Notions of simultaneity in strongly curved spacetime

[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).

 

[Austin0]

Quote by harrylin

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.

I don't know what Egan had to say but I think you are quite mistaken regarding Rindler coordinates and the horizon.
I don't think Rindler has anything to do with it. It is a coordinate artifact due to the dynamic metric in any accelerating system. This applies just as well to momentarily co-moving inertial frames. It happens because the distance to a point towards the rear shrinks due to contraction comparable to the increase in length due to system motion. SO the system asymptotically stops moving relative to points nearing the horizon as calculated . from a point within the system.
So harrylin is correct that Adam never crosses the horizon in any MCRF that Eve is at rest in.

 

[grav-universe]

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.

Right. Most of those invariants, some of which you mentioned in the other thread, don't really demonstrate much as far as I can see, as they are mostly measured externally to the horizon, so don't say a lot about what happens at the horizon itself. For instance, with the finite proper time of a freefalling clock, we could just as easily see that time showing the clock striking a point mass singularity in GUC as crossing a surface in SC.

Those invariants may make sense if taken all together as you said in the other thread, but on the flip side, there are also a few things that don't make sense, such as a clock traveling at c to a *hypothetical* static observer (which actually doesn't exist at the horizon, I know), infinite acceleration applied at a finite surface, and charts such as SC mapping out the physical space between the center and the horizon, but being unable to say anything at all about the spacetime there or events that occur there without referring to a different chart altogether. But these don't demonstrate anything definite either.

The one and only thing so far that I can see that does demonstrate anything substantial is what you just mentioned, the invariant locally measured surface area. In SC, there is only radial contraction and no tangent contraction of static rulers as inferred by a distant observer, so if the distant observer measures A = 4 pi r^2 = 4 pi (2 m)^2 = 16 pi m^2 at the horizon, then with no tangent contraction in SC all the way down to the horizon, so presumably at the horizon as well, a *hypothetical* static observer there should also. And regardless of how we change the coordinate system, that local measurement is invariant.

Even in GUC, the distant observer measures A = 0, but the tangent length contraction is 1 / (1 + 2 m / r1) = 0 at r1 = 0, so a static observer at the horizon measures A' = A (1 + 2 m / r1)^2 = 0 / 0^2 = any real number, including zero. Being invariant, however, it should agree with the local measurement made in SC, which is finite and non-zero. So there's that. Of course, however, since there can be no static observer at the horizon anyway, though, the surface cannot actually be measured there, which negates this result (lol jk). We could perhaps instead consider what an observer measures that just begins to freefall from rest (or near rest?) at the horizon, although that would already be assuming that a surface exists there that one could fall through, or if falling from rest just before the horizon, he could not reach it at less than c, so still very far from measuring its surface while static. Hmm, I'm actually not sure how that surface would be measured locally.

By the way, you said that some charts are not singular there, but how could that be? The local acceleration there is infinite, that is an invariant. Are you not defining a singularity as a place with infinite local acceleration? Also, due to the infinite acceleration, static observers cannot exist there, so that is also an invariant. Surely the chart you are referring to does not allow static observers there, right? Wouldn't that define the horizon, a place where static observers cannot exist and observers can never accelerate at a large enough rate to escape once there?

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.]

Um, wow, good post. Very detailed. That is a lot to look into. Thanks. :)

 

[PeterDonis]

It is a coordinate artifact due to the dynamic metric in any accelerating system. This applies just as well to momentarily co-moving inertial frames.

No, it doesn't; at least, not in flat spacetime. In flat spacetime, any inertial frame covers the entire spacetime, including the portion of Adam's worldline at and beyond the Rindler horizon. That's a basic fact about inertial frames in flat spacetime. An MCIF is an inertial frame, so this fact applies to MCIFs in flat spacetime. Another way of saying this is that in flat spacetime, every inertial frame is global.

