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What is wrong in the following? \begin{equation*} g_{\mu\nu}(x)g^{\mu\nu}(x)=D \\ \frac{\delta}{\delta g_{\alpha\beta}(x)}D=0 \\ \frac{\delta}{\delta g_{\alpha\beta}(x)}(g_{\mu\nu}g^{\mu\nu})=2g^{\alpha\beta}(x)=0 \end{equation*} The above equations are clearly inconsistent.

I am just trying to figure out what is the correct way of defining the correct functional derivative of $g_{\mu\nu}g^{\mu\nu}$ with respect to the components of $g$ itself.

Giuseppe
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    What's your question? –  Jun 02 '15 at 09:28
  • I am just trying to figure out what is the correct way of defining the correct functional derivative of gμνgμν with respect to the components of g itself. – Giuseppe Jun 02 '15 at 12:03

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Your question is posed rather weirdly...

And I am quite sure you cannot really differentiate it like that. Instead of attempting to calculate a Fréchet-derivative directly in a wrong way, try calculating a Gateux-derivative along an arbitrary curve in the space of possible metrics, eg. take $$ \left.\frac{d}{d\varepsilon}(g_{\mu\nu}(\varepsilon)g^{\mu\nu}(\varepsilon))\right|_{\varepsilon=0}, $$ where $g_{\mu\nu}(0)=g_{\mu\nu}$, with a bit of abuse of notation.

Edit: You seem to be rather confused about functional derivation.

First of all, the map $g_{\mu\nu}\mapsto g_{\mu\nu}g^{\mu\nu}$ is not really a functional in the sense that these type of maps lead into $C^\infty(M)$, not $\mathbb{R}$. However in this case it maps into a constant function, which can be viewed as an element of $\mathbb{R}$, thus it is a functional. It is also obvious that since $g^{\mu\nu}$ is not independent from $g_{\mu\nu}$, such expression for any sort of metric will always result in the same number, thus, the derivative must be zero, since the "functional" is constant.

Secondly, the way you tried to use chain rule here is just wrong, which is why you got nonzero result. In general, you cannot calculate a functional derivative "directly".

Most of the time in physics, these functionals will take the form $$ F[\phi]=\int L(\phi,\partial\phi,...,\partial^k\phi)\ \Omega ,$$ where $\Omega$ is some volume element. In this case, the functional derivative always exists (provided $L$ is a well behaved function), and a directional derivative at $\phi$ in the direction of some arbitrary function $\delta\phi$ (which we usually define as a tangent vector to some curve in the function space, eg. $\delta\phi=d\phi/d\varepsilon$) will always take the form $$ \frac{dF}{d\varepsilon}=\int\frac{\delta F}{\delta\phi}\delta\phi\ \Omega .$$ In this case we call $\delta F/\delta\phi$ the functional derivative of $F$, and obviously the above expression is linear in $\delta\phi$.

If the functional is NOT in this integral form, then we cannot really a define a functional derivative like that, rather, if the functional is differentiable, then there must exist some linear map from the space of functions to the reals, for which the directional derivative in the direction of an arbitrary function always equals the action of that linear map on the function.

Therefore, what you are really interested in is actually the expression $$ \left.\frac{d}{d\varepsilon}(g_{\mu\nu}(\varepsilon)g^{\mu\nu}(\varepsilon))\right|_{\varepsilon=0}, $$ where $g_{\mu\nu}(\varepsilon)$ is a family of metrics indexed by a real parameter in a smooth way, and the metric at the 0 parameter is the metric you want to perturb.

Can you take it from here?

Bence Racskó
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  • the problem seems to come from the constraint $g_{\mu\nu}g^{\mu\nu}=D$. As to the weird, these functional derivatives are quite common in theoretical physics. Only, I never thought of this one involving the metric tensor. The case with the electromagnetic tensor is simpler (no constraint). – Giuseppe Jun 02 '15 at 10:43
  • No, I mean, you asked the question weirdly. There is no problem, however, since the derivative is identically zero. Also the Einstein-Hilbert action involves functional derivatives involving the metric tensor. The problem comes from the fact that you cannot calculate a functional derivative the way you tried. – Bence Racskó Jun 02 '15 at 11:03
  • OK, I probably need to think some more on the subject :-) – Giuseppe Jun 02 '15 at 11:27
  • I am editing my answer – Bence Racskó Jun 02 '15 at 12:05
  • Ok, thanks!!! I was giving too much stuff for granted. – Giuseppe Jun 02 '15 at 12:29