GR coordinate acceleration problem

[PeterDonis]

So it does indeed work out the same, as an invariant, in both coordinate systems.

This can be confirmed in general, but unfortunately it's going to be disappointing when you see what the invariant actually is. In post #23, you defined $L$ and $L_t$ using the general form of the metric (you left out an $r^2$ in the formula I'm about to give, but I've put it back in since it's clear from your other posts that you intended it to be there):

$$
ds^2 = - z^2 dt^2 + dr^2 / L^2 + r^2 d\phi^2 / L_t^2
$$

Compare this with the line element in the equatorial plane for Schwarzschild coordinates:

$$
ds^2 = - \left( 1 - 2M / r \right) dt^2 + dr^2 / \left( 1 - 2M / r \right) + r^2 d\phi^2
$$

This gives $L = \sqrt{1 - 2M / r}$ and $L_t = 1$, so we have $L_t' = 0$ and your invariant $I = \left[ 1 - r \left( L_t' / L_t \right) \right] \left( L / L_t \right)$ just becomes $L$, i.e., $I = \sqrt{1 - 2M / r}$.

For isotropic coordinates (I'll use a capital $R$ for the radial coordinate here):

$$
ds^2 = - \left( 1 - M / 2R \right)^2 / \left( 1 + M / 2R \right)^2 dt^2 + \left( 1 + M / 2R \right)^4 \left( dR^2 + R^2 d\phi^2 \right)
$$

This gives $L = L_t = \left( 1 + M / 2R \right)^2$ and $L_t' = 2 \left( 1 + M / 2R \right) \left( - M / 2R^2 \right) = - \left( M / R^2 \right) \left( 1 + M / 2R \right)$. This gives for your invariant $I$:

$$
I = \left( 1 - R \frac{L_t'}{L_t} \right) \frac{L}{L_t} = 1 - R \frac{M}{R^2} \frac{1}{1 + M / 2R} = 1 - \frac{2M}{2R + M} = \frac{2R - M}{2R + M}
$$

If we substitute $r = R \left( 1 + M / 2R \right)^2$ into the Schwarzschild formula for $I$ that we obtained above, we get

$$
I = \sqrt{1 - \frac{2M}{R \left( 1 + M / 2R \right)^2}} = \sqrt{\frac{R \left( 1 + M / 2R \right)^2 - 2M}{R \left( 1 + M / 2R \right)^2}} = \sqrt{\frac{\left(2R + M \right)^2 - 4MR}{\left(2R + M \right)^2}} = \sqrt{\frac{\left( 2R - M \right)^2}{\left( 2R + M \right)^2}} = \frac{2R - M}{2R + M}
$$

So $I = \left[ 1 - r \left( L_t' / L_t \right) \right] \left( L / L_t \right)$ is indeed an invariant. But what is its physical meaning? Well, the formula for it in Schwarzschild coordinates should be a clue: the invariant $I$ is just $z$. Looking at the formula in isotropic coordinates, we see that this is indeed the case; we have

$$
z = \frac{1 - M / 2R}{1 + M / 2R} = \frac{2R - M}{2R + M}
$$

So unfortunately, all you've done is to rediscover the invariant $z$ that you already have. You haven't found any new invariants.

 

[PeterDonis]

You haven't found any new invariants.

As a more general comment, searching for invariants should not have to be hit-or-miss. We can look at the symmetries of the spacetime to help figure out what invariants we should expect.

For example, in any spacetime with a Killing vector field, the length of that KVF at a given event is an invariant. In Schwarzschild spacetime, the invariant associated with the "time translation" KVF (I put "time translation" in quotes because it's only timelike outside the horizon, but it's a valid KVF everywhere) is just your $z$. (To be precise, if we denote the KVF by $\xi^a$, then we have $\xi^a \xi_a = z^2$ everywhere in the spacetime. Outside the horizon, $z^2$ is positive and the KVF is timelike; on the horizon, $z^2 = 0$ and the KVF is null; inside the horizon, $z^2$ is negative and the KVF is spacelike.)

The other obvious invariant in Schwarzschild spacetime is $M$; and as far as I can tell, every other invariant can be expressed in terms of $M$ and $z$. For example, the proper acceleration $a$ of a "hovering" observer and the area $A$ of a 2-sphere at a given $z$ are given by

$$
a = \frac{\left( 1 - z^2 \right)^2}{4 M z}
$$

$$
A = \frac{16 \pi M^2}{\left( 1 - z^2 \right)^2}
$$

With the area of the 2-sphere we have also covered the spherical symmetry of the spacetime, and there are no other symmetries (no other KVFs), so we should not expect to find any other invariants based on the spacetime geometry alone.

