Black Holes - the two points of view.

[Mike Holland]

Roughly speaking, you can divide the spacetime up into three regions, the interior of the future light cone, the interior of the past light cone, and the exterior of the light cone. Events that will exist are in the first region, events that did exist are in the second, and all other events can be considered to exist "now", in a coordinate independent sense. In the Schwarzschild spacetime, the interior of the black hole is in the second and third regions, so it should be considered to exist now and in the future. If anything, you would say only that it did not exist in the past.

Slight confusion here. No light can leave the interior of a Black Hole, so how could any light from the interior be in my or your past light cone (second region)?

Similarly, you can choose the past light cone for any point in the third region, and nothing will have entered this cone from the inside of an event horizon.

So the inside of a black hole is not in the past light cone of any observer remote from the black hole.

The inside of a black hole is never in the past light cone of any observer, now or in a trillion years time, unless the observer falls into the black hole. The inside exists only for the inside. It does not exist for outside observers.

Mike

 

[Mike Holland]

But NOW does not have an invariant meaning; it depends on your choice of simultaneity convention. There are ways to choose a simultaneity convention so that the black hole at the center of the Milky Way galaxy exists NOW.

If the hole exists now, then it should be in our past light cone in about 27,000 years. Now you please tell me how long it takes for a photon to escape from the event horizon and become part of our past light cone. What simultaneity convention will get it here in 27,000 years?

Edit: I think you will find that as the supermassive object at the centre of the galaxy enters our past light cone, the cone becomes distorted - photons don't simply fly from there to here at c anymore, and photons from the EH never get here.

Mike

 

[Dale]

Slight confusion here. No light can leave the interior of a Black Hole, so how could any light from the interior be in my or your past light cone (second region)?

Oops, good catch. I messed up my own numbering. I meant that the interior was in the first and third regions (future and present), not second and third (past and present). I have corrected it above.

The inside exists only for the inside. It does not exist for outside observers.

No, remember your emphasis on the proper tense. For outside observers the inside exists in the present and will exist in the future. For outside observers it did not exist in the past.

 

[Dale]

If the hole exists now, then it should be in our past light cone in about 27,000 years.

No. Its existence or non existence "now" has nothing to do with some future events past light cone. It has only to do with whether or not it is in the manifold in the region which is space like separated from our present event. Chances are that we will not even exist in 27000 years to have a past light cone.

 

[Mike Holland]

No. Its existence or non existence "now" has nothing to do with some future events past light cone. It has only to do with whether or not it is in the manifold in the region which is space like separated from our present event. Chances are that we will not even exist in 27000 years to have a past light cone.

You are trying to ignore time and only look at space separation to decide whether something exists "now". But from this god-like perspective where time is just another dimension, all things exist in the eternal "now".

All the calculations show that there is a separation of infinite time in addition to the separation of finite space between an event horizon and its insides, and the remote observer. Because you insist on believing in Black Holes, you choose to ignore this. Do you not believe in gravitational time dilation?

As I have said before, I am looking at ther world from our perspective, and light cones are relevant in deciding what is in our past and what will be in our past. The insides of Black Holes willl never be in our past light cones (or anyone else's), wherever they might exist in the manifold, as long as we don't make the stupid mistake of falling into one.

Mike

 

[Dale]

You are trying to ignore time and only look at space separation to decide whether something exists "now". But from this god-like perspective where time is just another dimension, all things exist in the eternal "now".

This is incorrect, there is not just one region. As I explained before there are three regions, "future" (interior of future light cone), "past" (interior of past light cone), and "now" (exterior of future light cone). All other causal distinctions are coordinate-dependent.

All the calculations show that there is a separation of infinite time in addition to the separation of finite space between an event horizon and its insides, and the remote observer. Because you insist on believing in Black Holes, you choose to ignore this.

Only in Schwarzschild coordinates. Not in many other coordinate systems. I am not ignoring the infinities at the horizon in the Schwarzschild coordinates, merely pointing out that those infinities do not exist in other coordinates.

You state that "all the calculations show ...", and that is simply false. Only calculations based on Schwarzschild coordinates show what you claim. Calculations based on other coordinates show that the interior does exist now.

As I have said before, I am looking at ther world from our perspective, and light cones are relevant in deciding what is in our past and what will be in our past. The insides of Black Holes willl never be in our past light cones (or anyone else's), wherever they might exist in the manifold, as long as we don't make the stupid mistake of falling into one.

Agreed. Therefore as I said before, if you insist on talking with proper tenses then you can claim that the black hole did not exist in the past. But you cannot make the claim that it does not exist now and will not exist in the future.

 

[Mike Holland]

You state that "all the calculations show ...", and that is simply false. Only calculations based on Schwarzschild coordinates show what you claim. Calculations based on other coordinates show that the interior does exist now.

Hooray! This is what I have been asking for all along. Please direct me to some publications where these calculations for a remotely observed collapsing mass have been published. I am sick of being told by everyone that they exist, without any follow-up references.

Everywhere I've looked at other coordinate systems, they all seem to be the same as Schwarzschild coordinates outside the event horizon.

Mike

 

[Dale]

I am sick of being told by everyone that they exist, without any follow-up references.

Everywhere I've looked at other coordinate systems, they all seem to be the same as Schwarzschild coordinates outside the event horizon.

I linked you right to the Wikipedia pages on 4 alternative coordinate systems in the Schwarzschild spacetime back in post 16:
https://www.physicsforums.com/showpost.php?p=4029984&postcount=16

Don't try to claim that I have been holding out on you if you didn't bother to read the references I provided.

 

[PeterDonis]

If the hole exists now, then it should be in our past light cone in about 27,000 years.

This does not follow. In a flat spacetime it would; but our spacetime is not flat, it's curved, and it's curved in a way that makes the apparent logical implication in your statement invalid.

What *is* in our past light cone is the object (or objects) that originally collapsed to form the black hole. And what will be in our past light cone 27,000 years from now is a somewhat larger portion (i.e., closer to the horizon) of the history of that object (or objects) prior to it crossing the horizon.