In curved spacetime, there are *no* global inertial frames; *any* inertial frame can only cover a small patch of the spacetime. So in curved spacetime, you are correct that an MCIF at some event on an accelerated observer's worldline might not cover the horizon. But Egan's scenario is entirely set in flat spacetime, so the restrictions on inertial frames, including MCIF's, in curved spacetime doesn't apply.

Also, a word about "coordinate artifact". The fact that you can't assign a finite Rindler time coordinate to events at and beyond the Rindler horizon is an artifact of Rindler coordinates. But the fact that a light ray at the Rindler horizon will never intersect any of the "Rindler hyperbolas"--the curves with constant Rindler space coordinates--is not a coordinate artifact; you can express the same fact in any coordinate chart, because the curves themselves are geometric objects, not coordinate artifacts. So the existence of a "Rindler horizon" is not a coordinate artifact; there is something real and physical going on.

 

[PeterDonis]

with the finite proper time of a freefalling clock, we could just as easily see that time showing the clock striking a point mass singularity in GUC as crossing a surface in SC.

No, you can't, because the presence of a "point mass singularity", if it were true, would itself be an invariant; and computing invariants tells you that that whatever is there at that place in spacetime, it isn't a point mass singularity. See further comments below on the definition of a singularity.

Those invariants may make sense if taken all together as you said in the other thread, but on the flip side, there are also a few things that don't make sense, such as a clock traveling at c to a *hypothetical* static observer (which actually doesn't exist at the horizon, I know)

Not only does the hypothetical static observer at the horizon not exist; one *can't* exist, because the horizon is not a timelike surface. It's a null surface. That's the key fact you keep on missing, and it is one of the invariants I recommended that you compute. As I said, compute ds^2 for a line element at the horizon where dr = dtheta = dphi is zero. Technically you can't do it in SC coordinates because the line element is singular there, but PAllen said a while back that you can get around that even in SC coordinates by taking a limit as r -> 2m. Or you could do the computation in a chart that's not singular at the horizon, such as EIC. You will find that ds^2 = 0, and this is an invariant.

What is this invariant telling you? Well, look at similar line elements for r > 2m; i.e., pick some constant r > 2m, and plug in that r, plus dr = dtheta = dphi = 0, into the Schwarzschild line element. What do you get? You get ds^2 < 0 (with the usual sign convention), indicating that the line element is timelike; i.e., it's a possible worldline for an observer (a static observer, in this case). But when r = 2m, the corresponding line element is null; i.e., it's a possible worldline for a *light ray*, rather than a possible worldline for an observer.

That immediately tells us two things. First, it explains why the infalling observer moves at c relative to the horizon: the horizon is a light ray moving in the opposite direction to the observer (he's moving inward and the horizon is moving outward), so of course their relative velocity will be c. It's the *horizon* that's "moving at c", not the infalling observer; his worldline remains timelike, as it must.

Second, the fact that the horizon is null, rather than timelike, means that the horizon is not a "place" or "spatial location" in the way that the places occupied by static observers outside the horizon are. A "spatial location" requires a timelike curve going through it that has the same spatial coordinates everywhere. Curves of constant r > 2m (and constant theta, phi if we include the angular coordinates) meet that requirement; but a curve of constant r = 2m does not.

If you go back and look closely at the arguments you've made for why the infalling observer can't reach the horizon, you'll see that you were implicitly assuming that the horizon was a spatial location, a "place". It isn't. That's why your arguments don't show that an infalling observer can't reach the horizon.

infinite acceleration applied at a finite surface

There is no infinite acceleration, because there is no "place" where the infinite acceleration would exist. There is no "acceleration" along the path of a light ray. See above.

charts such as SC mapping out the physical space between the center and the horizon, but being unable to say anything at all about the spacetime there or events that occur there without referring to a different chart altogether.

I don't understand what you mean by this. The interior SC chart (i.e,. the SC chart with r < 2m) works perfectly well as a "map" of the interior of the black hole (the region inside the horizon). It's not a map that matches up with our intuitions very well, but so what? It's a perfectly valid map. It's also disconnected from the exterior SC chart, which maps the region outside the horizon, but again, so what? There's no requirement that any valid map has to continuously cover the entire spacetime.

