Existence of black holes

By Sean Hartnoll (Sah40) on Thursday, February 10, 2000 - 03:00 pm :

From an external perspective, the collapse of star to form a black hole never happens because as the collapsing star approaches the event horizon it goes slower and slower (as seen externally, for an infalling particle it all happens in a finite time).

How, then, can any black holes exist at all, because from our external viewpoint, they could not have formed yet given the finite age of the universe?

Sean

By Anton Ilderton (Abi20) on Friday, February 11, 2000 - 05:01 pm :

Go here:

NCSA Numerical Relativity

Anton.

By Michael Doré (P904) on textbfMonday, February 21, 2000 - 04:42 pm :

Hi!

I'm not at all sure about this but what about the following... You're saying that the matter in a collapsing star would move asymptotically closer to the "would-be" event horizon according to an external persperctive but it will never get there because of time dialation. If something is now dropped into the collapsing star then it too must also experience the same effects and stop outside the event horizon. So as far as someone outside the event horizon is aware the black hole is behaving exactly as black holes should - they are slowing down matter. The fact that the singularity will not form for an infinite time is possibly irrelevant - what do you think?

Thanks,

Michael

By Sean Hartnoll (Sah40) on Monday, February 21, 2000 - 05:33 pm :

Michael, I think your answer captures the point precisely. It is what I was starting to think, but you have expressed it particularly clearly. I had been thinking in terms of a photon emitted near the forming event horizon: it would be so redshifted that it would look black anyway.

The point is that a nearly black hole does behave like a black hole for an external observer, which, as you say, is what is relevant. It is sort of interesting though that "textbook" black holes don't strictly exist.

Sean

By Michael Doré (P904) on Wednesday, February 23, 2000 - 09:47 am :

Yes, I'm quite surprised that I hadn't heard this before. So what we are saying is that if you fall into a black hole then it doesn't become a black hole until you pass the event horizon. At this point your own time will have been dialated sufficiently to give the star a chance to collapse fully. Still I can't really expect to have much idea about what is really happening because I don't understand (or haven't yet learnt) any of the maths behind general relativity. Even so it's an interesting problem.

Michael

By Sean Hartnoll (Sah40) on Wednesday, February 23, 2000 - 01:26 pm :

The time effect isn't that complicated in itself, when you solve Einstein's equations for a spherical situation such as a star or black hole you get the Schwarzchild solution, which for a radially infalling photon (easy case) means

0 = ds² = (1 - r_s / r) dt² - (1 - r_s / r)^-1 dr²

So

dt/dr = (1 - r_s / r)

you can see that as r gets close to r_s (the Schwartzchild radius) this is going to blow up. And by integrating the equation we see that in fact it takes an infinite time to reach the horizon. However if we transform the equations into a frame falling into the hole the result turns out to be finite.

By Michael Doré (P904) on Saturday, February 26, 2000 - 05:42 pm :

Wow, that really is a simple result! I can understand why the time turns out to be finite from a free-falling frame. How drastically does this equation alter if the velocity is not entirely radial (I don't suppose you can consider each dimension separately...)? I've often wondered whether it would be possible to sling shot around a black hole through a region behind the event horizon. In terms of energy there is no particular reason that you cannot do this. If the universe is destined to expand forever and you actually tried to do this experiment then when you come out an infinite amount of time would have had to ellapsed for the rest of the universe! If the universe is closed then of course if you fall into a black hole you will simply be immediately at the big crunch (because while your time is dialated the rest of the matter in the universe will rush in).

Many thanks,

Michael

By Sean Hartnoll (Sah40) on Saturday, February 26, 2000 - 06:19 pm :

If the velocity is not radial there are more terms in the equation:

${ds}^{2} = (1 - r_{s} / r) {dt}^{2} - (1 - r_{s} / r)^{- 1} {dr}^{2} - r^{2} (d θ^{2} + \sin^{2} (θ) d ϕ^{2})$

Where $θ$ and $ϕ$ are angles. Incidently ${ds}^{2}$ is called an interval, and is a generalisation of the interval in special relativity:

${ds}^{2} = {dt}^{2} - {dx}^{2} - {dy}^{2} - {dz}^{2}$

(here I am using units in which $c = 1$ , this is standard, to avoid $c$ 's appearing everywhere).

The interval is fundamental as it is always the same for different observers. It is like a four-dimensional distance except that $t$ has a different sign than space. This is useful for thinking about it: it works in exactly the same way as the fact that the distance between two points in normal 3D space remains the same under a rotation.

We can find the time for infall in a similar way to above, but now the integation is much more complicated (normally you would do it by computer). You can also do it for massive particles, in which we don't have ${ds}^{2} = 0$ , but again it is more complicated. The essence is captured in the radially infalling photon case I gave you earlier.

