Black Holes - the two points of view.

[TrickyDicky]

You certainly didn't show that. To do that, you would have to write down the vector field and the metric in both coordinates, compute the Lie derivative of the metric wrt the vector field in each, and show that it is 0 in one and non-zero in the other.

I'm afraid you are not understanding what I'm saying, I have shown exactly what I said, not what you want me to show.
My point about coordinate-dependance was (as explained in a previous post) about the causal nature (the type if you will ) of the KVF, not about the existence or not of a KVF regardless of its causal character (timelike or spacelike).

 

[Dale]

My point about coordinate-dependance was (as explained in a previous post) about the causal nature (the type if you will ) of the KVF, not about the existence or not of a KVF regardless of its causal character (timelike or spacelike).

You didn't show that either. To do that, you would have to write down the vector field and the metric in both coordinates, compute the norm of the vector in each, and show that it has a different sign in one set of coordinates than in the other.

 

[TrickyDicky]

OK, and what is the point? Are you trying to claim that these different coordinate charts change the KVFs somehow? If so, simply citing them is insufficient.

No, as I said I'm only referring to the dependence of the causal character of a certain KVF on the coordinate chart.

 

[TrickyDicky]

You didn't show that either. To do that, you would have to write down the vector field and the metric in both coordinates, compute the norm of the vector in each, and show that it has a different sign in one set of coordinates than in the other.

For what I'm saying I think it is enough showing the dependence or lack of, of the metric coefficients of the line element (for coordinates covering the same region of the spacetime) on time.

 

[Dale]

Not if you are making a claim about the causal nature of the KVFs. Then you need to calculate the norm of the KVF in each coordinate system.

 

[TrickyDicky]

Maybe this reasoning helps to understand what I mean, of course if there is some flaw in it I'm sure you guys will point it out to me: different slicings of a spacetime ( in the case the spacetime is curved) cut the light cones in different ways altering the distribution of the inside and the outside of the cone and therefore of the vector field norms.
In the case the spacetime is flat like Minkowski's one has to chose patches from different regions of the spacetime to allow for non-staticity, see for instance the Milne universe that is a patch that corresponds to the interior of the Minkowski light cone and it is expanding.

But in the presence of curvature, one could (maybe not in all cases) pick the slicings that allow the same region of a spacetime to be static with one slicing and non-static with another.

 

[PAllen]

Maybe this reasoning helps to understand what I mean, of course if there is some flaw in it I'm sure you guys will point it out to me: different slicings of a spacetime ( in the case the spacetime is curved) cut the light cones in different ways altering the distribution of the inside and the outside of the cone and therefore of the vector field norms.
In the case the spacetime is flat like Minkowski's one has to chose patches from different regions of the spacetime to allow for non-staticity, see for instance the Milne universe that is a patch that corresponds to the interior of the Minkowski light cone and it is expanding.

But in the presence of curvature, one could (maybe not in all cases) pick the slicings that allow the same region of a spacetime to be static with one slicing and non-static with another.

No, I don't think any of this is true. A KVF is defined over spacetime. Different slicings and coordinates will relabel points, and modify the component expression of the KV at each point, but the timelike/spacelike nature of the KV at a particular event will never change.

 

[PAllen]

As the above statement indicates, there are also an infinite number of timelike KVFs on de Sitter spacetime, just as there are on Minkowski spacetime; but in de Sitter spacetime, each timelike KVF has its own "static region" in which it is timelike, and which is bounded by its own cosmological horizon on which the KVF becomes null (and then spacelike beyond the horizon).

Ok, now I am a bit confused. Both by definition (there exists a 'global' timelike KVF) and understanding (a timelike direction in which the metric doesn't change), any region that can be covered by these coordinates inside a cosmological horizon is static (since the KVF is also irrotational) , no scare quotes. The fact that some other coordinates don't manifest this in the metric expression shouldn't change this any more than Lemaitre coordinates cancel the static character of the exterior SC region.

 

[Dale]

different slicings of a spacetime ( in the case the spacetime is curved) cut the light cones in different ways

Yes, even in the case of a flat spacetime.

altering the distribution of the inside and the outside of the cone

No, the same events which are inside the cone remain inside the cone, as do the events which are outside. What changes is only which pairs of events are considered simultaneous or not.

and therefore of the vector field norms.

No, the norm of any given vector field at any given event will not change at all under any coordinate transform.

In the case the spacetime is flat like Minkowski's one has to chose patches from different regions of the spacetime to allow for non-staticity

Minkowski spacetime is static everywhere.

But in the presence of curvature, one could (maybe not in all cases) pick the slicings that allow the same region of a spacetime to be static with one slicing and non-static with another.

