Transformation of the Line-Element

[Ben Niehoff]

Coordinates do NOT have intrinsic meaning in general in GR (redundancy ftw). You can't define the r in the standard schwarzschild metric physically; it is not related to distance from the origin. How could you define the physical meaning of a coordinate when we don't have prescribed coordinate system when we solve the EFEs? I could just as easily transform to Eddington coordinates; how would you physically define the coordinates for those?

Actually, JDoolin has somewhat of a point. The coordinates in the standard Schwarzschild metric can be given physical meanings. The point that I meant is that if I write down any arbitrary metric in some collection of coordinates, call them say $(x, y, \eta, \Upsilon)$, the coordinates in general will not have any intrinsic meaning. Ultimately a coordinate chart is nothing but a way to label the points on the manifold in a way that is consistent with the manifold's topology (namely, continuity). For certain manifolds we can give logical meanings to some coordinate systems, but in general a coordinate system is just a collection of continuous maps from (some open region of) the manifold to (some open region of) $\mathbb{R}^n$.

Of course the coordinates have intrinsic meaning. If you're going to claim that the Schwarzschild metric or any other thing in general relativity has been experimentally verified, then the coordinates have to have intrinsic meaning, or you wouldn't have any experimental verification.

Angles, distances, and time intervals are what we measure physically, not coordinates. I can write the Schwarzschild solution in any funny coordinate system I like; they don't have to make sense to you. To find physically-observable quantities, I have to use the metric tensor to compute angles, distances, and time intervals.

In this case, I think Δr is the distance between two simultaneous events as measured by a ruler on (the non-rotating gravitational surface), while ΔR is the distance between the same two simultneous events as measured by a distant co-moving observer.

Wrong. In the standard Schwarzschild coordinates, r is a coordinate defined to be consistent with the circumference law of circles centered at the black hole. That is,

$$2\pi r = \oint_\mathcal{C} ds$$

where $\mathcal{C}$ is a circle centered at the singularity (at constant t and constant r). The coordinate r obeys the circumference law, but because the spacetime is curved, r does not represent distance from the singularity. To get the distance from the singularity, you must integrate ds along a radial path. $\Delta r$ does not represent any sort of distance at all. Infinitesimal radial distances in the Schwarzschild are represented by

$$\big( 1 - \frac{2m}{r} \big)^{-1} \; \Delta r$$

In the isotropic coordinates, R does not represent anything special, and neither do x, y, and z. The horizon is at $x = y = z = 0$, which is not a single point, but a whole 2-sphere worth of points. The isotropic coordinates are a coordinate chart that covers only the exterior region of the black hole.

Likewise, Δτ is the time measured between two events which occur at the same place, as measured by a stationary clock on the (non-rotating gravitational surface), while Δt is the time between the same two events as measured by a distant co-moving observer.

In both coordinate charts, t represents time measured by a stationary observer at infinity. But at any finite (constant) value of r, the time elapsed for a stationary observer is given by

$$\big( 1 - \frac{2m}{r} \big) \; \Delta t$$

But of course, a stationary observer at finite r is accelerating. An inertial observer (is this what you mean by "co-moving"?) would measure something else, given by integrating $ds$ along his worldline.

I'd be happy to be corrected based on something quantifiable*, but if you want to claim that Δr and ΔR have no intrinsic meaning, you must have a different idea of what "intrinsic" and "meaning" mean. If you want me to believe that these quantities are neither definable nor measurable, then what possible application can they have?

I do mean that 'r' and 'R' are not directly measurable. They do not represent measurable quantities! When one lays out measuring rods and clocks, one does not find his 'r' coordinate, nor even $\Delta r$ for nearby points.

One measures $\Delta s$ between nearby points. We can assign coordinates to the points however we please, so long as they are continuous and 1-to-1 in some open region of spacetime.

(*P.S. I see that you have given an explicit (quantifiabe) formulation relating r to R (what I've been calling r' and r), which I will verify and see if I agree with.)

(P.S.S. On second thought, I'm not sure if its worth bothering. How am I going to verify something that should have a serious problem at the schwarzschild radius based on a model that is only a Taylor series approximation? If we want to do the problem at all correctly, don't we have to go back to square one, and do it without a bunch of linear approximations?)

Not sure what you're getting at. No linear approximations have been used here whatsoever. The metric I gave in isotropic coordinates is an exact solution of the full Einstein field equations. In fact, it is just the Schwarzschild metric written in different coordinates. This is easy to verify: compute the Ricci tensor, observe that it vanishes, and observe that the metric is static and has spherical symmetry; uniqueness theorems tell you that it must be the Schwarzschild black hole.

 

[JDoolin]

Not sure what you're getting at. No linear approximations have been used here whatsoever. The metric I gave in isotropic coordinates is an exact solution of the full Einstein field equations.

I am not sure exactly how you derived it, but the original value of g_tt in the schwarzschild metric earlier is derived using several linear approximations, if you can see my derivation based on https://www.physicsforums.com/showpost.php?p=3415913&postcount=18

That took me several days to work through, but in the end, it made sense--but I was fully aware that it was a linear approximation of the actual solution.

I should ask, to be sure; did your calculation come from a serious "no approximations" version of the Schwarzchchild radius, or did it come from this more common linear-approximtion version? My expectation is that the exact version should have some hyperbolic geometry in it; and it shouldn't be some polynomial.

P.S. I will still address more of your post. This was just what I could answer quickly.

 

[WannabeNewton]

Actually, JDoolin has somewhat of a point. The coordinates in the standard Schwarzschild metric can be given physical meanings. .

I agree that certain coordinate functions in the standard metric can be made meaningful in your sense of the word such as $\theta$. What I was trying to say was that something like the radial coordinate doesn't really represent anything physically because if you find the radial distance between two points $$\Delta s = \int_{r_{1}}^{r_{2}}(1 - \frac{2GM}{r})^{-1/2}dr \approx \int_{r_{1}}^{r_{2}}(1 + \frac{GM}{r})dr = (r_{2} - r_{1}) + GMln(\frac{r2}{r1})$$, then, for GM << r, the distance is greater than that of (the meaningful notion of) the radius in euclidean space so you can't really give the circumferential radius in this metric any meaning.

 

[WannabeNewton]

I am not sure exactly how you derived it, but the original value of g_tt in the schwarzschild metric earlier is derived using several linear approximations

I am under the impression that you are thinking of a different form of the metric in question? You get the metric components simply by finding all components of $R_{\mu \nu } = 0$ and, of course, solving them for the metric with the spherical symmetry conditions that were placed and using classical parameters that correspond to parameters in the solution at infinity. Remember that by Birkhoff's Theorem if you ever solve the vacuum EFEs for a static and spherically symmetric space - time it will be the schwarzchild metric.

 

[Ben Niehoff]

I am not sure exactly how you derived it, but the original value of g_tt in the schwarzschild metric earlier is derived using several linear approximations, if you can see my derivation based on https://www.physicsforums.com/showpost.php?p=3415913&postcount=18

That took me several days to work through, but in the end, it made sense--but I was fully aware that it was a linear approximation of the actual solution.

I should ask, to be sure; did your calculation come from a serious "no approximations" version of the Schwarzchchild radius, or did it come from this more common linear-approximtion version? My expectation is that the exact version should have some hyperbolic geometry in it; and it shouldn't be some polynomial.

What a strange way to attack this problem. The original Schwarzschild metric does not require nor contain any approximations. I assure you, the full, exact solution really does just have polynomials. In fact we can write down analogues of Schwarzschild, Kerr, Reissner-Nordstrom, and Kerr-Newman black holes in d dimensions and they always just have polynomial coefficients.

I think you're making the problem harder than it is. You can easily get the Schwarzschild solution by starting with the ansatz

$$ds^2 = -f(r) dt^2 + \frac{1}{f(r)} dr^2 + r^2 (d\theta^2 + \sin^2 \theta \; d\phi^2),$$
computing its Ricci tensor, and finding the function $f(r)$ that makes the Ricci tensor vanish and satisfies the boundary condition (i.e., that $f(r) \rightarrow 1$ at infinity). I can do this in about 3 pages, it's not hard.

 

[Anamitra]

We may consider the the three metrics given below:

$${ds}^{2}{=}{dt}^{2}{-}{dx1}^{2}{-}{dx2}^{2}{-}{dx3}^{2}$$ --------------- (1)
$${ds}^{2}{=}{dt}^{2}{-}{dr}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{Sin}^{2}{\theta}{d}{\phi}^{2}{)}$$------------ (2)
$${ds'}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{Sin}^{2}{\theta}{d}{\phi}^{2}{)}$$ --------------------- (3)

Equations (1) and (2) represent flat spacetime metrics in the rectangular (t,x,y,z) and the spherical(t,r,theta,phi) systems[For flat spacetime the diagonal matrix of the form[1,-1,-1,-1]is valid only for the rectangular system of coordinates. The values of g(mu,nu) for flat spacetime will be different in the rectangular and the spherical systems but the Riemannian Curvatrure should have zero components in each system(for flat space time) ]
Equation three represents Schwarzschild’s Geometry in the spherical system[t,r,theta phi]system . Equations 2 and 3 use the same coordinate grid that is t,r,theta phi system.