What you say about coming out an infinite amount of time having passed is true. One way of travelling into the future would be to go up close to a black hole and then come back. Except that you must not go pass the horizon. Once you pass the event horizon there is no way back. This can be shown relatively straightforwardly from the equation above, but I think it is more useful if I show you a far deeper reason, which often isn't mentioned but is very fundamental:

in the equation for the interval, the $t$ -part has a different sign. However, note that when $r < r_{s}$ (i.e. when you are inside the event horizon) then the sign of the $t$ -part and the $r$ -part flip because $1 - r_{s} / r < 0$ . So now space will behave like time and time will behave like space. But time has the characteristic that it somehow "flows" forwards you can't go back. That is now happenning to space inside the black hole!! The SAME mysterious force that causes time to pass is now causing you to fall always inwards to the black hole, unable to turn back!! Think about this. It is very profound.

Sean

By Michael Doré (P904) on Saturday, February 26, 2000 - 08:43 pm :

Many thanks for enlightening me on this subject. I have found some info on this in one of my books: "Companion to the Cosmos" (by John Gribbin), page 371. It states that in 1965 Roger Penrose proved "that anything which falls into a non-rotating black hole must be crushed into the singularity at the centre of the black hole. There is no way for material to travel in an orbit that sends it looping past and around the singularity." (That was exactly what I was wondering by the way.) This leaves me confused - isn't the Schwarzchild solution (discovered in 1916) supposed to prove this?

I am about half way towards understanding your argument that nothing can escape from within the event horizon. Aren't the GR equations time symmetric? Also I don't understand the concept of time flowing forwards. It is true that entropy increases in one direction of time only but this is not due to any asymmetry in the laws of physics (I can show that the laws of Newtonian gravitation, Coulomb's electrostatics and classical magnetism are time symmetric. Apparently GR and QM are as well though of course I can't show this!)

The reason for the asymmetry is surely due to the universe having asymmetric boundary conditions. For some reason we know that 10-20 billion years ago the entropy of the universe was zero at the big bang. We have no such condition about any time in the future. Therefore statistically entropy must increase away from the big bang. This causes the perceived arrow of time.

So I am slightly confused by the "force that causes time to pass". Perhaps the argument is that while reversing the arrow of time requires an entropy decrease (which is statistically unlikely), here because space is acting like time and vice versa, reversing out of a black hole is equally unlikely (or impossible?).

Many thanks for your time and patience,

Michael

By Sean Hartnoll (Sah40) on Saturday, February 26, 2000 - 10:18 pm :

Your points are good. Some comments on them:

1) About the black hole. The wording from the book is slightly misleading. It is true that the Schwartzchild solution shows that you can't have orbits escaping from inside the event horizon. And this was known long before the Penrose result. I think the key idea in Penrose is that of "singularity". What Penrose proved (with Hawking) is, I think, that if you have something behaving like a black hole then you must have a singularity of spacetime inside. The Schwartzchild solution does have a singularity in it, but I think the Penrose result is more general.

2) About time symmetry. Actually I think the GR equations are only time symmetric under certain conditions. Also, although the Schrodinger equation is time symmetric, and this describes the evolution of the wavefunction in QM, what is not symmetric is the collapse of the wavefunction. So it is not really true to say that GR and QM are time symmetric, although this becomes quite a complicated issue.

3) I am sorry that my wording was quite bad and I should have used inverted commas to talk about "time flowing" and there certainly isn't a "force" in the usual sense of the word. I wasn't meaning to give a rigurous argument but just an idea. I don't think the question is properly understood anyway. The point is that if you just don't do anything time pases, you can't set up a coordinate system in which time stays still (unless you are a photon actually, travelling at the speed of light they have no time). Space on the other hand you can set up to be still. This is all a bit wolly but I think it is clear that there is a sense in which time flows and space doesn't. What I am saying is that once you pass the event horizon, these roles get inverted, and space flows, and it drags you into the hole in the same way time drags you into the future. Perhaps a proper discription of this process will require entropy, I don't know, and I don't think anyone does.

4) There is a book on special and general relativity, called "Introducing Einstein's Relativity" by Ray D'Inverno which is quite easy going and only assumes basic calculus and differential equations and developes all the maths need for GR in a straightforward way. If you have time, you might find it interesting. I think you would find it accessible, with some work of course. There is something nice about seing the math actually happen.

Sean

By Michael Doré ((P904) on Sunday, February 27, 2000 - 06:09 pm :

Thanks, Sean.

I hadn't thought about the collapse of the wavefunction before in the context of time symmetry. I'll take your word that this is not reversible (although I was sure I remember reading Feynman saying that the only law in physics that may not be time reversible is the weak force and beta decay - maybe he was ignoring the collapse).

I take your point about time not being able to stand still. It is certainly true that the world-line of all particles other than photons must occupy every point on the time axis once and only once. The same cannot be said of the space axes. I think that a full description would have to involve entropy, as the "direction" time occurs in is not well-defined. (Of course later events in time do happen later but this is just tautological.)

Thanks for your help (and I'm sorry it's you answering all the questions despite the fact that you started this discussion),

Michael

By Yvonne Teng (P2396) on Friday, April 21, 2000 - 10:28 am :

I think it depends on whether u believe it or not. If you all want a good book on time, energy,space, i suggest " A Wrinkle In Time", a Newberry award winner.