Notions of simultaneity in strongly curved spacetime

[zonde]

But in the static coordinate system (SC coordinates), the metric is diagonal; that means the surfaces of constant SC time are orthogonal to the worldlines of objects with static spatial coordinates. And *that* means the definition of "now" given by SC coordinates is the *same* as the definition of "now" given by the local inertial frames along the worldlines of objects with static spatial coordinates.

In the non-static coordinate system (GP coordinates), the metric is not diagonal; there is a dt dr "cross term" in the line element. That means the surfaces of constant GP time are *not* orthogonal to the worldlines of objects with static spatial coordinates. And that means the definition of "now" given by GP coordinates is *different* than the definition of "now" given by the local inertial frames along the worldlines of objects with static spatial coordinates.

So the sense in which the definition of "now" given by static (SC) coordinates is "unique" is that it is the only one that matches up with the definition of "now" in the local inertial frames of static observers.

So you are saying that the catch is that GP coordinates are non-static. But I don't see anything non-static about them. Slices of "now" are identical as we go along time coordinate.

I think that the catch is that in GP coordinates time coordinate is not orthogonal to space (radial) coordinate and in that sense they are not "right".

 

[zonde]

In my experience, only opinions about verifiable facts can be argued in a convincing way for those who are of a contrary opinion. Do you disagree?

Made me think once more about such question: what are observable differences between "frozen star" and "black hole" for distant observer? And I think that there are none.
But then there is an argument that "black hole" is somehow more correct extrapolation of known physical laws beyond limits of testability than "frozen star". This type of discussion is just empty.

Reminds me of Feynman's comments on the field of gravity, back in the 60's in his private letter to his wife.

I am learning nothing. Because there are no experiments this field is not an active one, so few of the best men are doing work in it. The result is that there are hosts of dopes here and it is not good for my blood pressure: such inane things are said and seriously discussed here that I get into arguments outside the formal sessions (say, at lunch) whenever anyone asks me a question or starts to tell me about his "work". The "work" is always: (1) completely un-understandable, (2) vague and indefinite, (3) something correct that is obvious and self evident, but a worked out by a long and difficult analysis, and presented as an important discovery, or, a (4) claim based on the stupidity of the author that some obvious and correct fact, accepted and checked for years, is, in fact, false (these are the worst: no argument will convince the idiot), (5) an attempt to do something probably impossible, but certainly of no utility, which it is finally revealed at the end, fails (dessert arrives and is eaten), or (6) just plain wrong. There is great deal of "activity in the field" these days, but this "activity" is mainly in showing that the previous "activity" of somebody else resulted in an error or in nothing useful or in nothing promising. It is like a lot of worms trying to get out of a bottle by crawling all over each other. It is not that the subject is hard; it is that the good men are occupied elsewhere. Remind me not to come to any more gravity conferences!

There has to be something to discuss that is within limits of testability.

 

[PeterDonis]

So you are saying that the catch is that GP coordinates are non-static. But I don't see anything non-static about them. Slices of "now" are identical as we go along time coordinate.

That's not enough for coordinates to be static. The slices of "now" also have to be orthogonal to the integral curves of the time coordinate, and as you observe, they're not.

 

[PeterDonis]

Made me think once more about such question: what are observable differences between "frozen star" and "black hole" for distant observer? And I think that there are none.

It depends on what you mean by "frozen star". Most people, when they use that term, really mean the *same* thing as other people mean by "black hole". In other words, they are using exactly the same spacetime and exactly the same solution of the EFE--their model of the physics is the same. They are just interpreting it differently. But since they're using the same model of the physics, they will make the same predictions for all observables. The difference is just a matter of interpretation.

(IMO, even the "difference in interpretation" is somewhat strained, since the "frozen star" people agree that an object falling into the hole/frozen star/whatever it is will experience only a finite amount of proper time to the horizon. That means that if we use coordinates that are not singular at the horizon, such as GP coordinates, we can assign a *finite* time to the event of any infalling object crossing the horizon. But I don't think we'll get any further with that discussion here.)

It might be possible to come up with a *different* model of a "frozen star", one which used a *different* spacetime and a *different* solution of the EFE, which actually made different physical predictions about what we would see because it was using a different model of the physics. But I've never seen one.

 

[zonde]

It depends on what you mean by "frozen star". Most people, when they use that term, really mean the *same* thing as other people mean by "black hole". In other words, they are using exactly the same spacetime and exactly the same solution of the EFE--their model of the physics is the same. They are just interpreting it differently. But since they're using the same model of the physics, they will make the same predictions for all observables. The difference is just a matter of interpretation.

Hmm, the way you say it gives me feeling that I am just fussing without much reason.

I guess it could be just enough to settle for meaning of "frozen star" as predictions of GR that are testable (falsifiable) by observations from Earth or at least without going on the suicide mission.

 

[harrylin]

Are you including the events that Eve calculates must exist, but can't receive light signals from (i.e,. events behind the Rindler horizon), in her "view of reality"?

Certainly not: "there is no time for Eve when, in her co-moving inertial reference frame, Adam passes through the horizon. In that sense, Eve could claim that Adam never reaches the horizon as far as she's concerned." It is specified that Eve uses a "co-moving inertial reference frame", which is only justified if she thinks that she is not accelerating. The author suggests further on in his discussion that she should adopt Adam's perception of reality. I agree that that is a more sensible approach, but my opinion is based on the fact that her "gravitational field" looks fictive to me: there is no physical cause that could allow for a difference from SR.

You're correct that Eve and Adam are in physically different states of motion. I'm not sure how that impacts their ability to have a "symmetrical interpretation". Both can make the same computations.

Perhaps we simply misunderstood each other on words; I'm not sure. Probably everyone proposes an asymmetry for such cases.
"They can make the same computations" is very much the "twin paradox". As Einstein explained, a symmetrical interpretation for an asymmetrical physical situation is incompatible with the foundations of GR. In fact, I don't know any theory of physics that violates that principle. In his discussion of the twin paradox (which was, I think, the first time that Einstein gave reason to doubt the reality of his "induced gravitational fields") he details how different the physical interpretation of a gravitational field is from that of acceleration; only the observable phenomena are held to be identical. And that brings me to a related point (I rearrange):

No, according to Adam, clocks at different locations in Eve's accelerating rocket are moving at different speeds. The clock at the nose of Eve's rocket is moving more slowly, according to Adam, than the clock at the tail of the rocket; so the clock at the nose will be ticking faster, according to Adam, than the clock at the tail (slower motion = less time dilation).

I did not say the contrary - and the point that I tried to make is obscured by your precision. I'll try again. In this illustration it is assumed that Adam uses his newly found rest frame as reference for physical reality. At the moment that Eve starts accelerating away, Adam ascribes the frequency difference that Eve observes to "classical" Doppler; according to him, her rocket has still negligible length contraction so that her clocks go at nearly equal rate. In contrast, Eve claims to be in rest and ascribes the frequency difference to the effect of a gravitational field which makes her clocks go at a different rate. This is just to illustrate how a different interpretation of gravitational fields and acceleration is both necessary and understood.

[..] the reason I brought it up was that it appeared that you were contradicting the equivalence principle). [..] What "inverse reasoning". Spell it out, please.

After some of you brought it up, I elaborated on the equivalence principle because you and several others seem to interpret it as requiring that we can make gravitational fields from matter "vanish" (which is simply wrong), and you seem to deny the physical reality of gravitational fields of matter. You thus claimed that in 1916GR, 'gravitation" can be turned into "acceleration" by changing coordinates'. That is the inverse of the equivalence principle that I have seen proposed in GR (of course, I may have just missed it; if so, please cite it!). GR is based on the assumption of physical reality of gravitational fields and the equivalence between acceleration and a homogeneous gravitational field. What Einstein originally denied was the physical reality of acceleration, which he thought could be "relativised" by pretending that instead a homogeneous gravitational field is induced. If we hold that GR was wrong on that last point, then that merely makes such induced fields "fictive" and acceleration "absolute". Einstein warned for a misconception that you seem to hold, and which I'll now cite it in full:

"From our consideration of the accelerated chest we see that a general theory of relativity must yield important results on the laws of gravitation. In point of fact, the systematic pursuit of the general idea of relativity has supplied the laws satisfied by the gravitational field. Before proceeding farther, however, I must warn the reader against a misconception suggested by these considerations. A gravitational field exists for the man in the chest, despite the fact that there was no such field for the co-ordinate system first chosen. Now we might easily suppose that the existence of a gravitational field is always only an apparent one. We might also think that, regardless of the kind of gravitational field which may be present, we could always choose another reference-body such that no gravitational field exists with reference to it. This is by no means true for all gravitational fields, but only for those of quite special form. It is, for instance, impossible to choose a body of reference such that, as judged from it, the gravitational field of the Earth (in its entirety) vanishes."