The one and only thing so far that I can see that does demonstrate anything substantial is what you just mentioned, the invariant locally measured surface area.

As I've shown above, the invariant ds^2 = 0 for a line element at the horizon demonstrates something substantial as well.

Even in GUC, the distant observer measures A = 0

No, the distant observer can't infer anything from this chart, because it's singular at the horizon--by which I mean *really* singular; you can't even compute the physical area of the 2-sphere at the horizon at all, because the line element is mathematically undefined. That doesn't allow you to conclude A = 0. It doesn't allow you to conclude anything.

Of course, however, since there can be no static observer at the horizon anyway, though, the surface cannot actually be measured there, which negates this result

It's true that one can't measure the area of the horizon in the obvious way, by having observers who are static at that radius lay down rulers. But one can measure it indirectly, by having static observers on 2-spheres closer and closer to the horizon measure areas, and taking the limit as r -> 2m. There may be other more ingenious ways of doing it as well. In any case, every coordinate chart which is not singular at the horizon will give you the same answer for the value of the invariant area of the horizon.

We could perhaps instead consider what an observer measures that just begins to freefall from rest (or near rest?) at the horizon

Near rest is the best you can do. The fact that the horizon is a null surface means that no observer can be at rest there even for an instant.

although that would already be assuming that a surface exists there that one could fall through, or if falling from rest just before the horizon, he could not reach it at less than c

See above for what the relative velocity of c actually means.

By the way, you said that some charts are not singular there, but how could that be? The local acceleration there is infinite, that is an invariant.

No, it isn't. There is no "local acceleration" at the horizon. The formula for "local acceleration" is only valid if the curve along which it is computed is timelike. As I showed above, the corresponding curve at the horizon is not timelike, it's null. So the formula fails. Again, there is *no* invariant that is not finite at the horizon.

Are you not defining a singularity as a place with infinite local acceleration?

No. The usual definition of a singularity is a place where the spacetime curvature becomes infinite. The only place in Schwarzschild spacetime where that happens is r = 0.

Strictly speaking, having *any* valid invariant (as I noted above, "local acceleration" isn't valid at the horizon because it only applies along timelike curves) become infinite is sufficient for a singularity, but when you work through the math you find that if any invariant is infinite, at least one of the invariants associated with curvature is infinite, so the usual definition in terms of curvature turns out to work fine.

Also, due to the infinite acceleration, static observers cannot exist there, so that is also an invariant.

You're correct that static observers can't exist at the horizon, and that's an invariant, but it's not due to "infinite acceleration". See above.

Surely the chart you are referring to does not allow static observers there, right?

Right.

Wouldn't that define the horizon, a place where static observers cannot exist and observers can never accelerate at a large enough rate to escape once there?

The usual definition is that the horizon is the surface at which radially outgoing light can no longer escape to infinity. But that also implies the things you state here, so they are valid ways of describing the horizon as well.

 

[grav-universe]

What is your definition of singular?

 

[PeterDonis]

What is your definition of singular?

We've been using the word in a couple of different senses:

(1) A line element is singular at a particular set of coordinate values if any of the coefficients is mathematically undefined for that set of coordinate values. For example, the SC line element is singular at r = 2m because the coefficient of dr^2 has (1 - 2m/r) in the denominator, which is mathematically undefined (you can't divide by zero).

(2) A spacetime is singular at a particular event if some invariant quantity is mathematically undefined at that event. For example, at r = 0 in Schwarzschild spacetime, the curvature is mathematically undefined; formulas for various curvature invariants have r in the denominator, so at r = 0 they are mathematically undefined (again, because you can't divide by zero).

People often use the term "goes to infinity" as a synonym for "mathematically undefined"; but that's just convenient (if sloppy) terminology. It doesn't imply that one can somehow evaluate singular quantities at the points where they are singular. One can try to take limits as the singular point is approached, but that only helps if the limit turns out to be finite; often it doesn't.