 

[Austin0]

Hooray! This is what I have been asking for all along. Please direct me to some publications where these calculations for a remotely observed collapsing mass have been published. I am sick of being told by everyone that they exist, without any follow-up references.

Everywhere I've looked at other coordinate systems, they all seem to be the same as Schwarzschild coordinates outside the event horizon.

Mike

Hi
I don't know enough GR to have an opinion one way or the other in this fascinating controversy but I do have a question.

Given an infalling atom. At the point where the metric indicates infinite dilation doesn't it also indicate infinite radial contraction? If this is the case then it would seem that the distinction between "at" the horizon and "in" the horizon becomes meaningless.
So larger structures would not maintain extent beyond the horizon but would, an atom at a time, become compacted to become effectively a dimensionless part of the horizon.

Based on your interpretation i get a picture of the horizon being an abstract geometric surface. A smooth, dimensionless and static boundary of absolute impenetrability.
While I can't argue against such a construct it doesn't seem as realistic as a more QM picture of a dynamic region with a certain degree of fuzzy dimensionality due to uncertainty,vacuum flux and reaction to the the changes occurring both within and without.
If there is no abstract dimensionless boundary then there is no abstract dimensionless mathematical point of infinite gamma but rather a fuzzy uncertain transition from outside to inside along the lines of electron tunneling. Just a thought.

 

[Mike Holland]

Austin0, from a remote point of view your picture is correct, but what is stopping collapse there is the stopping of time so that nothing can happen there. But remember that this is in a remote observer's timeframe. If you are falling into the collapsing mass that is becoming a Black Hole, you will not experience the time dilation because you are similarly dilated, and you will fall through the Event Horizon very quickly. The tricky part is getting your mind around time flowing very differently for different viewers, and both being equally valid.

Regarding the quantum fuzziness, I am stll waiting for a theory of quantum gravity to resolve this. Does the time dilation quell the fuzziness, or does the fuzziness make the location of the Event Horizon indeterminate? I don't know.


Dalespam, I have read the references you provided. Einstein-Finklestein coordinates are the same as Schwarzschild coordinates outside the Event Horizon, so they make no difference to the prediction that gravitational collapse will take an infinite time in a remote viewer's reference frame. The other systems just assign different scaling factors to space and time so that the numbers look different. It is like setting T = 10 + 1/t, so that a Black Hole forms at 10 o'clock, and my new "time" is continuous through 9, 10, 11 0'clock. They don't at any point show that the Schwarzschild calculations are incorrect. My T makes no difference to t being infinite.

Anyway, I was not asking for different coordinate systems to look at the situation. I was asking for actual calculations done using these coordinates to show that a Black Hole would form in a finite time, contrary to Schwarzschild. You have not provided this, and I don't believe any such calculation has been done.


And last, but not least, PeterDonis :
"This does not follow. In a flat spacetime it would; but our spacetime is not flat, it's curved, and it's curved in a way that makes the apparent logical implication in your statement invalid.
What *is* in our past light cone is the object (or objects) that originally collapsed to form the black hole. And what will be in our past light cone 27,000 years from now is a somewhat larger portion (i.e., closer to the horizon) of the history of that object (or objects) prior to it crossing the horizon."

That is more-or-less the point I was trying to make. If our "now" is defined by our past light-cone, then a Black Hole will never enter it, because of the time delay of photons escaping from near the Event Horizon. In this sense, Black Holes do not exist because we can never see them. But this ignores the added factor or time dilation which prevents the Black Hole from forming in a finite time anyway.

"But NOW does not have an invariant meaning; it depends on your choice of simultaneity convention. There are ways to choose a simultaneity convention so that the black hole at the center of the Milky Way galaxy exists NOW."

In simple, practical terms, "now" is my past light cone, but in a more theoretical sense it is a line drawn perpendicular to my world line in a space-time diagram. As you say, I need a simultaneity convention to determine what points/events lie on this line. Or in other words, I need to choose my coordinate system before I can draw my vertical line. But I challenge you to invent a convention that will change the infinity that comes out of Schwarzschild coordinates into a finite time. Multiply it be anything you like (except zero!), add or subtract anthing you like, and it will still be infinity.

Mike

 

[PeterDonis]

At the point where the metric indicates infinite dilation doesn't it also indicate infinite radial contraction?

No. But rather than use ambiguous English words, it's better to describe the math. In Schwarzschild coordinates, at the horizon, r = 2M, the metric coefficient g_tt goes to zero, and the metric coefficient g_rr goes to infinity.

"Infinite time dilation" refers to g_tt going to zero, but that's really a misnomer; a better description would be to say that at the horizon, a line element with only dt nonzero is not timelike any longer; it's null, or lightlike. (Strictly speaking, in Schwarzschild coordinates the line element is undefined, since g_rr is infinite, but there are other ways to describe "a line element on the horizon with only dt nonzero" that don't depend on Schwarzschild coordinates. One way is to look at the vector \partial_t, which can be defined as a vector field on the spacetime independent of coordinates, and see that that vector becomes null at the horizon, instead of timelike.) Since such a line element is no longer timelike at the horizon, it doesn't describe a "time interval" at all.

"Radial contraction" is also a misnomer; g_tt going to infinity does *not* mean that physical radial distances are "infinitely contracted". It means that, as you get closer and closer to the horizon, a given increment of radial coordinate dr corresponds to more and more physical radial distance. But it also means that, as you get closer and closer to the horizon, a line element with only dr nonzero gets closer and closer to being null instead of spacelike. (Or, equivalently, the vector \partial_r gets closer and closer to being null.) And once again, since such a line element is no longer spacelike at the horizon, it doesn't describe a "spatial length" at all.