We may think of a path that has the same coordinate description wrt to equations 2 and 3.The quantities t,r,theta and phi will be the same for both. Again dt,dr,d(theta),d(phi) will also be the same for both. But physical separations will be different.
In Schwarzschild’s Geometry the coordinate separation dr represents a physical distance of (1-2m/r)^(-1)dr.
The two manifolds cannot intersect physically but we may consider their intersection[or their existence for that matter] graphically for the purpose of calculations etc.
One may think of the example in post #76 . In presence of the spherical mass we have one manifold while in the absence of it we have the other on the same coordinate grid[t,r,theta,phi grid]. The existence or the non existence of the spherical mass is not possible simultaneously.
But graphically we may consider the simultaneously the two paths having the same coordinate description. For such a curve the coordinate separations between a pair of points should be the same but the physical separations would be different:
We may use the following notations:
$${dT}{=}{(}{1}{-}{\frac{2m}{r}}{)}{dt}$$ ------------- (4)
$${dR}{=}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}$$ ----------- (5)
The values of dr/dt, d(phi)/dt and d(theta)/dt are the same in both the manifolds
But the values dR/dT , d(phi)/dT and d(theta)/dT will be different in two manifolds
$$\frac{dR}{dT}{=}{(}{1}{-}\frac{2m}{r}{)}^{-2}\frac{dr}{dt}$$ ------------ (6)
$$\frac{{d}{\theta}}{dT}{=}{(}{1}{-}\frac{2m}{r}{)}^{-1}\frac{{d}{\theta}}{dt}$$ -------- (7)
$$\frac{{d}{\phi}}{dT}{=}{(}{1}{-}\frac{2m}{r}{)}^{-1}\frac{{d}{\phi}}{dt}$$ ----------------- (8)
Relative speed formula for Flat Space-time:
$${V}_{rel}{=}\frac{v2-v1}{{1}{-}\frac{v1v2}{{c}^{2}}}$$
Suppose motion is taking place in the radial direction[theta=const,phi=const]
[V1 and v2 are the coordinate speeds which should be the same in both the manifolds]
We may find the speed values [in the physical sense] in the other manifold[Schwarzschild’s Geometry] by using the following relations[from (6):
$${v1'}{=}{(}{1}{-}\frac{2m}{r}{)}^{-2}{v1}$$
$${v2'}{=}{(}{1}{-}\frac{2m}{r}{)}^{-2}{v2}$$
It is to be noted that “r” denotes the same quantity in both the manifolds[same with t,theta and phi]

In Flat Specetime the direction of relative velocity is unique.We simply carry this unit vector to the other manifold[Schwarzschild's Geometry for this case] to find the corresponding direction in curved spacetime
Points to Observe:
1. The following relation:
$${dx’}^{i}{=}\frac{\partial{x’}^{i}}{\partial{x}^{k}}{{d}{x}^{k}}$$
Is valid for coordinate transformation within the same manifold[ds^2 remains unchanged wrt coordinate transformations].We do not apply it in our case —we are interested here in in transitions between different manifolds[on the same coordinate grid ie, the (t,r,theta,phi) grid].

2. On dT
When we we say that the local speed of light is constant in vacuum we have in our mind some “time” wrt which this speed is defined. “dT” is a good option.
Lets take a metric:
$${ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}$$
$${ds}^{2}{=}{g}_{00}{dt}^{2}{-}{dL}^{2}$$
dL is the physical spatial element.
$${ds}^{2}{=}{dT}^{2}{-}{dL}^{2}$$ ------------ (9)
Where,
$${dT}{=}\sqrt{{g}_{00}}{dt}$$
For the null geodesic we have:
ds^2=0
From (9) we have
dL/dT= 1 [c=1 in the natural units]
Thus dT is a suitable candidate for the physical time interval for which the local speed of light is constant in vacuum.

 

[JDoolin]

Thank you, Ben, and WannabeNewton, for your thoughtful responses.

Actually, JDoolin has somewhat of a point. The coordinates in the standard Schwarzschild metric can be given physical meanings. The point that I meant is that if I write down any arbitrary metric in some collection of coordinates, call them say $(x, y, \eta, \Upsilon)$, the coordinates in general will not have any intrinsic meaning. Ultimately a coordinate chart is nothing but a way to label the points on the manifold in a way that is consistent with the manifold's topology (namely, continuity). For certain manifolds we can give logical meanings to some coordinate systems, but in general a coordinate system is just a collection of continuous maps from (some open region of) the manifold to (some open region of) $\mathbb{R}^n$.

It is one thing to say that in general, a coordinate chart only requires continuity. It is quite another to claim that a SPECIFIC mapping from one chart to another has no logical meaning. If I were to wad a paper into a little ball, it would preserve continuity, but the new distances between the points may be in some sense meaningless*. However, in the Schwarzschild metric case, there is unambiguous meaning.

(*no, I don't even want to play that game. Even the distances between points on a wadded piece of paper has meaning. It may be confusing, but it represents a specific configuration.)

Angles, distances, and time intervals are what we measure physically, not coordinates. I can write the Schwarzschild solution in any funny coordinate system I like; they don't have to make sense to you.

But once you do write the Schwarzschild solution in any funny coordinate system you like, I can apply an inverse transformation, and map them right back to (t,x,y,z).

In any case,
$$c^2 {d \tau}^{2} = \left(1 - \frac{2 G M}{r c^2} \right) c^2 dt^2 - \left(1-\frac{2 G M}{r c^2}\right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)$$

is not a particularly "funny" coordinate system, and it does make sense to me. Maybe not all the in's and out's but I feel I do have the general idea.

To find physically-observable quantities, I have to use the metric tensor to compute angles, distances, and time intervals.

Physically observable quantities such as angles are observer dependent. You can't claim to have a "correct" value for an angle. You can only say what it looks like from your point-of-view. Sure, you can use a metric tensor to figure out what it looks like from another point-of-view, (in general, the "local" point-of-view) but don't think just because it was measured locally, that represents the "true" value. It may represent the true age of the clock, or the true length of the stick, but my observations of it from afar are just as valid.

Especially with angles, is the angle of approach of a snowflake less real becuase you are riding a motorcycle? Is the true angle of approach of the snowflake falling straight down? No, neither one. The angle of approach is observer-dependent.

In this case, I think Δr is the distance between two simultaneous events as measured by a ruler on (the non-rotating gravitational surface), while ΔR is the distance between the same two simultaneous events as measured by a distant co-moving observer.

Wrong. In the standard Schwarzschild coordinates, r is a coordinate defined to be consistent with the circumference law of circles centered at the black hole. That is,

$$2\pi r = \oint_\mathcal{C} ds$$

I screwed up the variables. I should have said:

"ΔR is the distance between two simultaneous events as measured by a ruler on (the non-rotating gravitational surface), while Δr is the distance between the same two simultaneous events as measured by a distant co-moving observer."

where $\mathcal{C}$ is a circle centered at the singularity (at constant t and constant r). The coordinate r obeys the circumference law, but because the spacetime is curved, r does not represent distance from the singularity.

Here, I disagree. You are correct that r does not represent the distance from the singularity according to an internal observer, but it does represent the distance according to an external observer.

To get the distance from the singularity, you must integrate ds along a radial path. $\Delta r$ does not represent any sort of distance at all. Infinitesimal radial distances in the Schwarzschild are represented by

$$\big( 1 - \frac{2m}{r} \big)^{-1} \; \Delta r$$

How can you claim that Δr does not represent any distance at all, but after doing a path integral along a path involving Δr you have a distance? Doesn't the process of doing a path integral in some way imply that Δr is a coordinate of a path?

The (t,r,θ,Φ) coordinates represent a flat manifold in which the curved (tau,R,θ,Φ) coordinates are embedded.

In the isotropic coordinates, R does not represent anything special, and neither do x, y, and z.

Again, what an ambiguous statement! R does not represent anything special? Here, let me use WannabeNewton's post to help me explain. I think R represents something very special, (at least special in the sense that it is very interesting); but to define it, you need to do an integral, which means you have to determine an integration constant, and you can't be running that path integral down past the schwarzschild radius. WannabeNewton has graciously done the math for me.

I agree that certain coordinate functions in the standard metric can be made meaningful in your sense of the word such as $\theta$. What I was trying to say was that something like the radial coordinate doesn't really represent anything physically because if you find the radial distance between two points $$\Delta s = \int_{r_{1}}^{r_{2}}(1 - \frac{2GM}{r})^{-1/2}dr \approx \int_{r_{1}}^{r_{2}}(1 + \frac{GM}{r})dr = (r_{2} - r_{1}) + G M ln(\frac{r2}{r1})$$, then, for GM << r, the distance is greater than that of (the meaningful notion of) the radius in euclidean space so you can't really give the circumferential radius in this metric any meaning.

See; When you calculate the Δs between two simultaneous events of different r-values, but the same theta, and phi values, you are calculating the "local" distance, R, between those events.

R cannot be defined arbitrarily. You have to decide on a value of the lower radius r_1, and that becomes your zero. You also must set r₁ further out from the center than the Schwarzschild radius.

There's one little mistake in what WannabeNewton is saying here, and that is worrying that the value, ΔR

$$\Delta R = \Delta s|_{dt=d\theta=d\phi=0}$$

is too big to fit into Euclidean Space! It is NOT. The Euclidean space component is represented by the r-component here; not the R-component or the Δs value. The ΔR distance is the Non-Euclidean space that is embedded in the Euclidean Space. It's kind of like the Tardis on Dr. Who: It's bigger on the inside. But unlike the Tardis, you can see the whole thing from outside; you can see that it's bigger on the inside FROM the outside.