 

[PeterDonis]

Even in GUC, the distant observer measures A = 0

I forgot to comment further on this. Your coordinate transformation was (I'll write R instead of r1)

R = r(1 - 2m/r)

which is equivalent to

R = r - 2m

with the inverse

r = R(1 + 2m/R)

which is equivalent to

r = R + 2m

You wrote the line element

ds^2 = \frac{dt^2}{1 + 2m/R} - \left( 1 + \frac{2m}{R} \right) dR^2 - R^2 \left( 1 + \frac{2m}{R} \right) d\Omega^2

but I don't think this is quite correct. When I do the above transformation on the Schwarzschild line element, I get

ds^2 = \frac{dt^2}{1 + 2m/R} - \left( 1 + \frac{2m}{R} \right) dR^2 - R^2 \left( 1 + \frac{2m}{R} \right)^2 d\Omega^2

Note the (1 + 2m/R)^2 in the last term; your version didn't have the ^2 there, which may have been an inadvertent typo.

In any case, you were claiming that the area of the horizon is zero using this line element; but that's not correct; even though R^2 appears in the last term, and that is zero at R = 0, there is also the (1 + 2m/R)^2 factor, which goes to infinity (sloppy terminology, I know) at R = 0. Since both factors are squared, it's not obvious at first glance what really happens to the angular part of the line element at R = 0. But we can easily rewrite the line element so that the angular part isn't singular at all at R = 0:

ds^2 = \frac{dt^2}{1 + 2m/R} - \left( 1 + \frac{2m}{R} \right) dR^2 - \left( R + 2m \right)^2 d\Omega^2

The angular part now integrates easily at R = 0 to yield a horizon area of 16 \pi m^2. (Technically, we have to take a limit as R -> 0 to deal with the dR^2 term; we can rewrite the dt^2 term so it isn't singular at R = 0. But the limit of the dR^2 term as R -> 0 is zero if dR = 0, so that's not really an issue.)

 

[harrylin]

[rearrange]
I agree [with Adam and Eve's different views of physical reality], and would add that you don't need the qualifier "at the moment Adam falls away". [..]

Good - it allows for a point of agreement when I pick up that discussion in another thread.

[concerning "flat space-time":] [..] 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. [..]

Sorry, I stated it wrongly and agree with your last comment. I commented on your earlier statement that 'the scenario he decribes in flat spacetime is equivalent to "a first-order approximation of the Schwarzschild metric near a black hole's horizon".'

What I meant is that the Rindler horizon interpretation that Egan portrays does not exist in what I call "flat spacetime": in flat spacetime, Eve does not think that Adam never crosses the horizon.

[..] I don't think Rindler has anything to do with it. It is a coordinate artifact due to the dynamic metric in any accelerating system. This applies just as well to momentarily co-moving inertial frames. [..] harrylin is correct that Adam never crosses the horizon in any MCRF that Eve is at rest in.

Regretfully I said it wrongly, and I'm afraid that what I meant is just the inverse of what you mean...

 

[PeterDonis]

I commented on your earlier statement that 'the scenario he decribes in flat spacetime is equivalent to "a first-order approximation of the Schwarzschild metric near a black hole's horizon".'

What I meant is that the Rindler horizon interpretation that Egan portrays does not exist in what I call "flat spacetime": in flat spacetime, Eve does not think that Adam never crosses the horizon.

Then what's the difference in Schwarzschild spacetime? That's the whole point of the Rindler horizon analogy: that if Eve does not think Adam never crosses the horizon, Eve' who is hovering above a black hole horizon should not think that Adam', who drops off her spaceship and falls into the hole, never crosses the horizon either.

If you think there is a difference, what's the difference? Why can't Eve' reason the same way that Eve does, to conclude that Adam' does cross the horizon?

 

[grav-universe]

We've been using the word in a couple of different senses:

(1) A line element is singular at a particular set of coordinate values if any of the coefficients is mathematically undefined for that set of coordinate values. For example, the SC line element is singular at r = 2m because the coefficient of dr^2 has (1 - 2m/r) in the denominator, which is mathematically undefined (you can't divide by zero).

