Transformation of the Line-Element

[Dale]

Yes, you said that above. What do you mean by that? Please clarify, not repeat. Do you mean the coordinates in the new basis, or are you referring to the underlying geometrical object?

 

[Anamitra]

Yes, you said that above. What do you mean by that? Please clarify, not repeat. Do you mean the coordinates in the new basis, or are you referring to the underlying geometrical object?

I am not considering ds^2 as a tensor.
ds^2 is defined by the relation

$${ds}^{2}{=}{g}_{\mu\nu}{dx}^{\mu}{dx}^{\nu}$$
Now dx(i) are being transformed by the usual rules but the metric chosen is a new one such that it cannot be obtained from (1) by transformations where ds^2 is preserved

You may have a different interpretation of ds^2 and a new formulation.That is a different issue----there is no reason to mix up the two issues.

 

[Anamitra]

We may consider the following metric in the rectangular x,y,x,t system:

$${ds}^{2}{=}{dt}^{2}{-}{dx}^{2}{-}{dy}^{2}{-}{dz}^{2}$$ ---------------- (1)

The same metric in the r,theta ,phi,t reads

$${ds}^{2}{=}{dt}^{2}{-}{dr}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{sin}^{2}{\theta}{d}{\phi}^{2}{)}$$ --------------- (2)
ds^2 has not changed

Now in the same r,theta,phi system we may consider the metric:
$${ds'}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{sin}^{2}{\theta}{d}{\phi}^{2}{)}$$
--------------- (3)
ds^ 2 has now changed for identical values of dt,dr,d(theta) and d(phi)
The relationship between x,y,z,t and r,theta,phi,t is not changing when we are considering (1) and (2) or (1) and (3)

Each term on the RHS of (3) has a scale factor[=1 in some cases] wrt to each corresponding terms in (2)

 

[Dale]

I am not considering ds^2 as a tensor.
ds^2 is defined by the relation

$${ds}^{2}{=}{g}_{\mu\nu}{dx}^{\mu}{dx}^{\nu}$$

OK, thank you for the clarification, that is very helpful. Perhaps for clarity in this thread we can use ds² for the line element, a scalar, and g for the metric, a rank 2 tensor. If you will avoid calling ds² the metric then I will avoid interpreting it as a rank 2 tensor.

Now dx(i) are being transformed by the usual rules but the metric chosen is a new one such that it cannot be obtained from (1) by transformations where ds^2 is preserved

This is incorrect. The coordinate transformations listed are valid diffeomorphisms. ds² is unchanged under any diffeomorphism.

You may have a different interpretation of ds^2 and a new formulation.That is a different issue----there is no reason to mix up the two issues.

It is not my interpretation, it is the standard interpretation, see the Carroll notes. However, I am willing to use your non-standard interpretation in this thread as long as you do so consistently and do not call ds² the metric.

 

[Dale]

Now in the same r,theta,phi system we may consider the metric:
$${ds'}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{sin}^{2}{\theta}{d}{\phi}^{2}{)}$$
--------------- (3)
ds^ 2 has now changed for identical values of dt,dr,d(theta) and d(phi)

As discussed above ds² is not a metric, it is a scalar, and I assume that ds'² is also a scalar and not a rank 2 tensor (and therefore not a metric). However, the scalar ds'² is not equal to ds² in any coordinate system, they are unrelated scalars that have nothing to do with each other. You do not get ds'² from ds² by a change in coordinates.

Simply calling one quantity A and another quantity A' does not mean that they are in any way related to each other.

 

[Anamitra]

Suppose we have spherical body with its associated Schwarzschild's Geometry.Equation (3) in post #73 is valid and ds^2 is given by it.

Now due to some catastrophic effect the entire mass gets annihilated and the energy flies off to infinity. The present situation is described by the second or the first equation of the same posting. We have a new value of ds^2 on the same coordinate grid[t,r,theta,phi, system]

Thus a physical process can cause a change in the metric coefficients in the same coordinate grid leading to a change of the value ds^2.