 

[grav-universe]

This can be confirmed in general, but unfortunately it's going to be disappointing when you see what the invariant actually is. In post #23, you defined $L$ and $L_t$ using the general form of the metric (you left out an $r^2$ in the formula I'm about to give, but I've put it back in since it's clear from your other posts that you intended it to be there):

$$
ds^2 = - z^2 dt^2 + dr^2 / L^2 + r^2 d\phi^2 / L_t^2
$$

Compare this with the line element in the equatorial plane for Schwarzschild coordinates:

$$
ds^2 = - \left( 1 - 2M / r \right) dt^2 + dr^2 / \left( 1 - 2M / r \right) + r^2 d\phi^2
$$

This gives $L = \sqrt{1 - 2M / r}$ and $L_t = 1$, so we have $L_t' = 0$ and your invariant $I = \left[ 1 - r \left( L_t' / L_t \right) \right] \left( L / L_t \right)$ just becomes $L$, i.e., $I = \sqrt{1 - 2M / r}$.

For isotropic coordinates (I'll use a capital $R$ for the radial coordinate here):

$$
ds^2 = - \left( 1 - M / 2R \right)^2 / \left( 1 + M / 2R \right)^2 dt^2 + \left( 1 + M / 2R \right)^4 \left( dR^2 + R^2 d\phi^2 \right)
$$

This gives $L = L_t = \left( 1 + M / 2R \right)^2$ and $L_t' = 2 \left( 1 + M / 2R \right) \left( - M / 2R^2 \right) = - \left( M / R^2 \right) \left( 1 + M / 2R \right)$. This gives for your invariant $I$:

$$
I = \left( 1 - R \frac{L_t'}{L_t} \right) \frac{L}{L_t} = 1 - R \frac{M}{R^2} \frac{1}{1 + M / 2R} = 1 - \frac{2M}{2R + M} = \frac{2R - M}{2R + M}
$$

If we substitute $r = R \left( 1 + M / 2R \right)^2$ into the Schwarzschild formula for $I$ that we obtained above, we get

$$
I = \sqrt{1 - \frac{2M}{R \left( 1 + M / 2R \right)^2}} = \sqrt{\frac{R \left( 1 + M / 2R \right)^2 - 2M}{R \left( 1 + M / 2R \right)^2}} = \sqrt{\frac{\left(2R + M \right)^2 - 4MR}{\left(2R + M \right)^2}} = \sqrt{\frac{\left( 2R - M \right)^2}{\left( 2R + M \right)^2}} = \frac{2R - M}{2R + M}
$$

Thanks PeterDonis. Actually though, that should be

L = L_t = 1 / (1 + M / (2 R))^2, L_t' = M R / (R+ M / 2)^3

I = (1 - R L_t' / L_t) (L / L_t) = 1 - M R^2 (1 + M / (2 R))^3 / (R + M / 2)^2
= 1 - M (R + M / 2)^2 / (R + M / 2)^3 = 1 - 2 M / (2 R + M) = (2 R - M) / (2 R - M)

Strange how that works out the same either way though.

So $I = \left[ 1 - r \left( L_t' / L_t \right) \right] \left( L / L_t \right)$ is indeed an invariant. But what is its physical meaning? Well, the formula for it in Schwarzschild coordinates should be a clue: the invariant $I$ is just $z$. Looking at the formula in isotropic coordinates, we see that this is indeed the case; we have

$$
z = \frac{1 - M / 2R}{1 + M / 2R} = \frac{2R - M}{2R + M}
$$

So unfortunately, all you've done is to rediscover the invariant $z$ that you already have. You haven't found any new invariants.

