# Christoffel symbol with conformal time not equal to with cosmic time one when making a change of coordinates for d'Alembertian

I think I am having a misunderstanding that would be nice to clear up.

The covariant d'Alembertian $$\Box \phi = g^{\mu\nu}\nabla_\mu\partial_\nu \phi= \left(\partial^2 + \Gamma^\mu_{\mu\lambda}\partial^\lambda\right)\phi,$$ should be independent of coordinates since it is a Lorentz scalar. I am considering the two FLRW coordinate choices $$ds^2 = -dt^2+a^2\delta_{ij}x^ix^j,\\ d\tilde{s} =a^2\left(-d\eta^2+\delta_{ij}x^ix^j\right).$$ The Christoffel symbol for the one with cosmic time and the one with conformal $$\Gamma^\alpha_{t\alpha} = 3\frac{\dot a}{a} \equiv 3H,\\ \tilde\Gamma^\alpha_{\eta\alpha}= 4\dot a\equiv4\mathcal{H}.$$ We see $$\Box\phi = \partial^2\phi-3H\partial_t\phi,\\ \tilde\Box\phi = \tilde\partial^2\phi-4\frac{H}{a}\partial_\eta\phi = \partial^2\phi - 5H\partial_t\phi$$ where we transformed the coordinates of system 2 back into the first coordinate system using $$d\eta$$=dt/a. In other words, they are different. It seems like I have a sign error but I can't find it.

• I think it has to do with metric compatibility? It works out if I define $\Box = \nabla_\mu g^{\mu\nu} \partial_\nu$ Nov 13, 2022 at 9:59

Answer was that one needs to define $$\Box = \nabla_\mu g^{\mu\nu} \partial_\nu.$$ It does not matter in the case of the cosmic time coordinates since the metric commutes with the partial derivative, but it matters in the conformal time case. One can see in the derivation of the Euler Lagrange equation that the metric needs to be inside of the covariant derivative since the variation of the action happens with respect to the gradient of the field (which should have covariant index).
The d'Alembertian is defined as the divergence of the gradient, i.e $$\Box \phi = \nabla \vec \nabla \phi = \frac{1}{\sqrt{-g}}\partial_\mu \left( \sqrt{-g} g^{\mu\nu}\partial_\nu \phi \right) ,$$ where $$(\vec \nabla\phi)^\mu = g^{\mu\nu}\partial_\nu\phi \ \ , \ \ \nabla \vec F =\frac{1}{\sqrt{-g}} \partial_\mu \left(\sqrt{-g} F^\mu \right)$$
In terms of the set of coordinates $$(t,x,y,z)$$, with $$g_{\mu\nu}= diag(-1,a^2, a^2, a^2)$$, it is $$-\ddot\phi -\frac{3\dot a}{a} \dot\phi +\frac{1}{a^2} \Delta_x\phi$$ where, $$\dot A \equiv \frac{dA}{dt}$$, and in terms of the set of coordinates $$(\eta, x,y,z)$$, with $$\tilde g_{\mu\nu} =a^2 diag(-1,1,1,1)$$, it gives $$-\frac{1}{a^2} \phi'' - \frac{2a'}{a^3} \phi' + \frac{1}{a^2} \Delta_x\phi$$ where, $$A' \equiv \frac{dA}{d\eta}$$. Using the relation, $$A' = a \dot A$$, you can check that both expressions are equivalent.