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Juan Maldacena says here: "For a very massive and very compact object the deformation (or warping) of spacetime can have a big effect. For example, on the surface of a neutron star a clock runs slower, at 70 percent of the speed of a clock far away. In fact, you can have an object that is so massive that time comes to a complete standstill. These are black holes."

(from https://www.ias.edu/ideas/2011/maldacena-black-holes-string-theory)

It is very clear that he refers to time-dilation and that the clock runs at 70% of the speed of a clock in another reference frame. When the clock is at the event horizon though it stops. Obviously there is an infinity there. We can talk about relative speeds until we reach a 0%. "time comes to a complete standstill"

When I ask about black holes many people point to the central singularity. It seems though that for time to come to a halt at the event horizon we must also call this a singularity. Where am I wrong in my thinking?

For time to stop there at EH curvature must be already at maximum (if a maximum is possible). How can we proceed to think beyond this and why do so many people ignore the logical singularity at the horizon in favour of a central one?

I am not asking for opinions - what I mean specifically is why we still assume there is a central singularity when the metric arrives at zero long before that - why is that approach still taken?

and can a zero be not absolute when one considers quantum mechanics?

Clock
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It seems though that for time to come to a halt at the event horizon we must also call this a singularity.

Gravitational time dilation depends on the gravitational potential. A singularity (the type of singularity we mean when we talk about singularities in GR) is a certain type of misbehavior of the curvature, which causes geodesic incompleteness. Curvature and gravitational potential are two different things.

Any function can have a singularity, but this mathematical definition of a singularity is different from the very specialized usage in GR.

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The word "singularity" is already widely used in mathematics, before one looks at any examples in physics, so one has to be careful.

In the case of the spacetime at and around a black hole, there are two types of observation in which the word "singularity" is commonly used. In one observation, we adopt a certain coordinate system, called Schwarzschild coordinates, to map spacetime, and we find extreme behaviour at the horizon. But it turns out that the curvature of spacetime is not infinite at the horizon, whereas the curvature does tend to infinity at another location in spacetime, the location which is assigned the value $r=0$ in Schwarzschild coordinates. So the standard terminology is to say that a curvature singularity occurs at $r=0$ and a coordinate singularity occurs at $r=r_s$ in Schwarzschild coordinates. But one does not have to map spacetime using Schwarzschild coordinates. In anther coordinate system there need not be any coordinate singularity at the horizon, but there will always be a curvature singularity at the location of the curvature singularity, because curvature is absolute; it is a property of spacetime itself, not of coordinates.

These statements can be used to justify various physical predictions. One prediction is that if one were to lower a clock on a rope to a place just outside the horizon, and then raise the clock, then the total time elapsed on the clock will be much less than the time elapsed on a similar clock residing permanently at some high place well away from the horizon. Furthermore, the fraction of the overall evolution of the lowered and raised clock which takes place while it resides near the horizon tends to zero as that location approaches the horizon. Other types of experiment can also be used to confirm this. One can imagine that the clock stays near the horizon and sends out light signals, for example. So you are right to say that a certain kind of mathematical singularity is associated with observable physical effects connected to the horizon.

Nevertheless, the standard terminology in which we simply say "the singularity" when referring to the curvature singularity is a good terminology. It is a shorthand for "the curvature singularity".

Finally, you are right to suspect that quantum physics implies that statements about horizons and singularities made by classical general relativity do not capture all the relevant physics. In particular, black hole entropy and Hawking radiation is connected to this.