Actually, quite the contrary, thanks. If it can be demonstrated that I = z, then that is precisely what I need . With that we can solve for the unknowns very nicely. For instance, let's make the original coordinate choice, L_t = 1, whereby L_t' = 0. Then for the two relations

$z = (1 - r L_t' / L_t) (L / L_t)$ and

$m L_t^2 = z' L r^2$

we can reduce to just

$z = L$ and

$m = z' L r^2$

and putting those two together we get

$z z' = m / r^2$

and then solving for z using Wolfram, we find

$z = sqrt(c_1 - 2 m / r)$

where $c_1$ is a numerical constant. So since z must verge upon 1 for large r, $c_1$ must be 1, and since $L = z$, we therefore have

$L_t = 1, z = L = sqrt(1 - 2 m / r)$

 

[PeterDonis]

Actually though, that should be

L = L_t = 1 / (1 + M / (2 R))^2, L_t' = M R / (R+ M / 2)^3

I = (1 - R L_t' / L_t) (L / L_t) = 1 - M R^2 (1 + M / (2 R))^3 / (R + M / 2)^2
= 1 - M (R + M / 2)^2 / (R + M / 2)^3 = 1 - 2 M / (2 R + M) = (2 R - M) / (2 R - M)

Strange how that works out the same either way though.

Well, if we switch both definitions in concert, of course it doesn't change the invariant. Your $L$ and $L_t$ are just the reciprocals of mine, but since you took the reciprocal for both, the two changes cancel out.

(Using the reciprocals means you should also rewrite the line element; with your definitions here, $L$ and $L_t$ should appear in the numerator of their respective terms in the line element, but earlier you wrote them in the denominator, and that's the definition I used.)

Actually, quite the contrary

Contrary to what? Nothing you have done shows any new invariant other than $M$ and $z$. You've just shown how those invariants appear in the coordinate-dependent line element when you make particular coordinate choices.

 

[grav-universe]

Okay well, let's see. If

$z = (1 - r L_t' / L_t) (L / L_t)$

then from post #30 where we found that pervect's solution gives

$a' (r / L_t) / v_t'^2 = (1 - r L_t' / L_t) (L / L_t) / z$

the whole right side would reduce to unity, leaving just

$a' = v_t'^2 L_t / r$

and for the distant observer that becomes

$(a / (z^2 L)) = (v_t / (z L_t))^2 L_t / r$

$a = (v_t^2 / r) (L / L_t)$

cancelling out the $L / L_t$ I originally struggled with.

 

[grav-universe]

Well, if we switch both definitions in concert, of course it doesn't change the invariant. Your $L$ and $L_t$ are just the reciprocals of mine, but since you took the reciprocal for both, the two changes cancel out.

(Using the reciprocals means you should also rewrite the line element; with your definitions here, $L$ and $L_t$ should appear in the numerator of their respective terms in the line element, but earlier you wrote them in the denominator, and that's the definition I used.)

They are in the denominator, check the isotropic metric. That gets me a lot too , I have to double-check.

Contrary to what? Nothing you have done shows any new invariant other than $M$ and $z$. You've just shown how those invariants appear in the coordinate-dependent line element when you make particular coordinate choices.

I mean that if we can show that I = z by some simple method, we can solve for all of the unknowns, which is what I'm trying to do.

 

[PeterDonis]

$a = (v_t^2 / r) (L / L_t)$

Just for funsies, I'll express $v_t$ in terms of $M$ and $z$ as well. It's useful, first, to express $r$ that way; that's simple:

$$
r = \frac{2M}{1 - z^2}
$$

We have $v_t = \sqrt{M / \left( r - 2M \right)}$, which gives

$$
v_t = \sqrt{\frac{1 - z^2}{2 z^2}}
$$

It's interesting that $M$ does not appear explicitly in this formula.

 

[PeterDonis]

They are in the denominator, check the isotropic metric.

Ah, yes, I see; I was smart enough to make an even number of reciprocal errors, just as it's always good to make an even number of sign errors.

 

[grav-universe]

I can now visualize why there can be only one coordinate system where the coordinate acceleration is measured the same. Let's say we have such a coordinate system for the distant observer. Locally at r, the radial acceleration of a particle falling from rest is the same as that of a particle traveling tangently so that the equivalence principle will hold. As with the falling elevator example, the elevator falling from rest and a particle traveling tangently must accelerate radially together so that according to the elevator observer, the particle will travel straight across inertially, ignoring the gravitational gradient. The distant observer agrees that both the particle traveling tangently at r and another particle that falls from rest at the moment the first particle passes will both accelerate radially together, so having the same coordinate radial acceleration.

