Transformation of the Line-Element

[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<br /> <br /> d\tau = \frac{d\tau}{dt}\, dt<br /> <br />you have to use<br /> <br /> 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<br /> <br />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<br /> <br /> \frac{\partial \tau}{\partial r}= 0 \, \, \left( = \frac{\partial \tau}{\partial \phi} = \frac{\partial \tau}{\partial \theta} \right)<br /> <br />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<br /> <br /> d\tau = \frac{d\tau}{dt}\, dt<br /> <br />you have to use<br /> <br /> 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<br /> <br />

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<br /> <br /> \frac{\partial \tau}{\partial r}= 0 \, \, \left( = \frac{\partial \tau}{\partial \phi} = \frac{\partial \tau}{\partial \theta} \right)<br /> <br />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.

 

[Anamitra]

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.

Its quite different from what has has been claimed in the quoted portion.
Using a single parameter we may represent a curve--a space time one.
Ex: The one discussed in post #126

Thus we may come down to one dimension.The parameter t, takes care of the the geodesic or non geodesic nature of the curve in the spatial and in the temporal sense, simultaneously. The parameter reduces the curve a one dimensional one in R[1] .

The example in #126 is a good example. The curve is not a geodesic if one considers the temporal part.. The curve is given by:
r=kt,where k=Const,theta=const,phi=const.
The final values of R and T can be expressed in terms of t.

Link to Post #126: https://www.physicsforums.com/showpost.php?p=3458428&postcount=126

 

[JDoolin]

Post 204

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<br /> <br /> d\tau = \frac{d\tau}{dt}\, dt<br /> <br />you have to use<br /> <br /> 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<br /> <br />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<br /> <br /> \frac{\partial \tau}{\partial r}= 0 \, \, \left( = \frac{\partial \tau}{\partial \phi} = \frac{\partial \tau}{\partial \theta} \right)<br /> <br />but that condition is incompatible with your solution for \tau(t,r).

I can't speak for Anamitra right now, because he needs to get some sleep, and acknowledge his typo in post 184, and realize that r= k t does not specify an arbitrary path. As for me, I'm not trying to specify an arbitrary path. I am defining the relationship between two geometries by invoking the idea of a set of stationary clocks and rulers.

Please be aware that Anamitra's definition of d\tau and my definition of d\tau are different. It might be wise for one of us to switch variable names, (or, I would prefer, let us both define the family of space-time paths, r=r₀+k t and set k=0, so that he has the same paths I do!)

You raise an interesting question: Have I implicitly, or explicitly claimed that \frac{\partial \tau}{\partial r}= 0? Not that I know of.

Post 157

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.

Note the difference here. ∂R/∂τ = 0 claims that if we only let the time \tau change, the position of a clock does not change; this is true using my setup. ∂τ/∂R=0, on the other hand, claims that if we only let R change, the readings on the clocks do not change. I never claimed this. In fact, I am pretty sure that I pointed out that people observing clocks at different heights would be able to detect differences in the clocks.

Ahhh, here it is in post 159

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.

So, actually, I explicitly pointed out that ∂τ/∂r was a function of time.

 

[Anamitra]

Post 204


I can't speak for Anamitra right now, because he needs to get some sleep, and acknowledge his typo in post 184, and realize that r= k t does not specify an arbitrary path.

There is a typo in #184. But the Quoted portion in the same posting contains correct information[r=kt, where k=const , theta=const,phi=const].
Jdoolin missed it due to some haste,I believe.

Link for Post #184:

https://www.physicsforums.com/showpost.php?p=3462290&postcount=184

 

[JDoolin]

Anamitra, you said this in Post 132:

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

Why don't you use it?

Don't say r= k t; that's a worldline of a particle. (Specifically, a particle launched from the center at t=0)
Say r = r₀; that's the worldline of a coordinate. It's also what you have to do to get path 1.

If you want to map one space-time to another space-time, you need to find how their coordinates map, NOT their arbitrary space-time paths.

 

[Anamitra]

Anamitra, you said this in Post 132:

We can always do it in this way:

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

Why don't you use it?

Don't say r= k t; that's a worldline of a particle. (Specifically, a particle launched from the center at t=0)
Say r = r₀; that's the worldline of a coordinate. It's also what you have to do to get path 1.

