Notions of simultaneity in strongly curved spacetime

grav-universe

The equation of motion in GR, at least for radial freefall, as related to the time dilation is

sqrt(1 - (v'_r/c)^2) / z_r = K

where v'_r is the locally measured speed at r, z_r is the time dilation at r, and K is a constant of motion. With the initial condition of a particle freefalling from rest at infinity, for example, v' = 0 and z = 1, so K = 1. For a photon K = 0 and for any massive particle K > 0, depending upon the initial conditions for the state of motion of the particle, and K remains constant at all r whether the particle is 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.

PAllen

That is coordinate speed. Nobody measures coordinate speed. To get measured speed, you need to relate one 4-velocity (of measuring observer) to another 4-velociy (measured object). It is mathematically impossible for this to yield >=c.

PeterDonis

Once again, you stated it wrong. The correct statement is: "The equation of motion for radial freefall outside the horizon is..." The equation you wrote isn't valid at or inside the horizon, because what you are calling z_r is undefined there.

And what happens at or inside that limit?

Correct on both counts.

You keep on leaving out essential qualifiers. This time I've gone ahead and inserted the necessary qualifier in the quote above. With the qualifier inserted, your reasoning is no longer valid; the "locally measured speed" of an infalling particle, relative to an observer who is *not* static, does *not* necessarily approach c as you approach the horizon.

You can keep on trying at this, but I expect it to get monotonous. Let's try a different question: if a clock is released from rest at some radius r > 2m, with its clock time set to zero, and freely falls towards the horizon, what will the clock read at the instant it reaches the horizon? What do you think?

grav-universe

Right, by locally I mean measured by a static observer at that location. Okay well, let me ask you two these questions. A massive particle freefalls from rest at 4m toward the horizon and its speed is locally measured by static observers at each of their respective locations.

Can a massive particle cross the horizon before its locally measured speed reaches .99 c?

Can a massive particle cross the horizon before its locally measured speed reaches .9999999 c?

If no to both of these, is there any speed less than c that a massive particle can cross the horizon before reaching as measured locally by any static observer outside the horizon?

PeterDonis

No.

No.

No. So what? You didn't bother to make any argument that any of the above prevents the massive particle from crossing the horizon. Do you have such an argument? Bear in mind that your argument cannot assume that there is a static observer *at* the horizon, or that the concept of "locally measured speed" as you are using it is well-defined there, because there isn't one, and the concept isn't well-defined there.

grav-universe

z_r = 0 at the horizon. That is the time dilation.

That's the question, isn't it? Well, one could say it requires a different set of coordinates, or we can refer back to 2a. :)


I would have to see that.

It's finite, it stops at the horizon according to a distant observer. Your argument is that its proper time continues forward so that it passes the horizon. We could also switch to GR coordinates z = L = 1 / sqrt(1 + 2 m / r), another valid coordinate system but with no mapped interior coordinates and say that the clock strikes a point mass at that proper time. Or that the clock can never reach the horizon because its speed cannot reach c or that the time of the clock would freeze if it ever did reach c. So in a way it seems we're asking, if a clock did ever manage to reach c, even in a non-GR inertial system, would its proper time still continue?

grav-universe

Well, that is the argument. A massive particle cannot ever reach a locally measured speed of c, right? A static observer that lies outside the horizon will measure a speed that is less than c and a greater speed can always be achieved as it continues to approach the horizon, correct? Therefore, it never reaches the horizon at any speed less than c because any speed no matter how close to c would be measured outside the horizon and continue to increase, isn't that right?

And from your answers, if it cannot reach it before achieving a speed of .99 c or .9999999 c or any other speed at all that is less than c, then it can only reach it upon achieving c, correct?

PeterDonis

Oh, sorry, I mis-stated it. I should have said that, since z_r = 0 and it appears in the denominator of your equation of motion, your equation of motion is undefined at the horizon. I apologize for the error.

Or we could start talking about physics instead of coordinates. That is, we could start talking about actual physical observables *at* the horizon. I even gave you an example of one: the reading on a clock set to zero at the instant it is dropped from rest at a radius r > 2m, when it reaches the horizon. I see you gave an answer to that question; see below for further comments.

