Black Holes - the two points of view.

[PeterDonis]

Peter, the exponentials for the coefficients are used precisely to avoid that kind of dimensional swapping,IOW to distiguish clearlybetween the spacelike and the timelike coordinates, whose nature is determined a priori, not based on an arbitrary relabeling.

No, the nature of the spacelike and timelike coordinates is *not* determined a priori. That's the whole point. If you claim that they are, then please read the proof I posted and tell me which specific parts of it are not valid for r < 2m. If my proof is valid for r < 2m, then it is valid for t both timelike *and* spacelike (and for r both spacelike *and* timelike).

To make more precise what I was saying in #254, whether a coordinate is temporal or spatial is predefined for a given line element

Not for all charts. You can certainly *define* a coordinate in a way that requires it to be timelike or spacelike (or null), but there's nothing that requires you to. In the case we're discussing, the Schwarzschild r and t coordinates are *not* defined in a way that forces them to be timelike or spacelike or null. You have to *figure out* which they are by looking at the line element. And if the line element is valid in different regions which have different signs for the coefficients, then the coordinates can be timelike (or spacelike) in one region and spacelike (or timelike) in another.

When ambiguities can affect the results it is better to use the coefficients in exponential form, when MTW says that in the general case there is no such constraint on sign, it means exactly that, the general case, not the Schwarzschild's case, otherwise they could skip using the exponential form at all.

You are misreading MTW (and it doesn't appear that you've been reading my previous posts very carefully). MTW specifically use the exponential form in deriving Birkhoff's theorem, i.e., in discussing the *Schwarzschild vacuum* case. I have pointed you several times now at MTW's proof *and* at my own proof, which I've linked to; in both places it is explicitly stated that the proof applies for 0 < r < 2m (i.e., r timelike and t spacelike) as well as 2m < r < infinity (r spacelike and t timelike). And my proof is even written without the exponentials and *still* shows a change of sign in the metric coefficients for 0 < r < 2m. Either point out specifically where MTW's proof and my proof do not apply for 0 < r < 2m, or stop making these incorrect claims.

Try using the Schwarzschild metric in isotropic cordinates and you'll see the fact r is < or > than 2GM doesn't change the sign ofthe coefficients.

Yes, that's because isotropic coordinates do not cover the interior region; instead they double cover the exterior region. The range 0 < R < M/2 (R is the isotropic radial coordinate, and R = M/2 is the horizon) covers the *same* region of spacetime as M/2 < R < infinity. Which proves absolutely nothing about any other coordinate chart.

If the isotropic coordinates can be used in region II and t is still timelike, why use the coordinates that may produce far-fetched consequences?

See above. Isotropic coordinates do not cover Region II. The same comment applies to your further comments about isotropic coordinates. Others have made a similar comment; MTW is another reference that discusses this.

But there's an easy way to see it for yourself: pick two values of R, R1 and R2, such that R2 = M^2 / (4 * R1). For example, R1 = M and R2 = M/4 will work. Now compute the physical area of a 2-sphere at isotropic radial coordinates R1 and R2. You will find that they are the same; in other words, both R1 and R2 label the *same* 2-sphere, and it is a 2-sphere *outside* the horizon, since the physical area is greater than that of a 2-sphere at the horizon. You will *not* be able to find *any* value of R that labels a 2-sphere inside the horizon (i.e., with area less than the area of a 2-sphere at the horizon). This proves that isotropic coordinates only cover the exterior region.

Edit: I suppose I should explicitly write down the line element in the isotropic coordinates I was assuming above:

ds^2 = - \frac{\left( 1 - M / 2R \right)^2}{\left( 1 + M / 2R \right)^2} dt^2 + \left(1 + \frac{M}{2R} \right)^4 \left[ dR^2 + R^2 \left( d\theta^2 + sin^2 \theta d\phi^2 \right) \right]

Of course if one is convinced beforehand that inside the Schwarzschild radius there is no timelike KVF

I was not "convinced beforehand". I explicitly *proved* it, following the *proof* that MTW give. Either refute these proofs specifically or stop making this incorrect claim.

I find the attempt to derive something physical or geometrical from the Schwarzschild coordinate anomaly for r<2GM as misguided as trying to derive physical or geometrical consequences from the purely coordinate singularity at r=2GM. Guys, there is no real singularity there!

Yes, we all know that. So what? We're not deriving anything from the presence of the coordinate singularity; none of what we're saying depends on there being a coordinate singularity at r = 2m in Schwarzschild coordinates. Again, have you read MTW's proof, or the proof I posted?

