Black Holes - the two points of view.

[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 $x>ct$, and "below" the horizon are those points where $x < ct$. Obviously, if an object is just sitting around, at rest in the $x,t$ coordinate system, eventually it will "cross the horizon" so that $x < ct$.

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 $X > 0$, or alternatively, where $x > ct$. 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.

 

[PAllen]

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?

Actually, a free-falling initially locally spherical cluster of test bodies can never enclose the singularity. The definition of this procedure is that you start with a small enough ball of 'dust' that in some spatial slice it is geometrically arbitrarily close to a Euclidean 3-ball, then watch how it changes over a short period of time along the world lines. No such ball enclosing the singularity can ever have the geometry of a 3-ball. Further, the singularity is not even considered part of the manifold, so an alleged 3-ball enclosing it would not be a connected set.

 

[PAllen]

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.

Can you clarify what you mean? This makes no sense to me. A circular orbit world line advances steadily in proper time and in SC coordinate time. It is a helix in SC coordinates.

What Peter is saying is that a curve of constant (r,theta,phi) with t varying is not a geodesic - it has proper acceleration. A curve of constant (t,theta,phi) with varying r is also not a spacelike geodesic. Do you disagree with either of these statements?

 

[PeterDonis]

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.

I have no idea what this means, but whatever it means, it doesn't show that the worldline of an object in a free-fall orbit is an integral curve of any KVF, which was my point.

 

[PeterDonis]

Are you saying that having test particles enclosing the singularity implies there's no vacuum?

No vacuum somewhere inside the sphere, yes; in this case, no vacuum at the singularity at r = 0. But there is still vacuum everywhere else.

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?

You're confusing your spheres. The "spheres" referred to in the Baez web page you linked to are spheres of test particles; they can be anywhere you like, enclosing the singularity or not. The 2-spheres that foliate the spacetime according to its spherical symmetry enclose the singularity, yes, but those 2-spheres aren't the ones Baez is talking about.

[Edit: See my following post in response to PAllen; I was thinking of 2-spheres in the above, but I think PAllen is correct that Baez' actual argument refers to 3-balls, meaning 2-spheres plus their interiors. In that case a "sphere" enclosing the singularity doesn't meet Baez' specifications to begin with, so I was wrong to bring it up as a possible example.]

 

[PeterDonis]

Actually, a free-falling initially locally spherical cluster of test bodies can never enclose the singularity. The definition of this procedure is that you start with a small enough ball of 'dust' that in some spatial slice it is geometrically arbitrarily close to a Euclidean 3-ball, then watch how it changes over a short period of time along the world lines. No such ball enclosing the singularity can ever have the geometry of a 3-ball.

Hmm. I was thinking of the "sphere" as a 2-sphere of test particles, the 2-sphere that marks the boundary of a 3-ball, not the 3-ball itself. But re-reading the Baez web page, I think you're right, he means for it to be the full 3-ball, including its interior as well as its boundary. In that case, yes, you're right, no ball of test particles that encloses the singularity can meet his specifications.

Further, the singularity is not even considered part of the manifold, so an alleged 3-ball enclosing it would not be a connected set.

Yes, if the interior of the 3-ball has to be included as well as the boundary, you're right. The boundary itself, the 2-sphere, can still be a connected set, but its interior cannot.

 

[Dale]

What Peter is saying is that a curve of constant (r,theta,phi) with t varying is not a geodesic - it has proper acceleration. A curve of constant (t,theta,phi) with varying r is also not a spacelike geodesic. Do you disagree with either of these statements?

I agree with the first, but I question the second. Although I haven't solved the geodesic equation to prove it I think that the second would have to be a geodesic by symmetry.

 

[PeterDonis]

What Peter is saying is that a curve of constant (r,theta,phi) with t varying is not a geodesic - it has proper acceleration. A curve of constant (t,theta,phi) with varying r is also not a spacelike geodesic.

Those two statements are both true, but they're actually not what I was saying. What I was saying was that a curve of constant (t, r) but varying (theta, phi) cannot be a spacelike geodesic. Since any integral curve of one of the 3 KVFs arising from spherical symmetry must have constant (t, r) and varying (theta, phi), it follows that no integral curve of one of those KVFs can be a spacelike geodesic (contrary to what stevendaryl had said in the post I was responding to). The worldline of an object in a free-fall orbit can be a (timelike) geodesic, but such a worldline will not have constant t, only constant r (and of course it will have varying theta, phi), and it will not be the integral curve of any KVF.

