4
$\begingroup$

One thing I have been somewhat curious about is computing geodesic deviation for tidal effect. Specifically, given separation vector $\xi^a$ between two nearby timelike geodesics and $V^c$ the velocity vector fields tangent to the geodesics, we have \begin{equation} \frac{D\xi^a}{d\tau^2} = R^{a}_{\,\,bcd}V^bV^c\xi^d\,. \end{equation} Suppose I have a static spherically symmetric metric given by Schwarzschild metric, which only depends on radial coordinate $r$. The first step in computing tidal force, as is well-known, seems to be simplifying this expression. For example, if I wish to know how geodesics converge when two particles are placed side by side along $\phi$ direction, I expect that along this direction particles come closer together, as we expect from radial nature of gravity.

However, to do this explicitly, naively what I would do is to calculate for $\xi^\theta$, and in particular I choose $\xi^a$ such that initially only $\xi^\phi$ is nonzero. Then if pick $V^c$ such that the particles are initially at rest; this would lead to (?) \begin{equation} \frac{D\xi^\phi}{d\tau^2} = R^{\phi}_{\,\,tt\phi}\xi^\phi\,. \end{equation} Here is the problem: for Schwarzschild geometry, it can be shown that $R^{\phi}_{\,\,tt\phi}$ vanishes on the horizon of the black hole. This cannot be correct expression for tidal force, then, since it should be nonzero.

Curiously, repeating these calculations for weak-field limits (or far-distance limit of Schwarzschild black holes) do give the correct $r^{-3}$ dependence because the higher order terms can be neglected, and this expression seems to give reasonable tidal force if I am some distance away from the horizon (say, $2r_H$). This is true also for the "sphaghettification" scenario where instead one is interested in radial scenario; again, the same method (but now with different component of Riemann tensor) also leads to vanishing tidal force at the horizon, but good in weak field or some distance away from horizon. The problem arises because the component of Riemann tensor above also has a contribution at the order of $M^2/r^4$ which neatly cancels the usual "tidal" contribution at $r=r_H$. What is going on here?

I am aware of calculation methods using tetrads, but I am trying to stick this to elementary GR as much as I can.

Note: the (?) mark was meant to show that similar expression seems to appear if I do use tetrad formalism, and it might be possible that I am missing something, but I cannot tell why at the moment.

Update: I think I found the solution (thanks to comment from @A.V.S), but for personal reasons, I will write down the solution myself some time in April.

$\endgroup$

2 Answers 2

4
$\begingroup$

The problem is that you are trying to combine condition the particles are initially at rest with placement at the horizon. Massive object cannot be at rest at the horizon, since $\partial_t$ (in Schwarzschild coordinates) is null there. You need to include the radial component of the 4-velocity. Also note, that in Schwarzschild coordinates at the horizon the $g_{tt}$ metric component is zero, while $g_{rr}$ diverges, so be careful with that.

$\endgroup$
1
  • $\begingroup$ Thanks, I think I know how to fix the problem thanks to your reminder that at rest with the horizon is physically inconsistent. I will write the solution I found in April. $\endgroup$ Mar 15, 2019 at 19:58
0
$\begingroup$

I will answer this based on suggestion by A.V.S. in one of the answer posts. The problem is due to a common mistake in general relativity, namely $V^t = 1$ which is not correct unless one is in flat space. Taking this into account, the correct equation is \begin{equation} \frac{D^2\xi^\phi}{d\tau^2} = R^\phi_{\,\,tt\phi}V^tV^t\xi^\phi\,. \end{equation} Now, clearly this is only valid at $\tau=0$ when the four-velocity has no spatial component, but that is fine since it's only for this one time-slice. Crucially, $V^tV^t = 1/f(r)$ and combining this with the Riemann tensor gives \begin{equation} \frac{D^2\xi^\phi}{d\tau^2} = -\frac{GM}{r^3}\xi^\phi \end{equation} which is exact. Here $r$ is the fixed radius where the two geodesics begin.

One possible complication when one wants to compute tidal force is whether to take $\xi^\phi$ to measure proper length or coordinate length: as it is given, this equation will give tidal force based on coordinate separation, while possibly one is interested in proper separation. I think this can be handled at least for this case, recognizing that along $\phi$ direction, proper length and coordinate length only differ by a fixed constant prefactor dependent on $f(r)$, where $r$ is a fixed radius..

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.