Notions of simultaneity in strongly curved spacetimeby PAllen Tags: curved, notions, simultaneity, spacetime, strongly 

#163
Nov2812, 03:13 AM

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I just have time for this, as this subdiscussion took off:
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. 



#164
Nov2812, 09:10 AM

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(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. 



#165
Nov2812, 04:40 PM

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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 travelling 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? 



#166
Nov2812, 04:43 PM

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#167
Nov2812, 05:03 PM

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#168
Nov2812, 07:21 PM

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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 invariantsquantities 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. 



#169
Nov2812, 07:52 PM

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#170
Nov2812, 08:11 PM

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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 "Kruskaltype" 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 EddingtonFinkelstein 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 horizoninstead, as I've said before, it double covers the region outside the horizon. Another way of putting this is that the area of a 2sphere 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 2sphere (except at the horizon, r = m/2). This makes it pretty obvious that the EIC chart's coverage is incomplete: where are the 2spheres 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 2sphere 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.] 



#171
Nov2812, 09:03 PM

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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 FRWtype chart (the time reverse of the expanding FRWtype 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.




#172
Nov2912, 05:28 AM

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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: 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 "comoving inertial reference frame" is more precisely stated as "momentarily comoving inertial reference frame". Evans evidently means constantly comoving 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 comoving inertial reference frame, Adam passes through the horizon. 



#173
Nov2912, 08:44 AM

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



#174
Nov2912, 02:45 PM

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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: 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 offtopic in this thread; and now that I decided to start my own thread I'll include further elaboration there. 



#175
Nov2912, 03:06 PM

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#176
Nov2912, 05:29 PM

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Quote by harrylin
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 comoving 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. 



#177
Nov2912, 05:32 PM

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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 travelling 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 nonzero. 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? 



#178
Nov2912, 06:47 PM

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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 coordinatesis 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. 



#179
Nov2912, 07:24 PM

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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. 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. 



#180
Nov2912, 09:13 PM

P: 434

What is your definition of singular?



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