Transformation of the Line-Element

[Dale]

Now we may think of small portions of the above mentioned line which are open subsets.

No portion of the line is an open set in R(n). Please read the section I pointed you to in the Carroll notes. Pay attention carefully to the definition of an open set in R(n). An open set is constructed from a union of open balls. You cannot construct a line or a portion of a line from a union of open balls, therefore a line is not an open set in R(n).

 

[Anamitra]

You cannot construct a line or a portion of a line from a union of open balls, therefore a line is not an open set in R(n).

The highlighted portion is an incorrect statement--too remote from any rational interpretation

Regarding the open sphere:
Let X be a metric space with a metric d[the phrase metric d has been used in the sense of a line element here]. If x0 is a point is a point of X and r real positive number,then the open sphere[or open ball] S(r,x0) with cenre x0 and radius r is defined by:
S(r,x0)={x:d(x,x0)<r}
d(x,x0) : is the distance between the points x and x0.[x0 is an arbitrary point catering to the difference]From Carroll[page 36]:
Start with the notion of an open ball, which is the set of all points x in Rn such that
|x − y| < r for some fixed y ∈ Rn and r ∈ R, where |x − y| = [Summation(xi − yi)^2]^1/2. Note that
this is a strict inequality — the open ball is the interior of an n-sphere of radius r centered
at y.

x and y do not relate to the coordinate axis in this definition.They are simply points on R(n). In the case of the line I have considered both x and y belong to R1.

For R1 the open sphere is simply a straight line without its limit points[boundary points]--this is well with in Carroll's definition[it conforms to Carroll's definition] or any other relevant definition to be found in the texts.

[If one tries to construct a one to one map from a SPACE-CURVE[especially if it is a closed one] to R1 it is not possible.But we may take open subsets[open spheres to to be precise] and construct a map from each open subset on the curved line to R1. The concept of manifolds applies here--in the usual way. And I have tried to go against "this usual way" by the flattening of spheres " calculations/transformations and by considering open curves]
[For a straight line "i" has only one value since a straight line is one dimensional.["i" referred to here is in the formula in Carroll's definition]

 

[Dale]

x and y do not relate to the coordinate axis in this definition.They are simply points on R(n). In the case of the line I have considered both x and y belong to R1.

n is the dimensionality of the manifold. I.e. For spacetime n=4. You cannot construct a line out of open balls in R4.

Of course, you can embed a lower-dimensional manifold in a higher dimensional one and make coordinates in the lower dimensional manifold. However, in that case then the number of coordinates would be equal to the dimensionality of your lower-dimensional manifold. So, if you want to embed the line you are describing then you would get a single coordinate in R1, not 2 or 4. I have been under the assumption that we were interested in spacetime and therefore in 4-dimensional manifolds.

 

[Anamitra]

n is the dimensionality of the manifold. I.e. For spacetime n=4. You cannot construct a line out of open balls in R4.

Of course, you can embed a lower-dimensional manifold in a higher dimensional one and make coordinates in the lower dimensional manifold. However, in that case then the number of coordinates would be equal to the dimensionality of your lower-dimensional manifold. So, if you want to embed the line you are describing then you would get a single coordinate in R1, not 2 or 4. I have been under the assumption that we were interested in spacetime and therefore in 4-dimensional manifolds.

We can always parametrize a curve in four dimensions[or in any dimensions for that matter]
So it is a function of one variable.For any motion what we have in our minds is a path.So we have mapped a space curve to R1 to make the situation advantageous for us.

Now in the theory of manifolds we need to have open subsets for the purpose of creating local charts.These subsets can be of finite magnitude. At least from the theory we don't need to have infinitesimally small subsets.When we pass from the relevant area manifold to R(n) do we make sure that the line element squared has to be kept invariant in these transformations,keeping in mind that we may pass from a finite area of the manifold to the tangent plane[whose largeness/smallness is not restricted by theory]?

[Instead of taking tiny subsets on the manifold we can always take the largest possible ones for creating local charts]

 

[Dale]

We can always parametrize a curve in four dimensions[or in any dimensions for that matter]
So it is a function of one variable.

