The General Relativity Metric and Flat Spacetime

[Dickfore]

But, the original metric was:
$$ds^{2} = R^{2} \, \left(d\theta^{2} + \sin^{2}{\theta} \, d\phi^{2}\right)$$
Can't you see the difference?

 

[Anamitra]

The matter is quite simple. I have got the original metric from the flat spacetime metric using transformations. This is allowed.You may use the reverse transformations to pass from curved space to flat space.

It depends totally on the type of transformations you are using,what you would obtain finally.

We should always use a favorable type of transformations

 

[Dickfore]

You're a tool.

 

[Ben Niehoff]

Consider the metric:

$${ds}^{2}{=}{dx}^{2}{+}{dz}^{2}$$

Transformations:

$${z}{=}{R}{[}{1}{-}{cos}{\theta}{]}$$
And

$${x}{=}{R}{\phi}$$

$${dz}{=}{R}{sin}{\theta}{d}{\theta}$$
$${dx}{=}{R}{d}{\phi}$$

$${dx}^{2}{+}{dy}^{2}{=}{R}^{2}{sin}^{2}{\theta}{d}{\theta}^{2}{+}{R}^{2}{d}{\phi}^{2}$$
$${=}{R}^{2}{(}{d}{\phi}^{2}{+}{Sin}^{2}{\theta}{d}{\theta}^{2}{)}$$

But it is not the metric I asked for. The metric on a sphere is

$$ds^2 = d\theta^2 + \sin^2 \theta \; d\phi^2.$$

Notice the difference between this and yours.

 

[Anamitra]

The basic aim is to pass from curved spacetime to flat spacetime in a global manner--and that has been done[post #35].

ds^2 is invariant: but the metrics have different

$${ds}^{2}{=}{dx}^{2}{+}{dy}^{2}$$ \is a flat space metric,while

$${ds}^{2}{=}{R}^{2}{(}{d}{\phi}^{2}{+}{sin}^{2}{\theta}{d}{\theta}^{2}$$
represents curved space

We may pass from one metric to the other by Global Transformations. "ds^2" does not change in the process of transformations.
[It is important to observe that the transformations in #35 are of a global nature]

 

[Ben Niehoff]

The metric (coordinates re-labeled to avoid confusion with spherical metric)

$$ds^2 = R^2 (du^2 + \sin^2 v \; dv^2)$$

is NOT curved, as a relatively quick calculation of its curvature reveals. Try it.

 

[Anamitra]

So far as the metric coefficients are concerned it has to represent curved space[Hope v has been used as a variable]

 

[Ben Niehoff]

It's clear that you have some very deep-seated misunderstandings. I hope that upon further reading and reflection you will get over them. I suggest you practice actually computing some curvature tensors.

 

[Anamitra]

The metric (coordinates re-labeled to avoid confusion with spherical metric)

$$ds^2 = R^2 (du^2 + \sin^2 v \; dv^2)$$

is NOT curved, as a relatively quick calculation of its curvature reveals. Try it.

Take R as a fixed Schwarzschild Radius[coordinate value]
[You may think in parallel transporting a vector round the 45 degree latitude and see if it turns when it comes back to the original point--you will find curved space]

[On the same metric try out the transformations given in post #35]

 

[WannabeNewton]

Take R as a fixed Schwarzschild Radius[coordinate value]
[You may think in parallel transporting a vector round the 45 degree latitude and see if it turns when it comes back to the original point--you will find curved space]

[On the same metric try out the transformations given in post #35]

For that metric it is a rather trivial task to show that $R^{\alpha }_{\beta \mu \nu } = 0$ identically.

 

[Anamitra]

For that metric it is a rather trivial task to show that $R^{\alpha }_{\beta \mu \nu } = 0$ identically.

In such a situation a parallel-transported vector should not turn if it is taken round the 45 degree latitude and brought back to its initial position.
[The Christoffel Symbols work out to diffrent values in the Schwarzschild sphere and the ordinary sphere]

 

[TrickyDicky]

Anamitra, if you apply a global transformation to the metric of the ambient embedding space of the manifold you are considering you end up changing the original manifold, it is no longer the same spacetime, you just can't do that. You have a restriction equation that keeps you from doing it.
So, no you can't do the coordinate transformation in the OP and still have a curved manifold. Just like you can't do it in the case of the 2-sphere manifold acting from the flat ambient space, that 3 space is restrained to the 2-sphere surface and that restricts the kind of global transformations you may perform on it.
The paralled-transported vector turns just because you are restricting it explicitly with certain coordinate restricition, if you do a global transformation in the embedding space you may no longer keep on the 2-sphere surface and therefore it doesn't represent the curved space you think it's representing.