To actually find out what happens to objects that reach the horizon, or fall inside, you have to stop using Schwarzschild coordinates, and start using some method of description that doesn't have these issues at the horizon. One option that fixes the "spatial length" problem is Painleve coordinates: the Painleve radial coordinate is spacelike all the way down to r = 0. The only well-known chart I'm aware of that has a coordinate that's timelike all the way down to r = 0 is the Kruskal chart. But there are other ways of describing timelike vectors below the horizon.

 

[PeterDonis]

That is more-or-less the point I was trying to make. If our "now" is defined by our past light-cone

It is, in the sense that our "now" at any given event consists of those events that are *outside* our past light cone, and also outside our future light cone.

then a Black Hole will never enter it, because of the time delay of photons escaping from near the Event Horizon.

No, you're getting the definition of "now" mixed up. See above. The correct statement is that the region inside the horizon can never be in our past light cone if we are outside the horizon. A portion of the region inside the horizon is in our future light cone, at least as long as the BH is not close to finally evaporating; the rest of the region inside the horizon is in our "now".

In simple, practical terms, "now" is my past light cone

No, it isn't. See above.

But I challenge you to invent a convention that will change the infinity that comes out of Schwarzschild coordinates into a finite time.

I don't have to invent any; there are already at least three that I know of: ingoing Painleve coordinates, ingoing Eddington-Finkelstein coordinates, and Kruskal coordinates. All of those charts embody a simultaneity convention that removes the infinity at the horizon and includes many surfaces of simultaneity that cross the horizon.

 

[Austin0]

Austin0, from a remote point of view your picture is correct, but what is stopping collapse there is the stopping of time so that nothing can happen there. But remember that this is in a remote observer's timeframe. If you are falling into the collapsing mass that is becoming a Black Hole, you will not experience the time dilation because you are similarly dilated, and you will fall through the Event Horizon very quickly. The tricky part is getting your mind around time flowing very differently for different viewers, and both being equally valid.

Regarding the quantum fuzziness, I am stll waiting for a theory of quantum gravity to resolve this. Does the time dilation quell the fuzziness, or does the fuzziness make the location of the Event Horizon indeterminate? I don't know.

Mike

You seem to be contradicting yourself. On one hand you posit nothing can pass the horizon because time dilation becomes infinite and coordinate velocity becomes c which is zero at that point. Then you talk about about an observer passing through the horizon very quickly.

Assuming this observer was conscious he might not be aware of the time dilation ,but as it is a feature of the spacetime geometry at that location, it is not that it would not be in effect.

Infinite dilation. To me a single instant that is infinitely dilated, I.e extended in duration, is totally equivalent to a clock ticking away to an infinite reading. They both take effectively forever. Which is a longer infinity? , perhaps Cantor could say.

it seems to me that reaching this point is equivalent to actually reaching c in flat spacetime. Time no longer has any meaning. Whether or not this observer is moving or not would not be internally determinable if both his clock and brain activity had come to a stop.
So if we assume that motion stops at the horizon,collapsing matter etc. this would seem to have to include your observer. So we might say he is trapped there for eternity but when the end of days comes and he passes inside it will have happened quickly , having taken zero time by his stopped clock?
Or are you saying that matter does pass the horizon and black holes do form in a short finite time but that it is unobservable from the outside?
About your idea that time dilation would retard or prevent radial contraction at the boundary.
You may be right but I would think that both dilation and contraction were effects of the underlying geometry which is the cause. SO contraction would occur simply because of being at that location wrt the geometry and would not be subject to slowing through dilation which is just another independent effect of that geometry.

Well thanks for a very interesting and provocative thread.

I have no idea of the answer , my only belief is that there would be a singular dynamic geometric entity and so there would be a singular actuality independent of observation and coordinate choices.

 

[Mike Holland]

I have no idea of the answer , my only belief is that there would be a singular dynamic geometric entity and so there would be a singular actuality independent of observation and coordinate choices.

Sorry, but physics just doesn't work that way. Time passes differently for different observers. Our clocks run slow because of Earth's gravity. A clock in a high-altitude plane or a satellite runs faster. This has been measured. And this is what is predicted by Einstein's theories.

Mike

 

[Austin0]

Sorry, but physics just doesn't work that way. Time passes differently for different observers. Our clocks run slow because of Earth's gravity. A clock in a high-altitude plane or a satellite runs faster. This has been measured. And this is what is predicted by Einstein's theories.

Mike

`Yes of course this is simple fact. What could lead you to believe that I was not aware of or questioned this obvious reality.
Do you somehow question that this is the observable manifestation of a singular underlying geometry?
I am curious why you ignored everything in my post except a casual parting aside which you seem to have reinterpreted to have absurd implications.

 

[Dale]

Einstein-Finklestein coordinates are the same as Schwarzschild coordinates outside the Event Horizon

I don't know how you could possibly have actually read the link and gotten this so egregiously wrong.

T=t+R \; ln \left| \frac{r}{R}-1 \right| \neq t

The other systems just assign different scaling factors to space and time so that the numbers look different. It is like setting T = 10 + 1/t, so that a Black Hole forms at 10 o'clock, and my new "time" is continuous through 9, 10, 11 0'clock.

Yes, that is exactly the point. Your objection amounts to a complaint that some arbitrary number, t, goes to infinity, but that arbitrary number has no physical significance and can therefore be replaced by equally arbitrary number, T, which does not go to infinity.

They don't at any point show that the Schwarzschild calculations are incorrect. My T makes no difference to t being infinite.

The Schwarzschild calculations are correct, t is infinite. But t is a completely arbitrary label and has no physical significance.

Do you believe that changing coordinate systems can make something start or stop existing?

Anyway, I was not asking for different coordinate systems to look at the situation. I was asking for actual calculations done using these coordinates to show that a Black Hole would form in a finite time, contrary to Schwarzschild. You have not provided this, and I don't believe any such calculation has been done.

The Schwarzschild spacetime is static, representing a static eternal black hole, not the formation of a black hole. None of the coordinates listed, neither Schwarzschild nor the others, describe the formation.