Sure, but that does not make the parameter a coordinate of the higher-dimensional manifold. This is not my idea, this is standard Riemannian geometry. You are simply not defining a coordinate transformation in a 4D manifold if you are talking about a line.

Now in the theory of manifolds we need to have open subsets for the purpose of creating local charts.These subsets can be of finite magnitude.

Yes, and a line segment is not an open subset of R(4) because it cannot be constructed as a union of open balls in R(4).

 

[Anamitra]

A coordinate map, a coordinate chart, or simply a chart, of a manifold is an invertible map between a subset of the manifold and a simple space such that both the map and its inverse preserve the desired structure. For a topological manifold, the simple space is some Euclidean space Rn and interest focuses on the topological structure. This structure is preserved by homeomorphisms, invertible maps that are continuous in both directions.
The above portion is from a Wikipedia Link:
http://en.wikipedia.org/wiki/Manifold#Charts

If the simple space[highlighted] in the above definition is our tangent plane--Euclidean space Rn---the line element is not being preserved in the mapping.

Coordinate charts are created by Projection Transformations and not by Coordinate transformations
[These coordinate charts may not be confined to infinitesimal regions]

 

[JDoolin]

But r=kt

Anamitra, you have to be very careful to say something like this. That statement defines not just a worldline, but an infinite number of worldlines, all of them going in different radial directions. Actually, what you're describing is inertial (nonaccelerating) particles coming out of an explosion. With the schwarzschild metric, something like this equation might be useful to consider if you want to model a supernova explosion, except in that situation, k would not be a constant (because of acceleration).

Your definition for d\tau using this should be :

$$d\tau \equiv ds|_{d\theta=d\phi, dr=kdt}=\left ( \sqrt{1-\frac{G M}{r(t) c^2}} \right )dt$$

Contrast that with my definition:

First, a definition of their differentials:

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

(Edit: fixed from original post so second line is an integral with respect to dt instead of dr.)

My definition of \tau (dr=0) is different from Anamitra's (dr=k dt). My definition represents the time on stationary clocks in the gravitational field. Anamitra's definition represents the time on clocks that somehow manage to maintain a constant velocity in the gravitational field.

My point was that you could have chosen some other path, such as $r = kt$, and it would give you an entirely different definition of the new coordinate $\tau$. So it would be ambiguous to say

$$\tau = \int \sqrt{1-\frac{2m}{r}} \; dt$$
because no path has been specified. And it would be an outright lie to say

$$d\tau = \sqrt{1-\frac{2m}{r}} \; dt$$
because the right-hand side is not a total differential.

I could have chosen some other path, but I didn't. Notice my use of "≡" in my definition of d\tau. Under these circumstances, there is no ambiguity, but I can see how you would say that it is ambiguous in general, if you thought I meant r(t) were an arbitrary path.

Finally:

Something is wrong with your numerics. I double-checked my antiderivative in Mathematica. See the attached image of the output.

This is interesting. I can't figure out right exactly what the problem is. One possibility is since acosh is not a true function, that my calculator gave me the wrong part of the curve, or I flubbed it, entering it. If you plug in m=1/2, and r=1.1, into your formula do you get a negative value?

In any case Anamitra's equation from post 126 gave the same result as the numerical integration:

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

Edit: more info:


$$y=\cosh(\theta)=\frac{e^\theta + e^{-\theta}}{2}= \frac{e^{2\theta }+ 1}{2e^\theta}$$

reduces to a quadratic equation

$$(e^\theta)^2-2e^\theta y + 1 = 0$$

with solution

$$e^\theta=y\pm\sqrt{y^2-1}$$

So $$\theta = arccosh(y)=ln(y\pm \sqrt{y^2-1})$$

if that helps.

 

[Dale]

A coordinate map, a coordinate chart, or simply a chart, of a manifold is an invertible map between a subset of the manifold

Careful. An open subset of the manifold. r=kt defines a closed subset of the Schwarzschild space-time.

 

[Anamitra]

Anamitra, you have to be very careful to say something like this. That statement defines not just a worldline, but an infinite number of worldlines, all of them going in different radial directions. Actually, what you're describing is inertial (nonaccelerating) particles coming out of an explosion. With the schwarzschild metric, something like this equation might be useful to consider if you want to model a supernova explosion, except in that situation, k would not be a constant (because of acceleration).