 

[PeterDonis]

I only read the abstract where it says it admits any t and r, and by inspection it seemed the coordinates didn't have the problem with sign that the Schwarzschild have.

Did you actually read his *definition* of the coordinates, or look at the line element he wrote down in terms of those coordinates? That would be useful information.

 

[TrickyDicky]

What weird consequences are you concerned about here.

From wormholes to null surfaces, white holes etc. But usually GR texts use bigger adjectives.

This seems to be a complete mischaracterization. There is no need to be convinced beforehand of anything, simply take the KVFs and compute whether they are timelike or spacelike.

Well, sure, if you use the usual interpretation that the change of sign means that a temporal dimension suddenly becomes spatial, regardless of the fact that in Lorentzian manifolds they are not interchangeable(unless one takes the block universe interpretation seriously and wants to impose it).

 

[PeterDonis]

Well, sure, if you use the usual interpretation that the change of sign means that a temporal dimension suddenly becomes spatial, regardless of the fact that in Lorentzian manifolds they are not interchangeable

The change of sign in the metric coefficients for r < 2m does not mean any "dimension" has changed its nature. The integral curves of \partial / \partial t for r < 2m are *different* curves than the integral curves of \partial / \partial t for r > 2m. The same goes for integral curves of \partial / \partial r. The different sets of curves are not even parallel to each other.

Look on a Kruskal diagram like the one on the Wikipedia page:

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

The integral curves of \partial / \partial t in region I are the timelike hyperbolas curving from the lower right to the upper right. The integral curves of \partial / \partial t in region II are the spacelike hyperbolas curving from the upper left to the upper right. The integral curves of \partial / \partial r in region I are spacelike lines radiating from the central point out to the right towards spatial infinity. The integral curves of \partial / \partial r in region II are timelike lines radiating from the central point upward towards the singularity. Nothing in any of this requires any "dimension" to change nature or raises any issue with the Lorentzian nature of the manifold; the manifold is obviously Lorentzian everywhere, and the Kruskal coordinates used to make the diagram serve perfectly well as timelike and spacelike "dimensions" that never change.

 

[stevendaryl]

Given the weird consequences of taking seriously what looks like a purely coordinate artifact for r<2GM , if one doesn't want to use the coefficients in exponential form, the sensible thing to do is use a different set of coordinates, like the isotropic ones and check that with them there is no coordinate temporal or spatial nature swap.

I'm not sure what it is that you are claiming here, but the Schwarzschild line element
ds^2 = (1-2GM/(c^2 r)) dt^2 - 1/ (1-2GM/(c^2 r)) dr^2 - r^2 d\Omega^2

is a perfectly good solution to Einstein's field equations whether or not (1-2GM/(c^2 r)) is positive or negative. If it is positive, then the coordinate t is timelike and the coordinate r is spacelike. If it is negative, then the coordinate r is timelike and the coordinate t is spacelike.

There is no physical meaning to the claim that the coordinate t for the interior solution is "the same coordinate" as the coordinate t for the exterior solution, so the fact that t switches from timelike to spacelike is of no significance. What is of significance is that in the interior solution, there is no physically realizable worldline corresponding to constant r.

Of course if one is convinced beforehand that inside the Schwarzschild radius there is no timelike KVF,

What does "beforehand" me? Before what? It's a mathematical fact that there are 4 Killing vector fields that are all spacelike in the interior, one corresponding to translations in t, and the other 3 corresponding to the rotations. Whether there is another field that is timelike requires proof one way or the other.

...there is no way around it. But it looks rather arbitrary given the fact that the only reason to claim that is precisely the anomalous behaviour of the Schwarzschild coordinates for r<2GM when no constraint on sign is applied.

What are you claiming is anomalous about the region r < 2GM?

 

[Dale]

But you guys seem to have a different understanding of "existence", that if anything exists somewhere on a coordinate system, then it exists period! There is no becoming, no yesterday, today and tomorrow.

This is, unfortunately a slight limitation in the English language. There is no verb tense which could describe the concept of something existing at some instant in time without specifying whether that instant in time is future, past, or present.

The math of relativity, however, is clear that the tense of existence is not always something physical, but often is something which is merely an arbitrary human convention. The physical part is defined by the past and future light cones of the present event. Everything within the past light cone was past, and can affect the present event, and everything within the future light cone will be future, and can be affected by the present event.