Skyscrapers near the Schwarzschild radius would be only inches tall, looking from outside, but they would seem normal height to the internal observers. The strange thing is the theta and phi components are completely unaffected by the schwarzchild metric, so the local observers would be looking at very tall sky-scrapers with extremely narrow bottoms.

The horizon is at $x = y = z = 0$, which is not a single point, but a whole 2-sphere worth of points. The isotropic coordinates are a coordinate chart that covers only the exterior region of the black hole.

The horizon is at $r=\frac{G M}{ c^2}$
where
$$\frac{1}{1-\frac{G M}{r c^2}}\rightarrow \infty$$ where the internal coordinates descend for infinity in a finite amount of (external coordinate) space.

In both coordinate charts, t represents time measured by a stationary observer at infinity. But at any finite (constant) value of r, the time elapsed for a stationary observer is given by

$$\big( 1 - \frac{2m}{r} \big) \; \Delta t$$

But of course, a stationary observer at finite r is accelerating. An inertial observer (is this what you mean by "co-moving"?) would measure something else, given by integrating $ds$ along his worldline.

For an approximation of my meaning, imagine a stationary observer at a distance far enough away that the gravity is barely detectable.

Or for a more exact meaning, imagine replacing the entire scene with a hologram of the scene so the same images are seen, but there is no gravity. It's the same scene, but simulated, so there's no warping of space (you can walk right through the simulated schwarzschild radius, but there's no data coming from in there.) It's just a simulation of the events in the space, so the events are really happening in cartesian space.

I do mean that 'r' and 'R' are not directly measurable. They do not represent measurable quantities! When one lays out measuring rods and clocks, one does not find his 'r' coordinate, nor even $\Delta r$ for nearby points.

One measures $\Delta s$ between nearby points. We can assign coordinates to the points however we please, so long as they are continuous and 1-to-1 in some open region of spacetime.

But Δs does measure the laying out of measuring rods, and clocks. If Δs² is negative, it means you have a time-like interval, and you have the time between two events. If Δs² is positive, then you have measured the distance between two events. Sure, those events might not be perfectly aligned on the r-axis, but if they are aligned on the r-axis, then you are directly measuring ΔR.

There IS, of course, trouble in measuring (x,y,z) because of the difficulty of imagining "a stationary observer at a distance far enough away that the gravity is barely detectable." If the light is bending in the gravitational pull, you may have to do some work to establish where an event (t,x,y,z) actually happened.

(The hologram version is not really feasible, but it would have the advantage that light-rays wouldn't be bent.)

Not sure what you're getting at. No linear approximations have been used here whatsoever. The metric I gave in isotropic coordinates is an exact solution of the full Einstein field equations. In fact, it is just the Schwarzschild metric written in different coordinates. This is easy to verify: compute the Ricci tensor, observe that it vanishes, and observe that the metric is static and has spherical symmetry; uniqueness theorems tell you that it must be the Schwarzschild black hole.

Or, as WannabeNewton put it:

I am under the impression that you are thinking of a different form of the metric in question? You get the metric components simply by finding all components of $R_{\mu \nu } = 0$ and, of course, solving them for the metric with the spherical symmetry conditions that were placed and using classical parameters that correspond to parameters in the solution at infinity. Remember that by Birkhoff's Theorem if you ever solve the vacuum EFEs for a static and spherically symmetric space - time it will be the schwarzchild metric.

It seemed very surprising to me that a method using repeated approximations would yield the exact solution. Surprising, but not, I guess, unlikely, because if the exact solution is linear, of course, a linear approximation of a linear result is exact.

 

[Ben Niehoff]

It is one thing to say that in general, a coordinate chart only requires continuity. It is quite another to claim that a SPECIFIC mapping from one chart to another has no logical meaning. If I were to wad a paper into a little ball, it would preserve continuity, but the new distances between the points may be in some sense meaningless*. However, in the Schwarzschild metric case, there is unambiguous meaning.

(*no, I don't even want to play that game. Even the distances between points on a wadded piece of paper has meaning. It may be confusing, but it represents a specific configuration.)

OK, here's a metric:

$$ds^2 = \cos^4 \sqrt{u^2 + e^w} \; (\frac{du}{v} - \frac{dv}{\sqrt[3]{u}})^2 + \tanh^2 w (dv^2 + e^{uv} dw^2)$$
What do the coordinates (u, v, w) mean?

But once you do write the Schwarzschild solution in any funny coordinate system you like, I can apply an inverse transformation, and map them right back to (t,x,y,z).

Not if you don't know the coordinate transformation. And as should be clear in my example above, certainly not if you don't even know what the manifold is in the first place! I could very easily turn the Schwarzschild metric into something completely unrecognizable, and if I didn't outright tell you the coordinate transformation I used, you'd probably be helpless.

Physically observable quantities such as angles are observer dependent.

Of course they are. Physical measurements are always local. One lays out a local array of rods and clocks (called a "local inertial frame"), and uses them to measure distances, times, and angles. You can't measure the length of a distant object...that is, you can't lay out rods along its length, because you are here and it is over there. The best you can do is measure the angle between light rays from opposite ends of the object and try to make inferences. This is surveying, and astronomy.

"ΔR is the distance between two simultaneous events as measured by a ruler on (the non-rotating gravitational surface),

First of all, what do you mean by "gravitational surface"? I don't know of anything that goes by that name.

Second, rulers can only measure local lengths, so I'm not sure how you mean to compare rulers at different points. If I take a meter stick from R=200 to R=10, it is still a meter long. It doesn't make any sense to discuss "the length of a meter stick at R=10 as measured by an observer at R=200", because no such measurement can be made. If an observer at R=200 wants to make inferences about things at R=10, he'll have to look at light rays and compare the arrival times of photons.

while Δr is the distance between the same two simultaneous events as measured by a distant co-moving observer."

Again, same problem (unless you just mean events that are nearby to that observer!).

I can tell you this: In the asymptotic region, where spacetime approaches flatness, both coordinates 'r' and 'R' have the same interpretation, which is that dr and dR measure local distances in the radial direction, as in ordinary spherical coordinates. But this is only true in the asymptotic region. Closer to the horizon, 'r' and 'R' mean very different things, and neither dr nor dR gives a measure of local distance.

How can you claim that Δr does not represent any distance at all, but after doing a path integral along a path involving Δr you have a distance? Doesn't the process of doing a path integral in some way imply that Δr is a coordinate of a path?

Yes, of course 'r' is a coordinate along a path. I'm just saying that this is all 'r' is, intrinsically.

The (t,r,θ,Φ) coordinates represent a flat manifold in which the curved (tau,R,θ,Φ) coordinates are embedded.

What in the world are you saying here? There are so many things wrong here I don't know where to begin...

The (t,r,θ,Φ) coordinates represent a flat manifold...

The labels we choose to call the points on a manifold have nothing to do with whether the manifold is flat! I thought we went over this a million times in this thread...

...in which the curved (tau,R,θ,Φ) coordinates...

No one has used 'tau' as a coordinate in this thread, not even yourself. In your notation 'tau' is the proper time; that is $d\tau^2 = -ds^2$. 'tau' does not make a good coordinate; it can't be consistently extended to any finite neighborhood in the spacetime manifold, in general.

...are embedded.

You can't embed a 4-manifold within another 4-manifold, that makes no sense at all.

I think maybe you are confused by the definition of a coordinate chart, which is a differentiable, bijective map $\varphi : U \rightarrow V$ where U is an open region in M and V is an open region in $\mathbb{R}^n$. The map $\varphi$ is not an embedding. Furthermore, the coordinates in $V \in \mathbb{R}^n$ are not spherical coordinates on $\mathbb{R}^n$, even if that is their geometric meaning on M. In the case under discussion, the image of $\varphi$ in $V \in \mathbb{R}^4$ is actually a collection of Cartesian coordinates with the labels $(t,r,\theta,\phi)$.

There's one little mistake in what WannabeNewton is saying here, and that is worrying that the value, ΔR

$$\Delta R = \Delta s|_{dt=d\theta=d\phi=0}$$

is too big to fit into Euclidean Space! It is NOT. The Euclidean space component is represented by the r-component here; not the R-component or the Δs value. The ΔR distance is the Non-Euclidean space that is embedded in the Euclidean Space. It's kind of like the Tardis on Dr. Who: It's bigger on the inside. But unlike the Tardis, you can see the whole thing from outside; you can see that it's bigger on the inside FROM the outside.

First of all, WannabeNewton computed an integral for the 'r' coordinate of the standard Schwarzschild metric, so I'm not sure what you're going on about. It is a simple fact that the Schwarzschild geometry has "too much radial distance" within a circle of given 'r' (or 'R') coordinate; this tells you that the constant-time slices of the geometry are curved. This is an intrinsic part of the geometry; it does not depend on which coordinate system you use. Do the integral in both coordinate systems. You will get the same answer. That's the entire point of general covariance.

Skyscrapers near the Schwarzschild radius would be only inches tall, looking from outside, but they would seem normal height to the internal observers.

Again, such a statement makes no sense and reveals flaws in thinking. If you are at point A, you cannot measure distances at point B unless you physically go to point B! And when you get to point B, you will see that a meter is a meter and a skyscraper is quite tall.

