GR coordinate acceleration problem

[grav-universe]

Let's say we have a mass with an object orbitting with constant speed in a circular orbit and a distant observer Bob. According to Bob's coordinate system, the orbit is circular at a speed v and a constant inward coordinate acceleration a. The coordinate acceleration is just what is inferred according to Bob's coordinate system by the usual definitions of acceleration, s = vo t + 1/2 a t^2, 2 a s = vf^2 - vo^2, a = (vf - vo) / t, etc., where s is the distance traveled and vo and vf are the original and final velocities. The relativistic acceleration formulas also reduce to these when considering a body starting at rest and accelerating an infinitesimal distance over infinitesimal time, which we will be doing, so we need not worry with more complicated formulas for this.

So if the object is initially at coordinates x_o=r, y_o=0, and travels an infinitesimal distance y = v t in the tangent direction, it will have traveled a distance of x = r - sqrt(r^2 - y^2) in the radial direction toward the mass, since the orbit is perfectly circular. The distance being infinitesimal, we can drop higher orders and gain just x = r - r (1 - y^2 / (2 r^2)) = y^2 / (2 r) in the radial direction. We want the radial coordinate acceleration as inferred by Bob, so we can use s = vo t + 1/2 a t^2, where vo = 0 for the original velocity in the radial direction, so x = s = 1/2 a t^2.

So now we have x = 1/2 a t^2 = y^2 / (2 r), and since y = v t, this reduces to just

1/2 a t^2 = y^2 / (2 r)

a t^2 = (v^2 t^2) / r

a = v^2 / r

giving the usual acceleration formula for a circular orbit. Again, this is just the coordinate acceleration inferred by Bob's coordinate system. It is not saying anything about proper acceleration or anything else, just a coordinate effect. Does all of this look okay so far?

 

[Mentz114]

Your calculation looks right but your interpretation is not correct in the absence of gravity. In that case the acceleration is proper and the orbiting body will experience the force which keeps it in orbit.

 

[grav-universe]

Your calculation looks right but your interpretation is not correct in the absence of gravity. In that case the acceleration is proper and the orbiting body will experience the force which keeps it in orbit.

Okay good, thanks. I only want to measure the radial coordinate acceleration as inferred by Bob's coordinate system, independent of whatever the proper acceleration might be.

 

[grav-universe]

Okay, so next let's say we have a hovering observer Alice at r. At the moment the orbitting object passes Alice, Alice drops a ball from rest at r. We can say that the object and the ball coincide in the same place at that time. Now, according to the equivalence principle, the radial coordinate acceleration that Alice measures at that instant for both the object and the ball will be the same, correct? The object has no radial velocity, so it should be as if dropped from rest at that instant the same as the ball and accelerate radially with the ball for a moment, right? Although the object is also still traveling tangently.

As for the distant observer Bob, the same should be true according to his coordinate system, shouldn't it? He infers that the ball and the object both coincide at the same place at the same time and at that moment both coordinately accelerate radially from rest in the same way, right? Although of course, the radial coordinate acceleration Bob infers will be different from that which Alice measures locally.

 

[Mentz114]

I don't understand why the dropped object moves away from Alice. Is there a gravitational field ? You did use the word 'orbit' so I assume there must be.

In that case there is no proper acceleration to infer, and the distant observer is making an error.

 

[PeterDonis]

according to the equivalence principle, the radial coordinate acceleration that Alice measures at that instant for both the object and the ball will be the same, correct?

It depends on what spacetime you're in. If you're in flat spacetime, the radial coordinate acceleration of the dropped object will be zero in Alice's rest frame, because Alice will just be an inertial observer at rest in Bob's global inertial frame, and the dropped object is moving inertially and starts at rest (with Alice), so it just stays at rest. The ball, OTOH, if you're in flat spacetime, will have an inward coordinate acceleration of $v^2 / r$, as you calculated, because it has a force being exerted on it to hold it in a circular path.

If, OTOH, you are in Schwarzschild spacetime, at a sufficiently large radius (see below for why), then yes, the radial coordinate acceleration of both the dropped object and the ball, in Alice's momentarily comoving inertial frame, will be the same, because Alice is not an inertial observer; she must be firing rockets or otherwise be subject to a force that accelerates her radially outward. Both the ball and the dropped object are moving inertially, so their radial coordinate acceleration will be the same.

