Transformation of the Line-Element

[Anamitra]

It is important to observe that some parallel issues are going on in this thread----and all of them are interesting.

1. The issue of projecting one curved surface [manifold] on another,particularly on a flat surface. Here the entire surface is in consideration.The value of ds^2[line element squared] changes.
ex: posting #1 ,Posting #2...

An interesting type of such a projection would be the consideration of two manifolds on the same coordinate grid.[The other issues are very much alive]
[#73,# 83,#111]


2. Projecting a specific path from one manifold to another. Here we may not consider the projection of the entire surface. Projection in the vicinity/neighborhood of the curve is sufficient. "ds^2" that is the line element squared changes.
ex:postings #126,#132

Issues 1 and 2 have a strong correlation---they are allied issues.If two manifolds stand on the same coordinate grid--we are in effect projecting one manifold on the other. Any curve in one manifold will have its projection on the other.



[In posting #111 some square root signs were left out inadvertently. This was mentioned at the end of #126]

[It is also important to note that I have used ds^2 to mean the square of the line element.i would continue with this meaning.If I use it in a different way it will be specified separately.]

The next issue,that is the third one belongs to an entirely different category

3. Passing from one manifold to another keeping "ds^2" invariant. "Flattening of a sphere" belongs in this category.Rather one may define it in this way.

I have been currently trying my hands with this [the third issue]with the posting #147, so far as the current thread is concerned.[The other issues are very much alive]

 

[Anamitra]

The flattening of a sphere could become somewhat easier in view of the fact that each one of the coefficients of the metric tensor may not be functions of all the four variables involved.This can be an advantageous feature.We may take for example Schwarzschild's Geometry.

 

[JDoolin]

Again, it would be instructive for you to compute, and post here, the Schwarzschild metric in these new coordinates. I gave the solution to your first integral in my previous post, if you want to use it:

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

I've done this integral numerically, and your solution appears to be incorrect. For instance, when I set m=1/2, it gives a negative value at r=1.1.

Here is what I calculated at some specific radii. 1.1, 1.01, 1.001, 1.0001, 1.0001

If I could compute a closed-form solution for the integral, that would be excellent, but I don't think you have it.

Some Typos have been corrected from the original posting

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

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

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

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

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

 

[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<br /> \sqrt{r}}\right)}{\sqrt{r-2 m}+\sqrt{r}}-\frac{2 r \, dr -2 m\, dr}{8<br /> \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.

I am getting the value of dR as follows:

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

{dR}{=}\frac{1}{2}\frac{2r-2m}{\sqrt{{r}^{2}{-}{2m}}}{dr}{+}{2m}\frac{1}{\sqrt{r}{+}\sqrt{r-2m}}{[}\frac{1}{{2}\sqrt{r-2m}}{+}\frac{1}{{2}\sqrt{r}}{]}{dr}
{=}{[}\frac{r-m}{\sqrt{{r}^{2}{-}{2mr}}}{+}\frac{m}{\sqrt{{r}{(}{r}{-}{2mr}{)}}}{]}{dr}
{=}\frac{r}{\sqrt{{(}{r}^{2}{-}{2mr}{)}}}{dr}
{=}{(}{1}{-}\frac{2m}{r}{)}^{-1/2}{dr}
{=}\sqrt{{g}_{rr}}{dr}

 

[Anamitra]

OK, so now we are back to post 126? If so, please see my previous response.

For a Schwarzschild sphere of radius r=k and an ordinary sphere radius=r=k the line element on each sphere is the same in length.[t=constant for both the spheres;ie time slices are being considered]

In each case the line element is given by:

{ds}^{2}{=}{r}^{2}{(}{d}{\theta}^{2}{+}{sin}^{2}{(}{\theta}{)}{d}{\phi}^{2}{)}

In each case dr=0 and dt =0

The physical value of the radius of the Schwarzschild sphere is given by R in post #147

 

[Anamitra]

Thus in the last posting[post #155] we have flattened a sphere[r=constant ,t=constant].
One to one mapping from a Schwarzschild Sphere[ a sphere in curved space] to an ordinary sphere[a sphere in flat spacetime] exists for which the line element does not change in value.

 

[JDoolin]

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}

Ah. Interesting. You split the integral into two solvable integrals.