In the present context our transformation relates to the change of the metric coefficients wrt the same coordinate grid. This may be due to the redistribution of mass or due to annihilation.

 

[Dale]

I have no problem with that, but note that is a physical change, not merely a coordinate transformation. You have a new metric, a new manifold, and as I said above ds² is completely unrelated to ds'².

You can do physics in either manifold, but there is no way to transform the results of physics in one to get answers in the other since they are not related by a coordinate transformation. In particular, ds'² may tell you how fast a clock is ticking in the new manifold but it cannot tell you about gravitational time dilation in the previous manifold.

 

[Anamitra]

to understand let us consider a time dependent metric :

$${ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}$$ ----------- (1)

The coefficients, g(mu,nu) are time dependent ones.We assume that the metric coefficients remain time independent for one hour. Then they change for the next 10 minutes and again they become constant for the next one hour.[The next change may be of a different type]The process goes on repeating.
The same metric is taking care of different stationary manifolds --each manifold corresponding to the period when the metric coefficients are not changing. Several different manifolds are coming under the fold of the same metric
[The ds^2 values of the different stationary manifolds are related to each other by the general nature of the time dependent metric ,that is the overall metric given by (1)]

 

[Anamitra]

We may replace the variable time in the metric coefficients[of equation (1) in the last posting] by some variable lambda which is not time.This would give us an ensemble picture.[Of course dt^2 has to be maintained as before].
Different values of lambda would give us different stationary manifolds---having different values of ds^2

 

[Dale]

Sure. But since your metric is asymmetric wrt t or lambda you cannot take results from one t or lambda and use them for doing physics at another t or lambda. It is just not the same physical situation.

Similarly you cannot use results at one r to make predictions about another r in the Schwarzschild metric. E.g. you cannot say, a hovering observer at infinity feels no proper acceleration therefore a hovering observer at the event horizon feels no proper acceleration. The reason you cannot say this is the same as above, a lack of symmetry, in the radial direction instead of in time or lambda.

 

[Anamitra]

E.g. you cannot say, a hovering observer at infinity feels no proper acceleration therefore a hovering observer at the event horizon feels no proper acceleration. The reason you cannot say this is the same as above, a lack of symmetry, in the radial direction instead of in time or lambda.

The dependence of proper acceleration on the radial distance [in general on the spacetime coordinates]will be our law.

 

[Dale]

Yes, and in these other non-static spacetimes you describe it will depend on time as well.

 

[Anamitra]

A Thought Experiment to Perform
We are on initially flat spacetime and we undertake to perform experiments on geodesic motion, Maxwell's equations and other equations that may be expressed in tensor form.Then gravity is turned on and the field is increased in stages. Well the equations we were considering do remain unchanged so far as they are expressed in tensor form.
We also perform some other experiments simultaneously.
While in flat space-time we consider the length of an arrow lying on the ground. This arrow may represent a vector.We stick labels on its tip and tail.When the field is turned on the physical distance between the labels should change due to change in the values of g(mu,nu) .
One may consider a cuboidal box[on the ground] with the main diagonal representing a vector.
When the field[gravitational] is turned on the physical lengths of the components change.The norm as well as the orientation of the vector changes[in short, ds^2 changes].This is something similar to what is supposed to happen in projection geometry.

Actually the laws remain unchanged so long we are able to express them in tensor form.

Now let us think in the reverse direction.We are in curved space-time and the field is turned off gradually in stages. Finally we reach flat spacetime. The laws[in tensor form] remain unchanged at each and every stage of the process/experiment

Suppose there was a body at a distance when we were in curved spacetime. We consider a process where the field is reduced. In flat spacetime we can evaluate the velocity and then we may think of scaling up the velocity in regard of curved spacetime to evaluate the velocity of a body at a distance in curved spacetime.

 

[Dale]

Actually the laws remain unchanged so long we are able to express them in tensor form.

Certainly. Do you understand the difference between a law of physics and the boundary conditions?

Changing coordinates does not change the boundary conditions, changing the curvature does change the boundary conditions.