 

[Dickfore]

Would an OP please close this topic? It's excruciating to read this troll.

 

[Dale]

So far as the metric coefficients are concerned it has to represent curved space[Hope v has been used as a variable]

I just worked out the curvature. The above metric is flat, not curved. There is one non-zero Christoffel symbol, but no non-zero components to the Riemann curvature tensor. So it is apparently some sort of "polar-like" coordinates in a flat space.

Why do you think that "it has to represent curved space"?

 

[Dickfore]

Of course it is:
$$\sin^{2}{(\theta)} \, d\theta^{2} = \left(d(\cos{\theta})\right)^{2}$$

 

[TrickyDicky]

Would an OP please close this topic? It's excruciating to read this troll.

An OP? OP=Original Poster(or Post) He can't close it.
Besides I don't think he is trolling, he just got it wrong and it's one of the purposes of Forums like this to help him get it.

 

[Dale]

My experience with Anamitra is that he takes a long time to "get it", and fights you every step of the way, but eventually he comes around and understands. He is also eventually willing to work through the math and once he does so he tends to convince himself. I wouldn't recommend closing the thread.

 

[Dickfore]

Ok, I am sorry. Point taken. I lost my temper for awhile.

 

[Dale]

No problem, the way he fights you every step of the way can indeed be very frustrating. See: https://www.physicsforums.com/showthread.php?t=423334

 

[Anamitra]

OK, I think at this point you will not be convinced unless you actually do a computation. So let's simplify this as much as possible and do just 2 dimensions. Take the following metric for the sphere,

$$ds^2 = d\theta^2 + \sin^2 \theta \; d\phi^2,$$
and show us how to obtain a globally flat coordinate system, using your method.

This seriously heavy issue was raised by Ben Niehoff.[Post #18]

 

[Anamitra]

So, please eliminate one of them and express the metric in the form:

$$ds^{2} = A(x, y) \, dx^{2} + 2 \, B(x, y) \, dx \, dy + C(x, y) \, dy^{2}$$

What Dickfore said in post #24

 

[Anamitra]

In Relation to Post #46

Is DaleSpam ready to confirm that the Schwarzschild sphere[for a fixed coordinate r] is not a curved space?

I have made a claim[through #46] that it represents curved space. The answer from a Science Advisor would be crucial to the issue.

 

[Dale]

In Relation to Post #46

Is DaleSpam ready to confirm that the Schwarzschild sphere[for a fixed coordinate r] is not a curved space?

I have made a claim[through #46] that it represents curved space. The answer from a Science Advisor would be crucial to the issue.

This metric (not a sphere) is flat:
$$ds^2=R^2(d\phi^2+\sin^2(\theta) d\theta^2)$$


This metric (a sphere) is not flat:
$$ds^2=R^2(d\theta^2+\sin^2(\theta) d\phi^2)$$

 

[TrickyDicky]

In post number 35 you got this metric:
$$ds^2=R^2(d\phi^2+\sin^2(\theta) d\theta^2)$$

from:

$$ds^2=dx^2+dz^2$$

Wich is the metric of a flat surface.

It is a basic axiom of differential geometry that you cannot get from a flat surface to an intrinsically curved surface by any coordinate transformation.
Therefore the metric you get is still flat.

 

[WannabeNewton]

The paralled-transported vector turns just because you are restricting it explicitly with certain coordinate restricition, if you do a global transformation in the embedding space you may no longer keep on the 2-sphere surface and therefore it doesn't represent the curved space you think it's representing.