(The reason why we have to be at a sufficiently large radius is that the ball's orbital velocity must be much less than the speed of light; if it's a significant fraction of the speed of light, then the ball's radial coordinate acceleration will be larger than that of the dropped object because of relativistic effects--the same effects that make the bending of light by the Sun twice as large as a naive Newtonian calculation would predict.)

As for the distant observer Bob, the same should be true according to his coordinate system, shouldn't it?

Again, it depends on the spacetime. See above.

the coordinate acceleration Bob infers will be different from that which Alice measures locally.

This also depends on the spacetime; it's true in Schwarzschild spacetime, but false in flat spacetime: in flat spacetime, since Alice and Bob are both inertial observers and are at rest relative to each other, all coordinate accelerations are the same for both of them.

 

[grav-universe]

I don't understand why the dropped object moves away from Alice. Is there a gravitational field ?

Yes. There is a mass and Alice is a hovering observer at r. The object is orbitting in a perfectly circular orbit at r also. (r according to Bob's coordinate system)

 

[grav-universe]

If, OTOH, you are in Schwarzschild spacetime, at a sufficiently large radius (see below for why), then yes, the radial coordinate acceleration of both the dropped object and the ball, in Alice's momentarily comoving inertial frame, will be the same, because Alice is not an inertial observer; she must be firing rockets or otherwise be subject to a force that accelerates her radially outward. Both the ball and the dropped object are moving inertially, so their radial coordinate acceleration will be the same.

Right, this is what I am describing, although the ball is dropped and the object orbits :)

(The reason why we have to be at a sufficiently large radius is that the ball's orbital velocity must be much less than the speed of light; if it's a significant fraction of the speed of light, then the ball's radial coordinate acceleration will be larger than that of the dropped object because of relativistic effects--the same effects that make the bending of light by the Sun twice as large as a naive Newtonian calculation would predict.)

Okay, this is what I would be interested in. Before you said the coordinate acceleration at that instant when they coincide in the same place would be the same, but here it looks like something different. Let's throw in a relativitistic speed. I understand that the curvature of spacetime varies from Newtonian effects, giving twice the overall bending of light, but let's say for a moment that light passes tangently at the same moment that Carl falls from rest in an elevator at the same place as Alice. Wouldn't the coordinate acceleration as measured by Alice for both be the same at that moment? It would be the same as if Carl emitted light sideways in his elevator upon falling, would it not? So for a moment that light would continue to travel perpendicularly to Carl according to the equivalence principle, while both the elevator and the light coordinately accelerate radially at the same rate according to Alice, correct?

 

[grav-universe]

I will cut to the chase. The problem is I am still trying to find relationships and invariants that are independent of the coordinate system used, as we did here. There we found the relations

a' = G M L_t^2 / (z r^2) and a' = L z' c^2 / z

where at r, a' is the locally measured acceleration, z is the time dilation, L is the radial length contraction, and L_t is the length contraction in the tangent direction. From the derivatives of these two relations we can gain the R_00 tensor.

Okay, so now I want to look at what the distant observer infers and find relations and invariants from that. So far we have

a = v_t^2 / r

where v_t is the tangent velocity. From the equivalence principle, if the hovering observer Alice were at r in the same place as the object passes and Alice let's go of a ball at that moment, the ball and object should both accelerate toward the mass at the same rate at that instant. Although the object is traveling tangently, its initial radial velocity is zero, the same as the ball.

Alice measures the local acceleration to be a'. She measures the ball to fall some distance radially along the length of a short rod over some time. The distant observer Bob, however, infers that the time that passes for the ball to travel the length of the rod is 1/z longer and that the length of the rod itself is L shorter, so the coordinate acceleration that Bob measures is z^2 L smaller, whereby a = z^2 L a'. Likewise, the locally measured tangent velocity is v_t', and Bob infers 1/z greater time to travel a distance that is L_t shorter in the tangent direction, so v_t = z L_t v_t'. So now we have

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

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

Here's where the problem lies. Let's look at the invariants. L_t / r is an invariant, having the same value in any coordinate system. a' and v_t' are also invariants since they are what is measured locally, regardless of the distant observer's coordinate system. But that leaves

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

The right side is all invariant, so the left side should be too. But if the ratio of the radial to tangent length contraction is invariant, then there can be one and only one valid coordinate system. For instance, if one coordinate system gives some ratio L / L_t as an invariant for some spherical shell (with L being found between two very close shells), then the only way to change the coordinate system such that this ratio remains the same for that shell is to change the radius of all shells by the same ratio. But if we make them all .99 the original radiuses, then the distant observer's distance from the mass also changes by .99, whereas it should verge upon 1, so there can be only one coordinate system where this is also true.