Now let's change the coordinate system of the distant observer. Think of the space around the mass as composed of spherical shells at all r, each with their own values for z, L, and L_t, although L is measured between shells that are an infinitesimal distance apart. We will change the coordinate system with the simple coordinate transform r1 = r - 3 r_s. This simply pushes all spherical shells inward by the same amount, 3 r_s. The distances between shells, however, remains the same, since all of the shells move inward radially by the same amount, so the value of L also remains the same, since that is the length contraction of local radial rulers, which the distant observer measures the same as before. Of course the time dilation z is an invariant for each particular shell, so that remains the same as well. Since the shells are now smaller, though, the value for the length contraction in the tangent direction L_t will be smaller, so the distant observer will say that a ruler that is pointing tangently at each shell is now smaller by a factor of r1 / r = 1 - 3 r_s / r.

Okay, so the particle falling from rest will now fall over the same distance between shells in the same time, so that coordinate acceleration will remain the same as before. The orbitting particle however, traveling tangently at the same place as the particle dropped from rest, will now still travel the same angle in the same time around the orbit as before, but since the shell is now smaller, that means it will have traveled a lesser distance along the x and y directions than before, so falling inward at a lesser rate than the particle falling from rest by the same factor as that between the old and new shell radii. The radial coordinate acceleration of both particles will no longer be measured the same.

In order for both particles to still be measured to have the same coordinate radial acceleration according to the distant observer, L and L_t must change by the same amount, so the entire space around the mass can only be scaled up or down by some numerical ratio only, all shells at once. But any scaling of this nature, say by a factor of .99 for example, will also leave measurements at large radii working toward .99 rather than 1 as they should. So there is one and only one coordinate system where the coordinate acceleration is measured the same.

All observers agree upon events that coincide in the same place, such as the readings upon two clocks as they pass. It seems to me that as the two particles pass, coinciding in the same place, that if the local observer says that each accelerates radially at the same rate, then all observers should agree. Since acceleration is measured over a distance and time that the particles travel, however, there is some difference between that and the readings upon two clocks at the exact moment they pass. The distances and times the accelerations are measured over, however, are infinitesimal, so it would still seem that all observers should agree, but it turns out there would be only one coordinate system where that would be possible, an isotropic coordinate system it appears from what I have further worked out if I have done it correctly, where L_t = L.

 

[PeterDonis]

We will change the coordinate system with the simple coordinate transform r1 = r - 3 r_s. This simply pushes all spherical shells inward by the same amount, 3 r_s.

No, it doesn't. It *relabels* the shells; a shell which was labeled by $r$ is now labeled by $r_1 = r - 3 r_s$. So, for example, a shell which used to be labeled by $r = 6 r_s$ is now labeled by $r_1 = 3 r_s$.

And, obviously, a shell which used to be labeled by $r = 3 r_s$ is now labeled by $r_1 = 0$. And a shell which used to be labeled by $r = 2.5 r_s$ is now labeled by $r_1 = - 0.5 r_s$. And so on. (You should already see problems with your coordinate transformation from what I just said here, but I'll ignore those problems for the rest of this post.)

But note that, for example, the surface area of the shell now labeled by $r_1 = 0$ (and formerly labeled by $r = 3 r_s$) is $36 \pi r_s^2$, not zero. In other words, $r_1$ as a coordinate has no direct relationship to any actual geometric quantities. That means you need to be *very* careful about reasoning from it.

The distances between shells, however, remains the same, since all of the shells move inward radially by the same amount

No, the distances between the shells remain the same because relabeling the shells doesn't change the distance between them (obviously). However, it's also true that, since $dr_1 = dr$, the distance between two neighboring shells in terms of $dr_1$ is the same as the distance in terms of $dr$.

so the value of L also remains the same

Yes, because $dr_1 = dr$, and $L$ appears in the $dr^2$ metric coefficient only, which is unchanged by your coordinate transformation.

Since the shells are now smaller

No, they aren't. Relabeling the shells doesn't change their surface area. See above.

the value for the length contraction in the tangent direction L_t will be smaller

Will it? Let's see. Your coordinate transformation means that the tangential part of the line element will now be $\left( r_1 + 3 r_s \right)^2 d \Omega^2$. So $L_t$ is actually unchanged, numerically; the only thing that changes is how it's expressed in terms of your new coordinate. In other words, once again, relabeling the shells doesn't change any of their geometric properties.

so the distant observer will say that a ruler that is pointing tangently at each shell is now smaller by a factor of r1 / r = 1 - 3 r_s / r.

No, he won't. See above.