The remaining part of spacetime, the part that is neither in the past nor the future light cone, is ambiguous. Different arbitrary choices will make some event in that region be today, while other equally valid choices will make the same event be yesterday or tomorrow. Realizing that this temporal ordering is arbitrary and subject to human choice, not physics, is one of the most difficult concepts in relativity.

If I see a supermassive star collapsing, then you say the black hole exists, where I would say it is forming and is going to exist.

If you said that then you would be making a mistake. If you see it collapsing then its collapsing was at some time in the past. So what you should say is that it was forming.

Similarly, I cannot accept PeterDonis' idea of "now". He uses the word to cover the whole area between my past light cone and my future light cone. So a guy could live his whole life "now" on a planet 100 LY from me. He is born now, he is dead now. He is conceived now. This doesn't make sense to me. I would choose a simple method of drawing a line vertical to my world line in my space-time diagram, and say that defines my "now",

There are several problems with this.

The first is its arbitrariness, why should we pick your speed? Why should we pick a hyperplane orthogonal to your worldline? There is nothing in the laws of physics which constrains either choice.

The second problem is that you are non-inertial, so as you accelerate the plane orthogonal to your worldline changes orientation, which could make time go backwards for the alien guy. Why should he die before he was born simply because you drove to the store?

The third problem is that it only works in flat spacetime. In curved spacetime you can no longer do this at all. In curved spacetime, you have to simply pick some arbitrary convention. When you do that, you begin to realize that you have, in fact, been picking arbitrary conventions all along. You only thought that your previous conventions had some physical meaning.

If I hover near an Event Horizon and then return here, my clock will be retarded relative to yours. We can set our clocks side by side, and anyone in any coordinate system who can see them will agree that yours is ahead of mine.

Exactly, anyone in any coordinate system will agree that yours is ahead of mine. Which means that this experiment cannot distinguish between any coordinate system, further emphasizing the arbitrary nature of the coordinate systems. Both coordinate systems where the black hole exists now and coordinate systems where the black hole is collapsing now agree on the result of the experiment.

You lot seem so lost in your abstract theories that you forget there is a real world where things come into existence (like the flowers in my garden - it is Spring in Sydney) and then disappear, and where scientists actually measure time dilation and find that the facts agree with the theory.

Yes, the scientists do measure things which agree with the theory, and the theory says that the existence of some event is a topological feature of the manifold, not a feature of the coordinate system. You are the one who is ignoring the implications of the theory which agrees with the experiments.

 

[Dale]

I only read the abstract where it says it admits any t and r

Just because some coordinate ranges over -∞ to +∞ doesn't mean that the chart covers the entire manifold.

If the metric is the one given in the Wikipedia article then it doesn't cover the interior. If it isn't then it would be helpful to post it and the coordinate transform.

 

[Dale]

From wormholes to null surfaces, white holes etc. But usually GR texts use bigger adjectives.

Wormholes and white holes are certainly weird to me also, but they are not consequences of using the standard Schwarzschild coordinate chart on the interior region. Neither are null surfaces, but I don't find them weird at all.

Well, sure, if you use the usual interpretation that the change of sign means that a temporal dimension suddenly becomes spatial, regardless of the fact that in Lorentzian manifolds they are not interchangeable

Temporal coordinate, not temporal dimension. And temporal and spatial coordinates are interchangeable in Lorentzian manifolds.

Also, the KVF is a symmetry of the manifold, not the coordinate chart. You can express the KVF in any chart, although the one in question will be messier. In any coordinate system it will be timelike outside the EH, and spacelike inside.

 

[PeterDonis]

However, simply changing one's position on the time stream of one coordinate system can! There are things which exist today that did not exist yesterday.

If you use "exist" in the sense you're using it, yes. But that's not the only possible sense of "exist".

But you guys seem to have a different understanding of "existence", that if anything exists somewhere on a coordinate system, then it exists period!

In the sense we're using the word "exist", yes, that's true. This is a different sense of the word "exist" than the one you're using.

(Warning: Clintonesque discussion on what the meaning of "is" is follows. )

You appear to be treating "existence" as a property that something can have. Logically speaking, though, "existence" is a quantifier, not a property. Saying that something exists is not the same, logically, as saying that something is red or is large or has mass m. The two types of statements have a different logical structure.

The statement "this black hole has mass m" is expressed, logically, as Bx = mx, where "Bx" is "x is this black hole", mx is "x has mass m", and "=" means logical equivalence.