[...] Do you really not understand what the physical meaning of "proper time" is? It's at the foundation of the physical interpretation of relativity.

I wonder what you mean with "physical"; certainly nothing measurable! But now that also you talk of "the foundation of the physical interpretation of relativity": I may have overlooked it but I do not find the word "proper" in either "The Foundation of the Generalised Theory of Relativity" or "Relativity: The Special and General Theory". One should expect something that is at the foundation of the physical interpretation of relativity to be easy to find. So: reference please!

Do you think "spacetime" itself is a "mere mathematical term"?

"Mere mathematical" in the sense of Applied Mathematics? Not only I do I think so, GR is based on such thinking:

"The non-mathematician is seized by a mysterious shuddering when he hears of "four-dimensional" things, by a feeling not unlike that awakened by thoughts of the occult. [..] the world of physical phenomena which was briefly called "world" by Minkowski is naturally four dimensional in the space-time sense. For it is composed of individual events, each of which is described by four numbers" - Relativity:The Special and General Theory

A quick comment: do you see how this statement of Einstein's makes an implicit assumption that it is *possible* for a clock to be "kept at this place" (i.e., at the horizon). Have you considered what happens if that assumption is false--i.e., if a clock *cannot* be "kept" at the horizon (because it would have to move at the speed of light to do so, and no clock can move at the speed of light)?

That is completely wrong: he makes no such implicit assumption. Following your misunderstanding, Einstein would have meant with "For v=c all moving objects—viewed from the “stationary” system—shrivel up into plane figures" that it is *possible* for an object to move at c.

[..] I thought that by "Kruskal observer" you meant "someone calculating things using the Kruskal chart". [..] If you mean "coordinate chart", then say "coordinate chart". "Observer" does not mean "coordinate chart". [..]

PAllen introduced the Kruskal chart as giving a picture that differs from the equations of Oppenheimer. However I did not mean "coordinate chart", as I distinguish a chart from the opinion of the user of such a chart - PAllen suggested that the user of a Kruskal chart interprets the inside area as physical reality. Charts do not catch the topic of this thread which concerns human notions. However:

it will become more evident that many of the things you are saying are dependent on which chart you use, meaning that they're not statements about actual physics, just about coordinate charts.

The discussion of this thread appears to be about metaphysics. In order for the mentors not to close it, we should see if there is anything left related to this topic that is either physical in a verifiable sense, or pertaining to official GR theory. But even then, it may be better to start a fresh thread on that, as this thread is getting rather long.

 

[PeterDonis]

It is specified that Eve uses a "co-moving inertial reference frame", which is only justified if she thinks that she is not accelerating.

Eve *is* accelerating; she feels a nonzero acceleration, an accelerometer attached to her reads nonzero, if she stood on a scale it would register weight, etc. There is no way in which she "thinks she is not accelerating".

The term "co-moving inertial reference frame" is more precisely stated as "momentarily co-moving inertial reference frame" (MCIF). You can construct such a frame centered on any event on Eve's worldline; it will be the inertial frame in which Eve is momentarily at rest at the chosen event. This does not mean that Eve is not accelerating; it just means that it is easier to do a lot of the math in an inertial reference frame. The disadvantage of doing this is that, as I said, Eve is only at rest momentarily in any such frame; so if you want to look at the physics along Eve's worldline for any significant period of time, you can't use a single MCIF to do it. So the MCIF isn't a good representation of "Eve's point of view" over any significant length of Eve's time.

An MCIF is also *not* the same as Rindler coordinates; those are non-inertial. The advantage of Rindler coordinates is that they can cover all of Eve's worldline with a single coordinate chart in which Eve is "at rest" (her spatial coordinates in this chart don't change). This makes Rindler coordinates a better candidate for representing "Eve's point of view". But since the coordinates are non-inertial, they behave differently from the inertial coordinates you're used to in SR. For one thing, as has been noted before, Rindler coordinates don't cover the entire spacetime, so if Rindler coordinates describe Eve's "point of view", then as I said before, Eve's point of view is necessarily limited in a way that Adam's is not.

The author suggests further on in his discussion that she should adopt Adam's perception of reality. I agree that that is a more sensible approach, but my opinion is based on the fact that her "gravitational field" looks fictive to me: there is no physical cause that could allow for a difference from SR.

There is no "difference from SR". Rindler coordinates and Adam's coordinates (Minkowski coordinates) both describe the same spacetime; they describe the same geometric object (or at least a portion of it, in the case of Rindler coordinates). That spacetime is flat, so there is no spacetime curvature present. Whether that counts as a "fictive gravitational field", or no "gravitational field" at all, depends on how you define the term "gravitational field". The physics is the same either way.

This brings up a general comment: you are insisting on centering your reasoning around terms like "gravitational field" that are simply not fundamental according to GR. That is why you are having all these problems trying to interpret what's going on. There is no single consistent interpretation of the term "gravitational field" that matches all the physics. There just isn't. To find an interpretation that matches all the physics, you have to give up the term "gravitational field" and change your set of concepts, to include things like "spacetime curvature", "stress-energy", "Einstein Field Equation", etc. instead.

"They can make the same computations" is very much the "twin paradox". As Einstein explained, a symmetrical interpretation for an asymmetrical physical situation is incompatible with the foundations of GR.

How does making the same computations require a "symmetrical interpretation"? In the case of the twin paradox, both twins can compute that the traveling twin will have aged less when they meet up again. In other words, both twins compute an asymmetrical result. The same is true here; both observers (Eve and Adam) compute that Adam's point of view is not limited, while Eve's point of view is. What's the problem?

In this illustration it is assumed that Adam uses his newly found rest frame as reference for physical reality. At the moment that Eve starts accelerating away, Adam ascribes the frequency difference that Eve observes to "classical" Doppler; according to him, her rocket has still negligible length contraction so that her clocks go at nearly equal rate.

I didn't say Adam attributed the frequency difference to "length contraction". I said he attributed it to the fact that the nose of Eve's rocket is moving more slowly than the tail, in Adam's rest frame. A more precise description would actually be pretty much identical to "classical Doppler":

(1) A light beam emitted from the nose of Eve's rocket to the tail will look blueshifted at the tail, because the tail is accelerating towards it (in Adam's rest frame).

(2) A light beam emitted from the tail to the nose will look redshifted at the nose, because the nose is accelerating away from it (in Adam's rest frame).

Nowhere in any of this does "length contraction" appear; the reasoning applies equally well at the moment Adam drops off the rocket and at a later time when the rocket is moving at nearly the speed of light relative to Adam. If you work through the math, the observed blueshift/redshift depends only on Eve's proper acceleration; it does not depend on her instantaneous velocity relative to Adam. So it does not depend on the absolute value of her "length contraction" or "time dilation"; it only depends on the *change* in those values during the time of flight of a light beam across her rocket, which depends on her acceleration.

In contrast, Eve claims to be in rest and ascribes the frequency difference to the effect of a gravitational field which makes her clocks go at a different rate. This is just to illustrate how a different interpretation of gravitational fields and acceleration is both necessary and understood.

Yes, no problem here.

After some of you brought it up, I elaborated on the equivalence principle because you and several others seem to interpret it as requiring that we can make gravitational fields from matter "vanish" (which is simply wrong)

Agreed; the quote you gave from Einstein on this was apposite. I wasn't intending to argue about that.

and you seem to deny the physical reality of gravitational fields of matter.