Your definition for d\tau using this should be :

$$d\tau \equiv ds|_{d\theta=d\phi, dr=kdt}=\left ( \sqrt{1-\frac{G M}{r(t) c^2}} \right )dt$$

Actually in my definition I have said :

r=kt; where r=constant
theta= constant;
phi=constant;

The integration s requested in posting # 124Let us consider a path:
Theta=const ,phi =const; r=kt where k is a constant
[An arbitrary path may not correspond to a geodesic]

That selects a single radial line instead of several ones.

 

[Anamitra]

Careful. An open subset of the manifold. r=kt defines a closed subset of the Schwarzschild space-time.

You may consider r=kt on the open interval(t0,t) for the purpose of investigating the motion .

The motion physically may extend over the closed interval [t0,t] but we may select the open interval (t0,t) for the purpose of investigating the motion.

[If one considers a set of open subsets, each subset is open and closed at the same time.A subset is said to be closed if its complement is open. For each subset making up the manifold the complement[which is again a collection of open subsets ] is open.Such a subset is open and closed at the same time. ]

 

[Dale]

It is still not an open subset of R(4) even so. Do you understand the concept of open sets in topological spaces?

 

[Anamitra]

It is still not an open subset of R(4) even so. Do you understand the concept of open sets in topological spaces?

In the particular problem concerned[#126] I am not interested in the entire manifold.
In a problem the entire manifold may not be a matter of concern to us--for instance radial motion.

Again for investigating planetary motion we might be interested in a particular plane and not the full manifold[this is of course a different problem]

The basic interest here is to make as much simplification as we can on the space relevant to the problem.

 

[Anamitra]

In general the motion of a particle may be will be along a complicated spacetime curve on a 4D manifold[the spatial part may be a curve in a 3D region] . Such a curve may be characterized by a single parameter.The idea is to stretch this curve on R[1] treating it[ie, the curve] as a 1D manifold.

 

[Dale]

In the particular problem concerned[#126] I am not interested in the entire manifold.

I am not insisting that your coordinate chart cover the entire manifold, only that it be a valid coordinate chart.

 

[Anamitra]

I am not insisting that your coordinate chart cover the entire manifold, only that it be a valid coordinate chart.

You mean to say that we cannot or should not have a coordinate chart on a 1D manifold[ex: a 1D curve in lying in a 4D manifold] even if it is of interest to us?

[ I am treating the mentioned 1D curve as my manifold]

 

[Dale]

You can certainly have a coordinate chart on lower dimensional manifold which is embedded in a higher dimensional manifold. However:
1) the coordinate chart on the embedded manifold has a number of coordinates equal to the dimensionality of the embedded manifold, not the original manifold
2) the coordinate system of the embedded manifold is not a coordinate system for the original manifold
3) the curvature of the embedded manifold is not the same as the curvature of the original manifold

 

[Anamitra]

You can certainly have a coordinate chart on lower dimensional manifold which is embedded in a higher dimensional manifold. However:
1) the coordinate chart on the embedded manifold has a number of coordinates equal to the dimensionality of the embedded manifold, not the original manifold
2) the coordinate system of the embedded manifold is not a coordinate system for the original manifold
3) the curvature of the embedded manifold is not the same as the curvature of the original manifold

I am interested in the path concerned for example a radial path.The calculations in #126 relate to such an issue.

 

[Anamitra]

It is essential to remind the audience that there are several parallel issues in this thread[ and all of them are important]

The issues are:
1. Concentrating on the path of the particle instead of the entire manifold

2. Establishing one-to-one correspondence between a manifold and the corresponding Euclidean plane [of the same dimension].We may have mathematical theorems that do not allow the projection of the entire manifold on the corresponding R[n]. But is there any restriction that we cannot do it in the partial sense covering large non-local areas? In case there is a restriction we are basically doing projection transformations instead of coordinate transformations [as indicated in post #181] in the formation of local charts[ which may not be of microscopic nature]

3. Projection issues.

 

[Dale]

I am interested in the path concerned for example a radial path.

I am not. 1D manifolds are not very interesting. E.g. they can never be curved.

The calculations in #126 relate to such an issue.

