# Velocity inside black hole

1. Oct 30, 2015

### zylon

Consider a black hole with a radius of one light hour (as measured from outside, stationary observer), non rotating, having singularity in center, is radiating hawking radiation, and has no outside interference. Assume standard mathematical theory.
Traveller A is falling towards hole from infinite distance. Traveller B begins freefall at event horizon. In each case below, time is reckoned from the traveller’s point of view.
Question one: How long will it take traveller A to get from event horizon to the singularity?
Question two: How long will it take traveller B to get from event horizon to singularity?
Thank you.

2. Oct 30, 2015

### Staff: Mentor

Is there a particular reason you included this? Hawking radiation is going to be completely negligible for a hole of this size, and it's not relevant to any of the questions you're asking.

Strictly speaking, this is not possible. No observer can be at rest at the horizon, even for an instant. The best you can do is to have Traveler B start his free fall just slightly above the horizon.

However, for purposes of calculation, it turns out we can get away with plugging $r = 2M$ into the formula for Traveler B without causing any mathematical difficulties; so we can just consider the value we get as the limit of the values we would find if Traveler B started closer and closer to the horizon.

There is a simple formula for the time it takes Traveler A, by his clock, to fall from any radius $r$. The formula is:

$$\frac{\tau_A}{2M} = \left( \frac{r}{2M} \right)^{\frac{3}{2}}$$

Since you're asking about the time to fall from the horizon, $r = 2M$, this formula reduces to

$$\tau_A = 2M$$

So since $2M$ is one light-hour, it will take Traveler A one hour, by his clock, to fall from the horizon to the singularity.

The formula for this case is very similar to the above:

$$\frac{\tau_B}{2M} = \left( \frac{r}{2M} \right)^{\frac{3}{2}} \frac{\pi}{2}$$

For $r = 2M$, this reduces to

$$\tau_B = \pi M$$

3. Oct 31, 2015

### zylon

I mentioned Hawking radiation because , although I know that far from the hole (external time) the lifetime of the hole is practically eternal, I do not know how time behaves inside the hole. For all I know, the internal hour it takes for the traveler to reach the singularity could be longer than the external googol years the hole will live. In fact, as far as I know, it may be more than an external infinite time!
Therefore, until I understand how time is distorted within the event horizon, I cannot take anything for granted, concerning the comparison of times inside the horizon vs. far from the hole.

4. Oct 31, 2015

### m4r35n357

Like Peter I wondered why you mentioned Hawking radiation. Now you have answered that question I should point out that we don't have any closed form models of black holes that include time variation of its properties.

That said, if we restrict ourselves to a static solution such as Schwartzschild for example, all the mathematics that you will need to calculate coordinate and proper times is contained in this excellent article.

5. Oct 31, 2015

### Staff: Mentor

Hawking radiation has nothing to do with this.

The correct answer is that there is no well-defined way to compare the two. This is not just a feature of black holes; it is a general feature of curved spacetimes that there is no invariant way to compare the times of events that are spatially separated. We can adopt various conventions, but that's all they are, conventions. There is no physical fact of the matter. The only invariant fact of the matter is the time elapsed on the free-faller's clock; that's why I phrased my answer that way.

6. Oct 31, 2015

### zylon

Unless I am overlooking something, the way to compare spatially separated times is by comparing mutual events. In this case, the mutual event is the absence of the black hole, as what the situation would eventually become if hawking radiation continues until all the mass of the hole is depleted. As seen from the outside, this depletion, in its simplest mathematical description, would look like an explosion, followed by the absence of the hole. From the inside, the time which corresponds "the absence of the hole" can be matched by no longer being surrounded by an event horizon (or, if annihilated, the lack of any future events). For example, if the journey to the singularity takes longer than the lifetime of the hole, the traveler simply would never experience the event of reaching the singularity. My addition of Hawking radiation was just to create a (finite) time when the hole will no longer exist, which is a time event which both outside and inside observers could match (ignoring the time interval involved, which I know is a problem).
My thinking now is that relativistic time distortions are caused by acceleration against gravity (or inertia), and since free fall equates to a lack of gravity, perhaps there is no relativistic time distortion at all between observer and traveler, thus the condition of the hole after one hour of fall of the traveler is the same as the condition of the hole after one hour of watch by the outside observer, i.e. if the singularity still exists for the observer after one hour, the singularity would still exist for the traveler after one hour, and the far future when hawking radiation evaporates the hole can be ignored.
Am I on the right track??

7. Oct 31, 2015

### pervect

Staff Emeritus
We can (*) computed the times for someone to fall into an idealized black hole with a Schwarzschild geometry. Such a black hole does not evaporate, it's not part of the model.