The statement "this black hole exists" is expressed, logically, as Ex: Bx, where "Ex" is the existential quantifier. But in order to use a quantifier, you have to decide over what universal set you are going to quantify; different universal sets give different meanings for the quantifier, and therefore for the word "exist".

You are implicitly quantifying over the universal set: all events that lie in some particular spacelike hypersurface, which is labeled "now".

We (DaleSpam, PAllen, and I) are implicitly quantifying over the universal set: all events in the spacetime.

So we aren't really contradicting each other: we're just saying different things, using different logical quantifiers.

Similarly, I cannot accept PeterDonis' idea of "now". He uses the word to cover the whole area between my past light cone and my future light cone. So a guy could live his whole life "now" on a planet 100 LY from me. He is born now, he is dead now. He is conceived now. This doesn't make sense to me.

Of course it doesn't. But that doesn't make my definition of "now" logically contradictory, or even physically useless. It just means your intuition doesn't like it. Once again, this is just a third possible universal set over which to quantify when making statements about what is happening "now". For some purposes, such as figuring out what events I can and cannot be causally connected to, it is quite useful.

I would choose a simple method of drawing a line vertical to my world line in my space-time diagram, and say that defines my "now", while my past light cone defines an effective "now" because that is what I see now. A blind person might have a different view based on the speed of sound, but would agree with my geometric method.

If he had the same purpose in mind for using the definition, sure. This definition works fine for many purposes. But it's still a definition, and not the only possible one. Saying you don't like the others is not sufficient for maintaining that they are "wrong".

How would these clock differences occur if time dilation is just a coordinate thing?

"Time dilation" as it is usually defined is coordinate-dependent. But the differences in clock readings you are referring to are not. They occur because the length of a worldline in a curved spacetime is path-dependent: two worldlines between the same pair of events, but taking different paths, can have different lengths ("length" for a worldline meaning "proper time elapsed"). It's simple geometry.

You lot seem so lost in your abstract theories that you forget there is a real world where things come into existence (like the flowers in my garden - it is Spring in Sydney) and then disappear, and where scientists actually measure time dilation and find that the facts agree with the theory.

Your claim about what the "real world" is "really like" is not justified by the evidence you are citing. Once again, if you *choose* to describe the world that way, the description works, at least over a wide domain. But that does not show that your description is the only possible one, or even the only one that gives correct answers in that domain.

Nobody is disputing that there is a "real world", and that we can't change its nature just by changing the way we think about it. But accepting that is not sufficient to show that the real world *must* be described in terms of things coming into existence and then disappearing.

 

[stevendaryl]

I missed this key post.

Isn't a test particle orbiting a BH at a constant r (circular orbit, in a spherically symmetric vacuum setting this is possible unlike the real universe case) outside the EH also free-falling, that is, describing a timelike geodesic?

That sounds correct. So I guess there are geodesics that are integrals of Killing vectors. When I said "free-falling" I was thinking of a geodesic in which r is changing, but you're right that there are geodesics with constant r.

 

[PeterDonis]

That sounds correct. So I guess there are geodesics that are integrals of Killing vectors. When I said "free-falling" I was thinking of a geodesic in which r is changing, but you're right that there are geodesics with constant r.

But not constant r *and* t. The worldlines of particles in free-fall orbits around the BH are *not* integral curves of a KVF. Integral curves of the 3 KVFs that arise from the spherical symmetry are spacelike curves, not timelike ones, and those spacelike curves are *not* geodesics of the spacetime. (They are geodesics of the submanifold consisting of the particular 2-sphere they are in, but that's not the same thing.) And integral curves of the 4th KVF, \partial / \partial t, are worldlines of "hovering" observers (outside the horizon, of course--inside the horizon these curves are also spacelike, as has been noted many times in this thread), with zero angular momentum, not observers in free-fall orbits.

 

[Dale]

Integral curves of the 3 KVFs that arise from the spherical symmetry are spacelike curves, not timelike ones, and those spacelike curves are *not* geodesics of the spacetime. (They are geodesics of the submanifold consisting of the particular 2-sphere they are in, but that's not the same thing.)

I thought that the 3 KVFs represented rotations about 3 orthogonal axes, in which case their integral curves wouldn't be goedesics of the submanifold either, except on the "equator" of each rotation.

The KVFs really have nothing to do with geodesics. I don't know why they keep getting mentioned together.

 

[PeterDonis]

I thought that the 3 KVFs represented rotations about 3 orthogonal axes, in which case their integral curves wouldn't be goedesics of the submanifold either, except on the "equator" of each rotation.