If you want to map one space-time to another space-time, you need to find how their coordinates map, NOT their arbitrary space-time paths.

Actually Path 1,Path 2,Path 3 and Path 4 are infinitesimal paths emanating from the same point A. They have been used for the evaluation of the metric coefficients at A in post 132.
[Link for Post 132: https://www.physicsforums.com/showpost.php?p=3459224&postcount=132]

But the path in Post #126 is a spacetime path[World line] of finite length. Spatially it is a radial line. The point A can always lie on such a path and we may evaluate the values of the metric coefficients on the spacetime path in #126 by considering a point like A[ and considering 4 infinitesimal paths as defined in post#132] at each point[spacetime point ]of the line defined in #126[r=kt; k= const,theta= const and phi=constant]

Link to Post #126: https://www.physicsforums.com/showpost.php?p=3458428&postcount=126

 

[JDoolin]

Actually Path 1,Path 2,Path 3 and Path 4 are infinitesimal paths emanating from the same point A. They have been used for the evaluation of the metric coefficients at A in post 132.
[Link for Post 132: https://www.physicsforums.com/showpost.php?p=3459224&postcount=132]

But the path in Post #126 is a spacetime path[World line] of finite length. Spatially it is a radial line. The point A can always lie on such a path and we may evaluate the values of the metric coefficients on the spacetime path in #126 by considering a point like A[ and considering 4 infinitesimal paths as defined in post#132] at each point[spacetime point ]of the line defined in #126[r=kt; k= const,theta= const and phi=constant]



Link to Post #126: https://www.physicsforums.com/showpost.php?p=3458428&postcount=126

I think our problem is mostly notational. I am changing notation here and there, and you are changing notation here and there. You are using the same variables R and T to denote a space-time path, and (R,T), a coordinate. We need to use different variables to represent different things, and agree to what they mean. I see now, you agree that there are coordinates in post 132, and my post 133 came almost the same time with my definition of the coordinates. They are the same definition but different letters.


Post 132:

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:

For convenience, I am re-posting this in full, with equations fixed, outside of quotation.

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
​

-------------------------------------- (End of Anamitra's Post 132)

Please be aware that Anamitra's definition of d\tau and my definition of d\tau are different. It might be wise for one of us to switch variable names, (or, I would prefer, let us both define the family of space-time paths, r=r₀+k t and set k=0, so that he has the same paths I do!)

I would be happy to yield, and use (T,R,P,Q) for the coordinates, since Anamitra has already distinguished the coordinates from the spacetime path, defining (T,R,P,Q) as the coordinates, whereas I was calling them (\tau,R,\theta&#039;,\phi&#039;).

However, Anamitra, I would highly recommend that you use a different variable than R and T to describe the spacetime path of a particle following the path r=kt. The definition of T is different in post 132 and 126.

Post 126:

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.]

In post 132, you use dr=0 in your definition for T. In post 126, I think you use dr=k dt in your definition of T.

We can do it either way you like, but use one set of variables to represent the coordinates as defined in post 132, and another set of variables to represent the space-time path(s) you have in post 126.

Also, finish what you started in post 132. Do the integrals, and define the coordinates (R,T,P,Q) and see if you get something consistent with my definition of (R,τ,θ,Φ) in post 133. (fill in any missing factors of 2, of course.)

 

[JDoolin]

We can do it either way you like, but use one set of variables to represent the coordinates as defined in post 132, and another set of variables to represent the space-time path(s) you have in post 126.

Let me suggest a possible way of naming the coordinates and variables which I hope is simple and clear.

(t,r,\theta,\phi) represent the coordinates of events in the external frame. (The comoving Euclidean coordinates.)

(T(r,t),R(r),P(\theta)=\theta,Q(\phi)=\phi) represent the coordinates in the internal frame. (The Schwarzschild stationary coordinates)

(T_p,R_p,P_p,Q_p) represent an event on the space-time path of a particular particle, p. (Each of these should be a function of t or T.

