# Linearising the Geodesic Deviation Equation

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1. Nov 24, 2014

### Xander314

1. The problem statement, all variables and given/known data
Write down the Newtonian approximation to the geodesic deviation equation for a family of geodesics. $V^\mu$ is the particle 4-velocity and $Y^\mu$ is the deviation vector.

2. Relevant equations
$$D = V^\mu\nabla_\mu \\ V^\mu\approx(1,0,0,0) \\ g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$$

The geodesic deviation equation and the desired linearisation are
$$D^2 Y^\mu = R^\mu{}_{\rho\lambda\nu} \\ \partial_t^2 Y^i = -\partial_i\partial_j\phi Y^j$$

3. The attempt at a solution
Apologies if this is not the correct way to post in this subforum, but I'm not going to write up all of what I have so far as I don't think it's relevant to my question. My issue is with the left hand side, which with the approximation for the 4-velocity becomes
$D^2 Y^\mu = V^\mu\nabla_\mu V^\nu\nabla_\nu Y^\mu = \nabla_t^2 Y^\mu$
Everywhere I have looked, the next step is to replace this by $\partial_t^2 Y^\mu$, but I don't follow this logic. As far as I can see, the covariant and partial derivatives differ by Christoffel symbols containing terms O(h), which should enter into the linearisation. What am I missing? Is Y also O(h)?

2. Nov 27, 2014

### stevendaryl

Staff Emeritus
That equation $D^2 Y^\mu = R^\mu{}_{\rho\lambda\nu}$ can't possibly be right, because the left-hand side has only one index, $\mu$, while the right side has indices $\mu, \rho, \lambda, \nu$. I think it's supposed to be something like:

$D^2 Y^\mu = R^\mu{}_{\rho\lambda\nu} V^\rho V^\lambda Y^\nu$