Hm, good point. When I was visualizing it I was implicitly visualizing the equatorial rotation, but you're right, that's a special case even in the submanifold.

 

[TrickyDicky]

But not constant r *and* t.

Not constant t? You seem to be ignoring time-invariance.

 

[Mike Holland]

If you have a rocket ship accelerating at rate g in a straight line, then you can set up a coordinate system X, T using on-board clocks and rulers that are related to the usual Minkowsky coordinates
In terms of the coordinate system X, T, things behave as follows:

1. Clocks that are "higher up" (larger value of X) tick faster.
2. Clocks that are "lower" (smaller value of X) tick slower.
3. There is an "event horizon" at X = 0 such that clocks at this location have rate 0; they are slowed to a stop.
4. An object in freefall will drift closer to the event horizon as T → ∞, but never reach it.

Note that effects 1-4 are exactly analogous to the event horizon of a black hole. Yet, in this case, we know that the conclusion that an observer can never cross the "event horizon" is baloney. So of course it's possible to cross the event horizon. But the event of the object crossing the event horizon doesn't show up in the X,T coordinate system, unless you allow T = ∞.

Yes and No! You are not telling me anything I didn't know. I have said from the start that a falling observer will cross the event horizon in a very finite time in HIS timeframe, but that it will take an infinite time in OUR timeframe. Please go and re-read my very first post.

Just as, in your example, the situation is just a consequence of the remote observer's acceleration, so in my original description the identical situation was just a consequence of the presence of a black hole. In each situation we have the "baloney" situation that the faller cannot cross the EH in the remote observer's timeframe, and the "real" sutuation that he can in his own timeframe. Except that I don't believe in any "preferred" timeframes. They are equally real.

Mike

 

[Austin0]

In thinking about the "reality" of an in-falling observer crossing the event horizon, I think it's instructive to look at an analogous situation, which is the case of an accelerated observer in flat spacetime.

If you have a rocket ship accelerating at rate g in a straight line, then you can set up a coordinate system X, T using on-board clocks and rulers that are related to the usual Minkowsky coordinates x,t through

x = X cosh(gT)
t = X sinh(gT)

(For simplicity, I'm only considering one spatial coordinate. Also, I'm ignoring factors of c to make the equations simpler to write; it's easy enough to put them back in.)

In terms of the coordinate system X, T, things behave as follows:

1. Clocks that are "higher up" (larger value of X) tick faster.
2. Clocks that are "lower" (smaller value of X) tick slower.
3. There is an "event horizon" at X = 0 such that clocks at this location have rate 0; they are slowed to a stop.
4. An object in freefall will drift closer to the event horizon as T → ∞, but never reach it.

Note that effects 1-4 are exactly analogous to the event horizon of a black hole. Yet, in this case, we know that the conclusion that an observer can never cross the "event horizon" is baloney. The horizon at X=0 in the accelerated coordinates of the rocket corresponds to the points x=ct in the usual Minkowsky coordinates. "Above" the horizon are those points where x&gt;ct, and "below" the horizon are those points where x &lt; ct. Obviously, if an object is just sitting around, at rest in the x,t coordinate system, eventually it will "cross the horizon" so that x &lt; ct.

So of course it's possible to cross the event horizon. But the event of the object crossing the event horizon doesn't show up in the X,T coordinate system, unless you allow T = ∞. The existence of the event horizon, and the fact that apparently nothing can cross it, is an artifact of the accelerated coordinate system of the rocket; that coordinate system is only good for describing events "above" the horizon, where X &gt; 0, or alternatively, where x &gt; ct. The region at and below the horizon is just not adequately described by the coordinate system X,T.

I am not following your analogy. What does the fact that the clocks are increasingly dilated as they are located closer to X=0 have to do with an object crossing the horizon??
Hypothetical virtual Rindler observers at those points would agree the object was moving with increasing velocity (dilated clocks would increase the velocity measurement ) and would all agree the object crossed the horizon at X=0 ??

As far as direct observation by the actual Rindler observers; they would see the object fading off in the distance .Eventually disappearing far behind the horizon.
You can't say they would see it cross the horizon because of course they can't see the horizon. But you can say they see it at distances far beyond the calculated horizon.

So i don't understand why a stopped clock would in any way imply a stopped object
and it seems that both direct observation and the coordinate calculations would find the object sailing right through horizon.
So am I missing something here?