However, what you can show in the Schwarzschild spacetime is that a small amount of matter can indeed fall in and cross the EH. If you use Schwarzschild coordinates in the Schwarzschild spacetime you find that t is infinite. In the other coordinate systems you find that T is finite. In all coordinate systems you find that τ is finite.

 

[Mike Holland]

So if we assume that motion stops at the horizon,collapsing matter etc. this would seem to have to include your observer. So we might say he is trapped there for eternity but when the end of days comes and he passes inside it will have happened quickly , having taken zero time by his stopped clock?
Or are you saying that matter does pass the horizon and black holes do form in a short finite time but that it is unobservable from the outside?

Going back to your earlier post, my answer is YES to all those points. Yes, the falling observer would take an infinite time to reach the event horizon - in OUR reference frame. Yes, he wouild fall through very quickly, in HIS reference frame. We "see" him trapped there for eternity, ever edging closer to the EH. But he does not then fall through quickly as our clocks tick over to infinity. That is meaningless. His clock is never stopped, in either reference frame, because our clocks never "reach" infinity. It is like the old quandary "Do parallel lines never meet, or do they meet at infinity?". I'll know the answer when I get to infinity.

Mike

 

[Mike Holland]

It is, in the sense that our "now" at any given event consists of those events that are *outside* our past light cone, and also outside our future light cone.

OK, I see how you are using the word "now", to include ALL events that COULD be "now" for us, depending on our chosen coordinate system. I would have called them "indeterminate". As for the BH being there, I am still convinced that the eternally collapsing object is there, but not the BH yet.

A portion of the region inside the horizon is in our future light cone, at least as long as the BH is not close to finally evaporating; the rest of the region inside the horizon is in our "now".

I see that a portion of the BH can be in our future light cone - I could chuck something into it. But good old time dilation still comes into the picture, and just as our past light cone is distorted by the event horizon, so is our future one. The edge of my future light cone will approach the EH asymptotically (MY time frame, remember) and only reach it at t = infinity.

I don't have to invent any; there are already at least three that I know of: ingoing Painleve coordinates, ingoing Eddington-Finkelstein coordinates, and Kruskal coordinates. All of those charts embody a simultaneity convention that removes the infinity at the horizon and includes many surfaces of simultaneity that cross the horizon.

Painleve coords are based on a moving reference frame (the raindrop), and that is not MY coordinate system. They do not show the Schwarzschild coordinate system calculations for a remote observer to be false.

Look at the lines of equal Schwarzschild time in the Eddington -Finkelstein diagram, and you will see they all well up asymptotically from the event horizon in the infinite past. You will have to crossd an infinity of Schwarzschild time lines to get to the EH.

So none of these reference frames contradict calculations based on Schwarzschild coords. The only question is, what time coordinates is my wristwatch based on? V = root(r/2GM - 1)? Or those of the falling raindrop?

If you think something can fall through the event horizon in a finite time, how does it avoid the time dilation? Does it go round the back of the BH where we distant observers can't see it fall in? :)

Mike

 

[PAllen]

Mike, do you insist that seeing an Einstein ring means stars must be considered smeared into a ring by an observer that sees this? The effect BH on freezing light is physically the same phenomenon - just gravity bending or freezing light. There is no physical basis at all for interpreting this frozen light as 'reality' at all, let alone the only plausible reality for distant observers. Especially since even in SR (let alone GR) a basic understanding is that any choice of simultaneity is a pure convention, restricted only by the requirement that a surface of simultaneity be spacelike. Given this, there are uncountable infinite valid choices for simultaneity for distant observers which provide a specific time time for the formation of the event horizon and the singularity.

 

[TrickyDicky]

One thing that confuses me and is not often mentioned in discussions about BHs and EHs, is the spacelike nature of the infaller observer's worldline, when it crosses the EH, for most people this seems to be something natural, but I thought it was generally assumed that physical observers are always timelike, their worldlines can never become null-like, unless they are massless like photons and much less become spacelike. Now since the EH is not considered a true singularity, one would tend to think physics at the interior side of the EH should not be that radically different than physics at the outside.

Since this is a physics forum rather than just a mathematical one I thought this might be relevant in a physics discussion. Maybe it is not.

 

[Dale]

One thing that confuses me and is not often mentioned in discussions about BHs and EHs, is the spacelike nature of the infaller observer's worldline, when it crosses the EH

An in falling observers worldline is timelike at all points. Remember, a free faller follows a geodesic, and a geodesic parallel transports its own tangent vector, and parallel transport preserves the norm. So if the tangent vector is timelike outside the EH then it remains timelike inside the EH.

 

[Mike Holland]

Mike, do you insist that seeing an Einstein ring means stars must be considered smeared into a ring by an observer that sees this? The effect BH on freezing light is physically the same phenomenon - just gravity bending or freezing light. There is no physical basis at all for interpreting this frozen light as 'reality' at all, let alone the only plausible reality for distant observers.

Not true. In addition to the red shift "freezing " of light, there is time dilation. Time dilation is real. We have measured the effect on clocks in high-flying aircraft and in orbit. If I go and hang out near a Black Hole for a while, and then return, I will have aged less than you, and my clock will be retarded. These effects are real, even if they are different for different observers. They have nothing to do with light paths. They have everything to do with the passage of time in a gravity field. I accept that there are light delay effects superimposed on this in some situations, and you are welcome to call these effects illusions.

Reality depends on your point of view. But that doesn't make it any less real. Points of view is all we have.


Mike

EDIT: Sorry, let me rephrase that. All we have is points of view and Einsteins theories to try and make sense of them.

 

[PAllen]

Not true. In addition to the red shift "freezing " of light, there is time dilation. Time dilation is real. We have measured the effect on clocks in high-flying aircraft and in orbit. If I go and hang out near a Black Hole for a while, and then return, I will have aged less than you, and my clock will be retarded. These effects are real, even if they are different for different observers. They have nothing to do with light paths. They have everything to do with the passage of time in a gravity field. I accept that there are light delay effects superimposed on this in some situations, and you are welcome to call these effects illusions.