I certainly didn't intend to deny that matter causes gravity; I was only bringing up issues relative to the term "gravitational field" and what it means. See my comments above.

You thus claimed that in 1916GR, 'gravitation" can be turned into "acceleration" by changing coordinates'.

I suppose I should have included the qualifier "locally", since I was really just trying to affirm the equivalence principle, and the EP only says that you can do this locally (i.e., in a small patch of spacetime centered on a particular chosen event).

That is the inverse of the equivalence principle that I have seen proposed in GR (of course, I may have just missed it; if so, please cite it!).

There are a number of different ways of stating the EP; the Wikipedia page gives a decent overview:

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

The key thing I was trying to focus in on is that, in GR, you can always set up a local inertial frame centered on a particular event, in which "the acceleration due to gravity" vanishes. More precisely, you can always set up a local inertial frame centered on a particular event in which the following is true:

(1) The metric in the local inertial frame, at the chosen event, is the Minkowski metric; i.e., it is

$$ds^2 = - dt^2 + dx^2 + dy^2 + dz^2$$

where t, x, y, z are the local coordinates in the local inertial frame, whose origin (0, 0, 0, 0) is the chosen event.

(2) The first derivatives of all the metric coefficients are zero at the chosen event; this means that the metric coefficients are the Minkowski ones not just at the chosen event, but in the entire local inertial frame.

The second condition is what ensures that there is no "apparent gravitational field" in the local inertial frame; i.e., that the worldlines of inertial objects (i.e., freely falling objects) are straight lines in the local inertial frame. But this also means that the worldlines of accelerated objects--for example, the worldlines of objects at rest on the surface of the Earth, if we set up a local inertial frame centered on some event on the Earth's surface--are *not* straight lines in the local inertial frame: in fact they are hyperbolas, just like Eve's worldline in Adam's frame. This is the sense in which, locally, we can "make gravity look like acceleration"; we are making objects that are static in the local gravitational field look like accelerated objects in flat spacetime [Edit: and we are also making objects that are freely falling in the local gravitational field, and hence are "accelerating" from the viewpoint of an observer static in the field, look like objects at rest in an inertial frame in flat spacetime.]

But the local inertial frame only covers a small piece of spacetime around the chosen event; how small depends on how curved the spacetime is and how accurate our measurements of tidal gravity are. Spacetime curvature, i.e., tidal gravity, depends on the *second* derivatives of the metric coefficients, and those *cannot* all be set to zero by any choice of coordinates if the spacetime is curved. This is the sense in which we *cannot* "make gravity vanish" by choosing coordinates; the curvature will always be there, as in Einstein's example of being unable to make the gravitational field of the Earth vanish in its entirety by any choice of coordinates.

GR is based on the assumption of physical reality of gravitational fields

I would say it is based on the assumption of the physical reality of *spacetime* as a dynamical object. As I said above, the term "gravitational field" is problematic.

and the equivalence between acceleration and a homogeneous gravitational field.

With the qualifier "locally", and subject to reservations about the term "gravitational field", yes, this is OK.

What Einstein originally denied was the physical reality of acceleration, which he thought could be "relativised" by pretending that instead a homogeneous gravitational field is induced.

This is only true if "acceleration" is interpreted to mean "coordinate acceleration". Einstein never, AFAIK, claimed that *proper* acceleration (i.e., feeling weight, registering nonzero on an accelerometer, etc.) could be relativised. With the proper terminology, Einstein was correct: coordinate acceleration *can* be relativised (again, with the qualifier "locally"), and proper acceleration cannot (which is good since it's a direct observable).

If we hold that GR was wrong on that last point

It wasn't and isn't. See above.

Einstein warned for a misconception that you seem to hold

I don't. See above.

I wonder what you mean with "physical"; certainly nothing measurable!

You don't think proper time is measurable? What do you think your watch measures?

But now that also you talk of "the foundation of the physical interpretation of relativity": I may have overlooked it but I do not find the word "proper" in either "The Foundation of the Generalised Theory of Relativity" or "Relativity: The Special and General Theory".

If you want to learn about how a physical theory actually works, you can't depend on popular books, even if they're written by the person who invented the theory.

One should expect something that is at the foundation of the physical interpretation of relativity to be easy to find. So: reference please!

Try any relativity textbook. MTW talks extensively about proper time. So does Taylor & Wheeler's Spacetime Physics, which may be a better starting point since it is only about the fundamentals of relativity; MTW has a *lot* of other material.

"Mere mathematical" in the sense of Applied Mathematics? Not only I do I think so, GR is based on such thinking:

Einstein used the term "the world of physical phenomena". He wasn't talking about a mathematical abstraction; he was saying that *the real, actual universe* is a four-dimensional thing, which we call "spacetime".

That is completely wrong: he makes no such implicit assumption. Following your misunderstanding, Einstein would have meant with "For v=c all moving objects—viewed from the “stationary” system—shrivel up into plane figures" that it is *possible* for an object to move at c.

I have no idea what you're trying to say here. What you are claiming "follows my misunderstanding" does not follow from what I said at all, as far as I can see.

Perhaps I should belabor this some more, since it is an important point. Here's what you quoted from Einstein: "a clock kept at this place [i.e., at the horizon] would go at rate zero". I can interpret this one of two ways:

(1) Einstein is claiming that a clock can be kept at the horizon, and saying that it would go at rate zero.

(2) Einstein is claiming that a clock *cannot* be kept at the horizon, because if it could, it would go at rate zero, and that doesn't make sense.

Are you interpreting him as saying #1 or #2? If it's #1, the refutation is pretty easy: the clock would have to go at the speed of light, and no clock can do that. So let's look at the other possibility; #2 leads to one of the following:

(2a) A clock can't be kept at the horizon, because if it could, it would go at rate zero, and that doesn't make sense. Therefore the horizon can't exist, and neither can any spacetime inside it.

(2b) A clock can't be kept at the horizon, because if it could, it would go at rate zero, and that doesn't make sense. Therefore the horizon (i.e., a curve of constant r = 2m) can't be a timelike curve, because if it were, a clock could follow it as a worldline. But the horizon could still exist if it were some other type of curve, such as a null curve; and if so, there could also be a region of spacetime inside the horizon, where curves of constant r < 2m are also not timelike.

Einstein, as far as I can tell, believed #2a; but "modern GR" says #2b. I can't tell for sure what Oppenheimer and Snyder thought, since they didn't address the question in their paper; but everyone who has extended their model has come up with #2b as well.

I'll put comments on the "metaphysical" aspects of all this in a separate post.

 

[grav-universe]

(1) Einstein is claiming that a clock can be kept at the horizon, and saying that it would go at rate zero.

(2) Einstein is claiming that a clock *cannot* be kept at the horizon, because if it could, it would go at rate zero, and that doesn't make sense.

Are you interpreting him as saying #1 or #2? If it's #1, the refutation is pretty easy: the clock would have to go at the speed of light, and no clock can do that.

Just thought I'd add my two cents, but if that is your refutation of #1, then we should also add

(3) A clock can't cross the horizon, because if it could, it would have to go at the speed of light, and no clock can do that.


Of course there's still a way out. Due to the extreme gravity at the horizon, all matter might be annihilated so that only massless and timeless particles actually cross, but personally I opt for #2a.

 

[PeterDonis]

(3) A clock can't cross the horizon, because if it could, it would have to go at the speed of light

You stated this wrong. The correct statement is "a clock can't cross the horizon moving outward, because to do so it would have to go at the speed of light." That in no way prevents the clock from crossing the horizon moving inward.

Of course there's still a way out. Due to the extreme gravity at the horizon, all matter might be annihilated so that only massless and timeless particles actually cross, but personally I opt for #2a.

Then you opt incorrectly.

 

[PeterDonis]

I did not mean "coordinate chart", as I distinguish a chart from the opinion of the user of such a chart

I don't get this at all. A coordinate chart is a well-defined mathematical entity, and I understand what it describes: it describes a spacetime, or a portion of one. I don't understand what an "opinion of the user of such a chart" is--at least, not as it relates to any sort of actual physics.