Andrew Steane
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  • @Clock “space-time does extend through” [the coordinate singularity] - Only if you extend it by hand, which is non physical. See next to the last paragraph here for details: https://www.mathpages.com/rr/s7-03/7-03.htm – safesphere Feb 12 '20 at 16:20
  • “In anther coordinate system there need not be any coordinate singularity at the horizon” - The coordinate singularity at the horizon exists in all coordinate systems, because the coordinate transformation from Schwarzschild to any “well behaved” coordinates is necessarily singular and mathematically forbidden. You can solve the equations directly in the Kruskal coordinates with no horizon singularity, but the transformation from Kruskal to the coordinates of any physical observer is singular and puts the horizon singularity right back. – safesphere Feb 12 '20 at 16:46
  • @safesphere What about the physical observer in a state of free-fall? He will find that the metric close to him, as indicated by rods and clocks in his position and falling with him, will be the Minkowski metric, no matter where or when he is, except at curvature singularities where the maths breaks down. – Andrew Steane Feb 12 '20 at 18:30
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    @Andrew In any external coordinates, nothing ever crosses the horizon. For example, if you fall feet forward, you don’t “see” your feet crossing the horizon until your head does. (Where by “see” I mean the coordinate position, not a visual impression.) This implies that your local Minkowski region does not cross the horizon. The closer you get to the horizon, the smaller your local region becomes and asymptotically vanishes at the horizon. So the local flatness argument does not remove the horizon singularity. – safesphere Feb 12 '20 at 18:50
  • @safesphere this point is interesting; I'll think about it some more – Andrew Steane Feb 12 '20 at 19:44
  • @safesphere "your local [Minkowski] region becomes smaller and ... vanishes" I think is untrue. For example, if I were to fall through the horizon of a BH with $r_s = 10^{10}$ m, then I think my experience would be ordinary as I fell. The horizon is a null surface. On a spacetime diagram in LIF coordinates it is at 45 degrees. Whenever my feet pass such a line, I will not see them until my eyes do, but this is true for all the null surfaces I am continually crossing; a horizon is not special in this respect. – Andrew Steane Feb 13 '20 at 14:18
  • In a hovering frame momentarily coinciding with your LIF, your feet never reach the horizon (as this frame easily transforms to Schwarzschild). From this frame we transform to your LIF by a velocity boost, which does not change the position of your feet relative to the horizon. So in the LIF of your head, the coordinates of your feet are above the horizon for as long as your head is above the horizon. Please let me know if this is not clear. Also, you are not crossing any null surfaces in the direction of the null vector. If you tried doing this by accelerating, you would see the same effect. – safesphere Feb 13 '20 at 18:22
  • To see the same differently, consider the interval $\Delta t$ between the events of your head and your feet crossing the same $r$. This interval is timelike outside (decreasing as you approach the horizon), null (zero) at the null horizon, and spacelike inside in any valid coordinate system. So both your head and your feet cross the horizon at the same proper time and at the same Schwarzschild coordinate time $r=r_s$ on the inside (while the outside time $t$ diverges to the “same” infinity). – safesphere Feb 13 '20 at 19:15
  • Another point is that the event horizon is a property of the global spacetime, not a local phenomenon. In contrast, the equivalence principle is strictly local and thus does not apply at the horizon. Intuitively, the horizon extends to the infinite time in any reference frame, even in the frame of a free falling observer while he is outside (and he has no frame at the horizon). So the popular statement that “nothing special happens at the horizon” is deeply flawed to say the very least. BTW, we’ve discussed this earlier, but you never replied: https://physics.stackexchange.com/questions/492477 – safesphere Feb 18 '20 at 17:24
  • @safesphere Yes I know; I've been thinking about it. I am quoting the standard statements one finds in textbooks etc; you are insisting that it is not quite as straightforward as these statements suggest. I think what I am saying is correct, and you are pointing out that it cannot be mapped in any non-singular way to observations (clocks and rods etc.) of an observer who stays outside the horizon. This is partly because even to say "stays outside the horizon" is itself to make a statement about infinite time. – Andrew Steane Feb 18 '20 at 17:32
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Singularities mean the value becomes infinite .( For example it happens for 1/r potentials when r=0.) You could call it a discontinuity. The metric is not zero at the black hole singularity, but infinite.

A gravitational singularity, spacetime singularity or simply singularity is a location in spacetime where the gravitational field of a celestial body is predicted to become infinite by general relativity in a way that does not depend on the coordinate system.

Italics mine.

The Schwarzschild metric has a singularity for $r=0$ which is an intrinsic curvature singularity. It also seems to have a singularity on the event horizon $ r=r{_{s}}$. Depending on the point of view, the metric is therefore defined only on the exterior region $ r>r{_ {s}}$, only on the interior region $ r<r{_{s}}$ or their disjoint union. However, the metric is actually non singular across the event horizon as one sees in suitable coordinates (see below).

One has to consider the metric, as shown in the link. There is though a discontinuity.

There is also this observation:

Solutions to the equations of general relativity or another theory of gravity (such as supergravity) often result in encountering points where the metric blows up to infinity. However, many of these points are completely regular, and the infinities are merely a result of using an inappropriate coordinate system at this point.

So one has to be wary of the mathematics, and which metric and what coordinate systems the statement "In fact, you can have an object that is so massive that time comes to a complete standstill" is made.

and can a zero be not absolute when one considers quantum mechanics?

Quantum mechanics introduces the Heisenberg uncertainty to all measurement results, sure.

Look how the Big Bang singularity is changed by the introduction of effective quantum mechanics at the singularity.

anna v
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    Singularities mean the value becomes infinite . If you're describing the mathematical definition of a singularity, then this actually isn't correct. The metric is not zero at the black hole singularity, but infinite. This is wrong. There is no particular condition on the components of the metric that defines a singularity. Such a condition is neither necessary nor sufficient. –  Jan 21 '20 at 17:53
  • @anna v - Thank you for pointing out a discontinuity/disjoint union – Clock Jan 22 '20 at 10:21