This is the way I would define the coordinates:

\begin{matrix} R(r)=\int_{2 G M/c^2}^{r}\frac{1}{\sqrt{1-\frac{2 G M}{\rho c^2}}}d\rho \to {\sqrt{{r}^{2}{-}{2mr}}}{+}{2m}{\ln}{[}\sqrt{r}{+}\sqrt{(}{r}{-}{2m}{)}{]}{+}{C}<br /> \\ T(t,r)=\int_{0}^{t}\sqrt{1-\frac{2 G M}{r c^2}}dt = \left (\sqrt{1-\frac{2 G M}{r c^2}} \right )t \\ P(\theta)=\int_{0}^{\theta} d\theta=\theta\\ Q(\phi)=\int_{0}^{\phi} d\phi=\phi \end{matrix}

Anamitra, in post 126, it seems like you are discussing a path (T_p,R_p,P_p,Q_p), whereas in post 132, it seems like you are discussing the coordinates (T,R,P,Q). However, when you are finding R_p (in post 126), it appears that you have a function of r, only. If this is a space-time path, it should be a function of t as well.

The calculation you have done for R_p in post 126 would be correct if you were calculating the coordinate, R.

 

[Dale]

We can always do it in this way:
...
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 and Anamitra, you two are free to discuss coordinates on 1D manifolds, there is nothing wrong with that although I consider it a waste of time.

However, it is not mathematically valid to use those coordinates to construct the metric tensor in a higher dimensional manifold. Any attempt to do so is contrary to mainstream GR and would qualify as overly-speculative.

 

[PAllen]

It's not just speculative, it is mathematically absurd:

1) It is an elementary fact of differential geometry that a 1-manifold has no curvature at all; all 1-D manifolds are 'flat' and geometrically identical. Intuitively, you can see this by noting that any string can be laid along any curve without stretching, compressing, or tearing. This is in contrast to a hemisphere (for example; a curved 2-manifold), which cannot be flattened without stretching. So 2-manifolds are the lowest dimensionality that actually has non-trivial geometry.

2) The geometry of a sequence of n-manifolds embedded in a higher dimensional manifold tells you essentially nothing about the geometry of the higher dimensional manifold. Consider that flat Euclidean 3-space can be filled (except for one point) with a family of concentric 2-spheres. This doesn't change the fact that the 2-spheres are not flat while the 3-space is. Similarly, regions of a curved n-manifold can be filled with a family of flat (n-1)-manifolds; the n-manifold is still curved everywhere, while the embedded manifolds are still flat everywhere.

The whole framework of this thread is to assume counter factual mathematics, leading to nonsense.

 

[Dale]

It's not just speculative, it is mathematically absurd:

Agreed. I just used the word speculative to refer to the forum rules.

 

[Anamitra]

Let us consider the equation:

{ds}^{2}{=}{f1}{\,}{dp}^{2}{+}{f2}{\,}{dq}^{2}
-------------- (1)
[f1,f2 and f3 are well behaved functions in respect of continuity,differentiability etc]

We use the following substitutions:

{p}{=}{F1}{(}{x}{,}{y}{)} ------------ (2)
{q}{=}{F2}{(}{x}{,}{y}{)} ------------- (3)
{dp}{=}{\frac{{\partial}{p}}{{\partial}{x}}}{dx}{+}{\frac{{\partial}{p}}{{\partial}{y}}}{dy} -------- (3)
{dq}{=}{\frac{{\partial}{q}}{{\partial}{x}}}{dx}{+}{\frac{{\partial}{q}}{{\partial}{y}}}{dy} --------- (4){ds}^{2}{=}{[}{f1}{(}\frac{\partial p}{\partial x}{)}^{2}{+}{f2}{(}\frac{\partial q}{\partial x}{)}^{2}{]}{dx}^{2}{+}{[}{f1}{(}\frac{\partial p}{\partial y}{)}^{2}{+}{f2}{(}\frac{\partial q}{\partial y}{)}^{2}{]}{dy}^{2}{+}{2}{[}{f1}\frac{\partial p}{\partial x}\frac{ \partial p}{\partial y}{+}{f2}\frac{\partial q}{\partial x}\frac{\partial q}{\partial y}{]}{dx}{dy} ----------- (5)
In relation to the above workings we consider the following differential equations:
{1}{=}{[}{f1}{(}\frac{{\partial}{p}}{{\partial}{x}}{)}^{2}{+}{f2}{(}\frac{{\partial}{q}}{{\partial}{x}}{)}^{2}{]} ----------- (6)
{1}{=}{[}{f1}{(}\frac{{\partial}{p}}{{\partial}{y}}{)}^{2}{+}{f2}{(}\frac{{\partial}{q}}{{\partial}{y}}{)}^{2}{]} ---------- (7)