 

[Ben Niehoff]

Suppose there was a body at a distance when we were in curved spacetime. We consider a process where the field is reduced. In flat spacetime we can evaluate the velocity and then we may think of scaling up the velocity in regard of curved spacetime to evaluate the velocity of a body at a distance in curved spacetime.

Here you've given a vague description of some kind of procedure. To understand your procedure better, you should do an explicit calculation on a specific example. I want you to do the following example, because it will reveal the error in your reasoning:

The metric of an ellipsoid is

$$\begin{align*}ds^2 &= \Big( (a^2 \cos^2 \phi + b^2 \sin^2 \phi) \; \cos^2 \theta + c^2 \sin^2 \theta \Big) \; d\theta^2 + (a^2 \sin^2 \phi + b^2 \cos^2 \phi) \; \sin^2 \theta \; d\phi^2 \\ &\qquad \phantom{a} + 2 (b - a) \cos \theta \sin \theta \cos \phi \sin \phi \; d\theta \; d\phi \end{align*}$$

where $a, b, c$ are some constants. Now, Alice is sitting at $\theta = \pi/4, \; \phi = 5\pi/6$ and has local velocity vector $2 \partial_\theta - \partial_\phi$ at that point. Bob is sitting at $\theta = \pi/2, \; \phi = \pi/4$ and has local velocity vector $2 \partial_\phi$ at that point.

What is the difference between their vectors? Remember that the vectors are at different points on the ellipsoid! Do the calculation explicitly, using the method you have described. If you do not attempt to do your calculation explicitly, I will assume you are not interested in learning why your reasoning is wrong.

Alternatively, if you do realize what is wrong with your reasoning, then explain.

 

[Anamitra]

So far as the metric in the last posting[#85] is concerned, proper speeds are defined terms--d(phi)/ds and d(theta)/ds.
Now for a null geodesic ds^2=0
How does the observer define the coordinate speed or physical speed of light unless one of the variables theta or phi is the time coordinate?This has not been clearly specified
The speed of light is independent of the motion of its source---What does he understand by the speed of light here unless one of the coordinates in the metric represents time.?
Ben Niehoff should clearly specify the time coordinate.

 

[Anamitra]

General Covariance/Diffeomprphism covariance:
[Link: http://en.wikipedia.org/wiki/General_covariance ]
In theoretical physics, general covariance (also known as diffeomorphism covariance or general invariance) is the invariance of the form of physical laws under arbitrary differentiable coordinate transformations.

It should be clear that the coordinate transformations involved may of projection type. One has to include transformations wherer ds^2 change, where this quantity is no longer an invariant[this has been indicated in post #83: The fundamental laws should have the same form in all manifolds---flat spacetime and the different varieties of distinct curved spacetimes]. Interestingly for such transformations, as we pass between different manifolds, g(mu,nu) is not a tensor of rank two[since ds^2 is not preserved].Norms ,angles and dot products need not be preserved in these transformations[where ds^ changes].

 

[Ben Niehoff]

So far as the metric in the last posting[#85] is concerned, proper speeds are defined terms--d(phi)/ds and d(theta)/ds.
Now for a null geodesic ds^2=0
How does the observer define the coordinate speed or physical speed of light unless one of the variables theta or phi is the time coordinate?This has not been clearly specified
The speed of light is independent of the motion of its source---What does he understand by the speed of light here unless one of the coordinates in the metric represents time.?
Ben Niehoff should clearly specify the time coordinate.

There is no time coordinate. Erase the word "velocity" if it makes you feel better.

You claimed that you could define a canonical way to compare vectors at different points of a general curved manifold. So I am asking you to demonstrate your method on this manifold. Call Alice's vector A, and Bob's vector B. Is there any sensible way to define the vector $C = A - B$? If so, what is it? If not, why not? Explain, show work, etc.

I chose an ellipsoid because it has no continuous symmetries. Hopefully that will get you to think about it properly.

 

[Dale]

In theoretical physics, general covariance (also known as diffeomorphism covariance or general invariance) is the invariance of the form of physical laws under arbitrary differentiable coordinate transformations.