 

[TrickyDicky]

I don't see much benefit in a "yes, it is, no, it isn't" kind of exchange, I can see why you think my POV is wrong and I just presented it as my doubts, rather than in terms of right or wrong.

One of the things that concerns me about this is that for instance Weyl curvature is usually associated to static spacetimes, is there any example of non-static metrics with non-vanishing Weyl curvature?

This got lost in the debate. My question was trying to clarify how in the non-static(in this case contracting) metric inside the EH can we have a vanishing Ricci curvature when Ricci curvature describes (according to Baez at least:http://math.ucr.edu/home/baez/gr/ricci.weyl.html) change of volume, I though one of the hallmarks of expanding or contracting spaces was precisely change of volume.

 

[Mike Holland]

So i don't understand why a stopped clock would in any way imply a stopped object

A stopped clock does not mean a stopped object. Maybe it just needs winding, or a new battery. But stopped time most definitely does mean a stopped object.

When time stops (for any observer) then nothing happens (again, as far as that observer is concerned). There can be no before or after. All is one time.

As a falling object approaches the horizon, its speed is approaching c (in the local frame), and as the dilation seen by a distant observer aproaches infinity, the distance remaining to be traversed approaches zero. So it is by no means obvious which is going to win. But the calculations by Oppenheimer and others all show that the time dilation will win, and the object will only approach the horizon as the remote oberser's time approaches infinity.

Mike

 

[Austin0]

A stopped clock does not mean a stopped object. Maybe it just needs winding, or a new battery. But stopped time most definitely does mean a stopped object.

When time stops (for any observer) then nothing happens (again, as far as that observer is concerned). There can be no before or after. All is one time.

As a falling object approaches the horizon, its speed is approaching c (in the local frame), and as the dilation seen by a distant observer aproaches infinity, the distance remaining to be traversed approaches zero. So it is by no means obvious which is going to win. But the calculations by Oppenheimer and others all show that the time dilation will win, and the object will only approach the horizon as the remote oberser's time approaches infinity.

Mike

What are you going on about? My post was about a Rindler horizon not a BH (nary a mention)

At X=0 in an accelerating system there is no stopping of time. There is no possible acceleration or velocity that can produce infinite dilation anywhere in a real world system. SO no observer or clock could possible stop. Any dilation less than infinite would simply make the falling object have a greater coordinate velocity at that point.
So a point of actual infinite dilation,a BH EH would stop not only clocks but observers,,, but not an abstract coordinate projection which has no geometric or real world significance.
This is exactly why I thought his analogy wasn't really applicable.

 

[Mike Holland]

At X=0 in an accelerating system there is no stopping of time. There is no possible acceleration or velocity that can produce infinite dilation anywhere in a real world system. SO no observer or clock could possible stop. Any dilation less than infinite would simply make the falling object have a greater coordinate velocity at that point.
So a point of actual infinite dilation,a BH EH would stop not only clocks but observers,,, but not an abstract coordinate projection which has no geometric or real world significance.
This is exactly why I thought his analogy wasn't really applicable.

Light emitted from the Rindler horizon can never reach the accelerating observer. Light emitted very close to it will take an awful long time to catch up with him. So when looking back he will see a slowdown near the horizon, just like the slowdown near a gravitational event horizon.

In the case iof black holes I agree with you that nothing will actually reach infinite time dilation in the real world, which is why I don't think there are (quite) event horizons and black holes. But I haven't yet figured out why there shouldn't be a Rindler horizon in a finite time of accelerating. So I have a problem with my physics/philosophy.

Mike

Edit: Just had a thought while lying in the bath. All the Rindler diagrams I have seen, showing the acceleration curve approaching the light line asymptotically, do not show the time dilation of the accelerating guy as he approaches c. I need to take that into account and do some thinking/calculating to figure out whether he would ever see the event horizon forming behind him.
Off to bed. Mike

 

[PeterDonis]

Not constant t? You seem to be ignoring time-invariance.

I meant that there are no geodesics in Schwarzschild spacetime with constant r and t. Time-invariance has nothing to do with that.

 

[PeterDonis]

My question was trying to clarify how in the non-static(in this case contracting) metric inside the EH can we have a vanishing Ricci curvature when Ricci curvature describes (according to Baez at least:http://math.ucr.edu/home/baez/gr/ricci.weyl.html) change of volume, I though one of the hallmarks of expanding or contracting spaces was precisely change of volume.