 

[grav-universe]

Right, it's remapping the shells according to the distant observer's new coordinate system. The local values stay the same, nothing changes locally such as surface areas and so forth. With the coordinate system I changed to, L and z remains the same, z1 = z and L1 = L, but L_t1 changes only as it is expressed in terms of the new coordinate system, as you said. The locally measured radial acceleration of both particles is still the same, but are now different as they are inferred by the distant observer using the new coordinate system.

Also, the region within 3 r_s falls to a point with the coordinate system I gave, yes, off the map sort of speak, which I considered mentioning, but that is normal when changing coordinate systems. The same thing happens when changing from Schwarzschild to Eddington's isotropic coordinates, for instance. r_s falls from r = 2 m to r1 = m / 2 and the region originally below r = 2 m can no longer be mapped out in Eddington's coordinates, so the region mapped out below the event horizon at r1 = m/2 in Eddington's coordinates are a completely different region than that mapped below the event horizon at r = 2 m in Schwarzschild.

 

[PeterDonis]

L_t1 changes only as it is expressed in terms of the new coordinate system, as you said.

But what does that mean for the distant observer? As far as I can see, it means nothing.

The locally measured radial acceleration of both particles is still the same, but are now different as they are inferred by the distant observer using the new coordinate system.

But I don't see how that actually happens. From what I can see, what is "inferred by the distant observer" doesn't change either, because the distant observer will have to conclude that $r_1$ doesn't have any direct meaning. What has direct meaning is the actual numerical value of $L_t$, not the formula for $L_t$ in terms of a coordinate.

Also, the region within 3 r_s falls to a point with the coordinate system I gave

No, it doesn't. "Falls to a point" would imply that the surface area of a 2-sphere at $r_1 = 0$ was zero. It isn't.

The same thing happens when changing from Schwarzschild to Eddington's isotropic coordinates

Not really. In isotropic coordinates, $r_s = M / 2$, yes, but the region $r < r_s$ is perfectly well-defined; it's just another copy of the region $r > r_s$ instead of a different region altogether.

With your transformation, the region $r < 3 r_s$, which corresponds to $r_1 < 0$, is not well-defined--at least, not if you treat $r_1$ as a normal radial coordinate, which has to be nonnegative.

 

[grav-universe]

But what does that mean for the distant observer? As far as I can see, it means nothing.

But I don't see how that actually happens. From what I can see, what is "inferred by the distant observer" doesn't change either, because the distant observer will have to conclude that $r_1$ doesn't have any direct meaning. What has direct meaning is the actual numerical value of $L_t$, not the formula for $L_t$ in terms of a coordinate.

Right, the value for L_t changes. Instead of L_t = 1 in Schwarzschild, at r = 10 m for example, it would now be L_t1 = 1 - 6 m / (10 m) = .4. It means that a local 1 meter ruler placed in the tangent direction at r will be inferred to be 1 meter in length in Schwarzschild, but .4 meters with the new coordinate system.

No, it doesn't. "Falls to a point" would imply that the surface area of a 2-sphere at $r_1 = 0$ was zero. It isn't.

It implies that at first sight, but you showed in another thread that it doesn't even at r1=0. I'll have to find that. The locally measured surface areas remain the same. At r = 10 m for example, the surface area inferred by the distant observer is 400 ∏ m^2. In Schwarzschild that's what it is locally also since L_t = 1 for all shells. For r1 = r - 6 m, the inferred surface area is 64 ∏ m^2, but since local tangent rulers are length contracted by a factor L_t1 from what the distant observer infers, the locally measured surface area is still 4 ∏ (r1 / L_t1)^2 = 4 ∏ (4 m / .4)^2 = 400 ∏ m^2, so it remains the same.

Not really. In isotropic coordinates, $r_s = M / 2$, yes, but the region $r < r_s$ is perfectly well-defined; it's just another copy of the region $r > r_s$ instead of a different region altogether.

Right, it doubles back, but we can't have two sets of coordinates to map out the same spacetime, so it is only meant for the region above the event horizon at r = 2 m and the region below the event horizon at r1 = m/2 in Eddington is something altogether different.

With your transformation, the region $r < 3 r_s$, which corresponds to $r_1 < 0$, is not well-defined--at least, not if you treat $r_1$ as a normal radial coordinate, which has to be nonnegative.

It works for any coordinate transformation of the form 1 + n m, where n has a numerical value. We could use r1 = r - m and only lose half the radius below the event horizon, or r1 = r - 2 m to bring the event horizon to a point, or we could use 1 + m and map some extra undefined region. Any transformation between coordinate systems will generally subtract and/or add some undefined region in this way, unless they are transformed between point horizons I think. Makes me wonder if there should be only one "correct" coordinate system after all.