It should be clear that the coordinate transformations involved may of projection type.

No, it should be clear that only coordinate transformations are allowed, not projections. Projections change the manifold, not just the coordinates. If you were talking about simple coordinate transformations then you would be able to provide those coordinate transformations explicitly, which you have never done.

I think it is time for you to do some homework problems:
1) the ellipsoid problem
2) write the coordinate transformations to "flatten" the Schwarzschild metric

 

[Anamitra]

No, it should be clear that only coordinate transformations are allowed, not projections. Projections change the manifold, not just the coordinates.

1.The geodesic equations remain invariant in all manifolds
2. Maxwell's equations preserve their form in all manifolds--they remain invariant in form on transition from one manifold to another.

When we use the term "invariant" we have in our view certain transformations.

What are these transformations in the above two cases?
Would you like to call them transformations where ds^2 is preserved?

 

[Dale]

1.The geodesic equations remain invariant in all manifolds
2. Maxwell's equations preserve their form in all manifolds--they remain invariant in form on transition from one manifold to another.

This is true, but not relevant to the definition of diffeomorphism covariance. Read the definition you cited above carefully. It says that diffeomorphism covariance means that the form of the laws are invariant under coordinate transformations. It doesn't say that all transformations where the form of the laws are unchanged are diffeomorphisms.

Would you like to call them transformations where ds^2 is preserved?

Diffeomorphisms, or coordinate transformations.

 

[PAllen]

1.The geodesic equations remain invariant in all manifolds
2. Maxwell's equations preserve their form in all manifolds--they remain invariant in form on transition from one manifold to another.

When we use the term "invariant" we have in our view certain transformations.

What are these transformations in the above two cases?
Would you like to call them transformations where ds^2 is preserved?

Consider an arbitrary smooth bijection (one - one, invertible, mapping) on a manifold. To talk about geometric objects after transformation, you need to ask how they are transformed by the given mapping. If you define that:

1) scalar function of mapped point equals scalar function of original point
2) Vectors, covectors, and tensors transform according the the standard rules (including the metric tensor, of course)

then the manifold is unchanged, geometrically, from the original. The mapping thus describes a coordinate transformation. If you follow some different rules for transforming vectors, tensors, etc. they you get a different manifold with arbitrarily different geometry. This is what you are doing. It has nothing to do with the generality of the point to point mapping, it has to do with how geometric objects are transformed.

A non-coordinate transform such as you are describing can, for example, change a manifold describing the geometry of a single star and one planet into one describing the geometry of 4 stars with 3 planets each. It is obviously true that since Einstein Field Equations can be valid on any manifold, you have the same laws of physics in both. However, no computation you make in the 4 star universe will tell you anything about what to expect in the 1 star universe. For GR, there is no value to such a non-coordinate transformation.

 

[WannabeNewton]

Would you like to call them transformations where ds^2 is preserved?

Remember that if you have a diffeomorphism $\phi :M \mapsto N$ then $\phi _{*}\mathbf{T}$ can be expressed as $T^{\mu ...}_{\nu ...} = \frac{\partial x^{\mu }}{\partial x^{\alpha }}...\frac{\partial x^{\beta }}{\partial x^{\nu }}...T^{\alpha ...}_{\beta ...}$. A map from from one manifold to another that doesn't preserve the structure of the original manifold doesn't need to keep $ds^2$ the same and arbitrary tensor components won't transform according to such a law.

 

[JDoolin]

No, it should be clear that only coordinate transformations are allowed, not projections. Projections change the manifold, not just the coordinates. If you were talking about simple coordinate transformations then you would be able to provide those coordinate transformations explicitly, which you have never done.

I think it is time for you to do some homework problems:
1) the ellipsoid problem
2) write the coordinate transformations to "flatten" the Schwarzschild metric

Consider the Schwarzschild metric:

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

where

$$r_s= \frac{2 G M}{c^2}$$

(2) The r, θ, and Φ coordinates of the Schwarzschild metric already indicate that those components are definable from an external reference frame comoving with the gravitational mass. Those coordinates can be easily translated into Cartesian coordinates in the external comoving reference frame.

