I am trying to calculate $\sqrt{-g}$ in terms of a background metric and metric perturbations, to second order in the perturbations. I know how to expand tensors that depend on the metric, but I don't know how to expand the metric determinant. Does anyone know how to to this or (even better) know of a source that does second order perturbation theory in GR?


Dear David, as Ted already wrote, but didn't optimize the answer for your problem, there is an identity $$\det g = \exp(\mbox{Tr}(\ln g))$$ You may easily take the square root of it; let us assume that $g$ is positive definite for a while; the resulting formulae will obviously work for all matrices. $$\sqrt{\det g} = \exp\left(\frac{1}{2}\mbox{Tr}(\ln g)\right)$$ However, as he also realized, the formula is not terribly useful for your purpose, anyway, because you would need to differentiate a logarithm of a matrix which is even harder.

It is better to differentiate the determinant directly with respect to the individual components of the metric tensor explicitly. The determinant of a $d \times d$ matrix is the sum of the $d!$ products over all permutations covering all columns and rows, with the sign counting the parity of the permutation. What is the following derivative? $$ \frac{\partial (\det g)}{\partial g_{\mu\nu}}$$ Well, it's not hard to calculate. Among the $d!$ terms in the determinant, you only pick those that contain $g_{\mu\nu}$. I will return to the evaluation after a paragraph about a technicality.

Just to be sure, $g_{\mu\nu}$ and $g_{\nu\mu}$ are treated as independent variables and the calculation is formally done for general asymmetric matrices. At the very end, we may realize that the metric is a symmetric tensor which simplifies some formulae and allows us to be more sloppy about the ordering. But at no point, we will actually be forced to distinguish the derivatives with respect to diagonal and off-diagonal elements of the metric.

OK, so the derivative of the determinant with respect to a particular matrix element is the corresponding subdeterminant of the $(d-1) \times (d-1)$ matrix in which the $\mu$-th row and $\nu$-th column were omitted. You know that the inverse matrix is defined using a similar subdeterminant - however, it must be divided by the whole determinant. It leads us to the identity: $$ \frac{\partial (\det g)}{\partial g_{\mu\nu}} = \pm \mbox{subdet} (g_{omitted\,\mu,\nu}) = (\det g) g^{\mu\nu}$$ The $\pm$ symbol indicates that I want to stay sloppy about the overall sign. However, the sign is fixed in the following step and regardless of $\mu\nu$, it is a plus sign. This is very clear if you choose e.g. $(\mu,\nu)=(1,1)$ - and the formula must have a nice form for all values of the indices. The determinant had to be reinserted to the right hand side because the inverse matrix elements are defined as ratios with the determinant in the denominator, so I had to multiply it back. If you wrote the determinant as the exponential my answer started with, the exponential - the determinant - would be recycled back to the expression because the derivative of an exponential is the exponential times the derivative of the argument.

In the $+---$ or $-+++$ signature, the determinant is negative. The right sign is still the same $$ \frac{\partial (\det g)}{\partial g_{\mu\nu}} = \pm \mbox{subdet} (g_{omitted\,\mu,\nu}) = (\det g) g^{\mu\nu}$$ because the previous formula really holds for any matrix, regardless of the signature. Now, write $$g_{\mu\nu} = \eta_{\mu\nu}+h_{\mu\nu}$$ where $\eta$ doesn't necessarily have to have the simple Minkowski form. You can Taylor-expand the determinant in $h$: $$\det(g_{\mu\nu}) = \det(\eta_{\mu\nu}) + h_{\mu\nu} \frac{\partial (\det g)}{\partial g_{\mu\nu}}|_{g=\eta} + \frac{1}{2!} h_{\kappa\lambda}h_{\mu\nu} \frac{\partial^2 (\det g)}{\partial g_{\kappa\lambda}\partial g_{\mu\nu}}|_{g=\eta} + \dots$$ I am convinced that now you can solve it. The second derivatives with respect to the metric are not too much harder; they're the first derivatives of the first derivative. However, you know how to differentiate metric components themselves with respect to $g_{\mu\nu}$; and you also know how to differentiate the determinant.

In fact, it's helpful to tell you how to differentiate the inverse metric components with respect to the original metric components, too: $$\frac{\partial g_{\kappa\lambda}}{\partial g_{\mu\nu}} = \delta^\mu_\kappa \delta^\nu_\lambda$$ That was the simple Kronecker delta differentiation. Now the more complex one, with the inverse metric $$\frac{\partial g^{\kappa\lambda}}{\partial g_{\mu\nu}} = -g^{\kappa\mu}g^{\lambda\nu}$$ This formula may be determined from the fact that the derivative of $\delta^\pi_\rho$, which may be obtained as the product of $g^{\pi\alpha}g_{\alpha\rho}$, with respect to any $g_{\delta\epsilon}$ vanishes. The Leibniz rules produces two terms for it, and that's why one of them has to have a minus sign.

