Supermassive black hole, surface gravity and tidal forces

[PeterDonis]

They are geodesics.

When I do the computation in Painleve coordinates, I don't get that result.

A purely radial curve in Painleve coordinates with unit tangent has tangent vector $U = \partial_r$ (since $g_{rr} = 1$), i.e., its components are $(0, 1, 0, 0)$. The path curvature of this curve is:

$$
a = \sqrt{g_{ab} a^a a^b}
$$

where $a^a = U^b \nabla_b U^a$ is the path curvature 4-vector. It turns out to have two nonzero components, which, taking into account that all partial derivatives of $U$ are zero and using the above components of $U$, are:

$$
a^t = \Gamma^t_{rr} = \frac{M}{r^2} \sqrt{\frac{r}{2M}}
$$

$$
a^r = \Gamma^r_{rr} = - \frac{M}{r^2}
$$

Plugging in these and the relevant metric coefficients into the formula above, I get

$$
a = \sqrt{\frac{M}{2r^3}}
$$

which is nonzero.

This is most easily seen by looking at the variation in Lemaitre coordinates

I'll take a look at this when I get a chance.

 

[PAllen]

This is most easily seen by looking at the variation in Lemaitre coordinates

I'll take a look at this when I get a chance.

I took another look. I mis-handled the time dependence of the Lemaitre metric. Done right, the variation says the constant raindrop time lines are not geodesics of the spacetime - they are, indeed, only geodesics of the time slice.

It is still true that can get a wide range of values from just above zero (by using geodesics that asymptotically approach a forward going radial inward light path) to measuring back to horizon start by choice of geodesic that intersects horizon. Presumably, one of these would produce r-R, but it would not likely have any other distinguishing property.

 

[PeterDonis]

they are, indeed, only geodesics of the time slice.

Ok, good, thanks for checking again!