# Notions of simultaneity in strongly curved spacetime

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P: 1,376
 Quote by PeterDonis 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.
P: 3,187
 Quote by PeterDonis 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

 Quote by PeterDonis 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.
 Quote by PeterDonis [..] 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.
Physics
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 Quote by harrylin 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.

 Quote by harrylin 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.

 Quote by harrylin "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?

 Quote by harrylin 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.

 Quote by harrylin 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.

 Quote by harrylin 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.

 Quote by harrylin 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.

 Quote by harrylin 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).

 Quote by harrylin 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.

 Quote by harrylin 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.

 Quote by harrylin 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.

 Quote by harrylin 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).

 Quote by harrylin If we hold that GR was wrong on that last point
It wasn't and isn't. See above.

 Quote by harrylin Einstein warned for a misconception that you seem to hold
I don't. See above.

 Quote by harrylin 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?

 Quote by harrylin 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.

 Quote by harrylin 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.

 Quote by harrylin "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".

 Quote by harrylin 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.
P: 429
 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.

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.
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 Quote by grav-universe (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.

 Quote by grav-universe 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.
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 Quote by harrylin 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.

 Quote by harrylin 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.

 Quote by harrylin 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.
P: 429
 Quote by PeterDonis 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
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 Quote by grav-universe 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.
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P: 5,059
 Quote by grav-universe 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).
 P: 429 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 travelling 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.
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P: 5,059
 Quote by 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 travelling 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.
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P: 6,148
 Quote by 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
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.

 Quote by grav-universe 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?

 Quote by grav-universe 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.

 Quote by grav-universe 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?
 P: 429 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?
Physics
PF Gold
P: 6,148
 Quote by grav-universe Can a massive particle cross the horizon before its locally measured speed reaches .99 c?
No.

 Quote by grav-universe Can a massive particle cross the horizon before its locally measured speed reaches .9999999 c?
No.

 Quote by grav-universe 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.
P: 429
 Quote by PeterDonis 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?
P: 429
 Quote by PeterDonis 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?
Physics
PF Gold
P: 6,148
 Quote by grav-universe 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.

 Quote by grav-universe 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.

 Quote by grav-universe 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.

 Quote by grav-universe It's finite
Correct.

 Quote by grav-universe 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.

 Quote by grav-universe 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.

 Quote by grav-universe 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.

 Quote by grav-universe 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.

 Quote by grav-universe 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.

 Quote by grav-universe or that the time of the clock would freeze if it ever did reach c.
Irrelevant since the clock never does reach c.

 Quote by grav-universe 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.
Physics
PF Gold
P: 6,148
 Quote by grav-universe 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.

 Quote by grav-universe 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.

 Quote by grav-universe 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.

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