Doesn't it stand to reason that the standard definition of the Schwarzschild metric describe exactly what you are saying, i.e., how to "flatten" the Schwarzschild metric?

 

[WannabeNewton]

You can, for example, find a transformation that takes the $t = t_{0}$, $\theta = \pi / 2$ submanifold, of the 4 - manifold normally representing schwarzschild space - time, with the metric $ds^{2} = (1 - \frac{2M}{r})dr^{2} + r^{2}d\phi ^{2}$ that embeds this 2 - manifold in a flat 3 - space with metric $ds^{2} = dr^{2} + r^{2}d\phi ^{2} + dz^{2}$ but this is NOT a coordinate transformation. If you take the embedding map and evaluate the pushforward on, for example, a vector you will see that the vector's components will NOT transform according to the vector transformation law under coordinate transformations. You cannot find a COORDINATE transformation for the schwazrschild metric that makes $R^{\alpha }_{\beta \mu \nu } = 0$ identically.

 

[Dale]

(2) The r, θ, and Φ coordinates of the Schwarzschild metric already indicate that those components are definable from an external reference frame comoving with the gravitational mass. Those coordinates can be easily translated into Cartesian coordinates in the external comoving reference frame.

Doesn't it stand to reason that the standard definition of the Schwarzschild metric describe exactly what you are saying, i.e., how to "flatten" the Schwarzschild metric?

I don't understand your point. The standard definition of the Schwarzschild metric is not flat.

 

[JDoolin]

I don't understand your point. The standard definition of the Schwarzschild metric is not flat.

I'm not entirely sure what you mean either. I am referring to

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

as the standard definition of the Schwarzschild metric.

I would think that if I can take the coordinates (t, r, θ, Φ) and map them to a cartesian coordinate system plus time (t,x,y,z) then I have effectively mapped the events into a flat space-time. Actually, I guess there is a two-step process, because you have to go from $$(\tau, r', \theta, \phi) \rightarrow (t,r,\theta,\phi)\rightarrow (t,x,y,z)$$
by means of inverse Schwartzchild metric and then by use of spherical coordinates.

The point is that by the definition of the Schwarzschild metric, and its use of spherical coordinates, all of the events $(\tau, r', \theta, \phi)$, (outside the schwarszchild radius) are embedded in a flat spacetime, and can be uniquely identified by values of (t,x,y,z).

You can, for example, find a transformation that takes the $t = t_{0}$, $\theta = \pi / 2$ submanifold, of the 4 - manifold normally representing schwarzschild space - time, with the metric $ds^{2} = (1 - \frac{2M}{r})dr^{2} + r^{2}d\phi ^{2}$ that embeds this 2 - manifold in a flat 3 - space with metric $ds^{2} = dr^{2} + r^{2}d\phi ^{2} + dz^{2}$ but this is NOT a coordinate transformation. If you take the embedding map and evaluate the pushforward on, for example, a vector you will see that the vector's components will NOT transform according to the vector transformation law under coordinate transformations. You cannot find a COORDINATE transformation for the schwazrschild metric that makes $R^{\alpha }_{\beta \mu \nu } = 0$ identically.

Why would I want to set $\theta = \pi / 2$? What is this requirement of $R^{\alpha }_{\beta \mu \nu } = 0$? What does that mean, qualitatively, and why is it important?

Are you saying that I can't embed $(\tau, r', \theta, \phi)$ into (t,x,y,z); that somehow there is no way to do it uniquely? I sincerely doubt that's true. You merely need to define a reference t=τ=0 event. Well, okay, I see a little bit of a problem since, for symmetry, you'd want to define that event at r=0; which is troublesome because that's under the event horizon--But that wouldn't matter to the person looking from the external frame.) Wouldn't you at least agree that every event outside the Schwarzschild radius can be mapped uniquely to an event (t,x,y,z) in cartesian coordinates?

