Show that for a scalar field the following relation is true The FLRW line element is $ds^2=dt^2 -a^2(t)[(dx^1)^2+(dx^2)^2+(dx^3)^2]$. For a homogeneous scalar field $\psi$ show that
$$g^{\mu\nu}\nabla_{\mu}\nabla_{\nu} \psi = \ddot{\psi} + 3\frac{\dot{a}}{a}\dot{\psi}$$
The dot denotes a time derivative, the Einstein summation convention applies. I think I get close but I'm going wrong somewhere! Here's what I've tried:
Expand the second covariant derivative
$$g^{\mu\nu}\nabla_{\mu}\nabla_{\nu} \psi = g^{\mu\nu}\nabla_{\mu}(\partial_{\nu}\psi)$$
$$=g^{\mu\nu}(\partial_{\mu}\partial_{\nu}\psi-\partial_{\sigma}\psi\Gamma^{\sigma}_{\mu\nu})$$
$$=g^{\mu\nu} \partial_{\mu}\partial_{\nu}\psi - g^{\mu\nu}\Gamma^{\sigma}_{\mu\nu}\partial_{\sigma}\psi$$
Homogeneous means $\psi$ is a function of $t$ only, and so for the first term to be non-zero sub in $\mu=\nu=0$. Because they're dummy indices that doesn't mean I also have to set them to zero in the second term. 
$$=g^{00}\frac{\partial ^2\psi}{\partial t^2} - g^{\mu\nu}\Gamma^{\sigma}_{\mu\nu}\partial_{\sigma}\psi$$
From the line element, $g^{00}=1$. Now looking at the second term, I can write out the Christoffel symbol:
$$g^{\mu\nu}\Gamma^{\sigma}_{\mu\nu} = \frac{1}{2}g^{\sigma\epsilon}g^{\mu\nu}\left(\frac{\partial g_{\mu\epsilon}}{\partial x^{\nu}}+\frac{\partial g_{\nu\epsilon}}{\partial x^{\mu}} - \frac{\partial g_{\mu\nu}}{\partial x^{\epsilon}}\right)$$
Contract with the metric:
$$\color{red}{\frac{1}{2}g^{\sigma\epsilon}\left(\frac{\partial g^{\nu}_{\epsilon}}{\partial x^{\nu}}+\frac{\partial g^{\mu}_{\epsilon}}{\partial x^{\mu}} - 0\right)}$$ 
Because the last term is either one or zero so the derivative is always zero. Contract again: 
$$\frac{1}{2}\left(\frac{\partial g^{\sigma\nu}}{\partial x^{\nu}}+\frac{\partial g^{\sigma\mu}}{\partial x^{\mu}}\right)$$
Relabel the dummy index and get 
$$g^{\mu\nu}\Gamma^{\sigma}_{\mu\nu}=\frac{\partial g^{\sigma\nu}}{\partial x^{\nu}}$$
If I substitute that back in to the original equation, 
$$\frac{\partial ^2\psi}{\partial t^2} - g^{\mu\nu}\Gamma^{\sigma}_{\mu\nu}\partial_{\sigma}\psi = \frac{\partial ^2\psi}{\partial t^2} - \frac{\partial g^{\sigma\nu}}{\partial x^{\nu}}\partial_{\sigma}\psi$$
Looking at that I would conclude that only $\sigma = 0$ in the second term is non-zero, because I'd have $\partial_{\sigma}\psi$ which is zero unless the derivative is with respect to $t$. But the metric is diagonal so having $g^{0\nu}$ would mean I can't ever have the $a^2$ term and I won't get the $\dot{a}/a$ that I need. Where have I gone wrong? 
 A: The mistake is that you raised the indices of the metric going through a derivative, i.e. you took $g^{\mu\beta}\frac{d}{dx^\alpha}[g_{\beta\nu}] = \frac{d}{dx^\alpha}[g^{\mu}_{\nu}]$ which is wrong. Instead we have $\frac{d}{dx^\alpha}[g^{\mu}_{\nu}] = g^{\mu\beta}\frac{d}{dx^\alpha}[g_{\beta\nu}] + g_{\beta\nu}\frac{d}{dx^\alpha}[g^{\mu\beta}]$ by the product rule of differentiation.
To compute the Christoffel symbols note that since $g$ only depends on time we need $\nu = 0$ in order for $\frac{dg^{\alpha\beta}}{dx^\nu}$ to be non-zero. Also since $g$ is diagonal we need $\mu=\nu$ in order for $g_{\mu\nu}$ to be non-zero. This leads to the first two terms in the expression for the Christoffel symbols to vanish (e.g. the non-zero terms in $g^{\sigma\epsilon}g^{\mu\nu}\frac{dg_{\mu\epsilon}}{dx^\nu}$ must have $\nu=0$ (derivative) and then $\mu = 0$ (the $g^{\mu\nu}$) and then $\epsilon = 0$ ($g_{\mu\epsilon}$) leaving us with $g^{0\sigma}g^{00}\frac{dg_{00}}{dt} = 0$ since $g_{00}$ does not depend on time) and we are left with
$$g^{\mu\nu}\Gamma^{\sigma}_{\mu\nu} = -\frac{1}{2}g^{\sigma0}g^{\mu\nu}\frac{\partial g_{\mu\nu}}{\partial x^{0}} = -\frac{1}{2}g^{\sigma 0}\cdot 3\cdot (-a^{-2})\cdot \frac{d}{dt}(-a^2) = -3g^{\sigma 0}\frac{\dot{a}}{a}$$
since only the 3 terms with $\mu=\nu=1,2,3$ gives a non-zero contribution. This leads to the correct expression for the d'Alembertian of $\psi$.