I hope that you know how to Taylor-expand the square root as well, and how to insert Taylor expansions into each other - derivatives of composite functions. All of it follows standard formulae. You may forget that you're dealing with matrices; you may simply imagine that all the expressions are functions of $d^2$ independent variables $g_{\mu\nu}$.

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    $\begingroup$ great! thanks for that. Since I already worked it out ted's way, i'll compare it to your way to make sure I got the right answer. $\endgroup$ – david Jan 25 '11 at 21:27
  • $\begingroup$ Good idea, David. It was a pleasure. $\endgroup$ – Luboš Motl Jan 26 '11 at 6:35
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    $\begingroup$ Very nice! I really enjoy these kind of answers having explicit calculations in it - it uses the mathematical language of physics. $\endgroup$ – Robert Filter Jan 26 '11 at 8:47
  • $\begingroup$ Considering that the metric is symmetric under index exchange, the result $\frac{\partial g_{\kappa\lambda}}{\partial g_{\mu\nu}} = \delta^\mu_\kappa \delta^\nu_\lambda$ looks suspicious. Shouldn't it rather be $\frac{\partial g_{\kappa\lambda}}{\partial g_{\mu\nu}} = \delta^\mu_\kappa \delta^\nu_\lambda + \delta^\nu_\kappa \delta^\mu_\lambda$ ? $\endgroup$ – Kagaratsch Dec 16 '14 at 13:22
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    $\begingroup$ Hi, one may do the calculations in both ways - but your way would have to be fixed by adding the factor of 1/2 on the right hand side. My way assumes that $g_{\mu\nu}$ is a general tensor, not necessarily symmetric, and only treats the symmetricity as a "random feature of the initial conditions". Your way would effectively differ from mine by defining $g$ as the symmetric part of "my" $g$ but one must be careful about such things. My way is surely the standard way how the derivatives are calculated. $\endgroup$ – Luboš Motl Dec 18 '14 at 7:35

Just for posterity, I thought I'd put up the answer I calculated using the helpful answers given above:

Write $$g_{\mu \nu}=b_{\mu \nu}+h_{\mu \nu}$$ where $b$ is the background metric and $h$ is the perturbation; and raise/lower indeces on $h$ using $b$. Then to second order in $h$ we have: $$\sqrt{-g}=\sqrt{-b}\left(1+\frac{1}{2}{h^{\mu}}_{\mu}+\frac{1}{8}{h^{\mu}}_{\mu}{h^{\nu}}_{\nu}-\frac{1}{4}{h^{\mu}}_{\nu}{h^{\nu}}_{\mu}\right)$$

I'm pretty sure this is correct as I checked it on Mathematica for a general metric.

One possible derivation goes as follows:

We will use the properties $$ \log\det = \operatorname{tr}\log\quad\text{and}\quad det{(b+h)} = \det{b}\det{(1+b^{-1}h)} $$ and the power series of $\log{(1+x)}$ and of $\exp(x)$.