(2) A spacetime is singular at a particular event if some invariant quantity is mathematically undefined at that event. For example, at r = 0 in Schwarzschild spacetime, the curvature is mathematically undefined; formulas for various curvature invariants have r in the denominator, so at r = 0 they are mathematically undefined (again, because you can't divide by zero).

People often use the term "goes to infinity" as a synonym for "mathematically undefined"; but that's just convenient (if sloppy) terminology. It doesn't imply that one can somehow evaluate singular quantities at the points where they are singular. One can try to take limits as the singular point is approached, but that only helps if the limit turns out to be finite; often it doesn't.

The time dilation is always singular there since it is an invariant for that shell, at least for a *hypothetical* static observer, or more simply put, directly applying the co-efficient in the metric for that r, although the straight-forward application of the metric would still apply to the clock of a static observer there, but anyway, I suppose it could only be the dr^2 component that can be made non-singular as in sense #1 since that is coordinate dependent, right? Along with the tangent component though, so both spatial components can be made non-singular, but never the time component, correct? That's interesting. What is a form of the metric (the transformation of co-efficients from SC) that would allow both spatial components to be non-singular?

Note the (1 + 2m/R)^2 in the last term; your version didn't have the ^2 there, which may have been an inadvertent typo.

I had the ^2 in post #165. I'm not sure if I posted it anywhere else, but I don't think I did.

The angular part now integrates easily at R = 0 to yield a horizon area of 16 \pi m^2.

Oh yeah, right. Good catch. :)

 

[PeterDonis]

The time dilation is always singular there since it is an invariant for that shell

No, it isn't, because the horizon is not timelike. You evidently don't realize how much of your reasoning is valid only for a timelike surface, or, to put it another way, it's valid only for an *actual* static observer, one that moves on a timelike worldline. The "hypothetical static observer" you keep referring to at the horizon does *not* move on a timelike worldline, so he can't exist, so you can't draw any deductions from his "hypothetical" existence. (This is another way of stating that the horizon is not a "place" the way locations with constant r > 2m are places.)

In the case of time dilation, it is true that there is an invariant involved: it is the contraction of the 4-velocity of a static observer with the 4-momentum of a radial light ray either being emitted or absorbed, which gives the energy of the light ray as measured by the observer. (Strictly speaking, we then have to use quantum mechanics to convert energy to frequency, and frequency to "rate of time flow" for the static observer, but that's a minor technical point for this discussion.)

However, at the horizon, there is no "4-velocity", because the horizon is null, not timelike. The 4-velocity of a static observer is a unit vector that is tangent to his worldline; but there is *no* unit vector that is tangent to a null curve, because a null curve, by definition, has a tangent vector with length zero. So the invariant in question can't even be defined at the horizon.

directly applying the co-efficient in the metric for that r, although the straight-forward application of the metric would still apply to the clock of a static observer there, but anyway, I suppose it could only be the dr^2 component that can be made non-singular as in sense #1 since that is coordinate dependent, right? Along with the tangent component though, so both spatial components can be made non-singular, but never the time component, correct? That's interesting. What is a form of the metric (the transformation of co-efficients from SC) that would allow both spatial components to be non-singular?

I think you're making it more difficult for yourself by focusing so much on the metric coefficients. Read again what I wrote above, about why the "time dilation invariant" can't be defined at the horizon. Did I mention anything about metric coefficients? Everything I said was stated in terms of coordinate-free concepts, like whether a particular curve (such as a curve of constant r, theta, phi) is timelike or null.

As far as coordinate charts that are non-singular at the horizon, I think I already listed some, but maybe it wasn't in this thread; there are quite a few on this general topic right now. However, I should amplify that somewhat, since whether a chart is non-singular depends on what aspect of the chart you're looking at.

