Notions of simultaneity in strongly curved spacetime

[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.