Then: $$ \begin{align} \sqrt{-\det{g}} = \sqrt{-\det{(b+h)}} &= \exp\log\sqrt{-\det{(b+h)}}\\ &= \exp{\frac{1}{2}\log{(-\det{(b+h)})}}\\ &= \exp{\frac{1}{2}\log{(-\det{b}\det{(1+b^{-1}h)})}}\\ &= \exp{\left[\log{\sqrt{-\det{b}}} + \frac{1}{2}\log{\det{(1+b^{-1}h)}}\right]}\\ &= \sqrt{-\det{b}}\exp{\left[\frac{1}{2}\log{\det{(1+b^{-1}h)}}\right]}\\ &= \sqrt{-\det{b}}\exp{\left[\frac{1}{2}\operatorname{tr}{\log{(1+b^{-1}h)}}\right]}\\ &= \sqrt{-\det{b}}\exp{\left[\frac{1}{2}\operatorname{tr}{\left[b^{-1}h-\frac{1}{2}(b^{-1}h)^2+\dots\right]}\right]}\\ &= \sqrt{-\det{b}}\exp{\left[\frac{1}{2}\operatorname{tr}(b^{-1}h)-\frac{1}{4}\operatorname{tr}{(b^{-1}h)^2}+\dots\right]}\\ &= \sqrt{-\det{b}}\left(1 + \frac{1}{2}\operatorname{tr}(b^{-1}h)-\frac{1}{4}\operatorname{tr}{(b^{-1}h)^2} + \frac{1}{2}\left(\frac{1}{2}\operatorname{tr}(b^{-1}h)-\frac{1}{4}\operatorname{tr}{(b^{-1}h)^2}\right)^2\right) + \mathcal O(h^3)\\ &= \sqrt{-\det{b}}\left(1 + \frac{1}{2}\operatorname{tr}(b^{-1}h)-\frac{1}{4}\operatorname{tr}{(b^{-1}h)^2} + \frac{1}{8}\operatorname{tr}^2{(b^{-1}h)}\right) + \mathcal O(h^3)\\ \end{align} $$ In physics notation this reads $$ \sqrt{-g} = \sqrt{-b}\left(1+\frac{1}{2}h^\mu_\mu-\frac{1}{4}h^{\mu\nu}h_{\mu\nu}+\frac{1}{8}(h^\mu_\mu)^2\right) + \mathcal O(h^3) $$ To get a result to higher powers of $h_{\mu\nu}$ one simply has to keep more terms in the series expansion of $\log$ and $\exp$.

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    $\begingroup$ I think its a good idea to give answers to your own questions summarizing the ideas of others (which would already be fine) and finaly give a result based on it. Nice work! Greets $\endgroup$ – Robert Filter Jan 26 '11 at 8:43

I confess that I haven't looked at this sort of calculation in ages, but I have done other calculations involving perturbative expansions of determinants, and in those contexts the identity $$ \ln(\det M) = {\rm Tr}(\ln M) $$ has been really useful. If $g=g^{(0)}+\epsilon g^{(1)}+\epsilon^2g^{(2)}\ldots$ for some small parameter $\epsilon$, you can write out the power series of $\ln g$ to any desired order, take the trace, and exponentiate.

There may be some reason I'm not thinking of why this doesn't work in this context, in which case never mind.

  • $\begingroup$ that might be just what I need. I'll see if it works. $\endgroup$ – david Jan 25 '11 at 20:38
  • $\begingroup$ wait... how to I calculate the power series of ln(g+h) (expanded in powers of h) for g and h both matrices? $\endgroup$ – david Jan 25 '11 at 20:49
  • $\begingroup$ I think it's like this: $${\rm Tr}[\ln(g+h)]={\rm Tr}[\ln[g(1+g^{-1}h)]]={\rm Tr}[\ln g+\ln(1+g^{-1}h)]$$. Now the usual Taylor series formula for $\ln(1+x)$ works. I'm not sure that's 100% right in detail, but I think the idea is something like that. $\endgroup$ – Ted Bunn Jan 25 '11 at 20:52
  • $\begingroup$ Basically, a number of things that look like cheating turn out to be OK, as long as you wrap a trace around them. Now I have to try to remember exactly what things those are. I'll think about it and try to be more precise. $\endgroup$ – Ted Bunn Jan 25 '11 at 21:02
  • $\begingroup$ Thanks, I think that helps. I can check my answer against a specific metric once I'm done to see if I messed up (hopefully). I think the only thing missing is how to interpret g^(-1)*h raised to some power, but I can use trial and error I guess. Thanks alot for your help! $\endgroup$ – david Jan 25 '11 at 21:09

The rest of these answers are fine, but another way of doing this computation is to write the metric determinant as $g =\frac{1}{4!} \epsilon^{abcd}\epsilon^{efgh}g_{ae}g_{bf}g_{cg}g_{dh}$,

where you define the Levi-Civita symbol $\epsilon^{abcd}$ by the conditions:

  • $\epsilon^{txyz}$ is equal to 1, $</li> <li>$\epsilon^{abcd}=-\epsilon^{bacd}=-\epsilon^{acdb}=-\epsilon^{abdc}$</li> <li>$\epsilon^{abcd}=0$ if any two indices take the same value

Then, along with some algebra, you can see that $g^{ab} = \frac{1}{3!g} \epsilon^{aecd}\epsilon^{bfgh}g_{ef}g_{cg}g_{dh}$, and then doing the variation is easy.


I guess you basically answered the question yourself. You might have to treat $$g = \det g_{\mu\nu}$$ and the negative squareroot of it as a function of the metric as you would do for a scalar (rank zero tensor).
The definition of the $\det$ can be found e.g. on wikipedia.



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