The only charts I'm aware of that are *completely* non-singular at the horizon, meaning we can express *any* invariant there in the chart, are the Kruskal and Penrose charts. The key feature of these charts is that, if you look at the line element, not only are none of the coefficients mathematically undefined (i.e., no zeros in the denominator), none of them are *zero* either. That means the inverse metric (what you get if you consider the metric as a matrix and invert it) is also well-defined. (Btw, this includes the "time component", so it's not true that there are no charts where the "time component" is completely non-singular.)

The Painleve chart and the Eddington-Finkelstein chart have non-singular line elements at the horizon, but they do have a coefficient that's zero there (the coefficient of dt^2), so the inverse metric is not well-defined. (These charts have the same issue with the "time component" that the SC chart does at the horizon.)

 

[HomogenousCow]

Can't we just define a three dimensional time slice through the manifold for each coordinate at each coordinate time, then simply say that events in that slice are simultaneous?

 

[PeterDonis]

Can't we just define a three dimensional time slice through the manifold for each coordinate at each coordinate time, then simply say that events in that slice are simultaneous?

Yes, but there are multiple ways of doing that, and some of them don't even cover the entire manifold.

In the case of Schwarzschild spacetime, for example, consider the following three slicings:

(1) The Schwarzschild slicing: slices of constant Schwarzschild coordinate time. Strictly speaking, this is *exterior* Schwarzschild coordinate time, since the "t" coordinate in the SC chart is not timelike for r <= 2m. This slicing only covers the region outside the horizon; the slices actually "converge" as you approach r = 2m, and at r = 2m they all intersect (at least, in the idealized, not physically reasonable case where there is vacuum everywhere--see below under the Kruskal slicing), so the slicing is no longer valid there (you can't have the same event on multiple slices). This is similar to the way Rindler coordinates break down at the Rindler horizon in flat Minkowski spacetime.

(2) The Painleve slicing: slices of constant Painleve coordinate time. This slicing covers the regions outside *and* inside the horizon.

(3) The Kruskal slicing: slices of constant Kruskal time. This slicing also covers the regions outside and inside the horizon; but in addition, it reveals two *other* regions (at least, it does for the idealized, not physically reasonable case where the spacetime is vacuum everywhere, i.e, there is no matter present--in any real spacetime, there would be matter present and the other two regions would not be there) that are not covered by any other slicing.

 

[harrylin]

Then what's the difference in Schwarzschild spacetime? That's the whole point of the Rindler horizon analogy: that if Eve does not think Adam never crosses the horizon, Eve' who is hovering above a black hole horizon should not think that Adam', who drops off her spaceship and falls into the hole, never crosses the horizon either.

If you think there is a difference, what's the difference? Why can't Eve' reason the same way that Eve does, to conclude that Adam' does cross the horizon?

See the new thread, https://www.physicsforums.com/showthread.php?p=4181348

 

[Austin0]

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.

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.

Quote by harrylin


I don't know what Egan had to say but I think you are quite mistaken regarding Rindler coordinates and the horizon.
I don't think Rindler has anything to do with it. It is a coordinate artifact due to the dynamic metric in any accelerating system. This applies just as well to momentarily co-moving inertial frames. It happens because the distance to a point towards the rear shrinks due to contraction comparable to the increase in length due to system motion. SO the system asymptotically stops moving relative to points nearing the horizon as calculated . from a point within the system.
So harrylin is correct that Adam never crosses the horizon in any MCRF that Eve is at rest in.

No, it doesn't; at least, not in flat spacetime. In flat spacetime, any inertial frame covers the entire spacetime, including the portion of Adam's worldline at and beyond the Rindler horizon. That's a basic fact about inertial frames in flat spacetime. An MCIF is an inertial frame, so this fact applies to MCIFs in flat spacetime. Another way of saying this is that in flat spacetime, every inertial frame is global.

In curved spacetime, there are *no* global inertial frames; *any* inertial frame can only cover a small patch of the spacetime. So in curved spacetime, you are correct that an MCIF at some event on an accelerated observer's worldline might not cover the horizon. But Egan's scenario is entirely set in flat spacetime, so the restrictions on inertial frames, including MCIF's, in curved spacetime doesn't apply.