* (with the adjustments already mentioned of taking a limiting process in case B which is strictly speaking impossible exactly as written except as the result of a limiting process)

To compute the times for a non-idealized non-Schwarzschild geometry, you'd need the interior metric for a realistic black hole with Hawking radiation. This is a matter of some debate in the literature (even without the Hawking radiation!). So the question for a realistic black hole depends on the model used, and while there has been work on the subject, there isn't any widely-agreed on defnitive metric for this case, esepcially if you insist on having Hawking radiation.

That can probably be done without the extreme complexity involved in coming up with a realistic evaporating black hole metric. If you imagine an observer at infinity sending light signals radially downwards at regular intervals, for defininteness one second apart, each of the two observer can count the number of signals they receive from the observer at infinity, starting at the point where they are both co-located at the event hole horizon. This can be done most conveniently using ingoing Eddignton-Finklestein coordinates, a null coordinate where each "flash" has a constant coordinate value. I don't know the exact numerical answer for this case offhand, but I know it's compuatble, and that it will be finite for each of the two observers. Basically, there will be a "last event" that they see while falling down the purely classical, non-radiating, interior Schwarzschild geometry, a finite number of flashes that they see up until the time they reach the singularity. Strictly speaking, this is another limit, the actual point of reaching the singularity is outside the manifold, but the limit exists.

The next thing you can ask is how much mass the black hole has lost in this finite amount of time, and see if it's significant. I can pretty much say that it won't be, a black hole of this size is extremely long lived - though I don't have an exact number.

I know there are papers and textbooks that discuss the exact problem you mentioned (minus the part about the hawking radiation) but I haven't attempted to track them down. If (and only if) this is of some interest (and laso if I have time), I'll attempt to find them. If it's felt that the exact details would be "too techinical", my default assumption, I won't bother to attempt to find them. I don't believe the references I recall seeing will compute the "number of flashes seen", I can find at least a web reference that mentions that there are a finite number of them. I also have, somewhere on PF, some notes/calculations about ingoing Eddingtoin-Finklestein coordinates and the finite frequency shift seen by an infalling observer, but the details there are not textbook results, just some calculations I made. The calculations I did were, as Irecall, only up to the event horizon (where there was a modest 2:1 redshift at the horizon), but I believe they can be extended to the interior region as well (as long as you use the interior Schwarzschild geometry, that is.)

8. Oct 31, 2015

### Staff: Mentor

There is no such thing, in the sense you mean. "Events" are points in spacetime; they are things that happen at a particular place at a particular time according to a clock located at that place.

The black hole is not a single point in space at a single moment of time, so it isn't an "event" (much less a "mutual event"), nor is its absence.

The black hole is a region of spacetime that can't send light signals out to infinity. So instead of thinking about "events", if you want to know whether a given traveler will enter that region of spacetime, you have to look at the global geometry of spacetime and how the traveler's worldline fits into that geometry. See below.

Ok, this gives me a better understanding of why you are thinking about Hawking radiation for this scenario.

There is a way in which this can happen, but it is much more fruitful to think of it in the way implied by my comments above. You have a region of spacetime, the black hole, that can't send light signals to infinity. You have a traveler who has a specific worldline, i.e., a curve in spacetime. So the question is, does the traveler's worldline, as a curve in spacetime, intersect the black hole region of spacetime?

This is a simple question of geometry, just like the question of whether the equator on Earth intersects a given continent, for example, and it is best investigated using geometrical tools. My favorite tool is the Penrose diagram, which is a way of illustrating the geometric structure of a spacetime (which curves can enter which regions) that is symmetric enough that we can reduce the number of dimensions we need to look at to two. Take a look at this page, for example:

http://physics.oregonstate.edu/coursewikis/GGR/book/ggr/penrose

Figures 3 and 4 on that page show two Penrose diagrams: Figure 3 is for a black hole that forms from a collapsing massive object like a star, and then remains forever (i.e., no Hawking radiation). Figure 4 shows a black hole that forms by collapse but then eventually disappears by emitting Hawking radiation. In both diagrams, the region above and to the left of the 45 degree line marked "r = 2m" is the black hole, and the horizontal line marked "r = 0" is the singularity.

In Figure 3, note that the upper right end of the horizon line meets the point marked $i^+$. That point is called "future timelike infinity"--it is where all timelike lines that extend infinitely far in the future end up on this diagram (which includes points and lines "at infinity" in order to make the geometric structure clearer). The fact that the horizon intersects the point at infinity means that any timelike worldline (meaning any observer) has only two choices: he can stay away from the hole forever (which means he has to accelerate somehow; he can't remain in free fall forever because he will then fall into the hole), or he can fall into the hole and be destroyed in the singularity.

In Figure 4, however, note that there is a vertical line marked "r = 0" above the upper right end of the horizon line. That means that an observer who waits long enough, by his own clock, can indeed free-fall inward and still avoid falling into the hole, by arriving at the "r = 0" line at some point above the upper right end of the "r = 2m" line (which is the point at which the hole disappears).