No, this is not the case. If there is a timelike KVF at a given event in one coordinate system then there is a timelike KVF at that event in all coordinate systems.

 

[Dale]

TrickyDicky, why don't we work through the following exercise. Please pick any spacetime with two coordinate systems which you believe illustrate your point in some region of the manifold. Then, let's take anyone timelike KVF in that region, write it in terms of both sets of coordinates, and calculate the Lie derivative and the norm in each set of coordinates.

 

[erasrot]

Ok, I will assume it is correct that in Schwarzschild spacetime the difference between the proper times of any two free falling observers is finite. Would it means that any such an observer sees that the collapse ends after a finite amount of his proper time?

 

[PAllen]

Ok, I will assume it is correct that in Schwarzschild spacetime the difference between the proper times of any two free falling observers is finite. Would it means that any such an observer sees that the collapse ends after a finite amount of his proper time?

Any radial free faller will, in finite proper time, pass the event horizon and see earlier bodies having crossed the horizon, and (in finite proper time) meet the end of the collapse in the singularity. You never 'see' the singularity before reaching it in the following sense: there are no future directed null paths from the singularity. Further, you never see light that was emitted from closer to the singularity than where you are now. You will actually see objects in the direction of the singularity, but the light you receive from them was emitted from when they were further away from the singularity than you are now. This seemingly impossible state of affairs occurs because such outward pointed light emitted within the horizon is still making its way toward the singularity, and you receive it (bump into it, as it were) as outgoing light (even though it was emitted from further away from the singularity than you are now).

 

[PeterDonis]

Maybe this reasoning helps to understand what I mean, of course if there is some flaw in it I'm sure you guys will point it out to me: different slicings of a spacetime ( in the case the spacetime is curved) cut the light cones in different ways altering the distribution of the inside and the outside of the cone and therefore of the vector field norms.

Changing the spacetime slicing does not change the light cones. More precisely, it does not change whether a particular event is inside, on, or outside the light cone of another particular event. The causal relationships between particular events are invariants. So are the norms of particular vectors at particular events.

If you're reading references that use the term "norm", you should be aware that that term has two meanings. One is the sense in which DaleSpam used it: the norm of a vector is its invariant length. But "norm" can also mean "a vector that is orthogonal to some other vector or surface". Changing the spacetime slicing can change which particular surface you are using to judge orthogonality, so it can change a "norm" in the second sense, but not in the first sense.

In the case the spacetime is flat like Minkowski's one has to chose patches from different regions of the spacetime to allow for non-staticity, see for instance the Milne universe that is a patch that corresponds to the interior of the Minkowski light cone and it is expanding.

It is only "expanding" in the sense that the family of timelike curves that pick out worldlines of observers "at rest" in Milne's coordinates has non-zero (positive) expansion. That has nothing to do with whether the spacetime itself has zero or non-zero curvature, or is or is not static, in a particular region. You need to really take some time to understand these distinctions.

But in the presence of curvature, one could (maybe not in all cases) pick the slicings that allow the same region of a spacetime to be static with one slicing and non-static with another.

Nope, can't be done. Please re-read the previous posts by me, DaleSpam, and PAllen, where we have repeatedly given you the proper definition of "static" for a region of spacetime, and shown how staticity by that definition is invariant, independent of coordinate charts, spacetime slicing, etc.

 

[PeterDonis]

Ok, now I am a bit confused. Both by definition (there exists a 'global' timelike KVF) and understanding (a timelike direction in which the metric doesn't change), any region that can be covered by these coordinates inside a cosmological horizon is static (since the KVF is also irrotational) , no scare quotes. The fact that some other coordinates don't manifest this in the metric expression shouldn't change this any more than Lemaitre coordinates cancel the static character of the exterior SC region.

Once you've picked out a particular timelike KVF, yes, all this is true. But Schwarzschild spacetime only has one KVF that is timelike anywhere, and only one region in which it is timelike. In spacetimes where there are multiple timelike KVFs, things get more complicated: each one has to be looked at separately to see in which region of the spacetime it is timelike, and the answer may be different for different ones.

For example, Minkowski spacetime (we'll stick with that comparison first since it's easier) has two infinite families of KVFs which can be timelike, and they have different behaviors:

(1) All of the timelike KVFs in the first family (the ones corresponding to inertial observers) are timelike everywhere in the spacetime. So the entire spacetime is static with respect to anyone of them.