If you want to talk about how things appear to a distant observer, that's a perfectly valid question, and to answer it you must follow the null geodesics. The distant observer is only capable of measuring the photons he receives. He will use his local definition of simultaneity and his local definition of angles, and will calculate an angle subtended by a distant skyscraper (and it will probably be a smaller angle than expected, yes). But that is the only physical information he has.

The horizon is at x=y=z=0, which is not a single point, but a whole 2-sphere worth of points. The isotropic coordinates are a coordinate chart that covers only the exterior region of the black hole.

The horizon is at $r=\frac{G M}{ c^2}$
where
$$\frac{1}{1-\frac{G M}{r c^2}}\rightarrow \infty$$ where the internal coordinates descend for infinity in a finite amount of (external coordinate) space.

Not sure what you mean here. The coordinate 'r' is perfectly finite at the horizon. It is 'r = 2m', as you wrote. In isotropic coordinates, this corresponds to 'R = 0', which is the same as 'x=y=z=0'. So in isotropic coordinates, the horizon is a single point in coordinate space, despite the fact that the horizon in reality is an entire sphere's worth of points. I pointed this out so that you could see that one has to be careful how one interprets coordinates. The isotropic coordinates are only good for 'R > 0', but at 'R = 0', they break down...in fact, they break down so much that they tell you lies about the very topology of the manifold! But this is ok, as coordinate charts are only meant to cover open regions, and 'R <= 0' would be including the boundary of the chart.

For an approximation of my meaning, imagine a stationary observer at a distance far enough away that the gravity is barely detectable.

Such an observer can only make local measurements! However, such measurements can include measuring photons that arrive at the observer's location, which is I think what you mean by your "hologram" idea. But an observer measuring photons from distant objects can only make inferential statements; in particular, he can't say anything about distances between those objects unless he has a decent model of the spacetime curvature between the objects and himself. Such statements are then model-dependent. I wouldn't properly call them "measurements", unless the model has independent verification.

This is part of what's so tricky about cosmology. Making accurate measurements of very distant objects requires extra, non-geometrical information. One very helpful such piece of information is spectroscopy, which came from chemistry. Since spectral lines come in fixed patterns, it is easy to measure how far they have shifted from their expected location, so effectively we can measure redshift. This simple fact caused a revolution in cosmology when it was discovered. But without spectroscopy we would not be able to measure redshift at all, since we would have no information about the light's initial state when it was emitted!

There IS, of course, trouble in measuring (x,y,z) because of the difficulty of imagining "a stationary observer at a distance far enough away that the gravity is barely detectable."

Why should that be hard to imagine? The Schwarzschild geometry is the Schwarzschild geometry, regardless of the coordinates used to label it. It has an asymptotic region where it approaches flat space.

 

[JDoolin]

OK, here's a metric:

$$ds^2 = \cos^4 \sqrt{u^2 + e^w} \; (\frac{du}{v} - \frac{dv}{\sqrt[3]{u}})^2 + \tanh^2 w (dv^2 + e^{uv} dw^2)$$
What do the coordinates (u, v, w) mean?



Not if you don't know the coordinate transformation. And as should be clear in my example above, certainly not if you don't even know what the manifold is in the first place! I could very easily turn the Schwarzschild metric into something completely unrecognizable, and if I didn't outright tell you the coordinate transformation I used, you'd probably be helpless.

Quite right. I'm not at all confident that you've even given me a one-to-one mapping. (For all I know, you didn't just wad the paper, but you may have created a copy on a piece of silly-putty, and stretched it, pulled it, and wadded it; in which case, there IS no inverse transform.)

There's nothing preventing you from doing that sort of thing, but if the inverse transformation exists, you can use it to get back the original coordinates. What you are essentially saying, though, is that it is possible for you to create a de-codable coded message, but without the cipher, we are pretty helpless to figure out the message.

That's not the case with the Schwarzschild metric. It's not some kind of mysterious coded message. It's fairly straightforward mapping between (r', \tau,\theta \phi) and (r,t,\theta,\phi).

Of course they are. Physical measurements are always local. One lays out a local array of rods and clocks (called a "local inertial frame"), and uses them to measure distances, times, and angles. You can't measure the length of a distant object...that is, you can't lay out rods along its length, because you are here and it is over there. The best you can do is measure the angle between light rays from opposite ends of the object and try to make inferences. This is surveying, and astronomy.

I don't know if you're aware of it, but you're being circular. You are insisting that the values of what we're measuring have no meaning. If I agreed with that conclusion, then I would also agree with yo that "you can't measure the length of a distant object."

First of all, what do you mean by "gravitational surface"? I don't know of anything that goes by that name.

Sitting on the ground, or on a table.

Second, rulers can only measure local lengths, so I'm not sure how you mean to compare rulers at different points. If I take a meter stick from R=200 to R=10, it is still a meter long. It doesn't make any sense to discuss "the length of a meter stick at R=10 as measured by an observer at R=200", because no such measurement can be made. If an observer at R=200 wants to make inferences about things at R=10, he'll have to look at light rays and compare the arrival times of photons.

An observer will have to make inferences, because he has to estimate how far away the object is. Unless there is something actually preventing him from seeing the object, he's going to see something, and he will have to interpret that image in some way. There are difficulties in measuring distant quantities, but that does not mean those quantities don't exist.


[QUOTE

Again, same problem (unless you just mean events that are nearby to that observer!).

[/QUOTE]

No. I don't mean events that are near the observer. I mean an observer that is faraway. In fact, if you're near the Schwarzschild radius, things would be even more confusing to you than if you are fa raway. Near the Schwarzschild radius, the events which happened immediately above and below you

I can tell you this: In the asymptotic region, where spacetime approaches flatness, both coordinates 'r' and 'R' have the same interpretation, which is that dr and dR measure local distances in the radial direction, as in ordinary spherical coordinates. But this is only true in the asymptotic region. Closer to the horizon, 'r' and 'R' mean very different things, and neither dr nor dR gives a measure of local distance.

I assume by the asymptotic region, you mean far away from the center, where $$\frac{G M}{r c^2}\approx 0$$. In that case, I agree that dr and dR are the same.

But to compare r and R as more global parameters, you need to do the integral

$$R=\int dR=\int ds|_{d\theta=d\phi=dt=0}= \int_{r_0} \sqrt{\frac{1}{1-\frac{G M}{r c^2}}}dr$$

And to do this integral, you need to define an arbitrary radius $$r_0>\frac{G M}{ c^2}$$

...
Yes, of course 'r' is a coordinate along a path. I'm just saying that this is all 'r' is, intrinsically.

The coordinate, r, has physical meaning. It is the coordinate of the overlying cartesian coordinate system.

What in the world are you saying here? There are so many things wrong here I don't know where to begin...



The labels we choose to call the points on a manifold have nothing to do with whether the manifold is flat! I thought we went over this a million times in this thread...

If you have the argument with clarity, can you point it out to me? I've really been trying to follow the thread, but it wasn't until someone made the definitively ridiculous challenge "Give me a metric that will flatten the Schwarzschild metric" when I hopped in.

In any case, the (t,r,theta,phi) variables of the Schwarzschild metric are flat, or at least can be made flat by mappng to (t,x,y,z). By what definition of flatness do you claim otherwise?

No one has used 'tau' as a coordinate in this thread, not even yourself. In your notation 'tau' is the proper time; that is $d\tau^2 = -ds^2$. 'tau' does not make a good coordinate; it can't be consistently extended to any finite neighborhood in the spacetime manifold, in general.

I've used τ several times. Unfortunately, this font just doesn't how up. If I haven't made any errors in the math, here is how I would define it. (Very similar to the definition of R, above)

$$\tau=\int d\tau=\int ds|_{d\theta=d\phi=dr=0}= \int_{t_0} \sqrt{1-\frac{G M}{r c^2}}dt$$

You can't embed a 4-manifold within another 4-manifold, that makes no sense at all.

The Schwarzschild metric embeds a geometry of shorter rulers and slower clocks into a global Cartesian Space + Time. I don't know how that's different from embedding one 4-manifold into another 4-manifold, but that's what it sounds like to me.

I'll have to leave it for now, and go to work.

 

[Dale]

Anamitra, one of the most frustrating things about trying to discuss topics with you is that you don't use the quote feature at all, so we never know if you believe that you are responding to us or just talking to yourself. It generally seems that you are talking to yourself since your responses almost never clearly answer direct questions asked of you and your responses often seem to ignore previous posts which have directly rebutted something you are trying to do. For instance:

We may use the following notations:
$${dT}{=}{(}{1}{-}{\frac{2m}{r}}{)}{dt}$$ ------------- (4)
$${dR}{=}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}$$ ----------- (5)

As we have discussed previously several times from several people, there are no solutions to these equations. In other words, there is no valid coordinate transform from r, t to R, T.

If you disagree then please post the coordinate transform as I specifically requested in post 89. Otherwise please don't continue to rewind the conversation to topics already covered in depth.

We may consider the the three metrics given below:

$${ds}^{2}{=}{dt}^{2}{-}{dx1}^{2}{-}{dx2}^{2}{-}{dx3}^{2}$$ --------------- (1)
$${ds}^{2}{=}{dt}^{2}{-}{dr}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{Sin}^{2}{\theta}{d}{\phi}^{2}{)}$$------------ (2)
$${ds'}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{Sin}^{2}{\theta}{d}{\phi}^{2}{)}$$ --------------------- (3)

...

We may think of a path that has the same coordinate description wrt to equations 2 and 3.The quantities t,r,theta and phi will be the same for both. Again dt,dr,d(theta),d(phi) will also be the same for both. But physical separations will be different.

