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?

  • $\begingroup$ The potential is usually assumed to be time independent, which gets rid of two of the terms. If we can show that we can choose $\xi^0=0$ for all times, that gets rid of the last one, and you get your desired expression. $\endgroup$
    – Javier
    Commented Oct 29, 2020 at 3:25
  • $\begingroup$ That sounds reasonable! Still not sure about the issue of d vs $\partial$ though...but we probably assume that $U$ (four-velocity of the geodesic) is dominated by $(1,0,0,0)$ so $U^\mu \partial_\mu \xi \equiv \frac{\mathrm{d} \xi}{\mathrm{d} t} = \gamma (v) \left( \frac{\partial \xi}{\partial t} + \vec{v} \cdot \vec{\nabla} \, \xi \right) \approx \frac{\partial \xi}{\partial t}$. $\endgroup$
    – user16320
    Commented Oct 29, 2020 at 3:52
  • $\begingroup$ Your mistake, I think, is that you assume thet $U=e_0$ . This is obviously wrong, since U should be a geodesics of the given metric. The geodesic deviation equation mesures the acceleretion (ie. second derivative) of the relative dispacement of 2 nearby geodesics (the U field). Since your assumed U is wrong, the equation does not hold. $\endgroup$
    – magma
    Commented Oct 29, 2020 at 11:47
  • $\begingroup$ You should first calculate the U field for your metric using the geodesics equation. then take 2 geodesics dispaced by a small amount initially and then see how that displacement changes with time. Many GR textbooks report this calculation for the weak field. However the easiest thing is to see the derivation of the geodesic deviation equation simbolically, to convince yourself that it is correct $\endgroup$
    – magma
    Commented Oct 29, 2020 at 11:58
  • $\begingroup$ But for a weak field and small velocity we would neglect the higher orders (i.e. $U^\beta U^\mu \xi^\nu$) anyway, what would the solution for U provide? Moreover, it would be formal at best, since $\phi (x)$ has only a general form at this point, just independent of time. Assuming that $U = e_0$ at the beginning means that the objects fall from a stationary initial condition. The expression for $\dot{U}$ contains $\Gamma$'s, which contain the field $\phi$ so plugging that in will kill these extra terms, anyway, I think, but to be sure I have to try. $\endgroup$
    – user16320
    Commented Oct 29, 2020 at 13:21

1 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.


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