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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.

Ismasou
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1 Answers1

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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.

NormalsNotFar
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