{0}{=}{[}{f1}\frac{\partial p}{\partial x}\frac{\partial p}{\partial y}{+}{f2}\frac{\partial q}{\partial x}\frac{\partial q}{\partial y}{]} ----------- (8)

In case these equations yield solutions to suitable boundary conditions, we should be able to "flatten a sphere" in the literal sense.[f1,f2,f3,F1 and,F2 are assumed to be well behaved functions in respect of continuity differentiability etc]
[The Jacobian of the transformation is assumed to be non-singular]

My Question goes to Pallen Dr Greg and Dalespam:
In what situation(s) are the above partial differential equations not supposed to have solutions[boundary conditions being given]?
This question relates to the existence of solutions and not to the methodology of the solving process.

 

[Anamitra]

The situation in the last post has improved considerably.It my request to the audience to consider the inconvenience caused [before editing].

 

[JDoolin]

The situation in the last post has improved considerably.It my request to the audience to consider the inconvenience caused [before editing].

{0}{=}{[}{f1}\frac{{\partial}{p}}{{\partial}{x}}\frac{{\partial}{p}}{{\partial}{y}}{+}{f2}\frac{{\partial}{q}}{{\partial}{x}}\frac{{\partial}{q}}{{\partial}{y}}{]} ----------- (8)

0=f1\frac{\partial p}{\partial x}\frac{\partial p}{\partial y}+{f2}\frac{\partial{q}}{\partial x}\frac{\partial{q}}{\partial y}
----------- (8)

I used notepad search-and-replace to get rid of all the extraneous {} signs. I'm not sure which {} signs are specifically causing the LaTeX compiler to trip up, but getting rid of all the extras seemed to work.

 

[Dale]

My Question goes to Pallen Dr Greg and Dalespam:
In what situation(s) are the above partial differential equations not supposed to have solutions[boundary conditions being given]?
This question relates to the existence of solutions and not to the methodology of the solving process.

You have two unknown functions, F1 and F2, and three first order PDE's that they must satisfy. Such a system is called overdetermined and generally doesn't have a solution.

 

[PAllen]

My Question goes to Pallen Dr Greg and Dalespam:
In what situation(s) are the above partial differential equations not supposed to have solutions[boundary conditions being given]?
This question relates to the existence of solutions and not to the methodology of the solving process.

To paraphrase, you are asking the equivalent of: under what circumstances can I make 1 + 1 = 0 instead of 2? The fact that valid coordinate transforms (including correct transform of tensors) cannot change anything about the intrinsic geometry of the manifold is the core starting point (really, almost more like a definition; that is definitions were constructed to ensure that this is the case) of differential geometry. You need to at least read some elementary intro to differential geometry before you proceed further.

 

[Anamitra]

You have two unknown functions, F1 and F2, and three first order PDE's that they must satisfy. Such a system is called overdetermined and generally doesn't have a solution.

From equations (6) and (7) of Post 221 we have,
Eqn (1) below
\frac{\partial q}{\partial x}{=}\sqrt{\frac{1}{f2}{[}{1}{-}{f1}{(}\frac{\partial p}{\partial x}{)}^{2}{]}}
Eqn (2) below:

\frac{\partial q}{\partial y}{=}\sqrt{\frac{1}{f2}{[}{1}{-}{f1}{(}\frac{\partial p}{\partial y}{)}^{2}{]}}

Using the above two results in eqn (8) of Post #221 and simplifying we have,
Equation (3):
{(}\frac{\partial p}{\partial x}{)}^{2}{+}{(}\frac{\partial p}{\partial y}{)}^{2}{=}\frac{1}{f1}
Squaring eqn (1) and using eqn(3) on the RHS of squared (1) we have,
Eqn (4) below,
{(}\frac{\partial q}{\partial x}{)}^{2}{=}\frac{f1}{f2}{(}\frac{\partial p}{\partial y}{)}^{2}
Squaring eqn (2) and using eqn(3) on the RHS of squared (2) we have[eqn (5) below],
{(}\frac{\partial q}{\partial y}{)}^{2}{=}\frac{f1}{f2}{(}\frac{\partial p}{\partial x}{)}^{2}
Adding (4) and (5) and using (3),we have equation(6) below,
{(}\frac{\partial q}{\partial x}{)}^{2}{+}{(}\frac{\partial q}{\partial y}{)}^{2}{=}\frac{1}{f2}

