Jacobi equation for the geodesic deviation in the weak field limit

Question

The geodesic deviation equation can be written in the following form $$ \nabla_U^2 \xi = R (U, \xi) U \tag{1} $$ where $R$ is the Ricci tensor. It can also be written component-wise using the Riemann tensor $$ (\nabla_U^2 \xi)^\alpha = R^\alpha_{\;\, \beta \mu \nu} \: U^\beta U^\mu \xi^\nu \tag{2} $$

I'm interested in this equation on a Riemannian manifold equipped with the following metric tensor $$ g = -(1+2\phi(x))\mathrm{d} t \otimes \mathrm{d} t + (1-2\phi (x))\left(\mathrm{d} x \otimes \mathrm{d} x + \mathrm{d} y \otimes \mathrm{d} y + \mathrm{d} z \otimes \mathrm{d} z \right) \tag{3} $$ with the standard torsion-free and metric-preserving connection $\nabla$. I'm only interested in the spatial part of the Jacobi equation, that starts from $U = e_0$ i.e. $$ (\nabla_t^2 \xi)^i = R^i_{\;\, 0 0 \nu} \,\xi^\nu \tag{4} $$

In textbooks it can be found that in the weak-field limit (keeping only the linear terms involving $\phi (x)$) this reduces to $$ \frac{\mathrm{d}^2 \xi^i}{\mathrm{d} t^2} = - \phi_{,ij} \, \xi^j \tag{5} $$ where comma indicates partial derivatives.

However, when I try this for $g$ above, I get a different result. First, the left-hand side of (4) is $$ \nabla_t \xi = \left( \xi^\mu_{,t} + \Gamma^\mu_{\;\; \nu t} \, \xi^\nu \right) e_\mu $$ $$ \nabla^2_t \xi = \left( \xi^\mu_{,t} + \Gamma^\mu_{\;\; \nu t} \, \xi^\nu \right)_{,t} e_\mu + \left( \xi^\mu_{,t} + \Gamma^\mu_{\;\; \nu t} \, \xi^\nu \right) \Gamma^\lambda_{\;\; \mu t} e_\lambda $$

The Christoffel symbols are in general $$ \Gamma^\alpha_{\;\; \mu \nu} = \frac{1}{2} g^{\alpha \lambda} \left( g_{\lambda \mu, \nu} + g_{\lambda \nu, \mu} - g_{\mu \nu, \lambda} \right) $$ so in our case $$ \Gamma^0_{\;\; \mu \nu} = \begin{pmatrix} \phi_{,0} & \phi_{,1} & \phi_{,2} & \phi_{,3} \\ \phi_{,1} & - \phi_{,0} & 0 & 0 \\ \phi_{,2} & 0 & - \phi_{,0} & 0 \\ \phi_{,3} & 0 & 0 & - \phi_{,0} \end{pmatrix} $$ $$ \Gamma^1_{\;\; \mu \nu} = \begin{pmatrix} \phi_{,1} & - \phi_{,0} & 0 & 0 \\ - \phi_{,0} & - \phi_{,1} & - \phi_{,2} & - \phi_{,3} \\ 0 & - \phi_{,2} & \phi_{,1} & 0 \\ 0 & - \phi_{,3} & 0 & \phi_{,1} \end{pmatrix} $$ $$ \Gamma^2_{\;\; \mu \nu} = \begin{pmatrix} \phi_{,2} & 0 & - \phi_{,0} & 0 \\ 0 & \phi_{,2} & - \phi_{,1} & 0 \\ - \phi_{,0} & - \phi_{,1} & - \phi_{,2} & - \phi_{,3} \\ 0 & 0 & - \phi_{,3} & \phi_{,2} \end{pmatrix} $$ $$ \Gamma^3_{\;\; \mu \nu} = \begin{pmatrix} \phi_{,3} & 0 & 0 & - \phi_{,0} \\ 0 & \phi_{,3} & 0 & - \phi_{,1} \\ 0 & 0 & \phi_{,3} & - \phi_{,2} \\ - \phi_{,0} & - \phi_{,1} & - \phi_{,2} & - \phi_{,3} \end{pmatrix} $$ which, for $(\nabla^2_t \xi)^i$ gives (to the linear order in $\phi$) $$ \xi^i_{\;\;, 00} + \phi_{, i0} \xi^0 - \phi_{,00} \xi^i + 2 \phi_{,i} \xi^0_{\;\;, 0} - 2 \phi_{,0} \xi^i_{\;\;,0} $$