 

[PAllen]

I agree with the first, but I question the second. Although I haven't solved the geodesic equation to prove it I think that the second would have to be a geodesic by symmetry.

Oops, you are right (I just did work out the geodesic equations). I had blithely assumed that since radial spacelike geodesics 4-orthogonal to various free-fallers are r as f(t), that that covered all radial spacelike geodesics. So: constant (t,theta,phi) varying r is a spacelike geodesic, and definitely satisfies the geodesic equation with a suitably chosen affine parameter (that obviously cannot be t).

[above analysis only done for exterior region].

 

[PeterDonis]

constant (t,theta,phi) varying r is a spacelike geodesic

Hmm, yes, I missed this one.

[above analysis only done for exterior region].

The geodesic equation is the same in the interior region, so constant (t, theta, phi) varying r should still be a geodesic there--a timelike one instead of a spacelike one.

 

[Dale]

Oops, you are right (I just did work out the geodesic equations).

Thanks for doing that! That is helpful.

 

[Dale]

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.

I don't think that the Schwarzschild spacetime is contracting inside the EH. Is that just an assumption, or do you have a reference or a calculation that supports that?

Also, I am kind of lost as to the purpose of the recent lines of discussion. We seem to agree that the interior of a black hole has a spacelike KVF, is therefore not static, and yet is vacuum. We also seem to agree that Birkhoff's theorem, properly stated, avoids claiming otherwise. We seem to agree that the standard Schwarzschild coordinates may not be the best ones to use inside the EH. So what remains? I see that something still bothers you, but the objections you are bringing up recently seem like just throwing out random concepts unrelated to any of the previous discussion.

 

[TrickyDicky]

Can you clarify what you mean? This makes no sense to me. A circular orbit world line advances steadily in proper time and in SC coordinate time. It is a helix in SC coordinates.

What Peter is saying is that a curve of constant (r,theta,phi) with t varying is not a geodesic - it has proper acceleration. A curve of constant (t,theta,phi) with varying r is also not a spacelike geodesic. Do you disagree with either of these statements?

Hmmm, do you agree that in Schwarzschild solution there are circular orbits(stable only up to 6GM and unstable from 6 to 3GM). These are usually considered geodesics, right? And according to stevendaryl having constant r, theta, phi they correspond to integral curves of timelike KVFs.

 

[Dale]

Hmmm, do you agree that in Schwarzschild solution there are circular orbits(stable only up to 6GM and unstable from 6 to 3GM). These are usually considered geodesics, right? And according to stevendaryl having constant r, theta, phi they correspond to integral curves of timelike KVFs.

A circular orbit is a geodesic and it does have constant r and theta, but phi and t vary. They are not integral curves of any KVF.

I really don't understand this desire to link geodesics with KVFs. They are unrelated. I guess this is the part that seems random to me. I just don't get it at all.

 

[TrickyDicky]

I don't think that the Schwarzschild spacetime is contracting inside the EH. Is that just an assumption, or do you have a reference or a calculation that supports that?

It is just the educated guess that all 4-spacetimes that are not stationary(not including the cosmological constant in the EFE) must have an either expanding or contracting 3-space volume. But maybe this is not the case here.

Also, I am kind of lost as to the purpose of the recent lines of discussion. We seem to agree that the interior of a black hole has a spacelike KVF, is therefore not static, and yet is vacuum. We also seem to agree that Birkhoff's theorem, properly stated, avoids claiming otherwise. We seem to agree that the standard Schwarzschild coordinates may not be the best ones to use inside the EH. So what remains? I see that something still bothers you, but the objections you are bringing up recently seem like just throwing out random concepts unrelated to any of the previous discussion.

I'm trying to tie up all the (for me) loose ends. That's all.

 

[PeterDonis]

Hmmm, do you agree that in Schwarzschild solution there are circular orbits(stable only up to 6GM and unstable from 6 to 3GM). These are usually considered geodesics, right?

Yes.

And according to stevendaryl having constant r, theta, phi they correspond to integral curves of timelike KVFs.

That post by stevendaryl was in error. See the whole series of recent posts from me, PAllen, and DaleSpam on that subject.