 

[grav-universe]

Let's go ahead and try this for the Schwarzschild coordinate system and see if we can pinpoint where the problem actually lies. What is the locally measured orbittal speed as measured by a hovering observer at r for an object in a circular orbit at r?

 

[grav-universe]

Okay well, looks like the orbittal speed the distant observer infers is just v_t^2 = G M / r and the local hovering observer measures v_t' = (G M / r) / (1 - 2 G M / (r c^2)). So starting with the local acceleration and backtracking, we can get

a' = G M L_t^2 / (z r^2)

a' = L_t^2 v_t^2 / (z r)

a' = L_t^2 (z L_t v_t')^2 / (z r)

and picking out the invariants, that gives

a' / [(v_t'^2) (L_t / r)] = L_t^3 z

which says that L_t^3 z is invariant. That doesn't look right either.

 

[Mentz114]

Let's go ahead and try this for the Schwarzschild coordinate system and see if we can pinpoint where the problem actually lies. What is the locally measured orbital speed as measured by a hovering observer at r for an object in a circular orbit at r?

When the two worldlines coincide they will measure a relative velocity of $\sqrt{\frac{m}{r-2\,m}}$ which is close to $\sqrt{\frac{m}{r}}$ if $r>>m$.

This comes from $-\gamma = V^aU_a$ where $U$ and $V$ are the respective worldlines.

$V^a= \frac{\sqrt{r}}{\sqrt{r-2\,m}}\partial_t$
$U_a=-\frac{r-2\,m}{\sqrt{r}\,\sqrt{r-3\,m}}dt + \frac{\sqrt{m}\,r}{\sqrt{r-3\,m}} d\phi$

 

[grav-universe]

Oh, z is also an invariant, so we are left with L_t^3. It will work out, however, if v_t^2 = L_t^3 G M / r, although I don't immediately see why that should be the case. Plugging that into the coordinate acceleration formula for the distant observer, we get

a = v_t^2 / r

a = G M L_t^3 / r^2

z^2 L a' = G M L_t^3 / r^2

and separating the invariants,

L / L_t = G M (L_t / r)^2 / z^2

so we are still left with L / L_t as an invariant. It would be an easy fix if a = z^2 L_t a' instead of a = z^2 L a', but the acceleration is in the radial direction, not the tangent direction.

 

[grav-universe]

As far as v_t^2 = L_t^3 G M / r, I can almost see it now with

v_t^2 = L_t^3 G M / r

z^2 L_t^2 v_t'^2 = L_t^3 G M / r

z^2 v_t'^2 = G M (L_t / r)

which carries only invariants with nothing left over as it should be. But I still don't yet see the reason for the L_t^3 there.

 

[Mentz114]

Sorry I interupted your flow there ...

 

[grav-universe]

Sorry I interupted your flow there ...

lol It needs to be sometimes :) Thanks for your post. Not sure I understand it though.

 

[Mentz114]

I'm pretty sure the velocity is correct.

I'm not following you, but it could help if you try to use Latex.

For instance this L / L_t = G M (L_t / r)^2 / z^2, if you wrap it in two double #'s looks like this.

$L / L_t = G M (L_t / r)^2 / z^2$

 

[grav-universe]

I'm pretty sure the velocity is correct.

I'm not following you, but it could help if you try to use Latex.

For instance this L / L_t = G M (L_t / r)^2 / z^2, if you wrap it in two double #'s looks like this.

$L / L_t = G M (L_t / r)^2 / z^2$

Cool, thanks :)

 

[pervect]

Let's go ahead and try this for the Schwarzschild coordinate system and see if we can pinpoint where the problem actually lies. What is the locally measured orbittal speed as measured by a hovering observer at r for an object in a circular orbit at r?

I'm not following what you did, but I did something similar but more general in https://www.physicsforums.com/showthread.php?t=686147#post4354272
where I calculated a bunch of things for various general spherically symmetric metrics, one of which was the orbital velocity as measured by a co-located static observer. And of course the Schwarzschild metric is spherically symmetric.

Post #12 is the corrected version with all of the known typos removed (but I might have missed a few anyway).