In other words, you haven't even bothered to learn the actual physical model you are criticizing. This is a simple calculation that is routinely assigned as a homework problem in relativity textbooks. I don't have time right now to post the details; I encourage you to look them up. The bottom line is that it is possible for two infalling observers to have any relative velocity less than c when they cross the horizon together, depending on the initial conditions of their infall.

Correct.

Incorrect. All that the distant observer can properly assert is that he will never receive a light signal that is emitted by the infalling clock at (or inside) the horizon. He cannot claim that this means the clock simply stops at the horizon; see below for why.

Yes. Do you understand why? It's because all physical quantities are finite there. There is nothing physically present at the horizon that would cause the infalling clock to stop there. So it doesn't. To claim otherwise is to claim that the laws of physics suddenly work differently at the horizon, for no apparent reason.

The more technical way of putting this is that the solution of the Einstein Field Equation is perfectly finite and continuous at the horizon; there is nothing in the solution leading up to the horizon that makes the horizon a place where the solution (i.e., spacetime) could just stop. It has to continue if the EFE is valid, and therefore the worldline of the clock has to continue as well. To claim otherwise is to claim that the EFE suddenly stops being valid at the horizon, for no apparent reason.

Do you mean isotropic coordinates? As in the ones described in the "alternative formulation" section here:

http://en.wikipedia.org/wiki/Schwarzschild_metric

If so, you are correct that these do not cover the interior region; that's because they double cover the exterior region (outside the horizon). The range 0 < R < m/2 (where R is the isotropic radial coordinate) covers the same set of events as the range m/2 < R < infinity; each event in the exterior region maps to *two* values of R, not one.

How do you figure that? Isotropic coordinates do not somehow make a point mass magically appear at the horizon.

It's true that the clock's speed can never "reach c", but false that that implies that it can't reach the horizon. Here's another way of looking at it: the clock is falling inward, and the horizon is moving outward. The reason the clock is "moving at c" relative to the horizon, when it crosses it, is that the *horizon* is a lightlike surface--*it* is a surface made up of outgoing light rays. So of course anything passing those light rays will be "moving at c" relative to the light rays, because light rays move at c relative to any timelike object.

Irrelevant since the clock never does reach c.

I'm not asking that; the clock never does "reach c", except in the sense I gave above, that it "moves at c" relative to the horizon because the horizon itself is made of outgoing light rays, and any massive object "moves at c" relative to light rays.

PeterDonis

Correct up to here, yes.

No. Go back through this chain of reasoning again, carefully, and make explicit all the unstated assumptions.

No again. Same advice as above.

I'll make again a suggestion I made in my immediately previous post, but I'll amplify it somewhat. Think of the horizon as a light ray moving outward, and the infalling object as a timelike object falling inward. In a local inertial frame centered on the event at which the infalling object crosses the horizon, the horizon will appear as a 45 degree line (because it's a light ray); we'll say that the positive "x" direction is radially outward, so the 45-degree horizon line goes up and to the right. The worldline of the infalling object is the "t" axis of the local inertial frame; i.e., it's a vertical line going through the origin.

In this local inertial frame, the worldlines of static observers at distances closer and closer to the horizon appear as segments of hyperbolas that cross the vertical axis at negative "t" values closer and closer to the origin (i.e., to t = 0). These hyperbolas asymptote to the horizon line (but we don't see the axis of the hyperbolas, where the two asymptotes cross--it's way down and to the left somewhere, off our diagram). Each of these segments is inclined closer and closer to 45 degrees, so their speed relative to the infalling observer (which is also the infalling observer's speed relative to them) gets closer and closer to c. However, that in no way prevents the infalling observer from reaching and crossing the horizon; from his point of view, he passes static observers moving outward at closer and closer to c, until finally he passes an outgoing light ray--the horizon, which is moving outward *at* c. To him the horizon is just the "limit point" of the static observers; but the static observers can never see that limit point because light emitted at the horizon stays at the horizon; it never gets to any larger radius.

harrylin

I just have time for this, as this sub-discussion took off:
None of them can be a correct interpretation, just as (as I mistakenly thought to have clarified,) none of the following can be a correct interpretation of "For v=c all moving objects—viewed from the “stationary” system—shrivel up into plane figures":

1. Einstein is claiming that it is *possible* for an object to move at c.
2. Einstein is claiming that an object cannot move at c, because if it could, it would shrivel up into a plane figure, and that doesn't make sense.
3. Einstein is claiming that an object cannot move at c, because if it could, it would have infinite energy, and that is impossible.
etc.