PAllen suggested that the user of a Kruskal chart interprets the inside area as physical reality.

This seems very confused to me. The Kruskal chart, like any chart, maps 4-tuples of numbers to points of a geometric object. If you are trying to say that that, in itself, is "just mathematics", and doesn't necessarily have any physical interpretation, I agree. But such mathematical objects certainly serve as "building blocks" out of which we construct models that *do* have a physical interpretation. For example, we can take a portion of the geometric object described by the Kruskal chart and "glue" it together with another geometric object described by a collapsing FRW chart. The physical interpretation of this model is a spacetime containing a collapsing object such as a star, plus the vacuum region surrounding it.

Of course such a model is idealized; so is every model we use in physics. But its physical interpretation is not a matter of "opinion". Whether or not it's a *valid* model, taking its idealizations into account, is a separate question of what the physical interpretation of the model is.

Charts do not catch the topic of this thread which concerns human notions.

But coordinate charts are how we express the particular human notions that we are talking about. At least, they're a very convenient way of doing so. If you would prefer another way of expressing those notions, fine, please propose one. But you can't just punt on using coordinate charts without giving some other way of making precise, unambiguous statements about the subject under discussion.

 

[grav-universe]

You stated this wrong. The correct statement is "a clock can't cross the horizon moving outward, because to do so it would have to go at the speed of light." That in no way prevents the clock from crossing the horizon moving inward.

How do you mean? The locally measured speed also becomes precisely c right at the horizon freefalling inward with any initial speed from any r.

Then you opt incorrectly.

lol

 

[PeterDonis]

How do you mean? The locally measured speed also becomes precisely c right at the horizon freefalling inward with any initial speed from any r.

No, it doesn't, because the concept of "locally measured speed", as you are using it, no longer makes sense at the horizon. "Locally measured", as you are using the term, means "measured by an observer who is static at a given radius", and there are no observers who are static at radius r = 2m, i.e., at the horizon. Any such observer would have to be moving outward at the speed of light, and no observer can do that.

Even if you try to adjust "locally measured" to mean "measured in a local inertial frame which is instantaneously at rest at the given radius", that doesn't work at the horizon either. A local inertial frame can't even be instantaneously at rest at r = 2m, because r = 2m is a null curve, not a timelike curve; i.e., it's the path of a light ray (a radially outgoing light ray). So in any local inertial frame centered on an event at the horizon, the horizon itself will look like a radially outgoing light ray. That means any object at rest in such a local inertial frame, even instantaneously, must be moving radially inward.

 

[PAllen]

How do you mean? The locally measured speed also becomes precisely c right at the horizon freefalling inward with any initial speed from any r.

What do you mean? Who measures in infalling body crossing the horizon to go at c?

- A hovering observer just outside the horizon, measures the infaller pass at < c. At and inside horizon, they can take no measurement.
- Any other infaller whose trajectory allows them to measure the given infaller, measures the given infaller going < c.
- There is no such thing as a horizon observer (no local frame corresponds to the horizon; same as talking about the frame of light.

I can make no sense out of your statement. It is in violation of the mathematical structure of GR, which says suffuciently locally, all physics is SR, which means there is never local motion >= c for a material body (even alcubierre drive never violates this).

 

[grav-universe]

The equation of motion in GR, at least for radial freefall, as related to the time dilation is

sqrt(1 - (v'_r/c)^2) / z_r = K

where v'_r is the locally measured speed at r, z_r is the time dilation at r, and K is a constant of motion. With the initial condition of a particle freefalling from rest at infinity, for example, v' = 0 and z = 1, so K = 1. For a photon K = 0 and for any massive particle K > 0, depending upon the initial conditions for the state of motion of the particle, and K remains constant at all r whether the particle is traveling inward or outward. For a particle starting outside the horizon and falling inward, at the event horizon, where z_r = 0, in order for K to remain a finite value, then sqrt(1 - (v'_r/c)^2) = 0 also, whereby v'_r = c at the horizon regardless of the initial state of motion.

That is the general math of it anyway, taken up to the mathematical limit at the horizon. You both appear to disagree, however, that this applies at the limit itself, or one might simply refute that no static observers can exist at the horizon, so let me put it another way. The locally measured speed of the particle cannot be less than c at the horizon because if either of you were to give me the initial conditions for the state of motion and any value for the locally measured speed of the particle that is less than c, I can find the radius where that particular speed would be measured, and it would always lie outside the horizon. No matter how close to c you can get, the radius will still lie outside, and a greater speed will always be achieved as it continues to freefall toward the horizon. Therefore, at no locally measured speed less than c can a massive particle cross the horizon.

 

[PAllen]

The equation of motion in GR, at least for radial freefall, as related to the time dilation is

sqrt(1 - (v'_r/c)^2) / z_r = K

where v'_r is the locally measured speed at r, z_r is the time dilation at r, and K is a constant of motion. With the initial condition of a particle freefalling from rest at infinity, for example, v' = 0 and z = 1, so K = 1. For a photon K = 0 and for any massive particle K > 0, depending upon the initial conditions for the state of motion of the particle, and K remains constant at all r whether the particle is traveling inward or outward.

For a particle starting outside the horizon and falling inward, at the event horizon, where z_r = 0, in order for K to remain a finite value, then sqrt(1 - (v'_r/c)^2) = 0 also, whereby v'_r = c at the horizon regardless of the initial state of motion.

That is the general math of it anyway, taken up to the mathematical limit at the horizon. You both appear to disagree, however, that this applies at the limit itself, or one might simply refute that no static observers can exist at the horizon, so let me put it another way. The locally measured speed of the particle cannot be less than c at the horizon because if either of you were to give me the initial conditions for the state of motion and any value for the locally measured speed of the particle that is less than c, I can find the radius where that particular speed would be measured, and it would always lie outside the horizon. No matter how close to c you can get, the radius will still lie outside. Therefore, at no locally measured speed less than c can a massive particle cross the horizon.

That is coordinate speed. Nobody measures coordinate speed. To get measured speed, you need to relate one 4-velocity (of measuring observer) to another 4-velociy (measured object). It is mathematically impossible for this to yield >=c.

 

[PeterDonis]

The equation of motion in GR, at least for radial freefall, as related to the time dilation is

sqrt(1 - (v'_r/c)^2) / z_r = K

Once again, you stated it wrong. The correct statement is: "The equation of motion for radial freefall outside the horizon is..." The equation you wrote isn't valid at or inside the horizon, because what you are calling z_r is undefined there.

That is the general math of it anyway, taken up to the mathematical limit at the horizon.

And what happens at or inside that limit?

You both appear to disagree, however, that this applies at the limit itself, or one might simply refute that no static observers can exist at the horizon

Correct on both counts.

The locally measured speed of the particle cannot be less than c at the horizon because if either of you were to give me the initial conditions for the state of motion and any value for the locally measured speed of the particle relative to a static observer that is less than c, I can find the radius where that particular speed would be measured, and it would always lie outside the horizon.

You keep on leaving out essential qualifiers. This time I've gone ahead and inserted the necessary qualifier in the quote above. With the qualifier inserted, your reasoning is no longer valid; the "locally measured speed" of an infalling particle, relative to an observer who is *not* static, does *not* necessarily approach c as you approach the horizon.

You can keep on trying at this, but I expect it to get monotonous. Let's try a different question: if a clock is released from rest at some radius r > 2m, with its clock time set to zero, and freely falls towards the horizon, what will the clock read at the instant it reaches the horizon? What do you think?

 

[grav-universe]

Right, by locally I mean measured by a static observer at that location. Okay well, let me ask you two these questions. A massive particle freefalls from rest at 4m toward the horizon and its speed is locally measured by static observers at each of their respective locations.

Can a massive particle cross the horizon before its locally measured speed reaches .99 c?

Can a massive particle cross the horizon before its locally measured speed reaches .9999999 c?

If no to both of these, is there any speed less than c that a massive particle can cross the horizon before reaching as measured locally by any static observer outside the horizon?

 

[PeterDonis]

Can a massive particle cross the horizon before its locally measured speed reaches .99 c?

No.