You can also think of a path that has the same coordinate description wrt to equations 1 and 2. The labels for the coordinates are completely arbitrary, so you can simply use t=x0, x=x1, y=x2, z=x3, and t=x0, r=x1, theta=x2, and phi=x3. Or you could use r=x0, phi=x1, t=x2, and theta=x3 and think of a path that has the same coordinate description wrt 1 and 2. Coordinates are simply maps to open subsets of R4, so you can certainly always specify the same path in R4 provided the path is covered by both. I don't see the physics utility in doing so, but yes you can do it.

 

[Anamitra]

$${dT}{=}{(}{1}{-}{\frac{2m}{r}}{)}{dt}$$ ------------- (4)
$${dR}{=}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}$$ ----------- (5)

As we have discussed previously several times from several people, there are no solutions to these equations. In other words, there is no valid coordinate transform from r, t to R, T.

The metric in GR is valid only if a path is specifiied
$${ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{22}{-}{g}_{33}{dx3}$$

The following posting deals with the issue:
https://www.physicsforums.com/showpost.php?p=3436925&postcount=21

Once the path is specified the equations do yield solutions

The issues in this posting[the one given by the link] have not been responded to by DaleSpam [or by any other Science advisor/mentor for that matter ] since it does not contain any quotation.

 

[Dale]

The metric in GR is valid only if a path is specifiied.

This is not correct at all. The metric is used validly in many situations where there is no path. For example, it is used directly to compute the curvature tensors without the specification of any path.

The issues in this posting[the one given by the link] have not been responded to by DaleSpam [or by any other Science advisor/mentor for that matter ] since it does not contain any quotation.

Given your complete unresponsiveness on pretty much every point that has been raised, your complaint here on one small point is incredibly hypocritical. Once you start responding regularly then you have a basis to complain. Until then, just be glad that the rest of us treat you with far more respect and attention than you treat us with.

I note that you have yet to post the explicit coordinate transform from t to T and r to R despite repeated requests to do so and your repeated assertions that such a transformation exists.

 

[Ben Niehoff]

OK, I don't have time for long posts like that again, so I'm going to pick just a few important pieces...

Quite right. I'm not at all confident that you've even given me a one-to-one mapping. (For all I know, you didn't just wad the paper, but you may have created a copy on a piece of silly-putty, and stretched it, pulled it, and wadded it; in which case, there IS no inverse transform.)

You seem to be under the impression that I'm required to begin with some well-known metric and then transform its coordinates. I am not under any such restriction. My metric in (u, v, w) is completely made-up. I don't know what manifold it might represent. But it is the metric of some manifold, at least in some coordinate patch. Everything is continuous; there is no ripping or tearing involved, at least in the region where these coordinates are defined.

I don't know if you're aware of it, but you're being circular. You are insisting that the values of what we're measuring have no meaning. If I agreed with that conclusion, then I would also agree with yo that "you can't measure the length of a distant object."

OK, describe to me then, if you don't a priori know the metric of spacetime, how do you measure the length of a distant object?

[a "gravitational surface" is] Sitting on the ground, or on a table.

This definition makes no sense. There is no ground or table. There is a black hole and we're floating in space. Can you give a more precise definition?

An observer will have to make inferences, because he has to estimate how far away the object is. Unless there is something actually preventing him from seeing the object, he's going to see something, and he will have to interpret that image in some way. There are difficulties in measuring distant quantities, but that does not mean those quantities don't exist.

I'm not saying that facts like "The height of a skyscraper at r=10" do not exist. I'm saying that facts like "The height of a skyscraper at r=10 as measured by an observer at r=200" do not exist. The observer at r=200 cannot measure skyscrapers at r=10, because he is not there! The observer at r=200 measures light rays that appear to come from r=10. So the kinds of facts he can know are of the form "Using some specific model of spacetime (c.f. the Schwarzschild metric), and these light rays that came from r=10, the height of that skyscraper I see at r=10 would be X if I were standing at r=10".

You seem to be using facts of the form "Using a flat model of spacetime, and these light rays that came from r=10, the height of that skyscraper I see at r=10 would be X if I were standing at r=10". I'm telling you that this sort of fact is model-dependent, and I don't think it's any more natural to assume flat spacetime than it is to assume Schwarzschild, or a C-metric, or whatever.

The observer needs additional information in order to decide what model of spacetime to use in his calculations. A flat metric is reasonable on cosmological scales (actually, a metric with cosmological constant is even better). A Schwarzschild metric is reasonable if the observer is near a massive object (a Kerr metric is even better, considering nearly all massive objects are spinning). If the observer is near two massive objects, he's SOL because we don't have an exact metric for that. He'll have to run his data carefully through numerical simulations in order to figure out what the height of that skyscraper in the distance is.

But to compare r and R as more global parameters, you need to do the integral

$$R=\int dR=\int ds|_{d\theta=d\phi=dt=0}= \int_{r_0} \sqrt{\frac{1}{1-\frac{G M}{r c^2}}}dr$$

And to do this integral, you need to define an arbitrary radius $$r_0>\frac{G M}{ c^2}$$

You're wrong here, the 0 points of r and R are already well-defined, so there is no ambiguity. Small r=0 at the singularity (or alternatively, r=2m at the horizon). Big R=0 at the horizon.

The coordinate, r, has physical meaning. It is the coordinate of the overlying cartesian coordinate system.

There is no overlaying cartesian coordinate system! Such a thing is impossible.

In any case, the (t,r,theta,phi) variables of the Schwarzschild metric are flat, or at least can be made flat by mappng to (t,x,y,z). By what definition of flatness do you claim otherwise?

OK, here's the Schwarzschild metric in (t, x, y, z) coordinates:

$$ds^2 = -\big(1 - \frac{2m}{x} \big) \; dy^2 + \big( 1 - \frac{2m}{x} \big)^{-1} \; dx^2 + x^2 \; dz^2 + x^2 \sin^2 z \; dt^2$$
Now what?

Coordinates do not have the property of being "flat" or "not flat". Coordinates do not carry any geometrical information at all! They are just labels. The metric carries the geometrical information. You must compute its Riemann curvature tensor to determine if it describes a curved manifold or a flat one.

I've used τ several times. Unfortunately, this font just doesn't how up. If I haven't made any errors in the math, here is how I would define it. (Very similar to the definition of R, above)

$$\tau=\int d\tau=\int ds|_{d\theta=d\phi=dr=0}= \int_{t_0} \sqrt{1-\frac{G M}{r c^2}}dt$$

I can tell when you've used the letter tau, no problem with fonts. What I'm telling you is that you haven't used it as a coordinate. The quantity tau that you define here is not a coordinate. Do you see why?

 

[Anamitra]

This is not correct at all. The metric is used validly in many situations where there is no path. For example, it is used directly to compute the curvature tensors without the specification of any path.

The issue of path dependence is central to the issue of integration raised by DaleSpam in the posting #115.

The relevant portion has been referred to in posting #116

 

[Dale]

The metric is not path dependent. Please re-read my rebuttal that you quoted.

 

[Anamitra]

This is not correct at all. The metric is used validly in many situations where there is no path. For example, it is used directly to compute the curvature tensors without the specification of any path.

I have referred to the quantity ds^2 --- a scalar product
This value is path dependent.

From a particular point one may take out different paths[infinitesimal ] in different directions their their length square would have different values.For finite paths path lengths would be different.

The issue of path dependence is central to the issue of integration referred to in posts #115 and #116

 

[Anamitra]

Points to Observe:

1.In the same space[manifold] I can draw millions of infinitesimal curves having different orientations and positions.

2.The nature of the space[manifold] is determined by the metric coefficients.

3.The value of ds^2 is given by the orientation and the position of the infinitesimal curve. The orientation and position of the curve selects the appropriate values of the metric coefficients, relevant to something called "path".

 

[JDoolin]

Correction:

But to compare r and R as more global parameters, you need to do the integral

$$R=\int dR=\int ds|_{d\theta=d\phi=dt=0}= \int_{r_0} \sqrt{\frac{1}{1-\frac{G M}{r c^2}}}dr$$

And to do this integral, you need to define an arbitrary radius $$r_0>\frac{G M}{ c^2}$$

At work, I have access to mathematica, and I find that this integral is NOT behaving quite how I expected. I expected an integral that actually started at the schwarzschild radius in r would give an infinite R.

I'll give you a few examples of the calculation results:

$$\int_{1}^{1.1}\sqrt{1 -\frac{1}{r}} = .6428 n \Rightarrow \frac{\Delta R}{\Delta r} = 6.428$$

$$\int_{1}^{1.01}\sqrt{1 -\frac{1}{r}} = .2003 \Rightarrow \frac{\Delta R}{\Delta r} = 20.03$$

$$\int_{1}^{1.001}\sqrt{1 -\frac{1}{r}} = .0632 \Rightarrow \frac{\Delta R}{\Delta r} = 63.2$$

$$\int_{1}^{1.0001}\sqrt{1 -\frac{1}{r}} = .0200 \Rightarrow \frac{\Delta R}{\Delta r} = 200$$

$$\int_{1}^{1.00001}\sqrt{1 -\frac{1}{r}} = .00632 \Rightarrow \frac{\Delta R}{\Delta r} = 632$$

So I was wrong: You CAN start at the Schwarzschild radius and get a finite value for R. I thought that since

$$\left (\overset {\lim} {r \rightarrow 1^+} \left (\frac{\Delta R}{\Delta r} \right ) \right ) \to \infty$$

I thought the integral of that sort of thing would also give you infinity. But you don't, so you also don't have to define an arbitrary starting radius.

