# BH fall time = pi*m. What is external time?

1. Sep 19, 2011

### keithdow

Problem 17.3 in the "problem book in relativity and gravitation" states:

"Show that once a rocket ship crosses the gravitational radius (horizon) of a Schwarzschild black hole, it will reach r=0 in a proper time tau less than or equal to pi*m, no matter how the engines are fired."

My question is the proper time is at most pi*m. What time will elapse for an external observer at infinity during the pi*m proper interval? How do I show that?

Thanks.

2. Sep 19, 2011

### pervect

Staff Emeritus
It should be pretty clear that the observer at infinity won't be able to synchronize their clock via the Einstein convention.

Simultaneity is relative, the Einstein clock synchronization convetion is the closest thing we have to a standard, and it can't be applied to this case.

So I don't think the question has any definite answer.

3. Sep 19, 2011

### keithdow

Actually the following article does have an approach for some problems.

http://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates

The time of travel for light shining inward from event horizon to the center of black hole can be obtained by integrating the equation for the velocity of light, The result is 0.77M.

So if they can do it for light, why can't it be done for other objects?

4. Sep 19, 2011

### Passionflower

Not sure if I follow you. What can't be done?

The formula to calculate the proper time for an observer from a given r value starting with zero local velocity is:

$$1/4\,\pi \,r\sqrt {2}\sqrt {{\frac {r}{m}}}$$

In case r = 2m (in the limit) then we have indeed:

$$m\pi$$

If the observer is already at escape velocity when at r then the formula becomes:

$$1/3\,m\sqrt {2} \left( {\frac {r}{m}} \right) ^{3/2}$$

In case r = 2m then we get:

$$4/3\,m$$

In this case the observer has to travel a proper distance of:

$$m\pi$$

Last edited: Sep 19, 2011
5. Sep 19, 2011

### keithdow

Can we agree that the external time will be greater than m*pi? The question then becomes how much more.

6. Sep 20, 2011

### pervect

Staff Emeritus
You seem to think there is some unique notion of "external time". But it's not clear how you define this notion operationally, (i.e. how would you go about measuring it?) or what characteristics it should have.

If you can define how you'd measure "external time", we might be able to shed some light on the question or maybe even aswer it. If you can't define how you'd measure it, but can give some idea of why you want to know it, what purpose it serves,, we might be able to do something as well.

WHen I/we say "simultaneity is relative", does that mean anything to you? I suspect from your questions that you don't understand that, and as a result we wind up talking past one anonther.

Because if you accept that simultaneity is relative, then it should be clear why there isn't necessarily a unique answer to your question, similarly to the reason there isn't a unique answer to the aging of the twins in the twin paradox when they are separated.

7. Sep 20, 2011

### pervect

Staff Emeritus
I'd focus on asking whether that particular approach was unique. As far as I know, there are an infinite number of coordinate systems you could adopt, all of which would give different answers. And I'm not aware of any compelling (or even non-compelling) reasons to chose one coordinate system over another.

Coordinate choices are like labels on a map. They're human inventions, without any physical significance. So you can label an event "B4' or "C7", the physics won't change.

8. Sep 20, 2011

### keithdow

Let me make the question even simpler. I am at rest 10,000 Schwarzschild radii away from the event horizon of a black hole. I have a reference clock. I release an object that falls to just above the event horizon and then stops there for an instant then falls in. If time starts when I drop the object, at what time does it reach the event horizon? At what time does it reach the singularity? All times are measured from my reference clock.

9. Sep 20, 2011

### PAllen

Do you understand that your reference clock only represents time in its vicinity? Your question cannot be given any unique meaning. I assume you are aware that you would never see the object you dropped cross the event horizon.

Let me give you an example of something that could be calculated, though there is nothing uniuqe about it:

A stationary observer 10,000 Schwarzschild radii away drops clocks at regular intervals. Independent of whether this observer can communicate with the clocks, he defines (arbitrarily) that each dropped clock's reading is simultaneous with the time it was dropped plus its own time interval since dropping. In this arbitrary coordinate scheme, it is then possible to state when each clock reaches the singularity. This would be done purely by calculation. There is no possible measurement consistent with this definition of simultaneity (e.g. across the event horizon).

10. Sep 20, 2011

### Passionflower

Correct, however what we can determine is how much proper time the clock takes to go from the stationary position 10,000 Schwarzschild radii away all the way to the EH and also from the EH to the singularity.

In this case we have resp 1,570,794.756 * rs and 1.570796327 * rs seconds

Last edited: Sep 20, 2011
11. Sep 22, 2011

### Phrak

I believe I understand what you are trying to saying.

When you say "All times are measured from my reference clock," you mean that you will map your local inerital coordinate system over all space time, or at least as much to include the black hole horizon. This is a commonly used mapping by cosmologists.

In this coordinate system, the answer to "how long it will take?" is 'forever', which is why the horizon is called a coordinate singularity in the first place. The relationship between metric intervals between these coordinate maps is infinite.

Also, in any well behaved coodinate system exteral the horizon, the round trip time is infinite. That is, there is no loop causal relationship between an external object and a black hole unless the blackhole has existed forever. This is a time interval somewhat longer than 15 billion years, Earth time.

So if cosmologists want to keep saying things about what this-or-that black holes is doing, they should either claim blackholes have existed longer than the age of the universe, or say they are deviating from their standard adopted clock and using one, instead, where the age of the universe is infinite.

At this point, objections might be raised about primordial black holes or black holes resultant from 'quantum fluxations'. Again, it takes infinite time, as measured on Earthly clocks, for them to grow to solar mass.