Finally we have a pair of PDEs[(3) and (6) :
They have been written below again
{(}\frac{\partial p}{\partial x}{)}^{2}{+}{(}\frac{\partial p}{\partial y}{)}^{2}{=}\frac{1}{f1}
{(}\frac{\partial q}{\partial x}{)}^{2}{+}{(}\frac{\partial q}{\partial y}{)}^{2}{=}\frac{1}{f2}

 

[Dale]

Anamitra, I think your best bet is to actually work it out for a couple of concrete examples. I would try polar coordinates on a plane and standard coordinates on a sphere.

 

[Anamitra]

We again consider the three equations:
{1}{=}{[}{f1}{(}\frac{\partial p}{\partial x}{)}^{2}{+}{f2}{(}\frac{\partial q}{\partial x}{)}^{2}{]}--------------------------------- (1)
{1}{=}{[}{f1}{(}\frac{\partial p}{\partial y}{)}^{2}{+}{f2}{(}\frac{\partial q}{\partial y}{)}^{2}{]}------------------------------- (2)
{0}{=}{[}{f1}\frac{\partial p}{\partial x}\frac{\partial p}{\partial y}{+}{f2}\frac{\partial q}{\partial x}\frac{\partial q}{\partial y}{]} ------------- (3)
We choose the following two numbers:
f1/(f1+f2) and f2/(f1+f2)
Their sum is 1.
We write:
{f1}{(}\frac{\partial p}{\partial x}{)}^{2}{=}\frac{f2}{f1+f2}
{f2}{(}\frac{\partial q}{\partial x}{)}^{2}{=}\frac{f1}{f1+f2}
{f1}{(}\frac{\partial p}{\partial y}{)}^{2}{=}\frac{f1}{f1+f2}
{f2}{(}\frac{\partial q}{\partial y}{)}^{2}{=}\frac{f2}{f1+f2}
Equations (1) and (2) are clearly satisfied.
Now we write:
\sqrt{f1}\frac{\partial p}{\partial x}{=}\sqrt{\frac{f2}{f1+f2}}
\sqrt{f2}\frac{\partial q}{\partial x}{=}\sqrt{\frac{f1}{f1+f2}}
\sqrt{f1}\frac{\partial p}{\partial y}{=}\sqrt{\frac{f1}{f1+f2}}
\sqrt{f2}\frac{\partial q}{\partial y}{=}{-}\sqrt{\frac{f2}{f1+f2}}
The above four equations satisfy (1),(2) and (3) simultaneously.

 

[Dale]

Again, I recommend working out a couple of concrete examples. Polar coordinates on a plane, where we know there is a solution, and standard coordinates on a sphere, where we know there is not a solution.

 

[Anamitra]

A Modification over my Previous post[Post #228]

We again consider the three equations:
{1}{=}{[}{f1}{(}\frac{\partial p}{\partial x}{)}^{2}{+}{f2}{(}\frac{\partial q}{\partial x}{)}^{2}{]}--------------------------------- (1)
{1}{=}{[}{f1}{(}\frac{\partial p}{\partial y}{)}^{2}{+}{f2}{(}\frac{\partial q}{\partial y}{)}^{2}{]}------------------------------- (2)
{0}{=}{[}{f1}\frac{\partial p}{\partial x}\frac{\partial p}{\partial y}{+}{f2}\frac{\partial q}{\partial x}\frac{\partial q}{\partial y}{]} ------------- (3)
We choose the following two functions F1 and F2 and consider the expressions:
F1/(F1+F2) and F2/(F1+F2)
Their sum is always 1.
We write:
{f1}{(}\frac{\partial p}{\partial x}{)}^{2}{=}\frac{F2}{F1+F2}
{f2}{(}\frac{\partial q}{\partial x}{)}^{2}{=}\frac{F1}{F1+F2}
{f1}{(}\frac{\partial p}{\partial y}{)}^{2}{=}\frac{F1}{F1+F2}
{f2}{(}\frac{\partial q}{\partial y}{)}^{2}{=}\frac{F2}{F1+F2}
Equations (1) and (2) are clearly satisfied.
Now we write:
\sqrt{f1}\frac{\partial p}{\partial x}{=}\sqrt{\frac{F2}{F1+F2}}
\sqrt{f2}\frac{\partial q}{\partial x}{=}\sqrt{\frac{F1}{F1+F2}}
\sqrt{f1}\frac{\partial p}{\partial y}{=}\sqrt{\frac{F1}{F1+F2}}
\sqrt{f2}\frac{\partial q}{\partial y}{=}{-}\sqrt{\frac{F2}{F1+F2}}
The above four equations satisfy (1),(2) and (3) simultaneously.
It is important that we take care of the following:
\frac{{\partial}^{2}{p}}{\partial x \partial y}{=}\frac{{\partial}^{2}{p}}{\partial y \partial x} ----------- (4)
\frac{{\partial}^{2}{q}}{\partial x \partial y}{=}\frac{{\partial}^{2}{q}}{\partial y \partial x} ----------- (5)