No they didn't. The calculations in #126 had two coordinates, so they relate to a 2D manifold. Even in that 2D manifold you cannot have both k a constant and r=kt.

If you want to work on a 1D manifold you can only use one coordinate. If you are using 2 coordinates then you are using a 2D manifold, your coordinates must be open sets in R(2), and you cannot have both k constant and r=kt.

 

[Anamitra]

The calculations in #126 had two coordinates, so they relate to a 2D manifold. Even in that 2D manifold you cannot have both k a constant and r=kt.

If you want to work on a 1D manifold you can only use one coordinate. If you are using 2 coordinates then you are using a 2D manifold, your coordinates must be open sets in R(2), and you cannot have both k constant and r=kt.

There is only one variable "t"--only one coordinate.
Both R and T can be represented in terms of t as you may see from the portion quoted below:

The integration s requested in posting # 124Let 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 the expression for R the we may write r=kt on the RHS.

 

[Anamitra]

The basic idea behind the formation of local coordinate charts is "Projection Transformation"
This is an accepted principle in GR

[The word local is not synonymous with the word microscopic or infinitesimally small]

 

[Ben Niehoff]

I could have chosen some other path, but I didn't. Notice my use of "≡" in my definition of d\tau. Under these circumstances, there is no ambiguity, but I can see how you would say that it is ambiguous in general, if you thought I meant r(t) were an arbitrary path.

I see no point in arguing this any further. It doesn't matter whether you write "equals" or "defined as". The line is still false.

Do continue to read Carroll. Good luck.

 

[Anamitra]

I see no point in arguing this any further. It doesn't matter whether you write "equals" or "defined as". The line is still false.

Do continue to read Carroll. Good luck.

HELLO!
Whose words are you referring to in your quotation in #197?

The statements in the quotation[in #197]cannot be identified!

 

[JDoolin]

HELLO!
Whose words are you referring to in your quotation in #197.

The statements in the quotation[in #197]cannot be identified!

= means "equals"
≡ means "defined as"

I think Ben's trying to say I'm not allowed to define the variable $\tau$ and claim it is a coordinate. I guess we're done arguing, but it is a perfectly valid coordinate, according to the definition of coordinate I quoted earlier from wikipedia.

If you change your mind, Ben, there are several issues you never addressed in my post 133, but I fully understand time constraints; as school is starting.

 

[Anamitra]

Now I understand that Ben Niehoff had quoted Jdoolin in #197 and wished him good-luck.
It was a foggy thing out there[in #197] for quite sometime--was the software misbehaving?
[It did not report the name of the person whose words were there in the quotation in #197]

 

[JDoolin]

Actually in my definition I have said :

r=kt; where r=constant
theta= constant;
phi=constant;


That selects a single radial line instead of several ones.

It's overspecified, I'm afraid. If you let t change, and say r is constant, then either k=0, or k=r/t. If you don't let t change, then you're only talking about a point; not a radial line.

 

[Anamitra]

It's overspecified, I'm afraid. If you let t change, and say r is constant, then either k=0, or k=r/t. If you don't let t change, then you're only talking about a point; not a radial line.

If "t" changes "r" will change----it is not supposed to stay fixed in such a situation.

[Incidentally I had written k = constant in post #126. The quotation is out there in post#195
Please do take a careful note of it[ and a very quick note of it] in case software damages it

 

[Anamitra]

The integration s requested in posting # 124


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.]

The posting #126 has been requoted

 

[DrGreg]

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

= means "equals"
≡ means "defined as"

I think Ben's trying to say I'm not allowed to define the variable $\tau$ and claim it is a coordinate. I guess we're done arguing, but it is a perfectly valid coordinate, according to the definition of coordinate I quoted earlier from wikipedia.

The problem is that your "definition" works only if you restrict yourself to a single curve of constant $\theta, \, \phi, \, r$. If you try to "glue together" lots of curves to fill a 4D space, you have to "glue the coordinates" in a way that is compatible with the metric. Essentially you have to decide how $\tau$ behaves when you move orthogonally (e.g.) to the curves specified. For such arbitrary movement you cannot use$$d\tau = \frac{d\tau}{dt}\, dt$$you have to use$$d\tau = \frac{\partial \tau}{\partial t}\, dt + \frac{\partial \tau}{\partial r}\, dr + \frac{\partial \tau}{\partial \phi}\, d\phi + \frac{\partial \tau}{\partial \theta}\, d\theta$$That's where your method falls over because it's only preserving the metric of each one-dimensional manifold and fails to preserve the metric of the 4D manifold.