Can a massive particle cross the horizon before its locally measured speed reaches .9999999 c?

No.

If no to both of these, is there any speed less than c that a massive particle can cross the horizon before reaching as measured locally by any static observer outside the horizon?

No. So what? You didn't bother to make any argument that any of the above prevents the massive particle from crossing the horizon. Do you have such an argument? Bear in mind that your argument cannot assume that there is a static observer *at* the horizon, or that the concept of "locally measured speed" as you are using it is well-defined there, because there isn't one, and the concept isn't well-defined there.

 

[grav-universe]

Once again, you stated it wrong. The correct statement is: "The equation of motion for radial freefall outside the horizon is..." The equation you wrote isn't valid at or inside the horizon, because what you are calling z_r is undefined there.

z_r = 0 at the horizon. That is the time dilation.

And what happens at or inside that limit?

That's the question, isn't it? Well, one could say it requires a different set of coordinates, or we can refer back to 2a. :)

With the qualifier inserted, your reasoning is no longer valid; the "locally measured speed" of an infalling particle, relative to an observer who is *not* static, does *not* necessarily approach c as you approach the horizon.

I would have to see that.

You can keep on trying at this, but I expect it to get monotonous. Let's try a different question: if a clock is released from rest at some radius r > 2m, with its clock time set to zero, and freely falls towards the horizon, what will the clock read at the instant it reaches the horizon? What do you think?

It's finite, it stops at the horizon according to a distant observer. Your argument is that its proper time continues forward so that it passes the horizon. We could also switch to GR coordinates z = L = 1 / sqrt(1 + 2 m / r), another valid coordinate system but with no mapped interior coordinates and say that the clock strikes a point mass at that proper time. Or that the clock can never reach the horizon because its speed cannot reach c or that the time of the clock would freeze if it ever did reach c. So in a way it seems we're asking, if a clock did ever manage to reach c, even in a non-GR inertial system, would its proper time still continue?

 

[grav-universe]

No.



No.



No. So what? You didn't bother to make any argument that any of the above prevents the massive particle from crossing the horizon. Do you have such an argument? Bear in mind that your argument cannot assume that there is a static observer *at* the horizon, or that the concept of "locally measured speed" as you are using it is well-defined there, because there isn't one, and the concept isn't well-defined there.

Well, that is the argument. A massive particle cannot ever reach a locally measured speed of c, right? A static observer that lies outside the horizon will measure a speed that is less than c and a greater speed can always be achieved as it continues to approach the horizon, correct? Therefore, it never reaches the horizon at any speed less than c because any speed no matter how close to c would be measured outside the horizon and continue to increase, isn't that right?

And from your answers, if it cannot reach it before achieving a speed of .99 c or .9999999 c or any other speed at all that is less than c, then it can only reach it upon achieving c, correct?

 

[PeterDonis]

z_r = 0 at the horizon. That is the time dilation.

Oh, sorry, I mis-stated it. I should have said that, since z_r = 0 and it appears in the denominator of your equation of motion, your equation of motion is undefined at the horizon. I apologize for the error.

That's the question, isn't it? Well, one could say it requires a different set of coordinates, or we can refer back to 2a. :)

Or we could start talking about physics instead of coordinates. That is, we could start talking about actual physical observables *at* the horizon. I even gave you an example of one: the reading on a clock set to zero at the instant it is dropped from rest at a radius r > 2m, when it reaches the horizon. I see you gave an answer to that question; see below for further comments.

I would have to see that.

In other words, you haven't even bothered to learn the actual physical model you are criticizing. This is a simple calculation that is routinely assigned as a homework problem in relativity textbooks. I don't have time right now to post the details; I encourage you to look them up. The bottom line is that it is possible for two infalling observers to have any relative velocity less than c when they cross the horizon together, depending on the initial conditions of their infall.

It's finite

Correct.

it stops at the horizon according to a distant observer.

Incorrect. All that the distant observer can properly assert is that he will never receive a light signal that is emitted by the infalling clock at (or inside) the horizon. He cannot claim that this means the clock simply stops at the horizon; see below for why.

Your argument is that its proper time continues forward so that it passes the horizon.

Yes. Do you understand why? It's because all physical quantities are finite there. There is nothing physically present at the horizon that would cause the infalling clock to stop there. So it doesn't. To claim otherwise is to claim that the laws of physics suddenly work differently at the horizon, for no apparent reason.

The more technical way of putting this is that the solution of the Einstein Field Equation is perfectly finite and continuous at the horizon; there is nothing in the solution leading up to the horizon that makes the horizon a place where the solution (i.e., spacetime) could just stop. It has to continue if the EFE is valid, and therefore the worldline of the clock has to continue as well. To claim otherwise is to claim that the EFE suddenly stops being valid at the horizon, for no apparent reason.

We could also switch to GR coordinates z = L = 1 / sqrt(1 + 2 m / r), another valid coordinate system but with no mapped interior coordinates

Do you mean isotropic coordinates? As in the ones described in the "alternative formulation" section here:

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

If so, you are correct that these do not cover the interior region; that's because they double cover the exterior region (outside the horizon). The range 0 < R < m/2 (where R is the isotropic radial coordinate) covers the same set of events as the range m/2 < R < infinity; each event in the exterior region maps to *two* values of R, not one.

and say that the clock strikes a point mass at that proper time.

How do you figure that? Isotropic coordinates do not somehow make a point mass magically appear at the horizon.

Or that the clock can never reach the horizon because its speed cannot reach c

It's true that the clock's speed can never "reach c", but false that that implies that it can't reach the horizon. Here's another way of looking at it: the clock is falling inward, and the horizon is moving outward. The reason the clock is "moving at c" relative to the horizon, when it crosses it, is that the *horizon* is a lightlike surface--*it* is a surface made up of outgoing light rays. So of course anything passing those light rays will be "moving at c" relative to the light rays, because light rays move at c relative to any timelike object.

or that the time of the clock would freeze if it ever did reach c.

Irrelevant since the clock never does reach c.

So in a way it seems we're asking, if a clock did ever manage to reach c, even in a non-GR inertial system, would its proper time still continue?

I'm not asking that; the clock never does "reach c", except in the sense I gave above, that it "moves at c" relative to the horizon because the horizon itself is made of outgoing light rays, and any massive object "moves at c" relative to light rays.

 

[PeterDonis]

Well, that is the argument. A massive particle cannot ever reach a locally measured speed of c, right? A static observer that lies outside the horizon will measure a speed that is less than c and a greater speed can always be achieved as it continues to approach the horizon, correct?

Correct up to here, yes.

Therefore, it never reaches the horizon at any speed less than c because any speed no matter how close to c would be measured outside the horizon and continue to increase, isn't that right?

No. Go back through this chain of reasoning again, carefully, and make explicit all the unstated assumptions.

And from your answers, if it cannot reach it before achieving a speed of .99 c or .9999999 c or any other speed at all that is less than c, then it can only reach it upon achieving c, correct?

No again. Same advice as above.

I'll make again a suggestion I made in my immediately previous post, but I'll amplify it somewhat. Think of the horizon as a light ray moving outward, and the infalling object as a timelike object falling inward. In a local inertial frame centered on the event at which the infalling object crosses the horizon, the horizon will appear as a 45 degree line (because it's a light ray); we'll say that the positive "x" direction is radially outward, so the 45-degree horizon line goes up and to the right. The worldline of the infalling object is the "t" axis of the local inertial frame; i.e., it's a vertical line going through the origin.

In this local inertial frame, the worldlines of static observers at distances closer and closer to the horizon appear as segments of hyperbolas that cross the vertical axis at negative "t" values closer and closer to the origin (i.e., to t = 0). These hyperbolas asymptote to the horizon line (but we don't see the axis of the hyperbolas, where the two asymptotes cross--it's way down and to the left somewhere, off our diagram). Each of these segments is inclined closer and closer to 45 degrees, so their speed relative to the infalling observer (which is also the infalling observer's speed relative to them) gets closer and closer to c. However, that in no way prevents the infalling observer from reaching and crossing the horizon; from his point of view, he passes static observers moving outward at closer and closer to c, until finally he passes an outgoing light ray--the horizon, which is moving outward *at* c. To him the horizon is just the "limit point" of the static observers; but the static observers can never see that limit point because light emitted at the horizon stays at the horizon; it never gets to any larger radius.