Instead, such statements simply refer to (unattainable) physical limits; and in both cases it takes infinite coordinate time to reach such limits. This is acceptable shorthand among physicists, but "forbidden" for mathematicians.

PeterDonis

So basically, you are saying that the correct interpretation of what Einstein said is simply:

(4) No object can reach the horizon, because that would take an infinite amount of coordinate time.

In other words, eliminate all mention of "moving at c", and just focus on the coordinate time. Correct? If so, I'm confused about where the phrase "a clock kept at this place would go at rate zero" fits in.

grav-universe

That particular coordinate system is the one I discovered and you and I discussed in the other thread. Rather than keep calling it the 1/sqrt(1 + 2m/r1) coordinate system, let's call it GU coordinates :) . It coordinately transforms from SC with

r1 = r (1 - 2m/r), r = r1 (1 + 2m/r1) with the metric

ds^2 = c^2 dt^2 / (1 + 2 m / r1) - dr1^2 (1 + 2 m / r1) - dθ^2 r1^2 (1 + 2 m / r1)^2

I like it because it completely eliminates the event horizon and interior spacetime altogether, leaving only what external observers observe. It shrinks the boundary of the event horizon to a point, so that from the perspective of external observers applying this coordinate system, the mass lies at a point singularity in the center just as in Newtonian with infinite acceleration there, no event horizon and no interior spacetime. A clock falling to the point mass will still do so in finite time. Proper distance measured to the point mass is also finite. But we would expect these when measuring the distance to a point or the time to fall to a point. It is just as valid as SC, and the EFE's valid also, being only a coordinate transformation, with all of the same external observables, but looking at it, one would not expect any more spacetime to exist within a point. From the perspective of this coordinate system, that would be like falling out of this universe altogether into some other dimension if there were interior spacetime within a point. If Schwarzschild had happened to come up with this coordinate system rather than the one he did, each just as likely to have been derived before the other, we might not even consider that any interior spacetime exists in the first place.

You also mention Eddington's isotropic coordinates. These are also valid. But as you said, with a one to one correspondence to SC coordinates, they only map some of the interior spacetime of SC, then double back. If one were to fall past the horizon and all the way to the center of EIC, then, when transformed back to SC, it would be like falling part way past the horizon, then doubling back and 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?

PeterDonis

Can you give a reference to the "other thread" you refer to? This does not look at all familiar to me, but I may just be failing to remember a previous discussion. I'll refrain from commenting on the rest of your post until I've got the context clear.

grav-universe

Sure, here it is. "Shrinking event horizon to point singularity"

PeterDonis

Ah, ok, thanks, that helps to jog my memory.

Pretty much everything I would say in response has already been said in the other thread, so I don't see much point in anything more than a quick recap (and what I'm saying applies just as well to Eddington isotropic coordinates as any other chart):

(1) You can't change the physics by changing coordinate charts. You can choose coordinates such that what used to be r = 2m is now r1 = 0; but you can't change the physical nature of the spacetime at what used to be r = 2m and is now r1 = 0. Just labeling it with r1 = 0 doesn't make it a point instead of a surface.

(2) To actually talk about the physics, you have to compute invariants--quantities that don't change when you change coordinate charts. If your chart is singular at a particular place, you can't compute invariants there using the chart, so you can't say anything about the physics there using the chart. Your chart is singular at r1 = 0, so it can't say anything about the physics at that location: in particular, you can't compute any invariant in your chart that shows that what you are labeling r1 = 0 is an actual, physical point, instead of, say, a surface that your coordinates don't cover well.

These points are basic facts of differential geometry as it is used in physics. They have been stated ad nauseam, and you don't seem to be accepting them. That means we really don't have a good basis for discussion.

grav-universe

Right, but that could go either way. Both being equally valid external coordinate systems, how do you know we're not making a surface out of a point? What I am asking, though, is what coordinate system you think accurately maps out the spacetime, no more and no less? You stated that EIC do not. Why not? Do SC? Why or why not?

PeterDonis

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.

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.]