The one on the left, you are doing substitution of

u=r^2-2 m r , du = 2 r - 2 m

So you simplify to
\int \frac{1}{2} u^{-1/2} du = \frac{1}{1/2}\frac{1}{2}u^{1/2} = (r^2- 2 m r)^{1/2}

I am not seeing how you got the other integral, though. [strike]And the final answer doesn't seem to square with the values I got by numerical integration.[/strike] (Edit: My mistake, both the numerical integration values, and this equation are correct.)

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 only thing I would mention here is that by assigning a constant radial velocity, r=k t, you are trying to create a "coordinate system" full of objects that are moving in different directions, (radially inward or outward) which I'm pretty sure doesn't make sense. (or at least introduces a lot of unnecessary confusion)

When you set r= constant, that simplifies your calculation a lot, and it then represents a coordinate system defined by the positions of stationary clocks; both in terms of ∂R/∂τ = 0 and ∂r/∂t=0, if I'm not mistaken.

 

[Anamitra]

Ah. Interesting. You split the integral into two solvable integrals.

The one on the left, you are doing substitution of

u=r^2-2 m r , du = 2 r - 2 m

So you simplify to
\int \frac{1}{2} u^{-1/2} du = \frac{1}{1/2}\frac{1}{2}u^{1/2} = (r^2- 2 m r)^{1/2}

I am not seeing how you got the other integral, though. And the final answer doesn't seem to square with the values I got by numerical integration.

Its quite simple. We can do it in this way:
\int\frac {dr}{\sqrt{{r}^{2}{-}{2mr}}}
{=}\int\frac {dr}{\sqrt{{r}{(}{r}{-}{2m}{)}}}

Let

{r}{=}{z}^{2}
{dr}{=}{2z}{dz}

Therefore,
{Integral}{=}\int\frac{{2z}{dz}}{\sqrt{{z}^{2}{(}{z}^{2}{-}{2m}{)}}}
{=}\int\frac{2dz}{\sqrt{{(}{z}^{2}{-}{2m}{)}}}
{=}{2}{\,}{\,}{ln}{(}{z}{+}\sqrt{{z}^{2}{-}{2m}}{)}
{=}{2}{\,}{ln}{(}\sqrt{r}{+}\sqrt{{r}{-}{2m}}{)}

 

[JDoolin]

\tau \; \; \overset{?}{=} \; \int_{t_0} \sqrt{1-\frac{2G M}{r c^2}} \; dt
does not give a unique answer, because the value of the integral depends on the path. This is because the quantity being integrated is not an exact differential. This is why I said earlier that \tau is not a good coordinate.

Ah, now this is different. Here you have specified a path along which to do the previous integral. You have chosen the path r = \mathrm{const}. That's fine, I just want you to realize that you had to make that choice, and you could have made it another way.

Again, it would be instructive for you to compute, and post here, the Schwarzschild metric in these new coordinates. I gave the solution to your first integral in my previous post, if you want to use it:

\int_{2m}^r \Big( 1 - \frac{2m}{r&#039;} \Big)^{-1/2} dr&#039; = \sqrt{r ( r - 2m)} + 2m \; \cosh^{-1} \sqrt{\frac{r}{2m}}

The correct term for this is a stationary observer. Note that not every spacetime has stationary observers! It turns out that the existence of a stationary observer is intimately connected to the existence of a timelike Killing vector (this is a fancy way of saying the spacetime has time translation symmetry). When there is a timelike Killing vector, we can call an observer stationary if his worldline is everywhere tangent to the timelike Killing vector.

I'm not sure of all of the implications of this, but it seems to me, the way I've defined things, ∂τ/∂t is a function of r, while ∂τ/∂r= is a function of r and t.

More explicitly, I may have to set up a fairly complicated synchronization convention to get all the clocks set to internal coordinate time τ=0 at the external coordinate time t=0. But before that instant, and after that instant, the clocks aren't synchronized. (unless they are at the same radius) The clocks further from the center will be going faster than the clocks nearer the center.

So as time passes, the value of ∂τ/∂r at any given region will shrink and shrink before the synchronization convention, and will grow an grow afterward. This is no big deal. Any event is still uniquely located by (R,τ,θ,Φ) just as it is uniquely located by (r,t,θ,Φ). Each clock only passes through each time only once. No clocks are caused to go backward in time, and repeat an event over again.