The right-hand side needs $R^i_{\;\; 00 \mu}$. First, due to the symmetries, $R^i_{\;\;000} = 0$, so we only need $$ R^i_{\;\; 00j} = - \phi_{, ij} - \phi_{,00} \, \delta_{ij} $$

Now put it all together $$ \frac{\partial^2 \xi^i}{\partial t^2} + \phi_{, i0} \xi^0 - \phi_{,00} \xi^i + 2 \phi_{,i} \xi^0_{\;\;, 0} - 2 \phi_{,0} \xi^i_{\;\;,0} = - \phi_{,ij} \xi^j - \phi_{,00} \xi^i $$

Cancel out common terms $$ \frac{\partial^2 \xi^i}{\partial t^2} + \phi_{, i0} \xi^0 + 2 \phi_{,i} \xi^0_{\;\;, 0} - 2 \phi_{,0} \xi^i_{\;\;,0} = - \phi_{,ij} \xi^j $$

There are problems with this expression. It does not match what the textbooks say.

Even if it did, how can I get "$\frac{\mathrm{d}^2 \xi^i}{\mathrm{d} t^2}$" instead of the partial derivative? Is $\frac{\mathrm{d}}{\mathrm{d} t}$ to be interpreted as $U^\mu \partial_\mu$? In that case if $U = e_0$, we would have $\frac{\mathrm{d}}{\mathrm{d} t} = \frac{\partial}{\partial t}$, but probably only at the initial point of our geodesic...or do we reparametrize the geodesic in terms of the time $t$ and then derivative w.r.t $t$ is actually derivative w.r.t. the curve parameter, so $\nabla_U^2 = \frac{\mathrm{d}^2}{\mathrm{d} \lambda^2} = \frac{\mathrm{d}^2}{\mathrm{d} t^2}$? That would certainly interpret the left-hand side of (4), but we would still have that weird term $-\phi_{,00} \xi^i$ on the right-hand side (unless we assume that $\phi$ does not depend on time, on the top of all that). But even if we do it, then $U$ is no longer in just zeroth direction so the right-hand side will feature two $U$ terms (even if we begin with $U = e_0$, can it stay that way for the entire geodesic? or do we assume small $v$, therefore the zeroth component dominates and since $U \cdot U = -1$, then $U^0 = 1$?)

There's a bunch of other terms that don't seem to arise in textbooks, but I never saw the expression (5) derived, only stated.

Is there a conceptual mistake in my thinking, or I miscalculated something?

Answer 1

Here's an answer that I argued for myself and am pretty satisfied with it. The caveat is actually in what people mean by writing $\frac{\mathrm{d}^2 \xi}{\mathrm{d} t^2}$. This term is more of a placeholder for $$ \nabla_U (\nabla_U \xi) = \nabla_U^2 \xi \equiv \frac{\mathrm{d}^2 \xi}{\mathrm{d} \lambda^2} $$ where $\lambda$ is the parameter along the curve. When I read the corresponding chapter from my differential geometry book I convinced myself that that's it, there's nothing more to the left-hand side. The only thing we need to argue is that for a weak field $\phi$ and two close, slowly moving observers starting from the rest and not going too far along the geodesic${}^1$, their four-velocity $U$ is dominated by $e_0$ and moreover, we can reparametrize the curve in terms of $\tau$, the proper time, which becomes just the frame time $t$, if the observers are falling slowly. Therefore, $$ \nabla_U (\nabla_U \xi) \approx \frac{\mathrm{d}^2 \xi}{\mathrm{d} t^2} $$

The right-hand side was correct; for a weak field, the relevant part of the Riemann tensor is $R^i_{\;\; 000j}$ which is equal to $- \phi_{,ij} - \phi_{,00} \delta_{ij}$. If we additionally assume that $\phi$ does not depend on time, then we get $$ \frac{\mathrm{d}^2 \xi^i}{\mathrm{d} t^2} = - \phi_{,ij} \xi^j $$

${}^1$Which is how I imagine we (naively) gauge the tides and such effects; we let go off a small, sensitive gauging device in the gravitational field, let it measure for a few seconds and catch it again. That gives us how $\xi$ (gauging device's spring length, sensor measuring distance or however you would "practically" gauge this) changes in time, shortly after being let go, moving slowly, and not too far along the geodesic so it does not acquire much speed.