Applying the results from that post:

The orbital velocity measured by a co-located static observer should be:

c \sqrt{ \frac{h (\frac{d f}{d r}) }{ f (\frac{d h}{d r})}}

where for the Schwarzschild metric
f=c^2(1 -\frac{2 G M}{c^2 r}) \quad g = 1 / (1 -\frac{2 G M}{c^2 r}) \quad h = r^2

This gives:
<br /> v = c \, \sqrt{\frac {1}{\frac{c^2\,r}{GM}-2}} <br />

at r=3GM/c^2, the orbital velocity is c as expected (this is the photon sphere).

 

[Mentz114]

This gives:
<br /> v = c \, \sqrt{\frac {1}{\frac{c^2\,r}{GM}-2}} <br />

at r=3GM/c^2, the orbital velocity is c as expected (this is the photon sphere).

This agrees with the calculation in my post#12, viz $\sqrt{\frac{m}{r-2\,m}}$ with $m=GM,\ c=1$.

 

[WannabeNewton]

I still have no idea what this thread is about O.O

 

[grav-universe]

Wow, pervect :) That page is awesome, just what I needed. I'll start working through it, thank you.

 

[grav-universe]

Okay, so far from pervect's post #12 in the other thread, converting to the units I am using, $z$ for time dilation, $L$ for radial length contraction, and $L_t$ for the tangent length contraction, changes the metric from
<br /> ds^2 = -f\, dt^2 + g\, dr^2 + h\, d\phi^2<br />
to
<br /> ds^2 = -z^2\, dt^2 + dr^2 / L^2 + d\phi^2 / L_t^2<br />

and then we have

$f = z^2 c^2 , g = 1 / L^2, h = r^2 / L_t^2$

and first derivatives with

$f' = 2 c^2 z z', g' = -2 L' / L^3, h' = 2 r (L_t - r L_t') / L_t^3$

For the proper acceleration of a hovering observer, or measured acceleration by the same observer for a ball dropped from rest, pervect's post gives

$a' = c^2 f' / (2 f sqrt(g)) = c^2 (2 c^2 z z') / [2 (z^2 c^2) / L) = c^2 z' L / z$

The coordinate acceleration for a distant observer becomes

$a = - f' / (2 g) = - (2 c^2 z z') / (2 / L^2) = - c^2 z z' L^2$

whereby $a = a' z^2 L$, so everything agrees so far with what I've got for this much.

 

[Mentz114]

I still have no idea what this thread is about O.O

Me neither. Looks better in Latex, though.

 

[grav-universe]

I still have no idea what this thread is about O.O

Me neither. Looks better in Latex, though.

This thread is a continuation of another thread here (post #4). In that thread, we could find through somewhat straight-forward and intuitive methods a couple of relationships for local acceleration.

$a' = G M L_t^2 / (z r^2)$ and $a' = c^2 z' L / z$

Combining those gives

$G M L_t^2 / r^2 = c^2 z' L$

and finding the derivatives of that gives the $R_{00}$ tensor.

Basically, what I am trying to do is bypass the Einstein field equations and find other relations that will solve for the unknowns, $z, L,$ and $L_t$, through natural relations that can be found here and there and/or working through the invariants. With what we have so far, we can make a coordinate choice for one and find the relation between the other two, but there is not enough information to solve for all three. I couldn't determine anything more about the local acceleration, so instead I am now attempting it by looking at what is inferred by a distant observer.

 

[PeterDonis]

finding the derivatives of that gives the $R_{00}$ tensor.

I'm not sure I understand. So far all of your calculations have been for the vacuum exterior Schwarzschild metric; but in vacuum, the Ricci tensor is zero.

(Also, $R_{00}$ is not a tensor; it's one component of the Ricci tensor.)

With what we have so far, we can make a coordinate choice for one and find the relation between the other two, but there is not enough information to solve for all three.

That's to be expected, since two of the quantities you listed are coordinate-dependent; only the locally measured proper acceleration is coordinate-independent. So I would not expect to find coordinate-independent relationships between all three.

I think it would help if you would take a step back and explain at a higher level what you are trying to accomplish.

