The General Relativity Metric and Flat Spacetime

[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:

<br /> ds^{2} = A(x, y) \, dx^{2} + 2 \, B(x, y) \, dx \, dy + C(x, y) \, dy^{2}<br />

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.

 

[Ben Niehoff]

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.

It's actually worse than that! This statement:

dq = R \sin \theta \; d\phi

is a lie! The right-hand side is not 'd' of anything, so it is incorrect to call this one-form 'dq'.

Remember that for any p-form \omega, we must have dd\omega = 0. Taking 'd' of both sides of Anamitra's equation yields

\begin{align*} ddq &amp;= R \; d ( \sin \theta \; d\phi) \\ 0 &amp;= R \cos \theta \; d\theta \wedge d\phi \end{align*}
which is clearly false.

Edit: I have deleted the rest of my post. It was a careful explanation of the issues Anamitra would face if he tried to follow the procedure he has outlined. I am offended, however, at the idea of doing his work for him.

 

[DrGreg]

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

Because we must havedq = \frac{\partial q}{\partial \theta} d\theta + \frac{\partial q}{\partial \phi} d\phiyour equation implies not only\frac{\partial q}{\partial \phi} = R \, \sin \, \thetabut also\frac{\partial q}{\partial \theta} = 0As others have pointed out in different ways, there is no simultaneous solution to both those equations.

 

[Dale]

The right-hand side is not 'd' of anything, so it is incorrect to call this one-form 'dq'.

there is no simultaneous solution to both those equations.

Thanks, I liked both of these explanations better than mine. I didn't realize the problem was that serious, but I guess I should have.

 

[JDoolin]

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.

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

With {x}^{2}{+}{y}^{2}{+}{z}^{2}{=}{R}^{2}{=}{const}

{x}{=}{R}{Sin}{\theta}{cos}{\phi}
{y}{=}{R}{Sin}{\theta}{sin}{\phi}
{z}{=}{R}{Cos}{\phi}

Putting on a constraint that r=1:

\begin{align*} ds^2 &amp;= dr^2 +r^2 d\theta^2+r^2 sin^2(\theta) d\phi^2\\ (ds|_{r=1})^2 &amp;= 0 + d \theta^2 + sin^2(\theta)d\phi^2 \end{align*}

Nobody disagrees that you can embed a 2D curved surface like a sphere into a flat 3D space. That doesn't make the embedded 2D space flat.

Use your procedure to "flatten" the sphere. Or if a sphere is too easy then use your procedure to flatten the Schwarzschild spacetime.

Non-local velocities[in cosmology or elsewhere] should not be a problem since we have a flat spacetime in the physical context. Parallel-Transport is not so serious an issue in flat spacetime.

We have this mapping of the surface of a sphere, r=1, and we're able to make this relationship of

ds^2 = d\theta^2 + \sin^2 \theta \; d\phi^2,

However, there are no distinguishing features of that sphere that would tell anyone the value of θ or Φ. As far as they are concerned, the physical distances at the pole are identical to the physical distances at the equator. If you told a person at the pole that he could only travel in the theta direction he would say you were crazy. Yet, that's exactly what this coordinate system in (θ,Φ) would say. If you told someone walking around the pole that he was traveling at the same speed (in dΦ/dt) as someone flying around the equator once every three seconds, he would think you are crazy. Yet, that's exactly what this coordinate system in (θ,Φ) would say.

But that is certainly not the units that people living in that coordinate system would use. They would use the units from the higher dimensional flat coordinate system.

(Edit: or perhaps, they would all imagine themselves to be traveling along an equator, so that sin(θ)=1 and ds² =dθ²+dΦ² = dx² + dy²)

 

[JDoolin]

I just came across page 64 of the http://www.blatword.co.uk/space-time/Carrol_GR_lectures.pdf" where some of the ambiguity is introduced.

Within this, it is made clear that via parallel transport, there is no "well defined notion of relative velocity" namely because there is no well-defined notion of relative direction. However, I think in a static model (such as the surface of a sphere, and (possibly?) the Schwarzschild metric) there should still be a well-defined notion of "speed." i.e. the non-directional relative motion.

When Carroll gets to the bottom of the page, though, he invokes a non-static model; "the metric of spacetime between us and the galaxies has changed", and "the universe has expanded," suggesting a scale factor changing with respect to time, a(t), Under these circumstances, even the speed would be ambiguous, I guess.

(Although I may be reading too much into it... Carroll does not specifically say the metric has changed "over time." He may just mean the change in the metric through space.)

 

[pervect]

There's a well defined notion of a static observer in a static space-time like the Scwarzschild space-time and there is an unambiguous notion of the speed of any object relative to the static observer that's co-located with him.

Cosmological space-times are for the most part not static, so this trick won't work for them.

 

[JDoolin]

Flatlanders living on a sphere:

"Which way is your home?

"That way." (pointng in two opposite directions)

"And what direction did you just come from?"

"Sorry, that direction does not exist here. I would require a third degree of freedom to tell you, explicitly."

 

[JDoolin]

There's a well defined notion of a static observer in a static space-time like the Scwarzschild space-time and there is an unambiguous notion of the speed of any object relative to the static observer that's co-located with him.

Cosmological space-times are for the most part not static, so this trick won't work for them.

I still have to read more. Maybe when you put enough static schwarzschild metrics together, and rub them against each other, it becomes a nonstatic space-time.

 

[Anamitra]

By holding the other coordinates constant, you are able to do the integral, sure. But when you take the exterior derivative of both sides, you obtain

dT = \frac{\partial}{\partial x^\alpha} \Big( \int^{x^0} \sqrt{g_{00}(t&#039;,x^1,x^2,x^3)} \; dt&#039; \Big) dx^\alpha

which is a closed 1-form, but it is not a 1-form that fits nicely into your metric! It has a bunch of dx^1,\ dx^2,\ dx^3 terms in addition to dx^0. You will run into a similar problem with the rest of your integrals.



You should see now why the above claim is false. The point is that you have a system of differential equations (i.e., a bunch of 1-forms which you assume can be written as dx^\alpha for some coordinate functions x^\alpha), and this system of differential equations is not integrable in any finite-sized open region, in general.

If the GR metric is valid only in a path dependent situation. I would request Ben Niehoff to go through the following posting:
https://www.physicsforums.com/showpost.php?p=3436925&postcount=21

Right here you've lied. By calling it dT, you are claiming that the one-form \sqrt{g_{00}} dt is closed (i.e., locally exact). But as others have pointed out, this is not true unless g_{00} happens to be a function of t alone.

The Science Advisor considers it essential to use impolite language to express himself---possibly he feels that it should provide a special parameter to the forum