(2) The KVFs in the second family (the ones corresponding to Rindler observers) are only timelike in particular regions of the spacetime, and the regions are different for different KVFs in the family. For example, consider the following two members of the family, described with respect to a global inertial frame (I'll describe their integral curves since that's easier to write down):

KVF A has integral curves which are hyperbolas satisfying x^2 - t^2 = K, where K is a constant ranging from minus infinity to infinity. Obviously if K > 0 and x > 0, this family of curves corresponds to the worldlines of Rindler observers in Region I of the spacetime, which I'll actually call Region I-A. But there are three other branches of these hyperbolas, corresponding to Regions II-A, III-A, and IV-A, and the KVF is timelike only in regions I-A and III-A; it is spacelike in regions II-A and IV-A, and null on the Rindler horizons, x = +/- t. So with respect to this KVF, the spacetime is static only in Regions I-A and III-A.

KVF B has integral curves which are hyperbolas satisfying x^2 - (t - 1)^2 = K, where K again ranges from minus infinity to infinity. Obviously these are similar to the above hyperbolas, but with the Rindler horizon shifted along the time axis by 1 unit, to x = +/- (t - 1). This shifts all four regions accordingly: the spacetime, with respect to *this* KVF, is now static only in Regions I-B and III-B, which are *different* regions than I-A and III-A.

Similar remarks apply to de Sitter spacetime; here we can choose any spatial point to be the origin, r = 0, of the static chart that TrickyDicky wrote down the line element for. Each possible origin corresponds to a different member of the corresponding family of KVFs on de Sitter spacetime, and each member of the family is timelike in a different region. This is why I say that this family of KVFs in de Sitter spacetime corresponds to the "Rindler family" of KVFs on Minkowski spacetime. Which of course leaves open the question of which family of KVFs on de Sitter spacetime corresponds to the "Minkowski family" of KVFs on Minkowski spacetime (the one corresponding to inertial observers).

 

[Dale]

For example, Minkowski spacetime (we'll stick with that comparison first since it's easier) has two infinite families of KVFs which can be timelike, and they have different behaviors:

(1) All of the timelike KVFs in the first family (the ones corresponding to inertial observers) are timelike everywhere in the spacetime. So the entire spacetime is static with respect to anyone of them.

(2) The KVFs in the second family (the ones corresponding to Rindler observers) are only timelike in particular regions of the spacetime, and the regions are different for different KVFs in the family. For example, consider the following two members of the family, described with respect to a global inertial frame (I'll describe their integral curves since that's easier to write down):

This is why I specified taking one KVF in my proposed exercise in post 360. That KVF will have the same norm in both coordinate systems and will be a KVF in both coordinate systems, even if there are other KVFs.

 

[erasrot]

Thank you. I reached the same conclusion just looking at the corresponding Penrose diagram. And I think it also leads to the conclusion that no observer (asymptotic or free falling) is able to actually see the formation of a black hole, even if the collapse happen in a finite amount of his proper time. Right?

 

[PAllen]

Thank you. I reached the same conclusion just looking at the corresponding Penrose diagram. And I think it also leads to the conclusion that no observer (asymptotic or free falling) is able to actually see the formation of a black hole, even if the collapse happen in a finite amount of his proper time. Right?

Well this all depends on what you mean by 'see the formation of a black hole'. What do you mean by it? For me, I would say that any static observer, or sufficiently distant free fall observer, or non-radial free fall observer, sees a black hole form in finite time:

They see a collapse, ending with a region blacker than CMB, surrounded by an Einstein ring, with no sign of the matter that collapsed. I would call this seeing a black hole form. Note that Hawking radiation doesn't change this at all - the Hawking temperature of stellar (or larger) black hole is lower than the CMB temperature. Black holes in our universe will not decay until the heat death of the universe lowers the CMB temperature to below their Hawking temperature. Prior to that, they are actually growing via absorbing more energy from CMB radiation than they emit in Hawking radiation (and recall that the bigger a black hole is, the lower its Hawking temperature).

As for a radial free faller, if they start from far enough away, they see all of the above. Then, when they get very close to the horizon, it starts to brighten and they soon see objects that fell in shortly before them. They continue to see these objects appearing in front of them (if they are looking where the singularity is) until they splat on the singularity.

(Of course, I ignore tidal forces - I assume idealized point observers and point objects of negligible mass).

 

[erasrot]

Well this all depends on what you mean by 'see the formation of a black hole'. What do you mean by it? For me, I would say that any static observer, or sufficiently distant free fall observer, or non-radial free fall observer, sees a black hole form in finite time:

They see a collapse, ending with a region blacker than CMB, surrounded by an Einstein ring, with no sign of the matter that collapsed. I would call this seeing a black hole form. Note that Hawking radiation doesn't change this at all - the Hawking temperature of stellar (or larger) black hole is lower than the CMB temperature. Black holes in our universe will not decay until the heat death of the universe lowers the CMB temperature to below their Hawking temperature. Prior to that, they are actually growing via absorbing more energy from CMB radiation than they emit in Hawking radiation (and recall that the bigger a black hole is, the lower its Hawking temperature).