I don't see how the transported vector turns though under the metric $ds^{2} = d\phi ^{2} + sin^{2}(\theta) d\theta ^{2}$. If for example you choose to parallel transport the vector $\mathbf{v}$ around a circle of latitude $\theta = \theta _{0}$ then you could set up the parametric equation for the circle as $u^{A}(t) = t\delta ^{A}_{1} + \theta _{0}\delta ^{A}_{2}$ and the tangent vector would be $\dot{u^{A}} = \delta ^{A}_{1}$ and the parallel transport of the original vector , $\bigtriangledown _{\dot{\mathbf{u}}}\mathbf{v} = 0$, comes to $\frac{\partial v^{A}}{\partial \phi } + \Gamma ^{A}_{B1}v^{B} = 0$ and since for that metric that christoffel symbol vanishes I just get $\frac{\partial v^{A}}{\partial \phi }= 0$. If the vector's direction really does change maybe I interpreted the bases wrong; I interpreted them as they are on a 2 - sphere.

 

[Ben Niehoff]

I don't see how the transported vector turns though under the metric $ds^{2} = d\phi ^{2} + sin^{2}(\theta) d\theta ^{2}$. If for example you choose to parallel transport the vector $\mathbf{v}$ around a circle of latitude $\theta = \theta _{0}$ then you could set up the parametric equation for the circle as $u^{A}(t) = t\delta ^{A}_{1} + \theta _{0}\delta ^{A}_{2}$ and the tangent vector would be $\dot{u^{A}} = \delta ^{A}_{1}$ and the parallel transport of the original vector , $\bigtriangledown _{\dot{\mathbf{u}}}\mathbf{v} = 0$, comes to $\frac{\partial v^{A}}{\partial \phi } + \Gamma ^{A}_{B1}v^{B} = 0$ and since for that metric that christoffel symbol vanishes I just get $\frac{\partial v^{A}}{\partial \phi }= 0$. If the vector's direction really does change maybe I interpreted the bases wrong; I interpreted them as they are on a 2 - sphere.

The metric $ds^{2} = d\phi ^{2} + \sin^{2} \theta \; d\theta ^{2}$ is not even a metric for a 2-sphere, so it is silly to interpret either $\phi$ or $\theta$ as "latitude". This metric can be rewritten

$$ds^2 = dx^2 + dy^2$$

where

$$x = \phi, \qquad y = \cos \theta$$

and so it is a funny coordinate chart on the flat plane, covering only the horizontal strip where $-1 \le y \le 1$.

The metric for the 2-sphere is $ds^2 = d\phi^2 + \sin^2 \phi \; d\theta^2$. Anamitra is not reading carefully and has failed to notice the difference (on several occasions now).

 

[TrickyDicky]

I don't see how the transported vector turns though under the metric $ds^{2} = d\phi ^{2} + sin^{2}(\theta) d\theta ^{2}$. If for example you choose to parallel transport the vector $\mathbf{v}$ around a circle of latitude $\theta = \theta _{0}$ then you could set up the parametric equation for the circle as $u^{A}(t) = t\delta ^{A}_{1} + \theta _{0}\delta ^{A}_{2}$ and the tangent vector would be $\dot{u^{A}} = \delta ^{A}_{1}$ and the parallel transport of the original vector , $\bigtriangledown _{\dot{\mathbf{u}}}\mathbf{v} = 0$, comes to $\frac{\partial v^{A}}{\partial \phi } + \Gamma ^{A}_{B1}v^{B} = 0$ and since for that metric that christoffel symbol vanishes I just get $\frac{\partial v^{A}}{\partial \phi }= 0$. If the vector's direction really does change maybe I interpreted the bases wrong; I interpreted them as they are on a 2 - sphere.

I should have explained better that bit, it was referring to a previous 3 dimensional metric Anamitra wrote with a sphere restriction equation (post #19), not actually to the 2 dimensional last one.

 

[Anamitra]

Given metric:
$${ds}^{2}{=}{R}^{2}{[}{d}{\theta}^{2}{+}{Sin}^{2}{\theta}{d}{\phi}^{2}{]}$$

We start from a fixed point theta0, phi0 on the surface of a sphere of radius R
To locate an arbitrary point on the sphere first we move along a line of latitude,theta=theta0

$${X}{=}{R}{Sin}{\theta}_{0}{*}{\phi}$$
Then we move along a meridian
$${Z}{=}{R}{*}{\theta}$$