I'm not an expert on Killing vectors, but I'm not sure you'd have translation symmetry in time. If you move 10 seconds into the future or the past, the clocks below you would lose time, and the clocks above you would gain time. So you could easily calculate how long until, or how long since the synchronization convention, based on looking at nearby clocks. But the relative rate between them would still be the same--so there wouldn't be any difference in the physics.

 

[Dale]

I am getting the value of dR as follows: ...
{=}\sqrt{{g}_{rr}}{dr}

You are correct, I must have made a typo entering the equation in yesterday .

I will continue working with the rest of the transformation later today. It will not change the end result that the coordinate transformation cannot flatten a curved metric.

 

[JDoolin]

Its quite simple. We can do it in this way:

Excellent. Thank you. Now I realized that my spreadsheet defaults to Log base 10, rather than natural log. And I have verified that your equatin gives the same results as the numerical integration.

 

[PAllen]

Thus in the last posting[post #155] we have flattened a sphere[r=constant ,t=constant].
One to one mapping from a Schwarzschild Sphere[ a sphere in curved space] to an ordinary sphere[a sphere in flat spacetime] exists for which the line element does not change in value.

This may actually be true, but not very significant. Two surfaces of many geometries and topologies are embeddable in 3-space with the induced metric on them capturing geometry of a highly non-euclidean 2-manifold. I suspect that almost all two manifolds are embeddable in flat euclidean 4-space (certainly many that can't be embedded in flat 3-space can be embedded in flat 4-space). Thus the idea that pure 2-sphere (as a geometric as well topologic manifold) can be embedded in both Scwharzschild manifold and flat euclidean 4-space is a triviality with no significance the nature of mappings between the two 4-manifolds.

You haven't flattened anything. You've just established a well known feature of embedding: a flat space can embed lower dimensional curved manifolds; a curved manifold of one geometry can embed lower dimensional manifolds of all different geometries.

 

[JDoolin]

You haven't flattened anything. You've just established a well known feature of embedding: a flat space can embed lower dimensional curved manifolds; a curved manifold of one geometry can embed lower dimensional manifolds of all different geometries.

I don't think that's Anamitra's intention, or at least not mine, anyway. The point is that a coordinate system defined by stationary clocks in the schwarzschild coordinates can be mapped unambiguously to a coordinate system of stationary clocks in a cartesian coordinate system; event for event, a 1 to 1 relationship.

This is an embedding, or, if you prefer, a projection from one 4D space-time to another 4D spacetime, which could easily be co-located; for instance, simply by having the stationary clocks in the region each keep track of two different times.

 

[Anamitra]

Another Example of Flattening
a Curved Space-time Object
[-----The r-Slices "Flattened"]

We start with the Schwarzschild 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}{)} --------------- (1)
For Constant r we have:

{ds}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{Sin}^{2}{(}{\theta}{)}{d}{\phi}^{2}{)} --------------- (2)

For the surface r=Constant,we use the substitution:

{dT}{=}{(}{1}{-}\frac{2m}{r}{)}^{1/2}{dt}

{T}{=}\int{(}{1}{-}\frac{2m}{r}{)}^{1/2}{dt}
the limits of integration extending from to t0 to t. This eliminates the constant of integration.
{T}{=}{(}{1}{-}\frac{2m}{r}{)}{(}{t-t0}{)}For r=Const we have,the metric has the form:
{ds}^{2}{=}{dT}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{Sin}^{2}{(}{\theta}{)}{d}{\phi}^{2}{)} ----------- (3)

This is again a flat Spacetime metric[the above one--relation (3)]. "ds" remains invariant in the transformation used.
This time we have flattened a three dimensional surface[holding r as constant leaves us three independent variables--t,theta and phi]

[We have not changed theta and phi in our transformation]

 

[Dale]

a coordinate system defined by stationary clocks in the schwarzschild coordinates can be mapped unambiguously to a coordinate system of stationary clocks in a cartesian coordinate system; event for event, a 1 to 1 relationship.

Sure. That is what any coordinate chart does. They all perform an unambiguous 1 to 1 mapping from some open subset of the manifold to some open subset of R(N). That is part of the definition of a coordinate system, and is not in doubt.