 

[PAllen]

I'm not entirely sure what you mean either. I am referring to

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

as the standard definition of the Schwarzschild metric.

I would think that if I can take the coordinates (t, r, θ, Φ) and map them to a cartesian coordinate system plus time (t,x,y,z) then I have effectively mapped the events into a flat space-time. Actually, I guess there is a two-step process, because you have to go from $$(\tau, r', \theta, \phi) \rightarrow (t,r,\theta,\phi)\rightarrow (t,x,y,z)$$
by means of inverse Schwartzchild metric and then by use of spherical coordinates.

The point is that by the definition of the Schwarzschild metric, and its use of spherical coordinates, all of the events $(\tau, r', \theta, \phi)$, (outside the schwarszchild radius) are embedded in a flat spacetime, and can be uniquely identified by values of (t,x,y,z).




Why would I want to set $\theta = \pi / 2$? What is this requirement of $R^{\alpha }_{\beta \mu \nu } = 0$? What does that mean, qualitatively, and why is it important?

Are you saying that I can't embed $(\tau, r', \theta, \phi)$ into (t,x,y,z); that somehow there is no way to do it uniquely? I sincerely doubt that's true. You merely need to define a reference t=τ=0 event. Well, okay, I see a little bit of a problem since, for symmetry, you'd want to define that event at r=0; which is troublesome because that's under the event horizon--But that wouldn't matter to the person looking from the external frame.) Wouldn't you at least agree that every event outside the Schwarzschild radius can be mapped uniquely to an event (t,x,y,z) in cartesian coordinates?

You can map to x,y,z,t any way you want. However, given any such mapping, you transform the metric and vectors/covectors according to the standard formula. The result is the that ds for the x,y,z,t will not be cartesion (flat) for any possible mapping following these rules.

Put another way, using a hemisphere as an example, the rules of vector/tensor transformation ensure that your mapping constitutes a relabeling of points on the hemisphere. A mapping not accompanied by covariant/contraviariant transform of the metric allows for stretching the hemisphere such that it is flat. This is not a coordinate transform; it allows you the change the geometry (curvature) of the manifold. Dalespam's mention of the curvature tensor is just that this is the object that defines generalized curvature in n dimensions. For a valid coordinate transform, if it is non-zero for one coordinate system, it is non-zero for all. If you allow stretching rather then coordinate transform, then it can be made to vanish by geometry change.

 

[Ben Niehoff]

Here is another variation of the Schwarzschild metric in "isotropic coordinates":

$$ds^2 = - \frac{\big(1 - \frac{m}{2R} \big)^2}{\big(1 + \frac{m}{2R} \big)^2} \; dt^2 + \big(1 + \frac{m}{2R} \big)^4 \; (dx^2 + dy^2 + dz^2)$$
$$R = \sqrt{x^2 + y^2 + z^2}$$

This is the closest you can get to "cartesian coordinates" for the Schwarzschild geometry. This metric is not flat. The relation between the original Schwarzschild coordinates $(t, r, \theta, \phi)$ and the new coordinates $(t, x, y, z)$ is given by

$$\begin{align} r &= R \big(1 + \frac{m}{2R} \big)^2 \\ x &= R \sin \theta \cos \phi \\ y &= R \sin \theta \sin \phi \\ z &= R \cos \theta \end{align}$$

In general, however, you can't tell whether a metric is flat just by staring at it. In some cases it's obvious, such as Cartesian coordinates, but other cases might not be obvious. The only way to tell if a metric is flat is by computing its Riemann tensor. If (and only if) the Riemann tensor vanishes, then the metric is flat. The coordinates have no intrinsic meaning.

 

[Dale]

I'm not entirely sure what you mean either. I am referring to

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

as the standard definition of the Schwarzschild metric.

I would think that if I can take the coordinates (t, r, θ, Φ) and map them to a cartesian coordinate system plus time (t,x,y,z) then I have effectively mapped the events into a flat space-time. Actually, I guess there is a two-step process, because you have to go from $$(\tau, r', \theta, \phi) \rightarrow (t,r,\theta,\phi)\rightarrow (t,x,y,z)$$
by means of inverse Schwartzchild metric and then by use of spherical coordinates.