 

[grav-universe]

Okay, so the last equation in Pervect's post gives

v_t&#039;^2 = c^2 h f&#039; / (f h&#039;)

which becomes

v_t&#039;^2 = c^2 (r^2 / L_t^2) (2 c^2 z z&#039;) / [(z^2 c^2) (2 r (L_t - r L_t&#039;) / L_t^3)]

v_t&#039;^2 = c^2 r L_t z&#039; / (L_t - r L_t&#039;)

and we already have

a&#039; = c^2 z&#039; L / z

So now we can gain

a&#039; / v_t&#039;^2 = (L_t - r L_t&#039;) L / (z r L_t)

and bringing all of the obvious invariants to the left side, we are left with

a&#039; z (r / L_t) / v_t&#039;^2 = (L_t - r L_t&#039;) L / L_t^2 = (1 - r L_t&#039; / L_t) (L / L_t)

This means that $(1 - r L_t' / L_t) (L / L_t)$ is the invariant associated here while I had found just $L / L_t$. I'm not sure where the additional term of $(1 - r L_t' / L_t)$ comes from yet but I will continue to look into it.

 

[grav-universe]

Okay, so to verify that really quick, in Schwarzschild coordinates, we would have

L = sqrt(1 - 2 m / r) , L_t = 1 , L_t&#039; = 0

so for $r = 10 m$, the invariant

(1 - r L_t&#039; / L_t) (L / L_t)

would give

(1 - 0) (sqrt(1 - 2 m / (10m)) / 1) = .894427191

In Eddington's isotropic coordinates, we would have

r_1 = r / 2 - m / 2 + sqrt(r (r - 2 m) / 4) = 8.972135955 m

L_1 = L_{t1} = 1 / (1 + m / (2 * r_1))^2 = .8972135955

L_{t1}&#039; = m r_1 / (m / 2 + r_1)^3 = .01055728

giving

(1 - (8.972135955) (.01055728) / (.8972135955)) (1) = .894427191

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

 

[grav-universe]

I'm not sure I understand. So far all of your calculations have been for the vacuum exterior Schwarzschild metric; but in vacuum, the Ricci tensor is zero.

(Also, $R_{00}$ is not a tensor; it's one component of the Ricci tensor.)

Right. Here is post #5 from the other thread.

From the relationship we found before

m L_t^2 = (dz / dr) L r^2

which for convenience I will write dz / dr as just z' with second derivatives double primed.

We can re-arrange to gain

L_t^4 = z'^2 L^2 r^4 / m^2

The variables in the tensors are

A = 1 / L^2

B = - z^2

and B' = d(-z^2) = - 2 z z'

z' = - B' / (2 z)

z'^2 = B'^2 / (4 z^2) = - B'^2 / (4 B)

so we can rewrite the relationship once more to

L_t^4 = - B'^2 r^4 / (4 m^2 A B)

Finding the derivative for that using Wolfram, we get

d(L_t^4) = d[- B'(r)^2 r^4 / (4 m^2 A(r) B(r))] = r^3 B' (r B A' B' + A(r B'^2 - 2 B (r B" + 2 B')) / (4 m^2 A^2 B^2)

Now, if as a coordinate choice, we make L_t = 1, then its derivative is zero. So now we have

(0) (4 m^2 A^2 B^2) / (r^3 B') = 0 = r B A' B' + A(r B'^2 - 2 B (r B" + 2 B'))

r B A' B' + r A B'^2 - 2 r A B B" - 4 A B B' = 0

which when all terms are divided by 4 A^2 B r, gives

- B" / (2 A) + (B' / (4 A)) (A' / A + B' / B) - B' / (A r) = 0

This is the same as the R_00 Ricci tensor solution. Now I just need something similar to find the other two.

 

[grav-universe]

Oh wait. What I had originally would give

a = v_t^2 / r

z^2 L a&#039; = (z L_t v_t&#039;)^2 / r

L a&#039; = L_t v_t&#039;^2 / r

a&#039; (r / L_t) / v_t&#039;^2 = L_t / L

while pervect's post gives

a&#039; (r / L_t) / v_t&#039;^2 = (1 - r L_t&#039; / L_t) (L / L_t) / z

so that's quite different actually. Backtracking from what pervect has gives

a&#039; z = v_t&#039;^2 (1 - r L_t&#039; / L_t) L / r

[a / (z^2 L)] z = [v_t / (z L_t)]^2 (1 - r L_t&#039; / L_t) L / r

a z / L = v_t^2 (1 - r L_t&#039; / L_t) L / (L_t^2 r)

a = (v_t^2 / r) [(1 - r L_t&#039; / L_t) L^2 / (z L_t^2)]

That is very nonintuitive, so I'm not sure I can use it. I'll need something much more straight-forward.

 

[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&#039; = M R / (R+ M / 2)^3

I = (1 - R L_t&#039; / 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&#039; = M R / (R+ M / 2)^3

I = (1 - R L_t&#039; / 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.