# An identity between the d'Alembertian and the covariant derivative

Suppose $$f$$ is function which depends on $$\phi$$, $$f = f(\phi)$$; and $$\phi$$ is a scalar field. We define $$\square \equiv g^{\mu\nu} \nabla_\mu \nabla_\nu$$ and $$\nabla_\mu f(\phi) \equiv f_{;\nu}$$ Is this expression below correct?

$$-f_{;\mu\nu} + \square f g_{\mu\nu} = 0$$

I think it is correct since $$\nabla_\nu(\nabla_\mu f) = f_{;\mu\nu}$$ and $$(g^{\mu\nu} \nabla_\mu \nabla_\nu f) g_{\mu\nu} = \square f g_{\mu\nu} = (\nabla_\mu \nabla_\nu f)g^{\mu\nu}g_{\mu\nu} =\nabla_\mu \nabla_\nu f$$ but I doubt it, I need another point of view.

ps: for further, you can check arXiv:2009.11827. I try to prove equation (18).

• It is not correct. In your last line, you have mixed dummy indices (i.e. indices which are being summed over) with free indices. You should write $(g^{\alpha\beta} \nabla_\alpha \nabla_\beta f) g_{\mu\nu}$. Immediately it is possible to see that no simplification is possible. – Prahar Oct 19 at 12:25
• thank you, Prahar. I appreciate it – Adika Oct 19 at 14:16
• A counter-example is Minkowski space with $f = \frac{1}{2} (x^0)^2$. – Michael Seifert Oct 19 at 17:03

You are trying to prove eq (18) in 2009.11827. That has nothing to do with the identity you are asking about in the question (which is completely incorrect, btw). To derive (18) in the ref. we simply need to vary the action w.r.t. the metric ($$G_4$$ is a function only of $$\varphi$$) $$\delta \int \sqrt{-g} G_4 R = \int \left[ \delta \sqrt{-g} G_4R + \sqrt{-g} G_4\delta R \right]$$ Next, we use $$\delta \sqrt{-g} = - \frac{1}{2} \sqrt{-g} g_{\mu\nu} \delta g^{\mu\nu} , \qquad \delta R = R_{\mu\nu} \delta g^{\mu\nu} - \delta g^{\mu\nu}{}_{;\mu\nu} + g_{\mu\nu} \Box \delta g^{\mu\nu}$$ Then, \begin{align} \delta \int \sqrt{-g} G_4R &= \int \sqrt{-g} \left[ - \frac{1}{2} g_{\mu\nu} \delta g^{\mu\nu} G_4R + G_4 R_{\mu\nu} \delta g^{\mu\nu} - G_4 \delta g^{\mu\nu}{}_{;\mu\nu} + G_4 g_{\mu\nu} \Box \delta g^{\mu\nu} \right]\\ &= \int \sqrt{-g} \left[ G_4 E_{\mu\nu} - G_{4;\mu\nu} + g_{\mu\nu} \Box G_4 \right] \delta g^{\mu\nu} \end{align} Thus, the equations of motion are $${\cal E}_{\mu\nu}^{(4)} = G_4 E_{\mu\nu} - G_{4;\mu\nu} + g_{\mu\nu} \Box G_4$$ The identity in the question is not correct and is also not relevant to the calculation.

• Actually, I did the same way what you write above, but I simply didn't know that delta R has a form like that. I find only the first term (which contained Einstein tensor) so that I assumed the last two-term must be zero (that's why I wrote "I doubt it"). But, thank you very much. It is really helpful. – Adika Oct 19 at 15:42
• Given that you are reading the paper you mentioned, it seems that you should be able to derive $\delta R$. You shouldn't need to remember or know it. It would be good for you if you go through that derivation (for your own satisfaction). – Prahar Oct 19 at 15:44
• @Adika if you find an answer helpful accept the answer! This is indeed the derivation you're looking for! – ApolloRa Oct 19 at 16:31
• yeah, I should derive $\delta R$ by myself; for this case, I just using it (and don't really understand the derivation in the book). Thanks for the advice. – Adika Oct 20 at 3:48
• @ApolloRa Sorry. I am new at this, so I don't really know the rules. thanks for reminding me! – Adika Oct 20 at 3:53

This expression:

$$(g^{\mu\nu} \nabla_\mu \nabla_\nu f) g_{\mu\nu} = \square f g_{\mu\nu} = (\nabla_\mu \nabla_\nu f)g^{\mu\nu}g_{\mu\nu} =\nabla_\mu \nabla_\nu f,$$

is wrong. This does not hold. The Box operator is a scalar quantity. It is defined as:

$$\Box = g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}$$

and for a four dimensional diagonal metric this is:

$$\Box = g^{tt}\nabla_{t}\nabla_{t} + g^{xx}\nabla_{x}\nabla_{x} + g^{yy}\nabla_{y}\nabla_{y} + g^{zz}\nabla_{z}\nabla_{z}$$

You abused notation and thus you're not able to see this but the indices for the box term $$\mu, \nu$$ denote summation (dummy indices) while the indices for the metric denote components (free indices). A more correct way do write it would be:

$$\Box f g_{\mu\nu} = (g^{\alpha\beta}\nabla_{\alpha}\nabla_{\beta} f) g_{\mu\nu}$$

Also $$g_{\mu\nu}g^{\mu\nu} =d$$ where $$d$$ is the dimensionality of spacetime for a diagonal metric. I hope this helps.

• You're assuming here that the metric is diagonal, which it may not be. – Prahar Oct 19 at 12:26
• Oh you're right but it does not matter, ithink i've made my point. – ApolloRa Oct 19 at 12:27
• I corrected it though. – ApolloRa Oct 19 at 12:28
• yeah, I know your point. But, in the paper I read, it looks like the expression I wrote above must be true except my proving was wrong. – Adika Oct 19 at 13:18
• Could you give a reference? Add it in the question. – ApolloRa Oct 19 at 13:23

Your expression is correct if you're using a derivative that is compatible with the metric, so if $$\nabla_\mu g_{\rho\sigma}=0$$, otherwise the d'Alembertian would bring other terms too.

Edit: The reason why I think this is correct is

$$\Box f g_{\rho\sigma}=g^{\mu\nu}\nabla_\mu\nabla_\nu f g_{\rho\sigma}=\nabla^\nu\nabla_\nu f g_{\rho\sigma}$$
$$f_{,\rho\sigma}=\nabla_\rho\nabla_\sigma f =\nabla^\rho\nabla_\rho f g_{\rho\sigma}=\nabla^\nu\nabla_\nu f g_{\rho\nu}g^{\rho\nu}g_{\rho\sigma}=\nabla^\nu\nabla_\nu f g_{\rho\sigma}$$

I may be wrong, as @Prahar says, but I can't see where.

• thank you, Mauro. – Adika Oct 19 at 13:20
• This answer is incorrect. Kindly edit or delete. – Prahar Oct 19 at 14:27