 

[harrylin]

I just have time for this, as this sub-discussion took off:

["a clock kept at this place would go at the rate zero". - Einstein]

Quote by PeterDonis
"
(1) Einstein is claiming that a clock can be kept at the horizon, and saying that it would go at rate zero.

(2) Einstein is claiming that a clock *cannot* be kept at the horizon, because if it could, it would go at rate zero, and that doesn't make sense.

Are you interpreting him as saying #1 or #2? If it's #1, the refutation is pretty easy: the clock would have to go at the speed of light, and no clock can do that.
"

Just thought I'd add my two cents, but if that is your refutation of #1, then we should also add

(3) A clock can't cross the horizon, because if it could, it would have to go at the speed of light, and no clock can do that.
[...]

None of them can be a correct interpretation, just as (as I mistakenly thought to have clarified,) none of the following can be a correct interpretation of "For v=c all moving objects—viewed from the “stationary” system—shrivel up into plane figures":

1. Einstein is claiming that it is *possible* for an object to move at c.
2. Einstein is claiming that an object cannot move at c, because if it could, it would shrivel up into a plane figure, and that doesn't make sense.
3. Einstein is claiming that an object cannot move at c, because if it could, it would have infinite energy, and that is impossible.
etc.

Instead, such statements simply refer to (unattainable) physical limits; and in both cases it takes infinite coordinate time to reach such limits. This is acceptable shorthand among physicists, but "forbidden" for mathematicians.

 

[PeterDonis]

Instead, such statements simply refer to (unattainable) physical limits; and in both cases it takes infinite coordinate time to reach such limits.

So basically, you are saying that the correct interpretation of what Einstein said is simply:

(4) No object can reach the horizon, because that would take an infinite amount of coordinate time.

In other words, eliminate all mention of "moving at c", and just focus on the coordinate time. Correct? If so, I'm confused about where the phrase "a clock kept at this place would go at rate zero" fits in.

 

[grav-universe]

The more technical way of putting this is that the solution of the Einstein Field Equation is perfectly finite and continuous at the horizon; there is nothing in the solution leading up to the horizon that makes the horizon a place where the solution (i.e., spacetime) could just stop. It has to continue if the EFE is valid, and therefore the worldline of the clock has to continue as well. To claim otherwise is to claim that the EFE suddenly stops being valid at the horizon, for no apparent reason.

Do you mean isotropic coordinates? As in the ones described in the "alternative formulation" section here:

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

If so, you are correct that these do not cover the interior region; that's because they double cover the exterior region (outside the horizon). The range 0 < R < m/2 (where R is the isotropic radial coordinate) covers the same set of events as the range m/2 < R < infinity; each event in the exterior region maps to *two* values of R, not one.

How do you figure that? Isotropic coordinates do not somehow make a point mass magically appear at the horizon.

That particular coordinate system is the one I discovered and you and I discussed in the other thread. Rather than keep calling it the 1/sqrt(1 + 2m/r1) coordinate system, let's call it GU coordinates :) . It coordinately transforms from SC with

r1 = r (1 - 2m/r), r = r1 (1 + 2m/r1) with the metric

ds^2 = c^2 dt^2 / (1 + 2 m / r1) - dr1^2 (1 + 2 m / r1) - dθ^2 r1^2 (1 + 2 m / r1)^2

I like it because it completely eliminates the event horizon and interior spacetime altogether, leaving only what external observers observe. It shrinks the boundary of the event horizon to a point, so that from the perspective of external observers applying this coordinate system, the mass lies at a point singularity in the center just as in Newtonian with infinite acceleration there, no event horizon and no interior spacetime. A clock falling to the point mass will still do so in finite time. Proper distance measured to the point mass is also finite. But we would expect these when measuring the distance to a point or the time to fall to a point. It is just as valid as SC, and the EFE's valid also, being only a coordinate transformation, with all of the same external observables, but looking at it, one would not expect any more spacetime to exist within a point. From the perspective of this coordinate system, that would be like falling out of this universe altogether into some other dimension if there were interior spacetime within a point. If Schwarzschild had happened to come up with this coordinate system rather than the one he did, each just as likely to have been derived before the other, we might not even consider that any interior spacetime exists in the first place.

You also mention Eddington's isotropic coordinates. These are also valid. But as you said, with a one to one correspondence to SC coordinates, they only map some of the interior spacetime of SC, then double back. If one were to fall past the horizon and all the way to the center of EIC, then, when transformed back to SC, it would be like falling part way past the horizon, then doubling back and traveling back out of the horizon again. Likewise, I could find coordinate systems that have more spacetime than SC or even one that cuts out part of the exterior coordinates. So arbitrary coordinate systems may be valid, but obviously they are not equal. Some map out more or less spacetime than others, and some in ways that don't make sense, like the doubling back of EIC, although it would not actually double back in EIC itself. So how are we to know which one maps it out correctly? Personally I would go with the one I found, but if you insist that there must be interior spacetime, as I'm sure you do :) , then as you stated "you are correct that these do not cover the interior region" referring to EIC as compared to SC apparently, how do you know that they do not, or that SC does, with no more interior spacetime than actually exists and no less? SC is only the first coordinate system found. Since then, many others have been determined, and infinitely many are possible, all different in terms of how much spacetime is mapped, so statistically speaking, it is unlikely that SC maps it perfectly. How much spacetime is the right amount?

 

[PeterDonis]

That particular coordinate system is the one I discovered and you and I discussed in the other thread. Rather than keep calling it the 1/sqrt(1 + 2m/r1) coordinate system, let's call it GU coordinates :) . It coordinately transforms from SC with

r1 = r (1 - 2m/r), r = r1 (1 + 2m/r1) with the metric

ds^2 = c^2 dt^2 / (1 + 2 m / r1) - dr1^2 (1 + 2 m / r1) - dθ^2 r1^2 (1 + 2 m / r1)^2

Can you give a reference to the "other thread" you refer to? This does not look at all familiar to me, but I may just be failing to remember a previous discussion. I'll refrain from commenting on the rest of your post until I've got the context clear.

 

[grav-universe]

Can you give a reference to the "other thread" you refer to? This does not look at all familiar to me, but I may just be failing to remember a previous discussion. I'll refrain from commenting on the rest of your post until I've got the context clear.

Sure, here it is. "Shrinking event horizon to point singularity"

 

[PeterDonis]

Sure, here it is. "Shrinking event horizon to point singularity"

Ah, ok, thanks, that helps to jog my memory.

Pretty much everything I would say in response has already been said in the other thread, so I don't see much point in anything more than a quick recap (and what I'm saying applies just as well to Eddington isotropic coordinates as any other chart):

(1) You can't change the physics by changing coordinate charts. You can choose coordinates such that what used to be r = 2m is now r1 = 0; but you can't change the physical nature of the spacetime at what used to be r = 2m and is now r1 = 0. Just labeling it with r1 = 0 doesn't make it a point instead of a surface.

(2) To actually talk about the physics, you have to compute invariants--quantities that don't change when you change coordinate charts. If your chart is singular at a particular place, you can't compute invariants there using the chart, so you can't say anything about the physics there using the chart. Your chart is singular at r1 = 0, so it can't say anything about the physics at that location: in particular, you can't compute any invariant in your chart that shows that what you are labeling r1 = 0 is an actual, physical point, instead of, say, a surface that your coordinates don't cover well.

These points are basic facts of differential geometry as it is used in physics. They have been stated ad nauseam, and you don't seem to be accepting them. That means we really don't have a good basis for discussion.

 

[grav-universe]

Ah, ok, thanks, that helps to jog my memory.

Pretty much everything I would say in response has already been said in the other thread, so I don't see much point in anything more than a quick recap (and what I'm saying applies just as well to Eddington isotropic coordinates as any other chart):

(1) You can't change the physics by changing coordinate charts. You can choose coordinates such that what used to be r = 2m is now r1 = 0; but you can't change the physical nature of the spacetime at what used to be r = 2m and is now r1 = 0. Just labeling it with r1 = 0 doesn't make it a point instead of a surface.