As for a radial free faller, if they start from far enough away, they see all of the above. Then, when they get very close to the horizon, it starts to brighten and they soon see objects that fell in shortly before them. They continue to see these objects appearing in front of them (if they are looking where the singularity is) until they splat on the singularity.

(Of course, I ignore tidal forces - I assume idealized point observers and point objects of negligible mass).

Before answering your question, please, allow me one more: how does that picture is different from what happen for any other sufficiently compact (but not BH) object?

 

[PAllen]

Before answering your question, please, allow me one more: how does that picture is different from what happen for any other sufficiently compact (but not BH) object?

Any object that that looks like I describe is a black hole as I see it. If it has all the properties GR predicts for a black hole, what else do you want to call it? It doesn't have the properties of neutron star, for example. Within currently accepted theories there is only one type of object with these properties. Any stable collapsed object that is not a BH has (per currently accepted theories) very different properties than a BH.

[Note: if it is even 1% bigger than its SC radius, you could still bounce signals off it. If its mass consistent with a BH for its '(in)visible' horizon, by current theory, it is a BH. If you want to say: what if GR is wrong? That is a whole separate question. The question was: can you see a BH form? Since BH is an object of GR, I take that to mean can you see something form which GR says must be black hole, and has the right properties ? The answer is yes, as I see it.]

 

[TrickyDicky]

Yes, even in the case of a flat spacetime.

No, the same events which are inside the cone remain inside the cone, as do the events which are outside. What changes is only which pairs of events are considered simultaneous or not.

No, the norm of any given vector field at any given event will not change at all under any coordinate transform.

Minkowski spacetime is static everywhere.

No, this is not the case. If there is a timelike KVF at a given event in one coordinate system then there is a timelike KVF at that event in all coordinate systems.

The problem is I'm not disputing any of this. I'm not saying that an event changes in a coordinate transformation.
I'm talking about the obvious fact that one can use coordinates that pick different regions of a spacetime and also flat or curved spacelike foliations so that the coordinates are static or non-static.

 

[TrickyDicky]

Changing the spacetime slicing does not change the light cones. More precisely, it does not change whether a particular event is inside, on, or outside the light cone of another particular event. The causal relationships between particular events are invariants. So are the norms of particular vectors at particular events.

I am not saying otherwise.

It is only "expanding" in the sense that the family of timelike curves that pick out worldlines of observers "at rest" in Milne's coordinates has non-zero (positive) expansion. That has nothing to do with whether the spacetime itself has zero or non-zero curvature, or is or is not static, in a particular region. You need to really take some time to understand these distinctions.

The surfaces of constant t >0 are hyperboloids in Minkowski coordinates, all located within the forward lightcone. So the Milne coordinates cover only one quarter of Minkowski spacetime.
You seem more interested in repeating your mantra and proving me wrong than understanding what I'm saying. So be it.

 

[Dale]

The problem is I'm not disputing any of this. I'm not saying that an event changes in a coordinate transformation.
I'm talking about the obvious fact that one can use coordinates that pick different regions of a spacetime and also flat or curved spacelike foliations so that the coordinates are static or non-static.

Coordinates, yes. KVFs, no.

A KVF retains it's causal nature regardless of coordinate system.

 

[PeterDonis]

The problem is I'm not disputing any of this. I'm not saying that an event changes in a coordinate transformation.
I'm talking about the obvious fact that one can use coordinates that pick different regions of a spacetime and also flat or curved spacelike foliations so that the coordinates are static or non-static.

Ok, that helps to clarify things. I don't have a problem with any of this.

The surfaces of constant t >0 are hyperboloids in Minkowski coordinates, all located within the forward lightcone. So the Milne coordinates cover only one quarter of Minkowski spacetime.

Ok, this is fine too.

You seem more interested in repeating your mantra and proving me wrong than understanding what I'm saying. So be it.

I keep repeating the mantra because you keep appearing to contradict it, or at least to think there is some inconsistency or issue between what you're saying and what I've been saying. If all you were trying to say was the above, there's no inconsistency or issue at all that I can see. If you think you see one, then you're right, I don't understand what you're saying.

It might help to use standard terminology, or at least to be clear about what the terms you're using apply to: saying that a coordinate chart is static or non-static is very different from saying a region of spacetime is static or non-static, which is different again from saying that a family of timelike curves has zero or non-zero expansion. I have been quite confused trying to figure out which one you are talking about from post to post, since you have been using the same terms for all three concepts without any clarification. Similar remarks apply to your use of the word "norm", as I explained in a prior post.

