Notions of simultaneity in strongly curved spacetime

[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.