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

#145
Nov2712, 11:04 AM

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I guess it could be just enough to settle for meaning of "frozen star" as predictions of GR that are testable (falsifiable) by observations from Earth or at least without going on the suicide mission. 



#146
Nov2712, 02:26 PM

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"They can make the same computations" is very much the "twin paradox". As Einstein explained, a symmetrical interpretation for an asymmetrical physical situation is incompatible with the foundations of GR. In fact, I don't know any theory of physics that violates that principle. In his discussion of the twin paradox (which was, I think, the first time that Einstein gave reason to doubt the reality of his "induced gravitational fields") he details how different the physical interpretation of a gravitational field is from that of acceleration; only the observable phenomena are held to be identical. And that brings me to a related point (I rearrange): "From our consideration of the accelerated chest we see that a general theory of relativity must yield important results on the laws of gravitation. In point of fact, the systematic pursuit of the general idea of relativity has supplied the laws satisfied by the gravitational field. Before proceeding farther, however, I must warn the reader against a misconception suggested by these considerations. A gravitational field exists for the man in the chest, despite the fact that there was no such field for the coordinate system first chosen. Now we might easily suppose that the existence of a gravitational field is always only an apparent one. We might also think that, regardless of the kind of gravitational field which may be present, we could always choose another referencebody such that no gravitational field exists with reference to it. This is by no means true for all gravitational fields, but only for those of quite special form. It is, for instance, impossible to choose a body of reference such that, as judged from it, the gravitational field of the earth (in its entirety) vanishes." "The nonmathematician is seized by a mysterious shuddering when he hears of "fourdimensional" things, by a feeling not unlike that awakened by thoughts of the occult. [..] the world of physical phenomena which was briefly called "world" by Minkowski is naturally four dimensional in the spacetime sense. For it is composed of individual events, each of which is described by four numbers"  Relativity:The Special and General Theory 



#147
Nov2712, 04:23 PM

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The term "comoving inertial reference frame" is more precisely stated as "momentarily comoving inertial reference frame" (MCIF). You can construct such a frame centered on any event on Eve's worldline; it will be the inertial frame in which Eve is momentarily at rest at the chosen event. This does not mean that Eve is not accelerating; it just means that it is easier to do a lot of the math in an inertial reference frame. The disadvantage of doing this is that, as I said, Eve is only at rest momentarily in any such frame; so if you want to look at the physics along Eve's worldline for any significant period of time, you can't use a single MCIF to do it. So the MCIF isn't a good representation of "Eve's point of view" over any significant length of Eve's time. An MCIF is also *not* the same as Rindler coordinates; those are noninertial. The advantage of Rindler coordinates is that they can cover all of Eve's worldline with a single coordinate chart in which Eve is "at rest" (her spatial coordinates in this chart don't change). This makes Rindler coordinates a better candidate for representing "Eve's point of view". But since the coordinates are noninertial, they behave differently from the inertial coordinates you're used to in SR. For one thing, as has been noted before, Rindler coordinates don't cover the entire spacetime, so if Rindler coordinates describe Eve's "point of view", then as I said before, Eve's point of view is necessarily limited in a way that Adam's is not. This brings up a general comment: you are insisting on centering your reasoning around terms like "gravitational field" that are simply not fundamental according to GR. That is why you are having all these problems trying to interpret what's going on. There is no single consistent interpretation of the term "gravitational field" that matches all the physics. There just isn't. To find an interpretation that matches all the physics, you have to give up the term "gravitational field" and change your set of concepts, to include things like "spacetime curvature", "stressenergy", "Einstein Field Equation", etc. instead. (1) A light beam emitted from the nose of Eve's rocket to the tail will look blueshifted at the tail, because the tail is accelerating towards it (in Adam's rest frame). (2) A light beam emitted from the tail to the nose will look redshifted at the nose, because the nose is accelerating away from it (in Adam's rest frame). Nowhere in any of this does "length contraction" appear; the reasoning applies equally well at the moment Adam drops off the rocket and at a later time when the rocket is moving at nearly the speed of light relative to Adam. If you work through the math, the observed blueshift/redshift depends only on Eve's proper acceleration; it does not depend on her instantaneous velocity relative to Adam. So it does not depend on the absolute value of her "length contraction" or "time dilation"; it only depends on the *change* in those values during the time of flight of a light beam across her rocket, which depends on her acceleration. http://en.wikipedia.org/wiki/Equivalence_principle The key thing I was trying to focus in on is that, in GR, you can always set up a local inertial frame centered on a particular event, in which "the acceleration due to gravity" vanishes. More precisely, you can always set up a local inertial frame centered on a particular event in which the following is true: (1) The metric in the local inertial frame, at the chosen event, is the Minkowski metric; i.e., it is [tex]ds^2 =  dt^2 + dx^2 + dy^2 + dz^2[/tex] where t, x, y, z are the local coordinates in the local inertial frame, whose origin (0, 0, 0, 0) is the chosen event. (2) The first derivatives of all the metric coefficients are zero at the chosen event; this means that the metric coefficients are the Minkowski ones not just at the chosen event, but in the entire local inertial frame. The second condition is what ensures that there is no "apparent gravitational field" in the local inertial frame; i.e., that the worldlines of inertial objects (i.e., freely falling objects) are straight lines in the local inertial frame. But this also means that the worldlines of accelerated objectsfor example, the worldlines of objects at rest on the surface of the Earth, if we set up a local inertial frame centered on some event on the Earth's surfaceare *not* straight lines in the local inertial frame: in fact they are hyperbolas, just like Eve's worldline in Adam's frame. This is the sense in which, locally, we can "make gravity look like acceleration"; we are making objects that are static in the local gravitational field look like accelerated objects in flat spacetime [Edit: and we are also making objects that are freely falling in the local gravitational field, and hence are "accelerating" from the viewpoint of an observer static in the field, look like objects at rest in an inertial frame in flat spacetime.] But the local inertial frame only covers a small piece of spacetime around the chosen event; how small depends on how curved the spacetime is and how accurate our measurements of tidal gravity are. Spacetime curvature, i.e., tidal gravity, depends on the *second* derivatives of the metric coefficients, and those *cannot* all be set to zero by any choice of coordinates if the spacetime is curved. This is the sense in which we *cannot* "make gravity vanish" by choosing coordinates; the curvature will always be there, as in Einstein's example of being unable to make the gravitational field of the Earth vanish in its entirety by any choice of coordinates. Perhaps I should belabor this some more, since it is an important point. Here's what you quoted from Einstein: "a clock kept at this place [i.e., at the horizon] would go at rate zero". I can interpret this one of two ways: (1) Einstein is claiming that a clock can be kept at the horizon, and saying that it would go at rate zero. (2) Einstein is claiming that a clock *cannot* be kept at the horizon, because if it could, it would go at rate zero, and that doesn't make sense. Are you interpreting him as saying #1 or #2? If it's #1, the refutation is pretty easy: the clock would have to go at the speed of light, and no clock can do that. So let's look at the other possibility; #2 leads to one of the following: (2a) A clock can't be kept at the horizon, because if it could, it would go at rate zero, and that doesn't make sense. Therefore the horizon can't exist, and neither can any spacetime inside it. (2b) A clock can't be kept at the horizon, because if it could, it would go at rate zero, and that doesn't make sense. Therefore the horizon (i.e., a curve of constant r = 2m) can't be a timelike curve, because if it were, a clock could follow it as a worldline. But the horizon could still exist if it were some other type of curve, such as a null curve; and if so, there could also be a region of spacetime inside the horizon, where curves of constant r < 2m are also not timelike. Einstein, as far as I can tell, believed #2a; but "modern GR" says #2b. I can't tell for sure what Oppenheimer and Snyder thought, since they didn't address the question in their paper; but everyone who has extended their model has come up with #2b as well. I'll put comments on the "metaphysical" aspects of all this in a separate post. 