Reality depends on your point of view. But that doesn't make it any less real. Points of view is all we have.


Mike

EDIT: Sorry, let me rephrase that. All we have is points of view and Einsteins theories to try and make sense of them.

But the time dilation you refer to is not a function of position, but a function of path. If, instead of comparing a distant observer with minimal proper acceleration with one who experiences enormous proper acceleration, we compare them with one in free fall toward the EH, starting from the distant observer, there is no significant time dilation between the near horizon free faller and the distant stationary observer. This near horizon observer sees their clock and the distant clock going at essentially the same rate. Who appointed you arbiter of which near horizon observers define reality?

[edit: In a real sense, GR does not have much 'relativity' in it - it is theory of invariants. The geometry of the manifold is the invariant 'reality'. Different coordinate systems on it are conventions. You want to elevate one coordinate system as the definition of reality rather than looking at the geometry of the manifold in a coordinate independent way.]

 

[TrickyDicky]

An in falling observers worldline is timelike at all points. Remember, a free faller follows a geodesic, and a geodesic parallel transports its own tangent vector, and parallel transport preserves the norm. So if the tangent vector is timelike outside the EH then it remains timelike inside the EH.

That seems the logical thing to expect, but there is a switch from timelike geodesics to spacelike ones at the EH, at least that is stated in many descriptions of black holes.

 

[PAllen]

That seems the logical thing to expect, but there is a switch from timelike geodesics to spacelike ones at the EH, at least that is stated in many descriptions of black holes.

No, this is incorrect. Perhaps you are confusing that the t coordinate in SC coordinates changes from timelike to spacelike. However, that is just an artifact of a coordinate choice. It doesn't happen for some other coordinates.

 

[PeterDonis]

That seems the logical thing to expect, but there is a switch from timelike geodesics to spacelike ones at the EH, at least that is stated in many descriptions of black holes.

No, there isn't. The tangent vector to a given geodesic must be of the same causal nature--timelike, spacelike, or null--everywhere on the geodesic. That's an elementary consequence of the geodesic equation.

What happens at the EH is that a particular vector *field*, the "time translation" Killing vector field, changes from timelike (outside the EH) to null (on the EH) to spacelike (inside the EH). But first, the integral curves of that vector field are not geodesics, and second, as you move from outside the EH to on the EH to inside the EH, you are moving to *different* integral curves of the vector field.

(In Schwarzschild coordinates, the "radial" vector field, defined by \partial_r, also changes from spacelike outside the EH, to null on the EH, to timelike inside the EH. But that is only true in that particular chart; in other charts, such as Painleve, the "radial" vector field stays spacelike all the way down to r = 0. The statement I made above about the time translation vector field is a coordinate-independent statement.)

 

[Dale]

That seems the logical thing to expect, but there is a switch from timelike geodesics to spacelike ones at the EH, at least that is stated in many descriptions of black holes.

There is no such switch. A path which switches from timelike to spacelike can not be a geodesic.

Could you provide a reference to such a description?

 

[DrGreg]

That seems the logical thing to expect, but there is a switch from timelike geodesics to spacelike ones at the EH, at least that is stated in many descriptions of black holes.

No, that is a mis-stated effect. In external Schwarzschild coordinates (i.e. outside the horizon), the t_e coordinate is timelike. In internal Schwarzschild coordinates (i.e. inside the horizon) -- a different coordinate system -- the t_i coordinate is spacelike & the r_i coordinate is timelike. That's just a matter of labelling coordinates.

Outside the event horizon its possible for a massive object to maintain a constant r_e, impossible to maintain a constant t_e coordinate.

Inside the event horizon its possible for a massive object to maintain a constant t_i, impossible to maintain a constant r_i coordinate.

 

[PeterDonis]

OK, I see how you are using the word "now", to include ALL events that COULD be "now" for us, depending on our chosen coordinate system. I would have called them "indeterminate".

That's a matter of terminology, not physics. However, the real point I was making is that there is *no* reasonable definition of "now" such that "now" includes events inside your past light cone.

just as our past light cone is distorted by the event horizon, so is our future one. The edge of my future light cone will approach the EH asymptotically (MY time frame, remember) and only reach it at t = infinity.

This doesn't change the fact (which you have agreed with) that a portion of the BH is in your future light cone. It also doesn't change the fact that the proper time elapsed for an infalling object, from your radius to the horizon, is finite. See below.

So none of these reference frames contradict calculations based on Schwarzschild coords.

They don't contradict them, but they do show limitations in them. See below.

The only question is, what time coordinates is my wristwatch based on? V = root(r/2GM - 1)? Or those of the falling raindrop?

That's the only question if we're concerned about proper time elapsed along *your* worldline, yes. But it's *not* the only question if we're concerned about proper time elapsed along some *other* worldline, spatially separated from yours.

If you think something can fall through the event horizon in a finite time, how does it avoid the time dilation?

It doesn't have to. "Time dilation" is relative; it depends on your worldline.

Since you're so insistent on doing calculations in Schwarzschild coordinates, try this one: write down the equation defining the proper time of an object freely falling radially inward from a finite radius r = R > 2M, to radius r = 2M. Write it so that the proper time is a function of r only (this is straightforward because it's easy to derive an equation relating r and the Schwarzschild coordinate time t, so you can eliminate t from the equation). This equation will be a definite integral of some function of r, from r = R to r = 2M. Evaluate the integral; you will see that it gives a finite answer. Therefore, the proper time elapsed for an infalling object is finite, even according to Schwarzschild coordinates.

 

[Dale]

Mike Holland, I don't know if you are avoiding my post 117 or simply have not had time, but I would like a response, in particular, to the question I posed there:

Do you believe that changing coordinate systems can make something start or stop existing?

 

[TrickyDicky]

No, that is a mis-stated effect. In external Schwarzschild coordinates (i.e. outside the horizon), the t_e coordinate is timelike. In internal Schwarzschild coordinates (i.e. inside the horizon) -- a different coordinate system -- the t_i coordinate is spacelike & the r_i coordinate is timelike. That's just a matter of labelling coordinates.