 

[PeterDonis]

Right, the value for L_t changes. Instead of L_t = 1 in Schwarzschild, at r = 10 m for example, it would now be L_t1 = 1 - 6 m / (10 m) = .4.

No, that's not correct; you're confusing the re-labeling of the spheres with the value of $L_t$.

The formula is $L_t = \left( r_1 + 3 r_s \right)^2$, because you've re-labeled the spheres, so a sphere labeled by $r$ before is now labeled by $r_1 = r - 3 r_s$. So the sphere labeled by $r = 5 r_s$ before will now be labeled by $r_1 = 2 r_s$. But where before we had $L_t = r^2 = 25 r_s^2$, we now have $L_t = \left( 2 r_s + 3 r_s \right)^2 = 25 r_s^2$, so the value of $L_t$ on a given 2-sphere is unchanged; only the labeling of the 2-sphere changes.

It means that a local 1 meter ruler placed in the tangent direction at r will be inferred to be 1 meter in length in Schwarzschild, but .4 meters with the new coordinate system.

No, it doesn't. See above.

It implies that at first sight, but you showed in another thread that it doesn't even at r1=0.

You'll have to show me a link, because I don't remember ever saying anything that contradicts what you quoted, and I certainly don't remember ever saying anything that would imply that a point has nonzero surface area.

At r = 10 m for example, the surface area inferred by the distant observer is 400 ∏ m^2. In Schwarzschild that's what it is locally also since L_t = 1 for all shells.

No, $L_t = r^2$. At least, that's the definition of $L_t$ you have to use if you want to be consistent when you do coordinate transformations. Otherwise you will be doing things like trying to use formulas for $L_t$ in which $r$ and $r_1$ both appear, which is nonsense *unless* you eliminate one or the other (as I did above).

For r1 = r - 6 m, the inferred surface area is 64 ∏ m^2

Why? You haven't shown this at all. You've just assumed that the distant observer will "infer" a surface area equal to $4 \pi r_1^2$, instead of actually looking at what $r_1$ means or doesn't mean, physically. But that ignores the reason *why* the distant observer infers a surface area of $4 \pi r^2$ in the original Schwarzschild coordinates: because that's what the Schwarzschild line element *tells* him. He certainly doesn't make that inference simply because he labels the coordinate as $r$ and calls it "radial".

Right, it doubles back, but we can't have two sets of coordinates to map out the same spacetime

Sure you can. You can't treat the whole chart as a single coordinate patch, but that doesn't mean you can't treat both patches as mapping the same region of spacetime.

so it is only meant for the region above the event horizon at r = 2 m and the region below the event horizon at r1 = m/2 in Eddington is something altogether different.

What do you mean by "something altogether different"?

The standard interpretation, as far as I know, is that isotropic coordinates double-cover the exterior region of Schwarzschild spacetime: i.e. the patch $r < r_s$ is a "copy" of the patch $r > r_s$; both patches map the *same* set of physical events.

There is a second interpretation, according to which the patch $r > r_s$ maps the "ordinary" exterior region of the maximally extended Schwarzschild spacetime, and the patch $r < r_s$ maps the other exterior region--these two regions are the ones marked I (ordinary) and III (other) on the diagram of Kruskal coordinates on the Wikipedia page:

http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

If that's what you mean by "something altogether different", then yes, that's a possible interpretation, but since nobody, as far as I know, claims that region III actually exists in any real spacetime, it's an interpretation that's just a mathematical curiosity with no physical meaning.

It works for any coordinate transformation of the form 1 + n m, where n has a numerical value. We could use r1 = r - m and only lose half the radius below the event horizon, or r1 = r - 2 m to bring the event horizon to a point, or we could use 1 + m and map some extra undefined region. Any transformation between coordinate systems will generally subtract and/or add some undefined region in this way, unless they are transformed between point horizons I think. Makes me wonder if there should be only one "correct" coordinate system after all.

I don't understand where you're going with any of this. What is "it works" supposed to mean?

Also, you are still using the term "point" incorrectly: the transform $r_1 = r - 2m$ doesn't make $r_1 = 0$ a point, because the surface area of the 2-sphere there is still nonzero.