The question is whether you can flatten the metric by a coordinate transform, which you cannot. And which despite lots of posts Anamitra has never demonstrated. In fact, Anamitra has never even bothered to actually take one of his transforms and calculate the curvature tensor, always simply assuming flatness.

 

[Dale]

Hi Anamitra,

For:
T=c+\frac{\sqrt{k t (k t-2 m)}}{k}-\frac{2 m \log \left(\sqrt{t-\frac{2<br /> m}{k}}+\sqrt{t}\right)}{k}
and
r=k t
I am getting
\text{dT}=\frac{\frac{\text{dr} \left(4 m^2 t+m \left(2 r \sqrt{t} \sqrt{t-\frac{2 m<br /> t}{r}}-2 t \sqrt{r (r-2 m)}-2 r t\right)\right)+2 \text{dt} r (2 m-r) \left(\sqrt{r<br /> (r-2 m)}-m\right)}{2 m-r}+4 m (\text{dr} t-\text{dt} r) \log \left(\sqrt{t-\frac{2<br /> m t}{r}}+\sqrt{t}\right)}{2 r^2}

 

[Anamitra]

If the screen could be normalized to the correct width...This is a request.

Hi Anamitra,

For:
T=c+\frac{\sqrt{k t (k t-2 m)}}{k}-\frac{2 m \log \left(\sqrt{t-\frac{2<br /> m}{k}}+\sqrt{t}\right)}{k}
and
r=k t
I am getting ...

Lets do it:

{Integral}{=}{T}{=}\frac{1}{k}{\sqrt{{kt}{(}{kt}{-}{2m}{)}}}{-}\frac{2m}{k}{ln}{[}\sqrt{t}{+}\sqrt{(}{t}{-}{2m}{/}{k}{)}{]}{+}{C’}

Or,

{dT}{=}\frac{1}{k}\frac{1}{2}\frac{{2}{k}^{2}{t}{-}{2mk}}{\sqrt{{kt}{(}{kt}{-}{2m}{)}}}{dt}{-}\frac{2m}{k}\frac{\frac{1}{{2}\sqrt{t}}{+}\frac{1}{\sqrt{{t}{-}{2m}{/}{k}}}}{\sqrt{t}{+}\sqrt{{t}{-}{2m}{/}{k}}}{dt}

{dT}{=}\frac{kt-m}{\sqrt{{kt}{(}{kt}{-}{2m}}}{-}\frac{m}{k}\frac{1}{\sqrt{{t}{(}{t}{-}\frac{2m}{k}{)}}}

{dT}{=}\frac{kt-m}{\sqrt{{kt}{(}{kt}{-}{2m}}}{-}\frac{m}{\sqrt{{kt}{(}{kt}{-}{2m}{)}}}

We have,

{dT}{=}\frac{kt-2m}{\sqrt{{kt}{-}{2m}}}

{dT}{=}\sqrt{\frac{kt-2m}{kt}}

But r=kt

Therefore we have,finally,

{dT}{=}\sqrt{\frac{r-2m}{r}}{dt}

Or,
{dT}{=}{(}{1}{-}\frac{2m}{r}{)}^{1/2}{dt}

 

[Ben Niehoff]

\int_{2m}^r \Big( 1 - \frac{2m}{r&#039;} \Big)^{-1/2} dr&#039; = \sqrt{r ( r - 2m)} + 2m \; \cosh^{-1} \sqrt{\frac{r}{2m}}

I've done this integral numerically, and your solution appears to be incorrect. For instance, when I set m=1/2, it gives a negative value at r=1.1.

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

 

[Dale]

But r=kt

Therefore
dk = \frac{\text{dr} t-\text{dt} r}{t^2} \ne 0

 

[Ben Niehoff]

I'm not sure of all of the implications of this, but it seems to me, the way I've defined things, ∂τ/∂t is a function of r, while ∂τ/∂r= is a function of r and t.

More explicitly, I may have to set up a fairly complicated synchronization convention to get all the clocks set to internal coordinate time τ=0 at the external coordinate time t=0. But before that instant, and after that instant, the clocks aren't synchronized. (unless they are at the same radius) The clocks further from the center will be going faster than the clocks nearer the center.