"Change of volume" of a small sphere of test particles that are freely falling. Such a sphere does *not* change volume in Schwarzschild spacetime, *unless* it encloses the r = 0 singularity. Small spheres of freely falling test particles that do not enclose r = 0 change *shape* as they fall (due to Weyl curvature, or tidal gravity), but not volume. Spheres of freely falling test particles that do enclose r = 0 will change volume, but if they enclose r = 0 they are not entirely in vacuum because of the singularity.

 

[stevendaryl]

I am not following your analogy. What does the fact that the clocks are increasingly dilated as they are located closer to X=0 have to do with an object crossing the horizon??

The two effects go together. According to the X,T coordinate system, a "falling" clock takes an infinite amount of time to pass the "horizon" at X=0, but according to the clock, it only takes a finite amount of time. So those two facts imply infinite (or unbounded) dilation as the clock approaches X=0.

Hypothetical virtual Rindler observers at those points would agree the object was moving with increasing velocity (dilated clocks would increase the velocity measurement ) and would all agree the object crossed the horizon at X=0 ??

As I said, in the accelerated Rindler coordinate system, the clock never crosses the "horizon" at X=0. A clock in "freefall" from rest at X=L will approach the "horizon" according to the formula X = L/cosh(gT), which never goes to zero.

As far as direct observation by the actual Rindler observers; they would see the object fading off in the distance. Eventually disappearing far behind the horizon.

No, the clock never crosses the horizon, according to the Rindler coordinates (that is, the event at which it crosses the horizon is not covered by the coordinate system).

So i don't understand why a stopped clock would in any way imply a
stopped object

It's just a fact that in the Rindler coordinate system, an object in "freefall" will approach X=0, but never reach it.

 

[stevendaryl]

At X=0 in an accelerating system there is no stopping of time.

The phrase "stopping of time" doesn't really make any sense. But the significance of the event horizon (whether in a real black hole or in a Rindler accelerating coordinate system) is that the relationship between time on a clock and coordinate time becomes singular.

In particular, we have proper time \tau on a free-falling clock. We have coordinate time T in the coordinate system of the accelerating observer. The relationship between these are:

As T → ∞,
the clock approaches (but does not reach) the horizon,
the "clock rate" \dfrac{d \tau}{dT} → 0.

There is no possible acceleration or velocity that can produce infinite dilation anywhere in a real world system.

I'm not sure what you mean by that. In the Rindler coordinate system of an accelerated observer, "time dilation" defined as \dfrac{d \tau}{dT} goes to zero as the clock approaches X=0.

SO no observer or clock could possible stop. Any dilation less than infinite would simply make the falling object have a greater coordinate velocity at that point.

What are you talking about? In the Rindler coordinate system, the freefalling clock's coordinate velocity approaches ZERO as the clock approaches the Rindler horizon.

So a point of actual infinite dilation,a BH EH would stop not only clocks but observers,,, but not an abstract coordinate projection which has no geometric or real world significance.

I don't know what you mean by that, but in both the case of a real black hole, and in the case of a Rindler horizon, the infinite slowing of clocks as they approach the horizon are coordinate effects. They reveal limitations of the coordinate system being used, not anything physical going on with clocks.

This is exactly why I thought his analogy wasn't really applicable.

As I said, the main qualitative effects of an event horizon are the same for a Rindler event horizon and a Schwarzschild event horizon. The one difference is that in the Rindler case, it's not really possible for a rocket to remain "above" the horizon forever; eventually it will run out of fuel and fall below the horizon. In contrast, for a black hole, there are orbits around the black hole that allow an observer to stay forever outside the event horizon without expending fuel.

 

[Austin0]

quote austin0

As far as direct observation by the actual Rindler observers; they would see the object fading off in the distance. Eventually disappearing far behind the horizon.

No, the clock never crosses the horizon, according to the Rindler coordinates (that is, the event at which it crosses the horizon is not covered by the coordinate system).

.

Look. I was talking about actual observations. What people in an accelerating system would actually see with their eyes. Not what might be calculated.

I don't have time for more now. But am interested in the rest of your responce.

 

[stevendaryl]

quote austin0
Look. I was talking about actual observations. What people in an accelerating system would actually see with their eyes. Not what might be calculated.

Well, what the accelerated observer sees is very similar to what a "stationary" observer outside of an event horizon sees. A clock falling toward the event horizon will have its image red-shifted, so at some point, before it reaches the horizon, the clock will be too dim to see. The accelerated observer can never see the clock cross the event horizon.