 

[Dale]

I have referred to the quantity ds^2 --- a scalar product
This value is path dependent.

ds^2, a scalar product, is not the metric, a rank 2 tensor.

The metric is not path dependent, nor is a coordinate transform. The fact that the quantity you are referring to is path dependent should be a big hint to you that you are not doing a coordinate transform nor are you computing the metric.

Please explicitly write out the coordinate transform from t to T and r to R. If you cannot do that minimal step then none of the rest of what you are discussing has any relevance to anything. You have been asked to do this more than a half-dozen times now. Your inability to do this is a serious challenge to your method.

 

[Ben Niehoff]

But to compare r and R as more global parameters, you need to do the integral

$$R=\int dR=\int ds|_{d\theta=d\phi=dt=0}= \int_{r_0} \sqrt{\frac{1}{1-\frac{G M}{r c^2}}}dr$$

Incidentally, this line is wrong anyway, because

$$ds|_{d\theta = d\phi = dt = 0} = \Big( 1 + \frac{m}{2R} \Big)^2 dR \neq dR$$

As I told you earlier, the coordinate transformation between 'r' and 'R' is given by

$$r = R \Big( 1 + \frac{m}{2R} \Big)^2$$

One can transform to a radial coordinate that does represent distance from the horizon. This coordinate is given by

$$\rho = \int_{2m}^r \Big( 1 - \frac{2m}{r'} \Big)^{-1/2} dr' = \sqrt{r ( r - 2m)} + 2m \; \cosh^{-1} \sqrt{\frac{r}{2m}}$$

However, in order to carry out the transformation, you would have to invert this to get $r(\rho)$. Good luck.

 

[Anamitra]

The integration s requested in posting # 124

Please explicitly write out the coordinate transform from t to T and r to R...

Let us consider a path:
Theta=const ,phi =const; r=kt where k is a constant
[An arbitrary path may not correspond to a geodesic]

$${dR}=\frac{dr}{\sqrt{{1}{-}\frac{2m}{r}}}$$
$$\int{dR}{=}\int\frac{dr}{\sqrt{{1}{-}\frac{2m}{r}}}$$
$$\int\frac{r}{\sqrt{{r}{(}{r}{-}{2m}{)}}}{dr}$$
$${=}\int\frac{1}{2}\frac{2r-2m}{\sqrt{{r}^{2}{-}{2mr}}}{dr}{+}{m }\int\frac{1}{\sqrt{{r}^{2}{-}{2mr}}}{dr}$$
$${R}{=}{\sqrt{{r}^{2}{-}{2mr}}}{+}{2m}{ln}{[}\sqrt{r}{+}\sqrt{(}{r}{-}{2m}{)}{]}{+}{C}$$

Again,
$${dT}{=}\sqrt{(}{1}{-}\frac{2m}{r}{)}{dt}$$
$${=}\sqrt{\frac{{r}{-}{2m}}{r}}{dt}$$
$${=}\sqrt{\frac{{kt}{-}{2m}}{kt}}{dt}$$
$${=}\frac{{kt}{-}{2m}}{\sqrt{{kt}{(}{kt}{-}{2m}{)}}}{dt}$$
$${=}\frac{1}{2k}\frac{{2k}^{2}{t}{-}{2km}}{\sqrt{{k}^{2}{t}^{2}{-}{2ktm}}}{dt}{-}{m}\frac{1}{\sqrt{{kt}{(}{kt}{-}{2m}{)}}}{dt}$$
$${Integral}{=}{T}{=}\frac{1}{k}{\sqrt{{kt}{(}{kt}{-}{2m}{)}}}{-}\frac{2m}{k}{ln}{[}\sqrt{t}{+}\sqrt{(}{t}{-}{2m}{/}{k}{)}{]}{+}{C’}$$

The constants will be removed from initial conditions or by calculating definite integrals.

[In Post #111 the square root sign was inadvertently missed out in the expressions for dR and dT . I would request the audience to consider it.]

 

[Anamitra]

ds^2, a scalar product, is not the metric, a rank 2 tensor.

The metric is not path dependent, nor is a coordinate transform....

I would stress on the fact that the value of the line element in any manifold is a path dependent quantity.

Wherever I have used the term --the metric is path dependent---it should be interpreted in the sense that the value of the line element[represented by the scalar product ds^2] is path dependent. It is the path that selects out the appropriate values of the metric coefficients in the evaluation of the line element.

 

[JDoolin]

Incidentally, this line is wrong anyway, because

$$ds|_{d\theta = d\phi = dt = 0} = \Big( 1 + \frac{m}{2R} \Big)^2 dR \neq dR$$

As I told you earlier, the coordinate transformation between 'r' and 'R' is given by

$$r = R \Big( 1 + \frac{m}{2R} \Big)^2$$

I see now where you're getting that... From the "Alternative (isotropic) formulations of the Schwarzschild metric" on Wikipedia.

You do at least have to acknowledge that this is an alternative formulation. I am fairly sure the integration I did is correct, as is. If you want to add the complication suggested by Eddington, that is okay, but I don't think it represents the (t,x,y,z) variables that I was talking about.

It is also an extra level of encoding that I really don't want to deal with.

 

[Dale]

Wherever I have used the term --the metric is path dependent---it should be interpreted ...

You should stop using the term. It is factually incorrect. The metric is not path dependent.

You should say what you mean, that the line element depends on the path. I.e. The length of a line depends on the line being measured.

 

[Ben Niehoff]

I see now where you're getting that... From the "Alternative (isotropic) formulations of the Schwarzschild metric" on Wikipedia.

You do at least have to acknowledge that this is an alternative formulation.

It's another equally valid set of coordinates to describe the same geometry. In some situations, isotropic coordinates are actually the most convenient. In particular, it's very easy to find new solutions to Einstein's equations using isotropic coordinates.

I am fairly sure the integration I did is correct, as is. If you want to add the complication suggested by Eddington, that is okay, but I don't think it represents the (t,x,y,z) variables that I was talking about.

Then do tell us, what (t,x,y,z) variables are you talking about? Define them mathematically.

It is also an extra level of encoding that I really don't want to deal with.

There is no "extra level" of encoding. Every coordinate system is equally valid, so long as the metric still describes the same geometry. You seem to be clinging to Schwarzschild coordinates as though they are special; but I tell you, they are not special in any way. The following are all metrics that represent the Schwarzschild geometry:

$$ds^2 = - \Big( \frac{2r - m}{2r + m} \Big)^2 \; dt^2 + \Big(1 + \frac{m}{2r} \Big)^4 \; (dr^2 + r^2 \; d\theta^2 + r^2 \sin^2 \theta \; d\phi^2)$$

$$ds^2 = -dt^2 + \Big( \frac{4m}{3 (r - t)} \Big)^{2/3} \; dr^2 + m^{2/3} \Big( \frac{3}{\sqrt{2}} (r-t) \Big)^{4/3} \; (d\theta^2 + \sin^2 \theta \; d\phi^2)$$

$$ds^2 = \frac{16m^2}{u^2 - v^2} \left( \frac{\mathcal{W}(\tfrac{u^2 - v^2}{e})}{1 + \mathcal{W}(\tfrac{u^2 - v^2}{e})} \right) \; (du^2 - dv^2) + 4m^2 \Big(1 + \mathcal{W}(\tfrac{u^2 - v^2}{e}) \Big)^2 \; (d\theta^2 + \sin^2 \theta \; d\phi^2)$$

where $\mathcal{W}(x)$ is the Lambert W function and $e$ is the base of the natural logarithm.

 

[Dale]

$${R}{=}{\sqrt{{r}^{2}{-}{2mr}}}{+}{2m}{ln}{[}\sqrt{r}{+}\sqrt{(}{r}{-}{2m}{)}{]}{+}{C}$$$${T}{=}\frac{1}{k}{\sqrt{{kt}{(}{kt}{-}{2m}{)}}}{-}\frac{2m}{k}{ln}{[}\sqrt{t}{+}\sqrt{(}{t}{-}{2m}{/}{k}{)}{]}{+}{C’}$$

Excellent, thanks for that.

Now, the next step is to use these coordinate transformation equations to determine the metric (rank 2 tensor) in the new coordinates, and then to compute the curvature tensor (rank 4 tensor) in the new coordinates. Unfortunately, the given formulas are not analytically solvable for r, so you will have to do it numerically. Only then will you have demonstrated that your process generates a coordinate transform which flattens the metric.

 

[Anamitra]

Excellent, thanks for that.

Now, the next step is to use these coordinate transformation equations to determine the metric (rank 2 tensor) in the new coordinates, and then to compute the curvature tensor (rank 4 tensor) in the new coordinates. Unfortunately, the given formulas are not analytically solvable for r, so you will have to do it numerically. Only then will you have demonstrated that your process generates a coordinate transform which flattens the metric.