Just a matter of labelling coordinates? That was my initial idea but see for instance this quote by PeterDonis from a simultaneous similar thread:"The Killing vector field is not "swapped" by a coordinate transformation; it is a feature of the underlying geometry, independent of the choice of coordinates. So is the timelike, spacelike, or null nature of the Killing vector field at a particular event or within a particular region of the spacetime."
So then for instance the null-like condition of the EH is just a matter of labels or intrinsic to the geometry? And similarly what happens to the timelike or spacelike nature of geodesics outside and inside the EH, an arbitrary coordinate labelling issue? or a feature of the underlying geometry?

Inside the event horizon its possible for a massive object to maintain a constant t_i, impossible to maintain a constant r_i coordinate.

This doesn't seem a physical property for a timelike observer.

 

[DrGreg]

Just a matter of labelling coordinates? That was my initial idea but see for instance this quote by PeterDonis from a simultaneous similar thread:"The Killing vector field is not "swapped" by a coordinate transformation; it is a feature of the underlying geometry, independent of the choice of coordinates. So is the timelike, spacelike, or null nature of the Killing vector field at a particular event or within a particular region of the spacetime."
So then for instance the null-like condition of the EH is just a matter of labels or intrinsic to the geometry? And similarly what happens to the timelike or spacelike nature of geodesics outside and inside the EH, an arbitrary coordinate labelling issue? or a feature of the underlying geometry?

Whether a worldline is timelike, spacelike or null at a particular event is intrinsic to the geometry, and not dependent on a choice of coordinates. The scalar quantity given by the tensor expression g_{ab}U^aU^b, where U is a tangent vector to the worldline, is invariant, i.e. the same value no matter what coordinate system you use, and will be either positive, negative or zero. That determines whether the worldline is timelike, spacelike or null.

Particles with non-zero mass always travel along timelike worldlines (whether they are geodesics (in free fall) or not (acted on by a force)), everywhere, inside and outside the horizon. Particles with zero mass (e.g. photons) always travel along null worldlines.

 

[TrickyDicky]

The tangent vector to a given geodesic must be of the same causal nature--timelike, spacelike, or null--everywhere on the geodesic. That's an elementary consequence of the geodesic equation.

This is my understanding too.

What happens at the EH is that a particular vector *field*, the "time translation" Killing vector field, changes from timelike (outside the EH) to null (on the EH) to spacelike (inside the EH). But first, the integral curves of that vector field are not geodesics, and second, as you move from outside the EH to on the EH to inside the EH, you are moving to *different* integral curves of the vector field.

Are you sure?
I thought integral curves (flows) generated by Kiling vector fields in pseudoRiemannian manifolds were called geodesic flows because they were geodesic.

(In Schwarzschild coordinates, the "radial" vector field, defined by \partial_r, also changes from spacelike outside the EH, to null on the EH, to timelike inside the EH. But that is only true in that particular chart;

Didn't we agree that SC are valid only outside the EH?

The statement I made above about the time translation vector field is a coordinate-independent statement.)

And if so it must affect the geodesic flow consequently.

 

[TrickyDicky]

Whether a worldline is timelike, spacelike or null at a particular event is intrinsic to the geometry, and not dependent on a choice of coordinates. The scalar quantity given by the tensor expression g_{ab}U^aU^b, where U is a tangent vector to the worldline, is invariant, i.e. the same value no matter what coordinate system you use, and will be either positive, negative or zero. That determines whether the worldline is timelike, spacelike or null.

But that's what I said in my previous post and you replied that it was a matter of just labelling coordinates.

Particles with non-zero mass always travel along timelike worldlines (whether they are geodesics (in free fall) or not (acted on by a force)), everywhere, inside and outside the horizon. Particles with zero mass (e.g. photons) always travel along null worldlines.

This contradicts the quote from PeterDonis and your paragraph above, if you assert geometry determines whether a worldline is timelike, spacelike or null, you cannot say massive particles worldlines are always timelike independently of the intrinsic geometry.

 

[TrickyDicky]

There is no such switch. A path which switches from timelike to spacelike can not be a geodesic.

So I claim.

 

[TrickyDicky]

No, this is incorrect. Perhaps you are confusing that the t coordinate in SC coordinates changes from timelike to spacelike. However, that is just an artifact of a coordinate choice. It doesn't happen for some other coordinates.

I'm not talking about coordinates, I'm talking about Killing vector fields and the geodesic flows they generate in pseudoRiemannian manifolds.

 

[PeterDonis]

Are you sure?

Certainly. "Hovering" observers in Schwarzschild spacetime (observers who stay at a constant r > 2M for all time) travel along integral curves of the "time translation" Killing vector field. Those observers' worldlines certainly aren't geodesics; the observers have nonzero proper acceleration.

I thought integral curves (flows) generated by Kiling vector fields in pseudoRiemannian manifolds were called geodesic flows because they were geodesic.

They aren't. Where did you get the idea they were? [Edit: At least, they aren't in general. In some special cases, such as Minkowski spacetime, there are Killing vector fields that generate geodesic integral curves. I can't think of a case offhand of a curved spacetime where that's true, though.]

Didn't we agree that SC are valid only outside the EH?

No. Schwarzschild coordinates can be used inside the horizon, but that coordinate patch is disconnected from the patch that covers the exterior region (because of the coordinate singularity at r = 2M).

This contradicts the quote from PeterDonis and your paragraph above, if you assert geometry determines whether a worldline is timelike, spacelike or null, you cannot say massive particles worldlines are always timelike independently of the intrinsic geometry.

Yes, you can, because the law that "massive particle worldlines are always timelike" assumes that you already know the underlying geometry, so you know *which* curves are timelike. The law then just says that only those curves, the ones you already know are timelike, can be the worldlines of massive particles. (Similarly for massless particles like photons, with "timelike" replaced by "null".) Note that this goes for *any* curves, not just geodesics.