 

[PAllen]

Well, what the accelerated observer sees is very similar to what a "stationary" observer outside of an event horizon sees. A clock falling toward the event horizon will have its image red-shifted, so at some point, before it reaches the horizon, the clock will be too dim to see. The accelerated observer can never see the clock cross the event horizon.

Even more strongly: the event of the clock reaching the event horizon will never be in the past null cone of the accelerating observer, ever. Thus there is a true causal separation between 'half the universe' and uniformly accelerating observer. Causally speaking, the analogy between this situation and static SC observer watching free fall clock is identical.

In a post somewhere recently here, I went through the exercise of deriving the properties of a Rindler observer using Minkowski inertial coordinates. It was instructive how simple it was for the physics to emerge without using Rindler coordinates. All you do is take the envelope of past light cones along an accelerating world line written in Minkowski coordinates, and voila you have the Rindler horizon expressed in Minkowsi coordinates.

 

[TrickyDicky]

"Change of volume" of a small sphere of test particles that are freely falling. Such a sphere does *not* change volume in Schwarzschild spacetime, *unless* it encloses the r = 0 singularity. Small spheres of freely falling test particles that do not enclose r = 0 change *shape* as they fall (due to Weyl curvature, or tidal gravity), but not volume. Spheres of freely falling test particles that do enclose r = 0 will change volume, but if they enclose r = 0 they are not entirely in vacuum because of the singularity.

Are you saying that having test particles enclosing the singularity implies there's no vacuum?
But the singularity is at the center and the spacetime is foliated by 2-spheres that *enclose* the singularity so how could they not enclose it?

 

[TrickyDicky]

I meant that there are no geodesics in Schwarzschild spacetime with constant r and t. Time-invariance has nothing to do with that.

I meant that a time-invariant and time-symmetric system in a circular orbit behaves as if time is constant, that's why they are called static.

 

[PAllen]

I don't know that this post will convince anyone, but wrestling myself with the subtleties of the geometry of the interior region of the SC vacuum, I found it helpful to study Lemaitre coordinates, which I was not previously very familiar with (they seem least used of common coordinates for this geometry). Dalespam mentioned them in his list of coordinates much earlier in this thread:

http://en.wikipedia.org/wiki/Lemaitre_coordinates

I came to them via the following route:

It is clear that the core 'problem' with the SC t coordinate is that it is defined as parametrizing corresponding (same theta, phi) points on different 2-spheres of the same area, within spacetime. There must be a one parameter family of such spheres for all radii to properly cover spherically symmetric spacetime. For r > R (SC radius), the line connecting corresponding points on such spheres is timelike - representing the history of static observer. For r < R, there must still be a one parameter family of same area 2-spheres to properly fill spacetime. However, the feature that all timelike curves going through an r < R reach the singularity, means that the t parameter connecting same area 2-spheres for r < R must be spacelike. It is exclusively this definition of t that causes it's nature to flip at the horizon.

So I was thinking there ought to be a way chart SC spacetime such that you have timelike t coordinate everwhere, and a radially directed spatial coordinate everywhere, such that a surface of contant time coordinate is spatial 3-surface of concentric 2-spheres; and a surface of constant radial directed coordinate is time sequence of concentric 2-sheres down to the singularity. Accepting this feature of the radial directed coordinate (that held constant, it connects contracting spheres parametrized by time), makes it compatible with both the interior and exterior geometry. To validate this, I studied various timelike and spacelike curves in Kruskal coordinates and concluded that such thing should definitely be possible and cover all of region I and II (for example). Then I figured someone must have constructed such a thing before.

Lo and behold, that is exactly what Lemaitre coordinates accomplish! The metric smoothly covers regions I and II of the complete manifold, with a timelike time coordinate everywhere, and spacelike radial directed coordinate everywhere, with exactly the properties described above. In front of the surface angle portion of the metric is the area defined r value as function of the time and radially directed coordinates - as it must be to accommodate the non-stationary interior.

The horizon in these coordinates exists only as a derived feature, indistinguishable by any coordinate behavior. You can also directly see that a path connecting spheres of constant area is timelike >R, and spacelike < R, but this is unrelated to any change in coordinate nature.

The only downside I see to Lemaitre coordinates (besides that they don't cover regions III and IV - which presumably don't exist in the real world anyway), is that while the 3 rotational killing vectors are just as obvious as for the SC coordinates, the 'extra' killing vector field (that is timelike for r>R, and spacelike for r< R) is not manifest at all in these coordinates. It would have to be derived in some non-trivial way, or by converting to SC coordinates and then noting that a KVF is coordinate independent feature.