(2) To actually talk about the physics, you have to compute invariants--quantities that don't change when you change coordinate charts. If your chart is singular at a particular place, you can't compute invariants there using the chart, so you can't say anything about the physics there using the chart. Your chart is singular at r1 = 0, so it can't say anything about the physics at that location: in particular, you can't compute any invariant in your chart that shows that what you are labeling r1 = 0 is an actual, physical point, instead of, say, a surface that your coordinates don't cover well.

These points are basic facts of differential geometry as it is used in physics. They have been stated ad nauseam, and you don't seem to be accepting them. That means we really don't have a good basis for discussion.

Right, but that could go either way. Both being equally valid external coordinate systems, how do you know we're not making a surface out of a point? What I am asking, though, is what coordinate system you think accurately maps out the spacetime, no more and no less? You stated that EIC do not. Why not? Do SC? Why or why not?

 

[PeterDonis]

Right, but that could go either way. Both being equally valid external coordinate systems, maybe we're making a surface out of a point. How do you know that is not the case?

Because I have computed the invariants at r = 2m (using a chart that's not singular there), so I know what the actual physical quantities are there. That includes a computation of the physical area of a 2-sphere at r = 2m, *and* a computation of the causal nature of a curve with constant r = 2m (and constant theta, phi if we include the angular coordinates) to verify that it's a null curve, not a timelike curve.

What I am asking is what coordinate system would accurately map out the spacetime, no more and no less? You stated that EIC do not. Why not? Do SC? Why or why not?

If you want a map of the *entire* spacetime, including all regions that are mathematically possible according to the vacuum, spherically symmetric solution of the Einstein Field Equation, the only charts I'm aware of that cover it all are the Kruskal chart and the Penrose chart. (The technical term for the spacetime that the full Kruskal chart maps is the "maximally extended Schwarzschild spacetime".) However, as has been said before, nobody believes that this entire spacetime is physically reasonable, because it includes a white hole and a second exterior region.

If you want a map of a highly idealized spacetime consisting of a spherically symmetric region of collapsing matter with zero pressure, plus the vacuum region surrounding it, the only chart I'm aware of that covers it all with a single expression for the metric is the Penrose chart. There is a "Kruskal-type" chart for this spacetime, which covers it all, but the expression for the line element is different depending on whether you're in the vacuum region or the matter region. This spacetime is at least physically reasonable, though obviously it is highly idealized because of the exact spherical symmetry.

If you are willing to settle for a map that only covers the vacuum region exterior to a spherically symmetric collapsing body, there are two additional charts that will cover the entirety of this region: the ingoing Eddington-Finkelstein chart and the ingoing Painleve chart.

The common feature of all these charts is that they are nonsingular over the entire spacetime (or over the entire vacuum region, in the case of the last two), *and* the full range of their coordinates spans the full range of the region they cover. Both the SC chart and the EIC chart fail on at least one of these properties:

* The coordinate singularity at the horizon means that the SC chart can't accurately map the spacetime there, and it also means that the interior SC chart (with r < 2m) is a different, disconnected chart from the exterior SC chart (with r > 2m).

* The EIC chart is nonsingular at the horizon (actually, technically the inverse metric is singular there, but opinions differ on whether that counts as a "coordinate singularity" so I'm giving it the benefit of the doubt). However, the full range of the EIC "r" coordinate doesn't cover anything inside the horizon--instead, as I've said before, it double covers the region outside the horizon. Another way of putting this is that the area of a 2-sphere at radius "r" in EIC coordinates is not monotonic in r; it has a minimum at r = m/2, and increases both for r > m/2 *and* r < m/2. So there are two values of "r" that both map to the same physical 2-sphere (except at the horizon, r = m/2). This makes it pretty obvious that the EIC chart's coverage is incomplete: where are the 2-spheres with smaller area?

[Edit: btw, it's worth noting that the computation of invariants at the horizon that I referred to above can actually be done in the EIC chart, since the line element is not singular there. To compute the area of the 2-sphere at the horizon, plug in r = m/2 and dt = dr = 0, and integrate ds^2 over the full range of theta and phi. You should get 16 pi m^2. To compute the causal nature of a curve with constant r at the horizon, plug in r = m/2 and dr = dtheta = dphi = 0. You should find ds^2 = 0, indicating that a line element with constant r, theta, phi at the horizon (but nonzero dt) is null.]

 

[PeterDonis]

I realized I should add an additional comment to my last post about which charts cover which regions. In the case of a spherically symmetric collapsing body, one can cover the interior of the body (the region containing the matter) with a collapsing FRW-type chart (the time reverse of the expanding FRW-type chart that is used in cosmology to model the universe). MTW does this in their treatment of this model, for example. One can also construct a chart for the vacuum region that matches up with this chart at the boundary (the surface of the collapsing matter); IIRC this chart for the vacuum region is not the same as any of the ones I named. I believe the treatment of this model in MTW uses this type of chart for the vacuum region; if I get a chance I'll check my copy to see.

 

[harrylin]

So basically, you are saying that the correct interpretation of what Einstein said is simply:

(4) No object can reach the horizon, because that would take an infinite amount of coordinate time. In other words, eliminate all mention of "moving at c", and just focus on the coordinate time.

Correct? If so, I'm confused about where the phrase "a clock kept at this place would go at rate zero" fits in.

No. With that phrase he merely explains the meaning of Schwartzschild's solution (he even says so!). Here's a last attempt to clarify this.

Einstein: "a clock kept at this place would go at rate zero".
My translation attempt for mathematicians: setting dr/dt=0, dτ/dt->0 for r->μ/2

Compare Einstein in 1905: "For v=c all moving objects—viewed from the “stationary” system—shrivel up into plane figures".
My translation attempt for mathematicians: L->0 for v->c

And then I get to what might be "the mother" of all bugs (any further discussion on this topic is useless as long as this has not been fixed) - and this issue is perfectly on topic:

Eve *is* accelerating; she feels a nonzero acceleration, an accelerometer attached to her reads nonzero, if she stood on a scale it would register weight, etc. There is no way in which she "thinks she is not accelerating".

The term "co-moving inertial reference frame" is more precisely stated as "momentarily co-moving inertial reference frame" (MCIF).

What may be confusing is that Egan says to "use only SR", but as you see his discussion is an application of Einstein's equivalence principle and that is pure GR. For this illustration one only needs SR math. Now your claims:

1. Peter: There is no way in which Eve "thinks she is not accelerating", as "an accelerometer attached to her reads nonzero".

According to the Einstein Equivalence Principle, when she is sitting in her chair Eve can think that she is not accelerating; she may think that instead the force that she feels is due her being in rest in a gravitational field. As you noticed yourself, "if she stood on a scale it would register weight".

2. Peter: The term "co-moving inertial reference frame" is more precisely stated as "momentarily co-moving inertial reference frame".

Evans evidently means constantly co-moving inertial reference frame, and I will explain why. According to you, Evans means that according to Eve the force she feels is due to acceleration; so that she thinks that she is one moment at rest in one inertial frame, and the next moment she is at rest in a different inertial frame. Consequently she would use the same set of inertial frames as Adam - that is standard SR. In any such reference frame there is a time for Eve when Adam passes through the horizon. It would be just an SR simultaneity disagreement.

To the contrary, according to Egan there is no time for Eve when, in her co-moving inertial reference frame, Adam passes through the horizon.

 

[PeterDonis]

Einstein: "a clock kept at this place would go at rate zero".
My translation attempt for mathematicians: setting dr/dt=0, dτ/dt->0 for r->μ/2

Don't you mean r -> 2μ? (Or 2m in the more usual symbols.)

Anyway, I wasn't asking about how you would translate Einstein's statement into mathematics. I was asking what you thought it meant physically.

What may be confusing is that Egan says to "use only SR", but as you see his discussion is an application of Einstein's equivalence principle and that is pure GR.