Proceed with caution! Infinity is long time, and has been known to make many people very angry in these forums. Never use the synonym 'never' for an infinite time interval when discussing black holes among black hole enthusiasts.

Last edited: Sep 22, 2011
12. Sep 22, 2011

### keithdow

Let me make my question clearer.

I have put this problem on a computer that runs a simulation of two observers. The first observer is at rest far away from the black hole. The second observer is just above the event horizon of the black hole. Computers have to run simulations in lock step time or else you get unphysical phenomena. At computer time t=0 we release the plunging observer and set his clock to zero. Also the first observer has his clock set to zero. When the second observer in the simulation reaches the singularity his clock reads pi*m. We stop the simulation at that point. What time is on the first observers clock?

13. Sep 22, 2011

### PAllen

As has been explained several times, the first observer can never detect or be influenced by anything over the event horizon. You computer couldn't be built. If a wire is extended across the event horizon and one end anchored stationary, and assuming perfect ductility so it can stretch forever, it would be pulled thinner and thinner; no current or signal could pass up it across the horizon.

I can see there is a sort of inverse question that can be asked. If the crossing clock follows a path of maximum proper time to the singularity, starting from e.g. 1.5 time event radius, at which point we say it sees a reading of t=0 on a clock at 1000 radii, what is the last time it sees on the distant clock as it reaches the singularity? This is perfectly computable. I am not in a position to do it now, but maybe someone else will. This is the closest I can get to what you seem to be seeking.

[EDIT: minor correction: from 1.5 radii, there is no maximum proper time path because you could hover any amount of proper time before starting to fall. To make concretes, assume free fall from 1.5 radii, followed by max proper time from crossing of event horizon.]

Last edited: Sep 22, 2011
14. Sep 22, 2011

### keithdow

Computers to simulate black holes have been around for decades. Entering a black hole has been simulated for decades. I can put this problem on a computer and get an answer. Has anyone done it though?

15. Sep 22, 2011

### PAllen

You don't understand what the computers are simulating. This is an issue of understanding GR, not technical issues. In fact, it is not clear to me that you understand SR, because you have a concept of global simultaneity which has been rejected since 1905.

16. Sep 22, 2011

### DrGreg

In general relativity, there is no generally agreed definition of when two events, separated by a distance, occur simultaneously or not. You can choose a particular coordinate system, and then there will be an answer relative to those coordinates, but there is no way of saying one coordinate system is "right" and all the others are wrong. All the answers are equally right.

Having said all that, for a non-spinning black hole, the coordinates often used are Schwarzschild coordinates. If we choose to answer your question in those coordinates then what happens is this. It takes an infinite time, measured in external Schwarzschild coordinates, for your first observer to reach the horizon. So the answer is "more than infinity"! This is because external Schwarzschild coordinates don't extend inside the event horizon. Inside, you can use internal Schwarzschild coordinates, but they have no relation to the external coordinates.

17. Sep 22, 2011

### pervect

Staff Emeritus
Basically, if you can never receive a light signal from an event, you won't really know that it happened, and you'll have a great difficulty in assigning any meaingful time when it happened.

However, this doesn't mean that it didn't happen. The clearest example involves horizons, called Rindler horizons, similar to the black hole horizons, that occur for accelerating observers.

It's possible for someone to accelerate so that they won't see any events lager than 1 year after they started on their trip. At that time, the Earth falls behind their "Rindler horizon". But if such an observer were to leave the Earth, it wouldn't mean that time would stop in 1 year on Earth or anything. It would just mean that the accelerating observer wouldn't be able to see anything happen later than 1 year after their departure that occured here on Earth.

So, an observer outside the black hole would not ever see an event occur beyond the event horizon. However, said event would be quite real for the observer falling in, and would occur in a finite time for them.

18. Sep 22, 2011

### keithdow

Actually the idea of global simultaneity is fundamental to 3+1 formalism. Let me quote from "Introduction to 3+1 Numerical Relativity". Page 65.

"Let us start by considering a spacetime with metric gαβ. As already mentioned
in Chapter 1, we will always assume that the spacetimes of interest are
globally hyperbolic, that is, they have a Cauchy surface. Any globally hyperbolic
spacetime can be completely foliated (i.e. sliced into three-dimensional cuts) in
such a way that each three-dimensional slice is spacelike (see Figure 2.1). We
can identify the foliation with the level sets of a parameter t which can then be
considered a universal time function (but we should keep in mind that t will not
necessarily coincide with the proper time of any particular observer). Because
of this fact, such a foliation of spacetime into spatial hypersurfaces is often also
called a synchronization."

19. Sep 22, 2011

### keithdow

Thanks. It is obvious that I will have to write the computer program and run the simulation myself to get the answer. I can probably get a paper out of it.

20. Sep 22, 2011

### PAllen

Your quote of this shows your misunderstanding. As noted in the quote itself, this foliation does not represent global simultaneity for any observer, nor does it represent time according to any possible single clock (in general).

I did propose a way you could arbitrarily correlate the distant clock and the infalling clock from the point of view of the infalling clock. This is possible. It is not possible (in any meaningful way) to do it from the point of view of the distant clock. This asymmetry is coupled to the definition of an event horizon.

[EDIT: note that the described foliation is incredibly non-unique with no way to prefer one slicing over another. In any manifold where you can do this slicing, there are generally aleph-2 distinct ways to do it, with no way to prefer one over the other - because (unlike Lorentz simultaneity in flat spacetime) there is no physical basis to prefer one over the other.]