#148
Nov2712, 06:43 PM

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(3) A clock can't cross the horizon, because if it could, it would have to go at the speed of light, and no clock can do that. Of course there's still a way out. Due to the extreme gravity at the horizon, all matter might be annihilated so that only massless and timeless particles actually cross, but personally I opt for #2a. 



#149
Nov2712, 07:40 PM

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#150
Nov2712, 07:57 PM

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Of course such a model is idealized; so is every model we use in physics. But its physical interpretation is not a matter of "opinion". Whether or not it's a *valid* model, taking its idealizations into account, is a separate question of what the physical interpretation of the model is. 



#151
Nov2712, 08:05 PM

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#152
Nov2712, 08:44 PM

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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. 



#153
Nov2712, 08:45 PM

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 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). 



#154
Nov2712, 09:44 PM

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The equation of motion in GR, at least for radial freefall, as related to the time dilation is
sqrt(1  (v'_r/c)^2) / z_r = K where v'_r is the locally measured speed at r, z_r is the time dilation at r, and K is a constant of motion. With the initial condition of a particle freefalling from rest at infinity, for example, v' = 0 and z = 1, so K = 1. For a photon K = 0 and for any massive particle K > 0, depending upon the initial conditions for the state of motion of the particle, and K remains constant at all r whether the particle is travelling inward or outward. For a particle starting outside the horizon and falling inward, at the event horizon, where z_r = 0, in order for K to remain a finite value, then sqrt(1  (v'_r/c)^2) = 0 also, whereby v'_r = c at the horizon regardless of the initial state of motion. That is the general math of it anyway, taken up to the mathematical limit at the horizon. You both appear to disagree, however, that this applies at the limit itself, or one might simply refute that no static observers can exist at the horizon, so let me put it another way. The locally measured speed of the particle cannot be less than c at the horizon because if either of you were to give me the initial conditions for the state of motion and any value for the locally measured speed of the particle that is less than c, I can find the radius where that particular speed would be measured, and it would always lie outside the horizon. No matter how close to c you can get, the radius will still lie outside, and a greater speed will always be achieved as it continues to freefall toward the horizon. Therefore, at no locally measured speed less than c can a massive particle cross the horizon. 



#155
Nov2712, 09:49 PM

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#156
Nov2712, 09:56 PM

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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? 



#157
Nov2712, 10:16 PM

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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? 



#158
Nov2712, 10:23 PM

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#159
Nov2712, 10:35 PM

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#160
Nov2712, 10:45 PM

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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? 



#161
Nov2712, 10:59 PM

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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. 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. 



#162
Nov2712, 11:09 PM

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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 45degree 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 crossit'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 raythe horizon, which is moving outward *at* c. To him the horizon is just the "limit point" of the static observers; but the static observers can never see that limit point because light emitted at the horizon stays at the horizon; it never gets to any larger radius. 


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