For all movements we move first along the latitude. The value of $${sin}{\theta}_{0}$$ is an unchanging one for locating any arbitrary point on the sphere.
Now we have
$${ds}^{2}{=}{R}^{2}{[}{d}{\theta}^{2}{+}{sin}^{2}{\theta}{d}{\phi}^{2}{]}$$
Equivalent to,
$${ds}^{2}{=}{[}{d}{X}^{2}{+}{sin}^{2}{\theta}_{0}{d}{Z}^{2}{]}$$
The value,
$${Sin}{(}{\theta}_{0}{)}$$
Does not change for locating an arbitrary point since we are moving along the line of latitude first for locating any arbitrary point.
Better, we write $${k}{=}{Sin}{\theta}_{0}$$
Transformed metric:
$${ds}^{2}{=}{[}{d}{X}^{2}{+}{k}^{2}{d}{Z}^{2}{]}$$

Choose

$${Sin}^{2}{\theta}_{0}{=}{1}$$

You may take, theta0=pi/2
Finally we have,
$${ds}^{2}{=}{[}{d}{X}^{2}{+}{d}{Z}^{2}{]}$$

 

[Dale]

We start from a fixed point theta0

Yes, you can do that locally at any given point.

 

[Anamitra]

I am starting from a particular point[fixed one] theta0,phi0.
But the transformations I have given are not local. They have a global perspective.

$${X}{=}{R}{Sin}{\theta}_{0}{\phi}$$

$${Z}{=}{R}{*}{\theta}$$

Theta and phi are global---The only thing to keep in mind is that while creating the (X,Z) labels we have to move along the latitude starting from the theta0,phi point first to create the Xlabel and then we move along the meridian to get the Y label.

 

[TrickyDicky]

I am starting from a particular point[fixed one] theta0,phi0.
But the transformations I have given are not local. They have a global perspective.

$${X}{=}{R}{Sin}{\theta}_{0}{\phi}$$

$${Z}{=}{R}{*}{\theta}$$

Theta and phi are global---The only thing to keep in mind is that while creating the (X,Z) labels we have to move along the latitude starting from the theta0,phi point first to create the Xlabel and then we move along the meridian to get the Y label.

Theta and phi are not global in the 2-sphere manifold, they only cover a patch, you should read something about manifolds.

 

[Dale]

I am starting from a particular point[fixed one] theta0,phi0.
But the transformations I have given are not local. They have a global perspective.

Then your approximation $\sin(\theta) \approx \sin(\theta_0)$ doesn't hold. So instead we have:
$$ds^2=R^2(d\theta^2+\sin^2(\theta) \; d\phi^2)$$

$$x=R \sin(\theta_0) \; \phi$$
$$dx=R \sin(\theta_0) \; d\phi$$
$$z=R \; \theta$$
$$dz=R \; d\theta$$

$$ds^2 = dz^2 + \frac{1}{\sin^2(\theta_0)} \sin^2\left(\frac{z}{R}\right) dx^2$$

Which will not be flat.

 

[Anamitra]

Ok. Then let me put it in this way:

$${dp}{=}{R}{d}{\theta}$$

$${dq}{=}{R}{Sin}{(}{\theta}{)}{d}{\phi}$$
The above relation is integrable if the path is specified.From a reference point [theta0,phi0]we may take specified paths to all other points on the surface of the sphere.

Specification[of path] may be of the following type:along a meridian first and then along a line of latitude.One may apply the rule for points at finite or at infinitesimal separation.
Then one has to evaluate:

$${\int}{dq}$$ along the stated path on the sphere[between the two above stated points].
Finally we have:

$${ds}^{2}{=}{dx}^{2}{+}{dy}^{2}$$

 

[Anamitra]

In keeping with conventional ideas it is always possible to project the upper half of the a sphere on a tangent plane passing through the north pole.This covers a huge number of non-local points. May be the procedure does not give us a total global transformation--but it gives a semiglobal transformation that could simplify things for us.

In the above example a huge number of non-local points come under flat space time view.
We may do this[rather something similar] for the Schwarzschild sphere.

 

[Dale]

Ok. Then let me put it in this way:

$${dp}{=}{R}{d}{\theta}$$

$${dq}{=}{R}{Sin}{(}{\theta}{)}{d}{\phi}$$

That is not a coordinate transform. Once you integrate both sides to get p and q instead of dp and dq then you are going to have a mess. When you substitute that mess in it is not going to simplify the way you think it will. You will get a nice dp² term, a nice dq², and a messy dpdq term.