Egan's discussion is about a scenario in flat spacetime, which can be handled using only SR. That's why he says to "use only SR". The reason the scenario is relevant for our discussion here is that, as Egan says, the scenario he decribes in flat spacetime is equivalent to "a first-order approximation of the Schwarzschild metric near a black hole's horizon". Whether you call that "pure GR" or not is a matter of words, not physics.

1. Peter: There is no way in which Eve "thinks she is not accelerating", as "an accelerometer attached to her reads nonzero".

According to the Einstein Equivalence Principle, when she is sitting in her chair Eve can think that she is not accelerating; she may think that instead the force that she feels is due her being in rest in a gravitational field. As you noticed yourself, "if she stood on a scale it would register weight".

Sigh. I should have clarified (again) the distinction between coordinate acceleration and proper acceleration. Yes, Eve can think that she experiences no *coordinate* acceleration; she can view herself as at rest in a gravitational field. But she cannot think that she experiences no *proper* acceleration, because she feels weight, and that is the *definition* of proper acceleration. Proper acceleration is an invariant; it's a direct observable, so it's there regardless of which coordinates Eve uses (Rindler or Minkowski). Coordinate acceleration is *not* an invariant; Eve can make it disappear by viewing herself as at rest in Rindler coordinates (in which a "gravitational field" is present) rather than as accelerating in Minkowski coordinates (where there is no "gravitational field"). I thought you understood the difference between coordinate and proper acceleration, since it's been discussed enough times, but apparently not, so I'll try to be more careful about qualifying the term "acceleration", as in "feeling acceleration", which makes it clear that I'm referring to the direct physical observable, that Eve feels weight.

2. Peter: The term "co-moving inertial reference frame" is more precisely stated as "momentarily co-moving inertial reference frame".

Evans evidently means constantly co-moving inertial reference frame

If you left out the word "inertial", this would be fine. But with it included, it's false. There is no such thing as a "constantly co-moving inertial reference frame" for Eve (and Egan certainly isn't claiming any such thing). Eve feels proper acceleration; nobody at rest in an inertial frame (for more than an instant) can feel proper acceleration. That's the *definition* of an inertial frame: that any observer at rest in it (for more than an instant) is weightless, in free fall. Again, I thought you understood this, but apparently not. Sigh.

according to Eve the force she feels is due to acceleration; so that she thinks that she is one moment at rest in one inertial frame, and the next moment she is at rest in a different inertial frame.

This is not due to her "thinking" that the force is due to "acceleration": it's due to her actually *feeling* acceleration, i.e,. feeling weight.

Consequently she would use the same set of inertial frames as Adam - that is standard SR.

Adam only uses one inertial frame, since he is in free fall. Only Eve has to use a "set" of inertial frames if she wants to use inertial frames to describe her motion.

In any such reference frame there is a time for Eve when Adam passes through the horizon. It would be just an SR simultaneity disagreement.

And this is true; in any of the inertial frames in which Eve is momentarily at rest, there *is* a finite time at which Adam crosses the horizon. But Eve doesn't *stay* at rest in any of these frames, because she feels acceleration, i.e., she feels weight.

To the contrary, according to Egan there is no time for Eve when, in her co-moving inertial reference frame, Adam passes through the horizon.

As you state it, this is false; you need to leave out the phrase "in her co-moving inertial reference frame" (which Egan does *not* use, and your attributing it to him is mistaken). The "time for Eve" that Egan refers to is Rindler coordinate time, which is the same as proper time along her worldline. Since she feels acceleration, i.e., feels weight, that proper time is *not* the same as the time in *any* inertial frame, even inertial frames in which she is momentarily at rest. Egan's statement simply means that there is no Rindler coordinate time at which Adam crosses the horizon; it's not referring to the time in *any* inertial frame.

 

[harrylin]

Don't you mean r -> 2μ? (Or 2m in the more usual symbols.)
Anyway, I wasn't asking about how you would translate Einstein's statement into mathematics. I was asking what you thought it meant physically.

I directly used the notation of Einstein, in order not to mix up my own interpretation with my translation. And I already told you, it has no physical meaning without context, just as "v=c" has no physical meaning in itself. The context (incl. other papers) suggests to me that Einstein held both extremes for impossible in physical reality.

Egan's discussion is about a scenario in flat spacetime, which can be handled using only SR. That's why he says to "use only SR". The reason the scenario is relevant for our discussion here is that, as Egan says, the scenario he decribes in flat spacetime is equivalent to "a first-order approximation of the Schwarzschild metric near a black hole's horizon". Whether you call that "pure GR" or not is a matter of words, not physics.

I almost fully agree; but regretfully our differences are mostly a matter of words, which obscures eventual differences in physical models. For example, my notion of "flat space-time" means SR with reference systems that relate to each other by means of the Lorentz transformations - such a space-time lacks a Rindler horizon.

Apart of simple mistakes, we certainly come from different "schools" (even literally) that teach definitions which are incompatible with each other (Sigh indeed!). I will come back to the issue of definitions in a next thread.
Now, it will be a waste of time start a text exegesis of the meaning of words as used by Egan, with speculations of the school of thought that he is following; we don't really need him (except if we want to discuss "Egan's theory). And in the context of this thread you appear to agree with me on the simple point that I tried to make, and also the different "notions of simultaneity" are not in question:

[..]Yes, Eve can think that she experiences no *coordinate* acceleration; she can view herself as at rest in a gravitational field. [..]
in any of the inertial frames in which Eve is momentarily at rest, there *is* a finite time at which Adam crosses the horizon. [..]
Egan's statement simply means that there is no Rindler coordinate time at which Adam crosses the horizon [..]

That looks to me a reasonable summary of different "notions of simultaneity in strongly curved spacetime"; perhaps PAllen would like to clarify how his first post relates to this example (if indeed it does).

And from an earlier post, you seem to agree that at the moment that Adam "falls away" according to Eve, she ascribes the frequency difference from two clocks to the effect of a gravitational field which makes her clocks go at different rates; and that in contrast, for Adam the frequency difference that Eve observes is almost completely due to "classical" Doppler. My point was that in GR much more than in SR the different views relate to a disagreement about physical reality. However, that is a bit off-topic in this thread; and now that I decided to start my own thread I'll include further elaboration there.

 

[PeterDonis]

my notion of "flat space-time" means SR with reference systems that relate to each other by means of the Lorentz transformations - such a space-time lacks a Rindler horizon.

No, it doesn't. You may think it does, but that's because you don't fully understand the implications of "reference systems that relate to each other by means of the Lorentz transformations". Such a spacetime includes hyperbolas such as the worldline that Eve travels on, and it also includes the fact that a light ray emitted from the origin will never cross such a hyperbola (since the light ray is an asymptote of the hyperbola). That is the definition of a Rindler horizon, so your notion of flat space-time includes a Rindler horizon, whether you think so or not. If you didn't realize that all that was already included in your notion of a flat space-time, well, then you need to think more carefully about the implications of your notions.

Apart of simple mistakes, we certainly come from different "schools" (even literally) that teach definitions which are incompatible with each other (Sigh indeed!).

I don't know what "school" you come from, so I can't really evaluate this statement. I'm not aware of any definitions from my "school", i.e., standard GR, which are incompatible with each other. I don't really see a problem with incompatible definitions in our discussion; unclear definitions, yes, but that can be fixed by making them clear. When we've managed to do that, I don't see any incompatibility.

And from an earlier post, you seem to agree that at the moment that Adam "falls away" according to Eve, she ascribes the frequency difference from two clocks to the effect of a gravitational field which makes her clocks go at different rates; and that in contrast, for Adam the frequency difference that Eve observes is almost completely due to "classical" Doppler.

I agree, and would add that you don't need the qualifier "at the moment Adam falls away". The same reasoning, for both Eve and Adam, applies everywhere on Eve's worldline. The only potential issue with that is that Adam and Eve are spatially separated except at the moment Adam falls away; but since we're only talking about how Adam views what's happening on Eve's worldline, that isn't really an issue, since Adam's coordinates cover all of Eve's worldline, and Adam can receive light signals from any event on Eve's worldline (some of them he will receive after he crosses Eve's Rindler horizon, but he will receive them).