This all is neither addressing my statements nor correct.
A Mpmentarily Co-moving Inertial Frame is by difinition a limited slice of spacetime. A MOMENT of constant time in the chart of that frame. To say that a MCIF is global is simply false. As far as that goes neither is a Rindler chart global so in fact there is no global chart for an accelerating system in spite of the fact it is moving through flat spacetime. Or do you disagree?

So my statement:
" So harrylin is correct that Adam never crosses the horizon in any MCRF that Eve is at rest in"
unambiguously means that at the moment Eve is at rest in any frame, the charted position of Adam according to the instantaneous metric of this frame is inside the position of the horizon.
Do you still think this is incorrect??

Can you provide an example of a case where this would not apply?

Do you understand that the relevant question is not whether the chart covers the horizon but whether Adam's instantaneous position is inside the horizon's x coordinate or not at the time of evaluation?

Also, a word about "coordinate artifact". The fact that you can't assign a finite Rindler time coordinate to events at and beyond the Rindler horizon is an artifact of Rindler coordinates. But the fact that a light ray at the Rindler horizon will never intersect any of the "Rindler hyperbolas"--the curves with constant Rindler space coordinates--is not a coordinate artifact; you can express the same fact in any coordinate chart, because the curves themselves are geometric objects, not coordinate artifacts. So the existence of a "Rindler horizon" is not a coordinate artifact; there is something real and physical going on.

Your response here is not appropriate as I made no general statements about the horizon and was only talking within the limited context of the Adam and Eve example. Not related to light chasing an accelerating system. This phenomenon has nothing to do with coordinate systems (Rindler vs MCIF) per se and is just an empirical consequence of a finite light speed and a constantly accelerating system.

But as such still agrees with my statement that there is no significant effect due to Rindler coordinates as opposed to MCRFs . The effects are directly related to acceleration itself and are independent of coordinates.

 

[PeterDonis]

A Mpmentarily Co-moving Inertial Frame is by difinition a limited slice of spacetime.

In curved spacetime, yes. In flat spacetime, no. In flat spacetime, all inertial frames cover the entire spacetime. The MCIF is called "momentarily comoving" because an accelerated observer is only at rest in the MCIF for an instant; but that has nothing to do with how much of the spacetime the MCIF, or indeed any inertial frame, covers.

A MOMENT of constant time in the chart of that frame.

An inertial frame (momentarily comoving or not) is not the same thing as "a moment of time".

To say that a MCIF is global is simply false.

I disagree. See above.

as far as that goes neither is a Rindler chart global so in fact there is no global chart for an accelerating system in spite of the fact it is moving through flat spacetime. Or do you disagree?

It depends on what you mean by "a global chart for an accelerating system". If you mean a chart in which the accelerated object is at rest for more than an instant, then the most natural such chart, the Rindler chart, does not cover the entire spacetime. But there are other possible charts that could be used in which the accelerated object is at rest but the entire spacetime is still covered. In some recent thread or other, PAllen linked to a paper by Dolby and Gull that describes such a chart; if I can find the link I'll repost it here.

" So harrylin is correct that Adam never crosses the horizon in any MCRF that Eve is at rest in"
unambiguously means that at the moment Eve is at rest in any frame, the charted position of Adam according to the instantaneous metric of this frame is inside the position of the horizon.
Do you still think this is incorrect??

Yes, because any inertial frame, momentarily comoving with Eve or not, covers the entire spacetime, including the portion behind the horizon. The Rindler chart does not, but the Rindler chart is not an inertial frame.

The effects are directly related to acceleration itself and are independent of coordinates.

It depends on which "effects" you are talking about. The coordinates assigned to Adam are not "directly related to acceleration itself"; there is nothing requiring Eve to use the Rindler chart. Which light signals sent by Adam will intersect Eve's worldline *is* independent of coordinates.