...

9. Feb 24, 2013

### pleasehelpmeno

yeah but i read this:

http://en.wikipedia.org/wiki/Einstein–Hilbert_action

and the formula it uses is a trace but, are you saying that a trace would effectively just be a metric tensor $g^{\mu \nu} \delta g_{\mu \nu}$ anyway so it is in effect negated, also that minus sign is weird?

10. Feb 24, 2013

### Fredrik

Staff Emeritus
Do you know the definition of trace, and the definition of matrix multiplication?

Edit: OK, since there are some notational issues, I'll just explain what I'm thinking. Suppose that A is a matrix, and that we use the notation $A^{\mu\nu}$ for the component on row $\mu$, column $\nu$. Suppose that B is a matrix, and that we use the notation $B_{\mu\nu}$ for the component on row $\mu$, column $\nu$. If we use the notation $(AB)^\mu{}_\nu$ for the component on row $\mu$, column $\nu$ of AB, then by definition of trace and matrix multiplication, we have
$$\operatorname{Tr}(AB) =(AB)^\mu{}_\mu =A^{\mu\nu}B_{\nu\mu}.$$ If instead A is a (2,0) tensor, and B is a (0,2) tensor (like g), then
$$A^{\mu\nu}B_{\mu\nu} =A(e_\mu,e_\nu)B(e^\mu,e^\nu) =(A\otimes B)(e_\mu,e_\nu,e^\mu,e^\nu).$$ The right-hand side is a "contraction", and I suppose someone might want to use use the Tr notation for it.

Last edited: Feb 24, 2013
11. Feb 24, 2013

### Bill_K

gμνgμν ≡ 4, so vary this and get (δgμν)gμν + gμνδgμν = 0, or (δgμν)gμν = - gμνδgμν

12. Feb 25, 2013

### pleasehelpmeno

i see and understand al except still struggling to get my head around the trace. So simply $TRACE g^{\mu \nu} \delta g_{\mu \nu} = g^{\mu \nu} \delta g_{\mu \nu}$ or is the trace just not there in the first place? Thanks for all your help up to know

13. Feb 25, 2013

### Fredrik

Staff Emeritus
What sort of thing can you take the trace of? Matrices for sure. Maybe some sort of tensors and tensor fields, with an appropriate generalization of the definition of "trace". But $g^{\mu\nu}\delta g_{\mu\nu}$ is a scalar field. So the only way I can make sense of your left-hand side is to assume that it's a bizarre notation for a contraction of a tensor.

14. Feb 25, 2013

### pleasehelpmeno

15. Feb 25, 2013

### Fredrik

Staff Emeritus
Nothing like the notation $\mathrm{TRACE}(g^{\mu \nu} \delta g_{\mu \nu})$ appears in either of those articles. This is something you introduced. What you're doing is the equivalent of taking the equality I wrote as $\operatorname{Tr}(AB) =(AB)^\mu{}_\mu =A^{\mu\nu}B_{\nu\mu}$ and rewriting it in notation that doesn't make sense.

As opposed to? The components of the metric can of course be viewed as the components of a matrix.

16. Feb 25, 2013

### pleasehelpmeno

17. Feb 25, 2013

### Fredrik

Staff Emeritus
It's a trace, but there are no indices on the thing to the right of the Tr. For any matrix, the definition of trace is $\operatorname{Tr}A=\sum_i A_{ii}$. A notation like $Tr\left(\sum_{i}A_{ii}\right)$ would not make much sense. The only possible interpretation of it is that you're taking a trace of a 1×1 matrix. This will of course just give you the number inside the matrix.

18. Feb 25, 2013

### pleasehelpmeno

then how is $\delta \det g_{\mu \nu} = g g^{\mu \nu} \delta g_{\mu \nu}$

19. Feb 25, 2013

### andrien

20. Feb 25, 2013

### pleasehelpmeno

Thanks but shouldn't it be:

$\delta g = g Tr(g^{-1}. \delta g)$ ?

21. Feb 25, 2013

### Fredrik

Staff Emeritus
I'm not a fan of the notation $\det g_{\mu\nu}$ either, but I think it's clear that a person who uses it means "the determinant of the matrix that has $g_{\mu\nu}$ on row $\mu$, column $\nu$.

$g^{\mu \nu} \delta g_{\mu \nu}$ on the other hand can't be interpreted as a matrix because of the repeated indices, which imply summation. So it doesn't make sense to put "Tr" in front of it. It can however be interpreted as a trace of a product of two matrices. This follows immediately from the definition of trace and matrix multiplication. I showed the details in my first post in this thread.

22. Feb 25, 2013

### pleasehelpmeno

thank you, you have been helpful so just to be clear it should be: $TR(g^{-1}\delta g) = (g\delta g)^{\mu}_{\mu} = g^{\mu \nu} \delta g_{\mu \nu}$

23. Feb 25, 2013

### Fredrik

Staff Emeritus
Yes, that notation makes sense, assuming that $\delta g_{\mu\nu}$ means $(\delta g)_{\mu\nu}$ rather than $\delta(g_{\mu\nu})$.

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