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I have a Straightforward question. When the functional derivative of the Ricci scalar to get the GR field equations. As the derivative is done using the metric which is symmetric do I have to symmetrize my results.

On mathematica with the package xAct the "VarD" function always symmetrizes when I do a variation with a symmetric tensor.

For normal GR as I understand the symmetry pops out on its own, but as I'm working on other formulations of GR, sometimes this symmetry isn't a given.

Edit: As pointed in the comments by Qmechanic there is a similar answered question: Variation of the metric with respect to the metric I'm convinced now, as I expected, that I should symmetrize the results.

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  • $\begingroup$ Can you bit a more specific? Give some example $\endgroup$ – apt45 Aug 8 '18 at 9:34
  • $\begingroup$ Just in general, if we do the functional derivative of any arbitrary scalar with respect to a symmetric tensor do we have to take that into account? and do symmetrize the resulting tensor. $\endgroup$ – Ismasou Aug 8 '18 at 9:44
  • $\begingroup$ Maybe if we focus on clearness, and being concrete and rigorous the answer will lead to itself. I'm not sure about functional derivatives, a functional is a map from functions to scalar. I assume the functional derivative is like a derivative but with functionals. This seems analogous to how the regular derivative may need special treatment of curvilinear coordinate system and vectors. Basically how the derivative interacts with the basis in non Cartesian coordinates. $\endgroup$ – marshal craft Aug 8 '18 at 11:02
  • $\begingroup$ I assume here the functional is a matrix multiply with a vector to give a scalar. And the functional derivative, the generalized derivative. But in 3d space for example or manifold, you have a vector from $\Bbb R^3$ the point on the manifold. But then you also have the tangent space at that point describing the vector. For flat Cartesian the derivative is the same for each tangent space basis vectors . But for curved systems it depends on the point on the manifold. Maybe this is simarly related. $\endgroup$ – marshal craft Aug 8 '18 at 11:10
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/149066/2451 $\endgroup$ – Qmechanic Aug 8 '18 at 11:36
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Yes, since you are varying with respect to a symmetric tensor, only the symmetric part of the variation will survive on-shell.

To see this, suppose we are given a Lagrangian $\mathcal{L}=\mathcal{L}(T_{ab})$ which depends on a symmetric tensor field $T_{ab}$ (and perhaps some other independent fields) and an action $S=\int\text{d}^4x\mathcal{L}$. Then varying $S$ w.r.t $T^{ab}$ yields $$\delta S=\int\text{d}^4xG_{ab}\delta T^{ab}=\int\text{d}^4x\big(G_{(ab)}+G_{[ab]}\big)\delta T^{ab}=\int\text{d}^4xG_{(ab)}\delta T^{ab}=0$$ $$\implies G_{(ab)}=0$$ where I have denoted $G_{ab}:=\frac{\delta\mathcal{L}}{\delta T^{ab}}$ and have used the fact that $\delta T^{ab}=\delta T^{(ab)}$ is symmetric to kill the antisymmetric part of $G_{ab}$. This isn't very rigorous, but you get the idea.

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  • $\begingroup$ Thank you for your answer, my intuitions says so too. But can you link me to some kind of proof. Why would that be the case ? $\endgroup$ – Ismasou Aug 8 '18 at 11:31
  • $\begingroup$ No problem, I have updated my answer. $\endgroup$ – NormalsNotFar Aug 8 '18 at 11:45

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