So as time passes, the value of ∂τ/∂r at any given region will shrink and shrink before the synchronization convention, and will grow an grow afterward. This is no big deal. Any event is still uniquely located by (R,τ,θ,Φ) just as it is uniquely located by (r,t,θ,Φ). Each clock only passes through each time only once. No clocks are caused to go backward in time, and repeat an event over again.

I'm not sure if you read me very carefully. I agree with you that once you make a definite choice of path for the \tau integral, you end up with a valid coordinate system. In fact, the path r = \mathrm{const} is rather convenient, because then the \tau coordinate you get represents the elapsed proper time of stationary observers. Look up LeMaitre coordinates; he does something similar where he chooses the proper time of stationary observers as his time coordinate.

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'm not an expert on Killing vectors, but I'm not sure you'd have translation symmetry in time. If you move 10 seconds into the future or the past, the clocks below you would lose time, and the clocks above you would gain time. So you could easily calculate how long until, or how long since the synchronization convention, based on looking at nearby clocks. But the relative rate between them would still be the same--so there wouldn't be any difference in the physics.

If you don't understand the significance of Killing vectors, haven't computed a curvature tensor, and don't know how to solve "fancy differential equations" to obtain the easiest GR solutions (Schwarzschild and Reissner-Nordstrom), then I would say you have some woodshedding to do.

Please continue to study Carroll; it's an excellent book, and it will surely clear up all these misunderstandings.

 

[Anamitra]

Therefore
dk = \frac{\text{dr} t-\text{dt} r}{t^2} \ne 0

K is a constant--that was mentioned in the relevant posting

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]

You could view it in this manner also[remembering k=constant]:

r=kt
dr=kdt
tdr-rdt=tkdt-ktdt=0

 

[Dale]

K is a constant--...

r=kt

Since this is a coordinate transform you can't have it both ways. Either k is a constant or r=kt. Pick one, they contradict each other. I honestly don't care which you pick.

 

[Anamitra]

Since this is a coordinate transform you can't have it both ways. Either k is a constant or r=kt. Pick one, they contradict each other. I honestly don't care which you pick.

In Post #126 we were looking for a coordinate transformation along the path stated below[in relation to Schwarzschild's Geometry]:
r=kt; k: constant
theta=constant
phi=constant

There is absolutely no contradiction in having k as constant in the path r=kt

 

[Dale]

There is absolutely no contradiction in having k as constant in the path r=kt

Yes, there is. See the bottom figure on page 36 and the top figure on page 37 of the Carroll notes with the accompanying text.

A coordinate system is a one-to-one invertible map from open subsets of the manifold to open subsets of R(n). The set defined by r=kt is not an open subset for a constant k. Therefore, if you allow both k constant and r=kt then you do not have a coordinate transformation. This was already mentioned by Dr. Greg in 141, and emphasized by me in 144.

So as I said before, since this is a coordinate transform you cannot have it both ways, either k is a constant or r=kt. You must pick one because they contradict each other.

 

[Anamitra]

Yes, there is. See the bottom figure on page 36 and the top figure on page 37 of the Carroll notes with the accompanying text.

A coordinate system is a one-to-one invertible map from open subsets of the manifold to open subsets of R(n). The set defined by r=kt is not an open subset for a constant k. Therefore, if you allow both k constant and r=kt then you do not have a coordinate transformation. This was already mentioned by Dr. Greg in 141, and emphasized by me in 144.

So as I said before, since this is a coordinate transform you cannot have it both ways, either k is a constant or r=kt. You must pick one because they contradict each other.

The basic issue here is to have open subsets on the line in question[and the corresponding map should comprise open subsets]:The line has to be represented as the union of open subsets.

r=kt; where k is constant
theta=const
phi=const

Now we may think of small portions of the above mentioned line which are open subsets.These portions [small segments into which we will divide the line:These segments may overlap]should not include their end points[the limit points]

We simply express the the line as the union of open subsets.

In fact any open subset should not contain its limit points. The line I have defined falls in this category.

If an open subset is 3D region then the surface containing its limit points is two-dimensional
If an open subset is 2D region then the set containing its limit points is one-dimensional,that is a curve.
If an open subset is 1D region then the set containing its limit should contain two distinct points.

 

[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).