 

[Dale]

So I claim.

OK, so it sounds like you and I agree that an object which free falls across the EH must have a timelike worldline both inside and outside the EH. Then I guess it is just a mistake of one of the sources you have read that states otherwise. Could you please provide a link?

 

[TrickyDicky]

Certainly. "Hovering" observers in Schwarzschild spacetime (observers who stay at a constant r > 2M for all time) travel along integral curves of the "time translation" Killing vector field. Those observers' worldlines certainly aren't geodesics; the observers have nonzero proper acceleration.

These observers do not cross the EH so are not relevant to what I'm trying to get right.

They aren't. Where did you get the idea they were? [Edit: At least, they aren't in general. In some special cases, such as Minkowski spacetime, there are Killing vector fields that generate geodesic integral curves. I can't think of a case offhand of a curved spacetime where that's true, though.]

I think it is true of any spherically symmetric pseudoriemannian manifold. But let me find a good reference.

No. Schwarzschild coordinates can be used inside the horizon, but that coordinate patch is disconnected from the patch that covers the exterior region (because of the coordinate singularity at r = 2M).

Ah, ok, you just meant this.

Yes, you can, because the law that "massive particle worldlines are always timelike" assumes that you already know the underlying geometry, so you know *which* curves are timelike. The law then just says that only those curves, the ones you already know are timelike, can be the worldlines of massive particles. (Similarly for massless particles like photons, with "timelike" replaced by "null".) Note that this goes for *any* curves, not just geodesics.

Ok, so you are confirming that that law is subordinated to the geometry and therefore it should not contradict it.
So when DrGreg said that worldlines must obey the intrinsic geoemetry I uderstand that if the geoemetry says a worldline (geodesic or not) must be spacelike that means it cannot be followed by a massive observer, right?

 

[TrickyDicky]

OK, so it sounds like you and I agree that an object which free falls across the EH must have a timelike worldline both inside and outside the EH.

No, I agreed with what I quoted. I'm still trying to decipher how your statement here that "an object which free falls across the EH must have a timelike worldline both inside and outside the EH" does not contradict the intrinsic geometry of the K-S space.

 

[PeterDonis]

No, I agreed with what I quoted. I'm still trying to decipher how your statement here that "an object which free falls across the EH must have a timelike worldline both inside and outside the EH" does not contradict the intrinsic geometry of the K-S space.

Look at a Kruskal diagram, such as the one on the Wikipedia page:

http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

Any curve which is inclined to the vertical by less than 45 degrees on this diagram is timelike. There are many such curves that go from Region I to Region II; any such curve is a possible worldline for an infalling object. And of those curves, there are many that are geodesics.

If you want a mathematical description of such an infalling geodesic, it can be written in K-S coordinates, but it's kind of messy; it's easier to write it in ingoing Painleve coordinates for an ingoing timelike curve, or ingoing Eddington-Finkelstein coordinates for an ingoing null curve. For the timelike case, in Painleve coordinates, the ingoing geodesics are integral curves of the vector field \partial_T - \sqrt{2M / r} \partial_r, where T is the Painleve time coordinate. It is easy to show that this vector field is timelike everywhere.

 

[PeterDonis]

I think it is true of any spherically symmetric pseudoriemannian manifold. But let me find a good reference.

Please do, because Schwarzschild spacetime is a spherically symmetric pseudoriemannian manifold, and I gave an explicit example of a Killing vector field in it whose integral curves are not geodesics. So I'd be extremely surprised to find any reputable reference that claimed that the integral curves of a Killing vector field always *are* geodesics.

So when DrGreg said that worldlines must obey the intrinsic geoemetry I uderstand that if the geoemetry says a worldline (geodesic or not) must be spacelike that means it cannot be followed by a massive observer, right?

I would say "curve" instead of "worldline" for that precise reason; yes, a spacelike (or null) curve cannot be the worldline of any massive observer.

 

[Dale]

No, I agreed with what I quoted. I'm still trying to decipher how your statement here that "an object which free falls across the EH must have a timelike worldline both inside and outside the EH" does not contradict the intrinsic geometry of the K-S space.

What do you mean by "intrinsic geometry of the KS space"? I assume by KS you mean Kruskal–Szekeres coordinates, but that is a coordinate chart for a Schwarzschild spacetime and doesn't have any intrinsic geometry. The intrinsic geometry comes from the pseudo-Riemannian manifold representing the spacetime, not the coordinates.

In GR spacetime is represented by a pseudo-Riemannian manifold. The path of a massive particle is represented by a timelike curve in that manifold. So the worldline of any massive particle is always timelike regardless of whether it is geodesic or not and regardless of if it is inside or outside of an event horizon.

 

[Mike Holland]

Since you're so insistent on doing calculations in Schwarzschild coordinates, try this one: write down the equation defining the proper time of an object freely falling radially inward from a finite radius r = R > 2M, to radius r = 2M. Write it so that the proper time is a function of r only (this is straightforward because it's easy to derive an equation relating r and the Schwarzschild coordinate time t, so you can eliminate t from the equation). This equation will be a definite integral of some function of r, from r = R to r = 2M. Evaluate the integral; you will see that it gives a finite answer. Therefore, the proper time elapsed for an infalling object is finite, even according to Schwarzschild coordinates.

Peter, as I ubnderstand it, that is what Painleve did with his "raindrop" coordinates - provide a cioordinate frame that covers the object right through the falling process and beyond.

But I am at a loss to understand why there is an argument here. All along I have claimed that a falling observer will pass through the event horizon in his proper time, but that he won't in my proper time. This is what I understand from the calculations based on Schwarzschlild coordinates. So as long a I sit here in my armchair, I will see him hovering near the Evenyt Horizon forever edging closer and closer. As far as I am concerned, he hasn't fallen through it. But from his point of view in his spaceship, he just flies through and into the singularity. If the Black Hole is large enough, he won't even notice that he has gone through.

It is easy to fall into a Black Hole in a finite time, but not according to MY clock.