We can always do it in this way:

We have our definitions:
$${dT}{=}\sqrt{{g}_{tt}}{dt}$$
$${dR}{=}\sqrt{{g}_{rr}}{dr}$$
$${dP}{=}\sqrt{{g}_{\theta\theta}}{d}{\theta}$$
$${dQ}{=}\sqrt{{g}_{\phi\phi}}{d}{\phi}$$
Take any point A(t,r,theta,phi) in the manifold[it may be Schwarzschild's Geometry]
From this point take four distinct paths:
Path 1: r=const, theta=constant,phi=constant.
Path 2: t=const,theta=const, phi=const
Path 3: t=const,r=const,phi=const
Path 4: t=const,r=const,theta=const For Path 1.

$${dT}{=}\frac{{\partial}{T}}{{\partial}{t}}{dt}{+}\frac{{\partial}{T}}{{\partial}{r}}{dr}{+}\frac{{\partial}{T}}{{\partial}{\theta}}{{d}{\theta}}{+}\frac{{\partial}{T}}{{\partial}{\phi}}{d}{\phi}$$

For our path dr=0;d(theta)=0 ;d(phi)=0

Therefore,so far our path is concerned,we have

$$\frac{dT}{dt}{=}\frac{{\partial}{T}}{{\partial}{t}}{=}\sqrt{{g}_{tt}}$$

[How T changes along other paths is not our concern so far as the evaluation of g(tt) is our goal/objective]

For Path 2.

$${dR}{=}\frac{{\partial}{R}}{{\partial}{t}}{dt}{+}\frac{{\partial}{R}}{{\partial}{r}}{dr}{+}\frac{{\partial}{R}}{{\partial}{\theta}}{d}{\theta}{+}\frac{{\partial}{R}}{{\partial}{\phi}}{d}{\phi}$$

For our path dt=0;d(theta)=0 ;d(phi)=0

Therefore,so far our path is concerned,we have

$$\frac{dR}{dr}{=}\frac{{\partial}{R}}{{\partial}{r}}{=}\sqrt{{g}_{rr}}$$

[How R changes along other paths is not our concern so far as the evaluation of g(rr) at A is our goal/objective]

For Path 3.

$${dP}{=}\frac{{\partial}{P}}{{\partial}{t}}{dt}{+}\frac{{\partial}{P}}{{\partial}{r}}{dr}{+}\frac{{\partial}{P}}{{\partial}{\theta}}{{d}{\theta}}{+}\frac{{\partial}{P}}{{\partial}{\phi}}{d}{\phi}$$

For our path dt=0; dr=0;d(phi)=0

Therefore,so far our path is concerned,we have

$$\frac{dP}{{d}{\theta}}{=}\frac{{\partial}{P}}{{\partial}{\theta}}{=}\sqrt{{g}_{\theta\theta}}$$

[How P changes along other paths is not our concern so far as the evaluation of g(theta,theta) is our goal/objective]

For Path 4.

$${dQ}{=}\frac{{\partial}{Q}}{{\partial}{t}}{dt}{+}\frac{{\partial}{Q}}{{\partial}{r}}{dr}{+}\frac{{\partial}{ Q}}{{\partial}{\theta}}{d }{\theta}{+}\frac{{\partial}{Q}}{{\partial} {\phi}}{d}{\phi}$$

For our path dt=0;dr=0;d(theta)=0

Therefore,so far our path is concerned,we have

$$\frac{dQ}{{d}{\phi}}{=}\frac{{\partial}{Q}}{{\partial}{\phi}}{=}\sqrt{{g}_{\phi\phi}}$$

[How Q changes along other paths is not our concern so far as the evaluation of g(phi,phi) is our goal/objective]

Once we have the values of the metric coefficients at each point of the manifold we can always evaluate the Christoffel Symbols and the components of the Riemannian curvature tensor

 

[JDoolin]

You seem to be clinging to Schwarzschild coordinates as though they are special; but I tell you, they are not special in any way. The following are all metrics that represent the Schwarzschild geometry:

Oh, certainly I think that Schwarzschild coordinates are special, because I've seen a common-sense argument^{http://www.mathpages.com/rr/s8-09/8-09.htm"} showing that

$$\left (\frac{\partial \tau}{\partial r} \right )^2\approx 1-\frac{G M}{r c^2}$$

The fact is, I have worked through this on my own time, and I know exactly what the variables are referring to. So when you make the claim that these variables are meaningless, you're telling me something I know is not true; at least for the g₀₀ component.

On the other hand, I've not worked through any common-sense argument for the g₁₁ component of the Schwarzschild metric:
$$\left (\frac{\partial s}{\partial r} \right )^2\approx \frac{1}{1-\frac{G M}{r c^2}}$$

I can see there is a derivation in the http://www.blatword.co.uk/space-time/Carrol_GR_lectures.pdf" on pages 168-172. That remains a goal for me, to work through that derivation as well, but for now, I'm not comfortable with most of the concepts involved there. For instance, why does he start with the assumption that g₀₀ is the negative reciprocal of g₁₁?

But the point is, I didn't see the g₀₀ component by

• assuming $$g_{11}=-1/g_{00}$$
• finding the connection coefficients
• finding the nonvanishing components of the Reimann tensor
• taking the contraction (as usual?) to find the Ricci tensor
• setting all the components of the Ricci tensor to zero
• discovering that the g₀₀ and g₁₁ had to be functions of r, only,
• Setting R₀₀=R₁₁=0
• Doing some fancy differential equations with boundary conditions, and deriving the metric

Now, I am willing and (I think) able to go through all this; that's just a matter of time. But here's the point I want to make: If someone went through these steps, without going through the other derivation, they might be left with the impression that the coordinates involved were somehow, arbitrary, i.e. nonphysical. They might think that the physical interpretation of those variables were ambiguous, or even nonexistant. And that seems to be where you are coming from, Ben.

But where I'm coming from, you see, is the other derivation; where at all times, we're talking about clocks and rulers and measurements; figuring out how to modify the Rindler coordinates into the Schwarzschild coordinates. Even though I only figured out how to get the result for the time-time component, I trust the derivation to represent something meaningful.

There is no overlaying cartesian coordinate system! Such a thing is impossible.

I wish I could find some way to argue this point with you. Let me try a few things.

(1) If you could imagine yourself in a space-ship; reasonably far from a gravitational mass, could you go around it? In that asymptotic region, far from the system, where the schwarzschild metric approaches the flat space-time. Can't you go around the planet? Don't you have a fairly firm concept of how far you went? Don't base it on your path, but take someone further away, who can see both your starting position, and your ending position. He can see that you've moved from one side of the planet to the other side of the planet. And even with the planet there, he can describe your current and final position according to known distances, and known angles, effectively figured in an "overlying cartesion coordinate system."

(2) As the space-ship operator, do you feel that you just now went around a spatial anomoly. You may feel that since distances are changed inside that anomoly that you have gone an "unmeasurable" distance. However, couldn't you also take another approach? That spatial anomoly is CONTAINED in a region from your perspective. It is an unusual feature in an otherwise cartesian space. In that space, the volume where the black-hole, large planet, star, etc is a finite region; even a small and insignificant region if you're far enough away.

If you can localize a spatial anomoly to a given region; and if you can go AROUND a planet or star, I would say, you are operating with an overlying cartesian coordinate system. And the coordinates of that coordinate system run explicitly RIGHT THROUGH that spatial anomoly. They don't somehow take a break near the star and go all ambiguous on you. Every event that happens outside the schwarzchild radius, at least, is going to happen at a specific point in space and time in the overlying coordinate system.

I'm otherwise at a loss for how to explain this to you, but maybe you can identify what you think is in error.

Then do tell us, what (t,x,y,z) variables are you talking about? Define them mathematically.

If you're completely convinced that an overlying cartesian coordinate system is impossible then there's hardly a point to explain this to you, because that's my exact definition.

That being said, (t,x,y,z) represent the coordinates of events in the overlying cartesian coordinate system (plus time). Naturally, unless you've been at least somewhat swayed by the arguments above, that definition won't help. But it's all I've got.

Contrast that with your own explanation:

Coordinates do not carry any geometrical information at all! They are just labels. The metric carries the geometrical information. You must compute its Riemann curvature tensor to determine if it describes a curved manifold or a flat one.

(P.S. What criteria of the Riemann curvature determines whether a manifold is flat?)

In any case, it is my impression that coordinates DO carry geometrical information, and I'm rather at a loss for how you could argue otherwise.

There is no ground or table. There is a black hole and we're floating in space. Can you give a more precise definition?

Mathematically, the Schwarzschild metric identifies the transformation:

$$c^2 {d \tau}^{2} = \left(1 - \frac{2 G M}{r c^2} \right) c^2 dt^2 - \left(1-\frac{2 G M}{r c^2}\right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)$$

It's a differential equation, so you'll need some boundary values to define things explicitly.

But this isn't just a mapping of $$(dt,dr,d\theta,d\phi) \to ds$$

It is a mapping of $$(dt,dr,d\theta,d\phi) \to (d\tau,dR, d\theta',d\phi')$$

http://en.wikipedia.org/wiki/Coordinate_system In geometry, a coordinate system is a system which uses one or more numbers, or coordinates, to uniquely determine the position of a point or other geometric element.


My claim is that after applying boundary conditions, $(\tau, R, \theta', \phi'), (t,r,\theta,\phi), \, \mathrm{and}\, (t,x,y,z)$ all represent coordinates, in that they all uniquely determine the position of events.

It should be obvious already that $$(t,r,\theta,\phi)$$ represent coordinates. If not, please explain why.