Along the lines of the previous responses, you can certainly do a coordinate transform from (t,r,θ,Φ) to (t,x,y,z) but the metric expressed in the new coordinates will not be the usual flat spacetime metric regardless of the details of the transform.

Similarly, you can take the usual flat metric (t,x,y,z) and do a coordinate transform to (t,r,θ,Φ) but the metric expressed in the new coordinates will still be flat. It may not be obvious that it is flat, but if you compute the curvature tensor it will be zero.

 

[Anamitra]

In all the cases you have been referring to, the line element ds^2 has between taken as invariant. These are not projection transformations where the line element itself changes.
You consider a hemispherical surface and the flat surface it is resting on.The line element corresponding to the hemispherical surface represents curved space while the line element corresponding to the surface the hemisphere is resting on represents flat space. There is a correspondence between the two through the projection transformation------These is a one to one correspondence between each point on the hemispherical surface and the flat space it is resting on.

A practical example in relation to gravity has been provided in post # 83

 

[JDoolin]

Here is another variation of the Schwarzschild metric in "isotropic coordinates":

$$ds^2 = - \frac{\big(1 - \frac{m}{2R} \big)^2}{\big(1 + \frac{m}{2R} \big)^2} \; dt^2 + \big(1 + \frac{m}{2R} \big)^4 \; (dx^2 + dy^2 + dz^2)$$
$$R = \sqrt{x^2 + y^2 + z^2}$$

This is the closest you can get to "cartesian coordinates" for the Schwarzschild geometry. This metric is not flat. The relation between the original Schwarzschild coordinates $(t, r, \theta, \phi)$ and the new coordinates $(t, x, y, z)$ is given by

$$\begin{align} r &= R \big(1 + \frac{m}{2R} \big)^2 \\ x &= R \sin \theta \cos \phi \\ y &= R \sin \theta \sin \phi \\ z &= R \cos \theta \end{align}$$

In general, however, you can't tell whether a metric is flat just by staring at it. In some cases it's obvious, such as Cartesian coordinates, but other cases might not be obvious. The only way to tell if a metric is flat is by computing its Riemann tensor. If (and only if) the Riemann tensor vanishes, then the metric is flat. The coordinates have no intrinsic meaning.

Please don't argue your case based on the claim that "whatever I'm saying has no intrinsic meaning." That's nonsense.

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.

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. 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.

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?

(*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?)

 

[JDoolin]

Along the lines of the previous responses, you can certainly do a coordinate transform from (t,r,θ,Φ) to (t,x,y,z) but the metric expressed in the new coordinates will not be the usual flat spacetime metric regardless of the details of the transform.

Similarly, you can take the usual flat metric (t,x,y,z) and do a coordinate transform to (t,r,θ,Φ) but the metric expressed in the new coordinates will still be flat. It may not be obvious that it is flat, but if you compute the curvature tensor it will be zero.

I think you skipped a step. First do a coordinate transform from (t',r',θ,Φ) to (t,r,θ,Φ). Then do the transform from (t,r,θ,Φ) to (t,x,y,z). The two together create a mapping of the curved four-dimensional coordinates to a flat four-dimensional coordinates.

 

[Dale]

I think you skipped a step. First do a coordinate transform from (t',r',θ,Φ) to (t,r,θ,Φ). Then do the transform from (t,r,θ,Φ) to (t,x,y,z).

If there exists a diffeomorphism from coordinate system A to B and from B to C then there exists one directly from A to C. I wasn't suggesting skipping a step, but just using the direct transform for brevity.

The two together create a mapping of the curved four-dimensional coordinates to a flat four-dimensional coordinates.

No, there is nothing that you can do in two diffeomorphisms that cannot be done in one.

 

[WannabeNewton]

Please don't argue your case based on the claim that "whatever I'm saying has no intrinsic meaning." That's nonsense.

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?