Mike

 

[Mike Holland]

Mike Holland, I don't know if you are avoiding my post 117 or simply have not had time, but I would like a response, in particular, to the question I posed there:
Do you believe that changing coordinate systems can make something start or stop existing?

My apologies, DaleSpam, I have been very busy. Also, your posts are usually the most difficult ones to answer. I need time to think.

If two coordinate systems do not overlap, then things that exist in one will not exist in the other. Similarly, if two systems have common space coordinates but no overlap in time coordinate, then you cannot convert x,y,z,t in one system to x,y,z,t in the other, so you cannot have the same event in both systems. But there can be cases of partial overlap to complicate things.

In the case of an Event Horizon, we can have a common x,y,z, but t does not overlap between my coordinates and the local ones. Any time coordinate at the event horizon corresponds to t = infinity in my coords, so no event there can be mapped into an x,y,z,t in my system. Events there do not exist for me. I can change coordinate system by jumping into the forming Black Hole, and when I get there it will come into existence for me.

Does that answer your question?
Mike

Edit: I've been thinking a bit further. What I said about Event Horizons still applies, but my ideas don't cover an event that did occur in my past, or will occur in my future. Such events exist on the timeline, but don't exist NOW. Things come into existence as I travel along my timeline, and others cease to exist as they move into my past. But in another sense of the word "exist", they do have an existence in my timeframe which Black Holes don't.

 

[TrickyDicky]

What do you mean by "intrinsic geometry of the KS space"? I assume by KS you mean Kruskal–Szekeres coordinates, but that is a coordinate chart for a Schwarzschild spacetime and doesn't have any intrinsic geometry. The intrinsic geometry comes from the pseudo-Riemannian manifold representing the spacetime, not the coordinates.

I was referring to the maximally extended Schwarzschild vacuum spacetime solution. KS coordinates happen to cover the whole spacetime.

In GR spacetime is represented by a pseudo-Riemannian manifold. The path of a massive particle is represented by a timelike curve in that manifold. So the worldline of any massive particle is always timelike regardless of whether it is geodesic or not and regardless of if it is inside or outside of an event horizon.

This is my understanding too, but IMO this contradicts the geometrically intrinsic definition of EH that PeterDonis gave.
See my answer to his posts below.

 

[TrickyDicky]

Look at a Kruskal diagram, such as the one on the Wikipedia page:

http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

Any curve which is inclined to the vertical by less than 45 degrees on this diagram is timelike. There are many such curves that go from Region I to Region II; any such curve is a possible worldline for an infalling object. And of those curves, there are many that are geodesics.

If you want a mathematical description of such an infalling geodesic, it can be written in K-S coordinates, but it's kind of messy; it's easier to write it in ingoing Painleve coordinates for an ingoing timelike curve, or ingoing Eddington-Finkelstein coordinates for an ingoing null curve. For the timelike case, in Painleve coordinates, the ingoing geodesics are integral curves of the vector field \partial_T - \sqrt{2M / r} \partial_r, where T is the Painleve time coordinate. It is easy to show that this vector field is timelike everywhere.

Yes, and it's a Killing vector field too, as long as Birkhoff's theorem holds.

Please do, because Schwarzschild spacetime is a spherically symmetric pseudoriemannian manifold, and I gave an explicit example of a Killing vector field in it whose integral curves are not geodesics. So I'd be extremely surprised to find any reputable reference that claimed that the integral curves of a Killing vector field always *are* geodesics.

I didn't say always, we are talking about spherically symmetric vacuums only.
According to Birkhoff's theorem a spherically symmetric vacuum has a timelike ∂t Killing vector field, in this context, the integral curves of this KVF are timelike geodesics.
Yor example is of an ingoing geodesic, so I don't understand what you mean by "are not geodesics".

Let me try and clarify where I see a problem here.
You said in the other thread " the event horizon is the boundary of the region in which the Killing vector field of the "time translation" isometry of Schwarzschild spacetime is timelike--i.e., the EH is the Killing horizon associated with the "time translation" isometry."
But then how is this compatible with the fact that spherically symmetric vacuums have timelike Killing vector fields (or to use Carroll's words in his Notes on GR:"We have therefore proven a crucial result: any spherically symmetric vacuum metric possesses a timelike Killing vector"), the logical question here would be then, is the region inside the EH (region II) a vacuum or not? If it is it must have ∂t KVF and therefore your EH definition would not be valid.

 

[Dale]

My apologies, DaleSpam, I have been very busy. Also, your posts are usually the most difficult ones to answer. I need time to think.

No problem, I understand that it can be difficult to keep up with a multiple-poster thread like this.

Does that answer your question?

Yes. The short answer is that you do believe that changing coordinate systems can make something start or stop existing.

Unfortunately, that position becomes ridiculous pretty quickly. In GR you are free to choose your coordinates in almost any way you like, you don't even need three for space and one for time, they don't have to be orthogonal or normal nor cover the whole spacetime. So, I could choose coordinates that do not cover anything to my right. Then, according to you, that light to my right, from which I am receiving a signal, does not exist. A change in coordinates and suddenly it exists.

Carried to the extreme, I do not even need to use coordinates which include me. So I can make myself not exist. Which begs the question, if I don't exist when I choose coordinates that don't include me, then how can something which doesn't exist choose coordinates?

Hopefully you can now understand why the idea of existence being determined by coordinate systems is repugnant to many people who understand the flexibility of coordinate systems in GR.

 

[Dale]

I was referring to the maximally extended Schwarzschild vacuum spacetime solution. KS coordinates happen to cover the whole spacetime.

OK, thanks for the clarification.

This is my understanding too, but IMO this contradicts the geometrically intrinsic definition of EH that PeterDonis gave.
See my answer to his posts below.

Since you and I agree then I will stop arguing . I didn't see any conflict with PeterDonis' comments (particularly since Killing vectors don't generally represent worldlines nor geodesics), but rather than try to explain what I think he meant I will let him explain directly. Perhaps you picked up on something I missed.