 

[PAllen]

Any object that that looks like I describe is a black hole as I see it. If it has all the properties GR predicts for a black hole, what else do you want to call it? It doesn't have the properties of neutron star, for example. Within currently accepted theories there is only one type of object with these properties. Any stable collapsed object that is not a BH has (per currently accepted theories) very different properties than a BH.

[Note: if it is even 1% bigger than its SC radius, you could still bounce signals off it. If its mass consistent with a BH for its '(in)visible' horizon, by current theory, it is a BH. If you want to say: what if GR is wrong? That is a whole separate question. The question was: can you see a BH form? Since BH is an object of GR, I take that to mean can you see something form which GR says must be black hole, and has the right properties ? The answer is yes, as I see it.]

Adding to this, I bring up yet again super massive BH. At the point of horizon formation, density is quite modest (unlike stellar mass BH). So, if we observer (as is hoped over the next decade) a surface with all expected horizon properties for galactic supermassive BH, what alternative is there to GR model (at least up to horizon; at least macroscopically). Since the densities are far less than the outer layers of an ordinary star, it is really hard to conceive of a collapsed object alterntative to BH horizon.

 

[Mike Holland]

Austin0, I gather that I did not reply to this post of yours.

you say the falling observers clock is never stopped in either frame because the distant observers clock never reaches infinity.
I agree. but you seem to ignore the fact that this is only true in the region where the faller has NOT reached the singularity.
you then want to magically have the faller PASS the horizon without ever having reached it.

It appears you interpret time dilation in a way that creates alternate contradictory realities.
If your premise that reaching the horizon requires infinite coordinate time for the distant observer is correct, that means that at all points in that interval the times at the two locations will be related by the SC metric. Both observers will agree on these relative elapsed times and both observers will agree that the faller has not reached the horizon.

The answer here is that all points on the two time scales ARE related by the SC metric, all the points from 0 to infinity on the distant observer's clock are rerelated to the points from 0 to T on the faller's clock, where T is his local time when he gets to the horizon. Obviously it is not a linear relationship, more like a tangent graph where tangent goes to infinity as angle goes to 90 degrees, and so they don't agree on relative elapsed times. Each sees the other's clock ticking at a different rate to his own, an ever increasing difference.

 

[PeterDonis]

Each sees the other's clock ticking at a different rate to his own, an ever increasing difference

This isn't quite correct; SC coordinates can be thought of as the "natural" ones for the distant observer, but they are not the "natural" ones for the infalling observer. So it's not really correct to equate SC coordinate values to anything the infalling observer "sees".

 

[ianpaul12345]

i have a question, if protons are massless why in the event Horison are they unable to escape. what is grabbing hold of um, they have no mass? why can they no escape

 

[ianpaul12345]

it has mass or it or it don't pick one

 

[Nugatory]

i have a question, if protons are massless why in the event Horizon are they unable to escape. what is grabbing hold of um, they have no mass? why can they no escape

You said "proton", but I think that might be a typo and you meant to say "photon", as protons are most certainly not massless.

Assuming that you meant to say "photon"... In general relativity gravity is not treated as a force that grabs onto massive objects. Instead space and time are curved so that anything moving through them is affected by the curvature; whether massive or massless makes no distance. Below the event horizon, the curvature is such that no paths, even that of a massless photon moving at the speed of light, cross the event horizon into the space outside.

 

[Dale]

i have a question, if protons are massless why in the event Horison are they unable to escape. what is grabbing hold of um, they have no mass? why can they no escape

Nugatory gave a good response for GR, but even classical Newtonian gravity could in principle affect photons. The acceleration of gravity in Newtonian physics is independent of the mass, so even a photon would be accelerated the same.

 

[Saw]

... even classical Newtonian gravity could in principle affect photons. The acceleration of gravity in Newtonian physics is independent of the mass, so even a photon would be accelerated the same.

An acceleration, however, which would happen without a force, since the product M x m would be zero... (?)

 

[PAllen]

An acceleration, however, which would happen without a force, since the product M x m would be zero... (?)

For small test bodies, you normally divide out the the test body m anyway to get an acceleration formula rather than a force formula. For astronomy purposes, a many body approximation is often written in terms of acceleration rather than force. In any case, to take the limit as m/M->0, you just divide out m.

 

[Dale]

An acceleration, however, which would happen without a force, since the product M x m would be zero... (?)

Yes, of course. Consider Newtons 2nd law; how much force does it take to produce a finite acceleration on a massless object?