Now the functions F1 and F2 are of our choice and we do have a certain amount of freedom over them.If they cater to the requirements in (4) and (5) the process of integration would be favored.
Though we can exert our freedom of choice over F1 and F2 ,two distinct conditions (4) and (5) have to be satisfied.It might be difficult if F1 and F2 are considered as analytical functions.
But we may think in the direction of numerical integration

[What DaleSpam has advised in the previous posting[#229]---- to work out concrete examples is extremely important-- and I do have that in my mind. I have simply placed here what I have done uptill now.]

 

[Anamitra]

It is important to remember that there are several issues going on in the thread--and "flattening of the sphere" in the literal sense is one of them.I have been trying it out in #228,#230.

We should always be aware of the fact that we may have curved surfaces in flat space-time and flat surfaces in curved spacetime.[The planetary orbits lie on such planes in curved space.]
[We can always have a straight line[spatial] in curved spacetime: The only thing is that such a line,generally speaking, will not be a spatial geodesic.But in some cases straight lines may correspond to spatial geodesics: for ex: radial lines in Schwarzschild's Geometry]

One of our basic interests is to find a one to one correspondence between curved space time and flat space time,preserving the line element[ds^2 invariant]. I have some examples at following posts[#155 and #164]:

https://www.physicsforums.com/showpost.php?p=3460459&postcount=155

https://www.physicsforums.com/showpost.php?p=3460801&postcount=164
In the second one we have "flattened" r= constant slices in Schwarzschild's Geomtery. These are 3D surfaces which we have flattened

An arbitrary curve[world line or spacetime curve] will lie on some 3D surface embedded in 4D space. The same curve may be contained by several surfaces[the curve lying at the intersection of several surfaces] The curve itself seems to be of paramount importance]

 

[WannabeNewton]

One of our basic interests is to find a one to one correspondence between curved space time and flat space time,preserving the line element[ds^2 invariant]. I have some examples at following posts[#155 and #164]:

Your examples don't show that you have a one to one correspondence between curved and flat space. You can find an embedding map that takes a curved manifold into a submanifold of a flat one but an embedding map of such a kind isn't a diffeomorphism so if you have an embedding \phi that doesn't mean \phi ^{-1} exists. If the mapping isn't a diffeomorphism then ds^{2} doesn't have to remain invariant. Also, the metric for a 2 - sphere doesn't describe flat space; there is a non - vanishing component R_{\theta \phi \theta \phi }.

 

[Anamitra]

Further Calculations[in relation to Post #230]:

Metric[Line element]

{ds}^{2}{=}{d}{\theta}^{2}{+}{Sin}^{2}{(}{\theta}{)}{d}{\phi}^{2}
-------------- (1)
Equations:

\frac{\partial \theta}{\partial x}{=}{\psi} -------- (2)
\frac{\partial \theta}{\partial y}{=}{\chi} ------------- (3)
\frac{\partial \phi}{\partial x}{=}\frac{1}{{Sin}{(}{\theta}{)}}{\chi} ----- (4)
\frac{\partial \phi}{\partial y}{=}{-}\frac{1}{{Sin}{(}{\theta}{)}}{\psi} ----- (5)


{\psi}^{2}{+}{\chi}^{2}{=}{1} -------- (6)