The argument that you and Anamitra have used tries to glue the coordinates using$$\frac{\partial \tau}{\partial r}= 0 \, \, \left( = \frac{\partial \tau}{\partial \phi} = \frac{\partial \tau}{\partial \theta} \right)$$but that condition is incompatible with your solution for $\tau(t,r)$.

 

[Anamitra]

The problem is that your "definition" works only if you restrict yourself to a single curve of constant $\theta, \, \phi, \, r$. If you try to "glue together" lots of curves to fill a 4D space, you have to "glue the coordinates" in a way that is compatible with the metric. Essentially you have to decide how $\tau$ behaves when you move orthogonally (e.g.) to the curves specified. For such arbitrary movement you cannot use$$d\tau = \frac{d\tau}{dt}\, dt$$you have to use$$d\tau = \frac{\partial \tau}{\partial t}\, dt + \frac{\partial \tau}{\partial r}\, dr + \frac{\partial \tau}{\partial \phi}\, d\phi + \frac{\partial \tau}{\partial \theta}\, d\theta$$

In the above relation[the last one in the quotation] Dr Greg has assumed right at the outset that d(tau) is a perfect differential in respect of t,r,theta and phi. Thanks for that!
[The line element has been considered as a perfect differential--that is quite interesting]

That's where your method falls over because it's only preserving the metric of each one-dimensional manifold and fails to preserve the metric of the 4D manifold.

The argument that you and Anamitra have used tries to glue the coordinates using$$\frac{\partial \tau}{\partial r}= 0 \, \, \left( = \frac{\partial \tau}{\partial \phi} = \frac{\partial \tau}{\partial \theta} \right)$$but that condition is incompatible with your solution for $\tau(t,r)$.

I have never suggested the gluing of curves anywhere!
Rather I have suggested concentrating on a particular curve relevant to the motion in question.

 

[DrGreg]

I have never suggested the gluing of curves anywhere!
Rather I have suggested concentrating on a particular curve relevant to the motion in question.

But that doesn't tell you anything useful. If you restrict yourself to a one-dimensional manifold, you won't know anything about the 4D-manifold in which it is embedded. E.g. you can't decide if the curve is a geodesic with respect to the the 4D manifold unless you use the coords and metric of the 4D manifold. Within a 1D manifold, every curve is a geodesic, rather trivially.

 

[Anamitra]

Within a 1D manifold, every curve is a geodesic, rather trivially.

This is not correct. You are restricting yourself only the spatial part of the geodesic, completely ignoring the temporal aspect!

 

[Dale]

There is only one variable "t"--only one coordinate.

OK, with that I am done with this thread unless the conversation winds back to interesting topics. IMO, lines are not interesting manifolds since they are always intrinsically flat and so their flatness tells you nothing about any physics nor about the manifold in which it is embedded. I don't see any value in this approach at all.

 

[DrGreg]

This is not correct. You are restricting yourself only the spatial part of the geodesic, completely ignoring the temporal aspect!

If you want to separate space and time as two dimensions, you need a manifold with at least two dimensions and therefore a coordinate system that covers at least a two dimensional region of space time. If you restrict yourself to coordinate system that is defined only along a single spacetime worldline, that's a one-dimensional manifold and you can't define space and time within that manifold except by embedding it in at least two dimensions.

 

[Anamitra]

OK, with that I am done with this thread unless the conversation winds back to interesting topics. IMO, lines are not interesting manifolds since they are always intrinsically flat and so their flatness tells you nothing about any physics nor any manifold in which it is embedded. I don't see any value in this approach at all.

The radial lines[considered in the spacetime perspectivre,the spatial part is radial] mentioned in the problem [#126] are within the manifold. I would request the audience to go through posting #126.
[It is not necessary that paths in a manifold have to be geodesics]

We can have curved lines[spacetime ones] of a complicated nature that lie within the manifold.We can always try to stretch such lines on R1 to the maximum possible extent.