 

[grav-universe]

Not sure what you did in your last post. The tangent length contraction L_t has a numerical value with no units, as do the time dilation z and radial length contraction L. The invariants with changes in coordinate systems are

L_t / r = L_t1 / r1

L / dr = L1 / dr1

z = z1

For Schwarzschild we have L_t = 1. For the coordinate system r1 = r - 6 m, it is

L_t / r = L_t1 / r1

1 / (r1 + 6 m) = L_t1 / r1

L_t1 = r1 / (r1 + 6 m) = 1 / (1 + 6 m / r1)

or

1 / r = L_t1 / (r - 6 m)

L_t1 = (r - 6 m) / r = 1 - 6 m / r

Here's the post where you showed that the coordinate system for r1 = r - 2 m, where the event horizon is brought to a point according to the distant observer's coordinate system, still has a nonzero area as measured locally.

 

[PeterDonis]

The tangent length contraction L_t has a numerical value with no units, as do the time dilation z and radial length contraction L.

Yes, that's right. And I was calculating what that numerical value was in the different charts. However, it looks like you're using different definitions of $L_t$ and $L_{t1}$ than the ones I was using. That affects the numerical values of those quantities, but not the physics or the things I was saying about it. See below.

The invariants with changes in coordinate systems are

L_t / r = L_t1 / r1

L / dr = L1 / dr1

z = z1

First of all, $L_t / r$ and $L / dr$ are *not* invariants. (Nor are the correct quantities that should go in the equality; see below.) They are coordinate-dependent quantities. You can't even define them, for example, in the Painleve chart. The only invariant you give here is $z$.

Second, the equality you give above, $L_t / r = L_{t1} / r_1$, doesn't look right to me given the rest of your post. The correct equality, as far as I can tell, is $L_t r = L_{t1} r_1$. See further comments below.

For Schwarzschild we have L_t = 1.

In other words, the tangential part of the line element is supposed to be $L_t^2 r^2 d \Omega^2$ in Schwarzschild coordinates, correct? I'm guessing, then, that for your chart, the tangential part of the line element is supposed to be $L_{t1}^2 r_1^2 d \Omega^2$.

But if that's the case, then the coefficients of $d \Omega^2$ must be equal, so we have (taking the square root since everything is squared)

$$
L_{t1} r_1 = L_t r
$$

which gives

$$
L_{t1} = \frac{r}{r_1} = \frac{r_1 + 3 r_s}{r_1} = 1 + \frac{3 r_s}{r_1}
$$

since $L_t = 1$. So for $r = 5 r_s$, we have $r_1 = 2 r_s$, and $L_{t1} = 1 + 3 r_s / 2 r_s = 5 / 2$. So you're correct that $L_{t1} \neq L_t$, but you have the relationship between the two incorrect, if I'm interpreting your intent correctly above.

However, the appropriate quantity to use for tangential distances is the entire coefficient of $d \Omega^2$; you can't arbitrarily remove $r_1^2$ from the comparison just because you are calling $r_1$ a "radial coordinate". And the coefficient of $d \Omega^2$, of course, is the same for the Schwarzschild chart and your chart; that's guaranteed by the above.

Another way of putting this is, the physical quantity is the surface area of the 2-sphere, and changing the labeling of the 2-sphere doesn't change its surface area. There's no such thing as "local surface area" vs. "surface area as seen by the distant observer"; the spacetime is spherically symmetric, so surface areas of 2-spheres aren't affected by "looking at them from a distance".

Here's the post where you showed that the coordinate system for r1 = r - 2 m, where the event horizon is brought to a point according to the distant observer's coordinate system, still has a nonzero area as measured locally.

You're misrepresenting what I said. I never used the word "point". If I had used it in that post, it would have been to say that, *because* of the nonzero area, that location is *not* a point. Which is what I've been saying here. You can't use the word "point" to describe a 2-sphere with nonzero area.

 

[grav-universe]

I'm using the metric

ds^2 = -z^2\, dt^2 + dr^2 / L^2 + dΩ^2 r^2 / L_t^2

so it looks like we're using the inverses of each other for L and L_t. Sorry about that.

 

[PeterDonis]

I'm using the metric

ds^2 = -z^2\, dt^2 + dr^2 / L^2 + dΩ^2 r^2 / L_t^2

so it looks like we're using the inverses of each other for L and L_t. Sorry about that.

Ok, thanks. That means the formula for $L_{t1}$ is the reciprocal of the one I wrote down, and its numerical value changes accordingly. It doesn't change the rest of what I said.