These spherical coordinates can be mapped to cartesian coordinates in the standard way

$$\begin{matrix} t=t\\ z=r \cos(\theta)\\ x=r \sin(\theta)\cos(\phi)\\ y=r \sin(\theta)\sin(\phi) \end{matrix}$$

Which is also a unique mapping, except at r=0, there are several different values of phi all mapping to the same point.

The hard part is showing that $(\tau, R, \theta', \phi')$ are coordinates.

First, a definition of their differentials:

$$\begin{matrix} dR \equiv ds|_{d\theta=d\phi=dt=0}=\left ( \frac{1}{\sqrt{1-\frac{G M}{r c^2}}} \right )dr\\ d\tau \equiv ds|_{d\theta=d\phi=dr=0}=\left ( \sqrt{1-\frac{G M}{r c^2}} \right )dr\\ r d\theta' \equiv ds|_{dt=d\phi=dr=0} =r d\theta\\ r sin(\theta)d\phi' \equiv ds|_{d\theta=dt=dr=0} =r sin(\theta)d\phi \\ \end{matrix}$$

Since $$d\phi'=d\phi\; \mathrm{and}\; d\theta'=d\theta$$, I use them interchangably.

Using boundary conditions of $$\begin{matrix} R(r=\frac{G M}{ c^2})=0\\ \tau(t=0)=0 \end{matrix}$$

We can calculate the definite integrals:

$$\begin{matrix} R(r)=\int_{G M/c^2}^{r}\frac{1}{\sqrt{1-\frac{G M}{\rho c^2}}}d\rho\\ \tau(t,r)=\int_{0}^{t}\sqrt{1-\frac{G M}{r c^2}}dt = \left (\sqrt{1-\frac{G M}{r c^2}} \right )t \\ \theta'(\theta)=\int_{0}^{\theta} d\theta=\theta\\ \phi'(\phi)=\int_{0}^{\phi} d\phi=\phi \end{matrix}$$

There is no ground or table. There is a black hole and we're floating in space. Can you give a more precise definition?

If I am setting dr=0, I am saying, essentially there is no "drop" between the events. In a black hole, this might not be available; you'd have to be in a rocketship thrusting away from the center in order to maintain dr=0. But if you are on a solid planet, there is no problem. You just set your clock on a floor or table.

I can tell when you've used the letter tau, no problem with fonts. What I'm telling you is that you haven't used it as a coordinate. The quantity tau that you define here is not a coordinate. Do you see why?

No. I don't. Again, the definition I see says: "a coordinate system is a system which uses one or more numbers, or coordinates, to uniquely determine the position of a point or other geometric element."

It appears to me that tau is one of four coordinates that can uniquely determine the space-time location of an event.

 

[Dale]

We can always do it in this way:

We have our definitions:
$${dT}{=}\sqrt{{g}_{tt}}{dt}$$
$${dR}{=}\sqrt{{g}_{rr}}{dr}$$
$${dP}{=}\sqrt{{g}_{\theta\theta}}{d}{\theta}$$
$${dQ}{=}\sqrt{{g}_{\phi\phi}}{d}{\phi}$$

Unfortunately you get stuck even there. For the transforms you posted you do not get these expressions back. For example differentiating the R to r coordinate transform equation gives

$${R}{=}{\sqrt{{r}^{2}{-}{2mr}}}{+}{2m}{ln}{[}\sqrt{r}{+}\sqrt{(}{r}{-}{2m}{)}{]}{+}{C}$$
$$dR=\frac{2 m \left(\frac{dr}{2 \sqrt{r-2 m}}+\frac{dr}{2 \sqrt{r}}\right)}{\sqrt{r-2 m}+\sqrt{r}}-\frac{2 r \, dr -2 m\, dr}{8 \left(r^2-2 m r\right)^{3/2}}$$
$$dR=\frac{m (4 r (r-2 m)+1)-r}{4 (r (r-2 m))^{3/2}}dr \ne \sqrt{{g}_{rr}}{dr}$$

That is the problem with trying to solve equations that don't have a solution. If you try and get some result then when you plug it back into the original equations you don't get what you thought you would. Please go ahead and do the differentiation yourself to confirm that I have done it correctly.

 

[Anamitra]

Unfortunately you get stuck even there. For the transforms you posted you do not get these expressions back. For example differentiating the R to r coordinate transform equation gives

$${R}{=}{\sqrt{{r}^{2}{-}{2mr}}}{+}{2m}{ln}{[}\sqrt{r}{+}\sqrt{(}{r}{-}{2m}{)}{]}{+}{C}$$
$$dR=\frac{2 m \left(\frac{dr}{2 \sqrt{r-2 m}}+\frac{dr}{2 \sqrt{r}}\right)}{\sqrt{r-2 m}+\sqrt{r}}-\frac{2 r /, dr -2 m/, dr}{8 \left(r^2-2 m r\right)^{3/2}}$$
$$dR=\frac{m (4 r (r-2 m)+1)-r}{4 (r (r-2 m))^{3/2}}dr \ne \sqrt{{g}_{rr}}{dr}$$

That is the problem with trying to solve equations that don't have a solution. If you try and get some result then when you plug it back into the original equations you don't get what you thought you would. Please go ahead and do the differentiation yourself to confirm that I have done it correctly.

For the purpose of evaluating the metric coefficients we have to take convenient paths.
Paths have to be chosen with an aim to get the metric coefficients.I have stated such paths[Path 1,Path 2, Path 3and Path 4] in Posting #132.

The path r=kt, theta=cost and phi=const
[k=const] will not be convenient for such a purpose. We may need such a path for a specific problem--- but it will not be useful for the purpose of evaluating the values of g(mu,nu) at each point.
You must remember that our aim in posting 132 was to get the metric coefficients[so that we may evaluate the christoffel symbols and the curvature tensor ]at different points of the manifold and we are definitely at liberty to choose convenient paths for our work.

 

[Dale]

For the purpose of evaluating the metric coefficients we have to take convenient paths.

This is a lie as we have discussed over and over and over and over. See posts 117, 120, 124, and 129.

I use the word lie deliberately. A lie is something that you say which is false and which you know to be false. You have been corrected sufficiently in this thread that you are no longer mistakenly stating a falsehood. This is now deliberate, so it is a lie.

 

[JDoolin]

It appears that at least in some sense, Anamitra and I are in agreement:

Take any point A(t,r,theta,phi) in the manifold[it may be Schwarzschild's Geometry]
From this point take four distinct paths:
Path 1: r=const, theta=constant,phi=constant.
Path 2: t=const,theta=const, phi=const
Path 3: t=const,r=const,phi=const
Path 4: t=const,r=const,theta=const

That's essentially what I did here:

First, a definition of their differentials:

$$\begin{matrix} dR \equiv ds|_{d\theta=d\phi=dt=0}=\left ( \frac{1}{\sqrt{1-\frac{G M}{r c^2}}} \right )dr\\ d\tau \equiv ds|_{d\theta=d\phi=dr=0}=\left ( \sqrt{1-\frac{G M}{r c^2}} \right )dr\\ r d\theta' \equiv ds|_{dt=d\phi=dr=0} =r d\theta\\ r sin(\theta)d\phi' \equiv ds|_{d\theta=dt=dr=0} =r sin(\theta)d\phi \\ \end{matrix}$$

Since $d\phi'=d\phi\; \mathrm{and}\; d\theta'=d\theta$, I use them interchangably.

The metric is not path dependent. Please re-read my rebuttal that you quoted.

Hmmm. I can't imagine what you mean here, DaleSpam. Doesn't the metric essentially define a path integral along a path in spacetime?

$$s=\int ds=\int \sqrt{g_{00}dx_0^2 + g_{11}dx_1^2 + g_{22}dx_2^2+g_{33}dx_3^2}$$

How can a path integral not be path dependent?

 

[Anamitra]

This is a lie as we have discussed over and over and over and over. See posts 117, 120, 124, and 129.

I use the word lie deliberately. A lie is something that you say which is false and which you know to be false. You have been corrected sufficiently in this thread that you are no longer mistakenly stating a falsehood. This is now deliberate, so it is a lie.

The metric coefficients are point functions[ functions of the space-time points]. Suppose we choose two points A and B[close to each other].We may always access B from A along different paths and integrate ds.This will simply give different lengths along different routes and these lengths in general will involve all the metric coefficients.

But when you move along a particular axis only one metric coefficient get involved in the evaluation of ds---thats an advantage

The metric coefficients are definitely point functions. That does not contradict my method.
I have no reason to tell you a lie.

 

[Dale]

Hmmm. I can't imagine what you mean here, DaleSpam. Doesn't the metric essentially define a path integral along a path in spacetime?

$$s=\int ds=\int \sqrt{g_{00}dx_0^2 + g_{11}dx_1^2 + g_{22}dx_2^2+g_{33}dx_3^2}$$

How can a path integral not be path dependent?

The metric is a rank 2 tensor, a path integral is a scalar. Please see posts 117, 120, 124, and 129. I am not about to rehash it all with you too.

 

[Dale]

The metric coefficients are point functions[ functions of the space-time points].

And therefore do not depend on any path.

There is nothing ambiguous or confusing here. You can use the metric to calculate the length along a certain path, but you can also use it for many other things. The length along a path is clearly path dependent, but the metric is not. You and JDoolin are confusing two separate things, and I don't understand how either of you can continue to make such obvious mistakes when they have been pointed out clearly and repeatedly.

A path dependent scalar (a path integral) cannot possibly be the same as a path independent rank two tensor (the metric). Stop equating the two and saying lies.