Exact Differential Conditions:
1. \frac{\partial \psi}{\partial y}{=}\frac{\partial \chi}{\partial x} ------- (7)
2. \frac{\partial }{\partial y}{(}\frac{1}{Sin{(}{\theta}{)}}{\chi}{)}{=}{-}\frac{\partial }{\partial x}{(}\frac{1}{Sin{(}{\theta}{)}}{\psi}{)} -------- (8)

The first condition[in relation to exact differentials] implies:

{\psi}{=}\int\frac{\partial \chi}{\partial x}{dy}{+}{f}{(}{x}{)}
Or,
{\psi}{=}\int\frac{\partial }{\partial x}\sqrt{{1}{-}{\psi}^{2}}{dy}{+}{f}{(}{x}{)}
----------- (9)
If solutions exist for the above integral equation,we write,
\frac{\partial \theta}{\partial x}{=}{\psi} --------- (10)
\frac{\partial \theta}{\partial y}{=}{\chi} --------- (11)
The exact differential condition is satisfied for theta.

What about phi?
By the differentiation of Equation (8) we have,
{-}\frac{Cos\theta}{{sin}^{2}{\theta}}\frac{\partial \theta}{\partial y}{\chi}{+}\frac{1}{Sin \theta}\frac{\partial \chi}{\partial y}{=}\frac{Cos \theta}{{sin}^{2}\theta}\frac{\partial \theta}{\partial y}{\psi}{-}\frac{1}{{Sin}\theta}\frac{\partial \psi}{\partial y}
Or
\frac{{Cos}{(}{\theta}{)}}{{Sin}^{2}{(}{\theta}{)}}{(}{\chi}^{2}{+}{\psi}^{2}{)}{=}\frac{1}{{Sin}{(}{\theta}{)}}{(}\frac{\partial \psi}{\partial x} {+}\frac{\partial \chi}{\partial y}{)}
[(10) and (11) have been used in obtaining the above relation]
Finally:
{Cot}{(}{\theta}{)}{=}{(}\frac{\partial \psi}{\partial x} {+}\frac{\partial \chi}{\partial y}{)} ----- (12)
Or,
{Cot}{(}{\theta}{)}{=}{[}\frac{{\partial}^{2}{\theta}}{{\partial x}^{2}}{+}\frac{{\partial}^{2}{\theta}}{{\partial y}^{2}}{]} ----- (13)

On relation (13) we may apply the relations (10) ,(11) and (6) to go in the backward direction and arrive at the second exact differential condition that is, relation (8)

[The values of del(psi)/del(x) and del(chi)/del(y} as obtained from the results of the integral equation[ relation (9) should satisfy relation (12). This implies that at least certain solutions of (13) will cater to our requirement]]

 

[Dale]

This implies that at least certain solutions of (13) will cater to our requirement]]

Such as which solutions? Any one will do.

 

[Anamitra]

An Analytical Endeavor:
Equation:
{Cot}{(}{\theta}{)}{=}{[}\frac{{\partial}^{2}{\theta}}{{\partial x}^{2}}{+}\frac{{\partial}^{2}{\theta}}{{\partial y}^{2}}{]} ----- (1)

We look for a solution of the form:
\theta{=}{A}{(}{-}{ln}{Sin}{x}{-}\frac{1}{2}{x}^{2}{)}{+}{B}{(}{-}{ln}{Sin}{y}{-}\frac{1}{2}{y}^{2}{)}{+}{\lambda}{x}^{2}------------- (2)
A and B are constants Lambda is a parameter independent of x and y.

Using the above trial in relation (1) we have

{A}{Cot}^{2}{x}{+}{B}{cot}^{2}{y}{+}{2}{\lambda} {=}{Cot}\theta--- (3)

We may eliminate lambda between (2) and (3) to find Particular Integrals[surfaces theta=f(x,y)]
[Even if one has to use numerical methods, the equation will be an ordinary one instead of a“differential Equation”]
Instead of (2) one may use a trial of the form below, to obtain a greater variety of particular integrals:
\theta{=}{A}{f1}{(}{x}{)}{+}{B}{f2}{(}{y}{)}{+}{\lambda}{x}^{2}------------- (4)
f1 and f1 are known ,arbitrary functions--well behaved ones of course,in terms of continuity ,differentiability etc.

The homogeneous part of equation (1) is simply Laplace’s equation in two dimensions. We have the familiar solutions.

[This posting will be revised]

 

[Dale]

So plug your transformation equation back into the metric and calculate the curvature.