I Supermassive black hole, surface gravity and tidal forces

[Orodruin]

Gullestrand-Painleve coordinates use a time slicing good from approaching the singularity to infinity. Each slice includes different horizon events as desired. In this time slicing, any difference in areal radius r, is, in fact, a proper distance per the metric along the time slice.

It should not be surprising that such coordinates do exist. However, it should be noted that these time slices are not orthogonal to the time coordinate lines, which is a bit dissatisfying in itself.

 

[PAllen]

It should not be surprising that such coordinates do exist. However, it should be noted that these time slices are not orthogonal to the time coordinate lines, which is a bit dissatisfying in itself.

But these slices are orthogonal to the congruence of inertial free fallers from infinity (which is different from the Gullestrand-Painleve t coordinate lines; to see this you write out the rain drop world line tangents in GP coordinates). Who says the congruence of stationary observers is fundamentally preferred over the ‘rain drop’ congruence?

 

[Orodruin]

Who says the congruence of stationary observers is fundamentally preferred over the ‘rain drop’ congruence?

That depends to some extent on your preference. In a general spacetime there would be no preference for one set of observers over the other. However, in the case of a stationary spacetime, such as the exterior Schwarzschild solution, the existence of a timelike Killing field is related to a time translation symmetry of the spacetime itself, independent of any actual observers. That stationary observers are then defined based on this time translation symmetry is a different issue.

So I would say that if you are going to call any convention ”preferred”, the it better be based on the symmetries of the spacetime than an arbitrarily chosen set of observers. Thereby not saying that you definitely should do so.

 

[PAllen]

What the discussion of Lemaitre and Gullestrand-Painleve coordinates leads to is the following conclusion:

From any event outside an non-rotating uncharged BH, if you define distance to the horizon by the proper distance along the radial spacetime geodesic [edit: path of Gullestrand-Painleve simultaneity] through that event 4-orthogonal to a free faller with escape velocity toward the BH, then that distance is just r - R, where r is the areal radius, and R is the SC radius.

In contrast, the radial lines in SC coordinates are spacelike spacetime geodesics orthogonal to a hovering observer at a given event. Unfortunately, for a BH from collapse, these don’t intersect the horizon at all, as discussed in recent posts. Thus, IMO, it is useless to use them to discuss possible definitions of distance to the BH.

Of course, among all the radial spacelike geodesics from an event that do reach the horizon, there is no reason to favor one orthogonal to an escape velocity free faller. Thus you can justify a wide range of values greater than zero using proper distance along geodesics.

 

[Orodruin]

The question is how to relate all of this to the question of surface gravity as posed in the OP.

In respect to this I’d argue that sufficiently locally, so locally that tidal effects may be neglected, there is nothing separating the two black holes of different masses at the horizon. For the stationary observer hovering just above the horizon with constant proper acceleration, the event horizon is mapped to its Rindler horizon in the local Minkowski frame. Regardless of the size of the black hole.

 

[PeterDonis]

the radial spacetime geodesic through that event

I don't think those radial lines are spacetime geodesics. They are geodesics of the 3-surface of constant Painleve coordinate time that passes through the chosen event, but I believe those surfaces are not generated by radial spacetime geodesics. The radial spacetime geodesics lie in the surfaces of constant Schwarzschild coordinate time, which are different from the surfaces of constant Painleve coordinate time.

 

[PeterDonis]

In respect to this I’d argue that sufficiently locally, so locally that tidal effects may be neglected, there is nothing separating the two black holes of different masses at the horizon.

But if we adopt this approximation then there is no difference in their surface gravities either. The "surface gravity" of a Rindler horizon is zero.

In order to differentiate between surface gravities, we have to include the black hole masses in our model somehow.

 

[PAllen]

I don't think those radial lines are spacetime geodesics. They are geodesics of the 3-surface of constant Painleve coordinate time that passes through the chosen event, but I believe those surfaces are not generated by radial spacetime geodesics. The radial spacetime geodesics lie in the surfaces of constant Schwarzschild coordinate time, which are different from the surfaces of constant Painleve coordinate time.

They are geodesics. This is most easily seen by looking at the variation in Lemaitre coordinates, where the result is almost obvious. Note that through a point there are infinite radially directed geodesics, each with a different tangent. Similar to in x t Minkowsky plane, at any point there are infinite space like geodesics with -x component. Thus, the SC radial geodesic from an event is just one of infinitely many radial geodesics from that event.
[see later discussion].

 

[Orodruin]

But if we adopt this approximation then there is no difference in their surface gravities either. The "surface gravity" of a Rindler horizon is zero.

No, it is infinite as the proper acceleration of a hovering observer tends to infinity as you get closer to the horizon. My point is that locally there is no difference until you start feeling the tidal effects.

In order to differentiate between surface gravities, we have to include the black hole masses in our model somehow.

That’s assuming that whatever we want to call ”surface gravity” is different in the cases of different mass black holes.

 

[PeterDonis]

That’s assuming that whatever we want to call ”surface gravity” is different in the cases of different mass black holes.

The standard definition of surface gravity for a black hole (and indeed for a Killing horizon in general) is, as I said earlier in this thread, the redshifted proper acceleration at the horizon (or more precisely the appropriate limit as the horizon is approached). For a good (although A-level) treatment, see Wald's monograph Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics. This surface gravity is different for different black holes.

 

[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!