9. Nov 2, 2015

### zylon

First, the event of the outside observer saying "hey, the black hole isn't there anymore" (e.g. observed to have exploded) can be by definition matched to the event of the traveler first noting that he is no longer surrounded by an event horizon, thus connecting the first moments of the black hole's absence from both frames. The size of the time intervals are not relevant.

Second, my question was not about whether the traveler would enter the black hole; it was given that he did. It was only about whether or not he would reach the singularity of a black hole with a finite lifetime. My addition of hawking radiation just represented anything which would terminate the hole; I could have used the big crunch, but that would be even more problematic.

I realize now that any attempt to terminate the hole would create enormous mathematical complications, and would probably push the traveler even faster into the singularity so that event of being outside the event horizon again would never exist. But I am still very far from sure about this.

10. Nov 2, 2015

### Staff: Mentor

How does the infaller discover that he is not surrounded by an event horizon? That is, what has to happen along the traveler's worldline before the event "the traveler first noting that he is no longer surrounded by an event horizon"?

11. Nov 2, 2015

### Staff: Mentor

How? Please show your work. You appear to think this is a trivial point; it isn't. In fact, it's not even possible to "match" spatially separated events in an invariant manner.

I didn't gather that from your previous posts, since they were talking about Hawking radiation and other phenomena at the horizon, not phenomena at the singularity.

If the model of a black hole emitting Hawking radiation that is described in the Penrose diagram I linked to earlier is correct, then anyone who falls through the horizon hits the singularity. The only way to avoid the singularity is to avoid the horizon.

However, it's not at all clear that that model is correct. In fact, nobody knows what the correct model of an evaporating black hole is. In fact, if quantum gravity effects are taken into account, it's not even certain that a true black hole (i.e., an event horizon and a region inside it with a singularity) can exist. So we don't know for sure what the answer to your question is.

12. Nov 3, 2015

### zylon

My addition of Hawking radiation (HR) was only to transform the hole into a clock in case times in the far future were relevant. If so, both the outside observer and the traveler inside the hole could match time events by calculating the loss of the mass (or radius) of the hole due to HR. Only HR's effect of shrinking the hole was relevant; the radiation itself was not. My purpose was to make the hole mortal, to give me a better understanding of the time involved for the interval between the travelers being at the horizon and the singularity, to make sure no hidden infinities were involved. For this I will now need a new thread about relativistic speeds (Lorentz transformations) vs. free fall vs. escape velocities. Any suggestions?

Last edited: Nov 3, 2015
13. Nov 3, 2015

### Staff: Mentor

No, they can't, because the two observers will not see the same behavior of the mass. The outside observer will see the mass and radius (more precisely, the area of the horizon) of the hole slowly decrease. The infalling observer will see the mass as basically constant during his fall (meaning, between the horizon and the singularity), and he has no way of even measuring the radius of the hole (more precisely, the area of its horizon). The decrease in mass of the hole due to Hawking radiation is never seen by the infalling observer, because he hits the singularity and ceases to exist before any of that information can reach him.

14. Nov 3, 2015

### zylon

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The outside observer would also not see the hole's mass decrease, if he is destroyed in one hour (i.e. the time you said the traveler would exist in his own time frame inside the hole).

Since my entire question is based on "when" the traveler hits the singularity, saying that the reason the traveler cannot see the hole shrink is because he hits the singularity "too soon" is so circular that there seems to be a problem in communication here.

If the traveler calculates the same mass for the singularity after his hour of being in the hole that the outside observer does after his one hour after seeing the traveler approach close to the horizon, then time as reckoned by the "BH shrinkage clock" can be considered similar in both frames. Or at least there would be no contradiction if considered so.

Am I still in error?

15. Nov 3, 2015

### Staff: Mentor

One hour for the outside observer does not have any relationship to one hour for the infalling traveler. They are traveling on separate paths in spacetime and there is no invariant way to "match" times on one path with times on another.

That said, if the outside observer were destroyed after one hour of time by his own clock, yes, of course he wouldn't see the hole's mass decrease. But that's irrelevant, for the reason I just gave.

And the only answer to that question, which you don't seem to be getting, is that the question has no meaning in any absolute sense. The only "when" that has any meaningful answer at all is "when" according to the traveler's own clock, and that answer is only valid on the traveler's own worldline. It isn't valid anywhere else.

No, it isn't circular. But your insistence on phrasing the question in terms of "when", instead of looking at things geometrically, as I asked you to a number of posts ago, is hindering your understanding. Geometrically, the reason the infalling traveler cannot see the hole shrink is simple: no light signals from the region of spacetime where the hole is shrunk ever intersect his worldline. That is a geometric fact about the spacetime and curves within it. It has nothing to do with any "when".

No, it cannot. This has been pointed out to you multiple times. Please do not keep repeating this error.

Yes, there would. See above.

Yes. See above. And I strongly advise you to think hard about how to abolish the word "when" from your vocabulary, at least for this discussion, and learn how to view all this geometrically.