The General Relativity Metric and Flat Spacetime

[Anamitra]

Let us consider the General Relativity metric:
{ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{{dx}_{1}}^{2}{-}{g}_{22}{{dx}_{2}}^{2}{-}{{g}_{33}}{{dx}_{3}}^{2} ---------------- (1)
Using the substitutions:
{dT}{=}\sqrt{{g}_{00}}{dt}
{dX}_{1}{=}\sqrt{{g}_{11}}{dx}_{1}
{dX}_{2}{=}\sqrt{{g}_{22}}{dx}_{2}
{dX}_{3}{=}\sqrt{{g}_{33}}{dx}_{3}
We have,
{ds}^{2}{=}{dT}^{2}{-}{{dX}_{1}}^{2}{-}{{dX}_{2}}^{2}{-}{{dX}_{3}}^{2} ------------ (2)
The above metric corresponds to flat spacetime.
Now let us consider the following integrals:
{T}{-}{T}{0}{=}\int\sqrt{{g}_{00}}{dt}
[along lines for which coordinate values of x1,x2 and x3 are constant.
{X}_{1}{-}{(}{X}_{1}{)}_{0}{=}\int\sqrt{{g}_{11}}{dx}_{1}

[x2,x3 and t are held constant for the above integral]
{X}_{2}{-}{(}{X}_{2}{)}_{0}{=}\int\sqrt{{g}_{22}}{dx}_{2}
[x1,x3 and t are held constant for the above integral]
{X}_{3}{-}{(}{X}_{3}{)}_{0}{=}\int\sqrt{{g}_{33}}{dx}_{3}
[x1,x2 and t are held constant for the above evaluation]
[The previous four integral on the RHS are definite integrals having limits between t0 and t1,x1(0) and x1,x2(0) and x2,x3(0) and x3]
We are simply using physical distances between the coordinate labels to get our new coordinate system.
The flat spacetime metric given by relation (2) seems to be globally valid if the above integrals exist.We may describle spacetime globally with the variables T,X1,X2 and X3 having the metric equation(2) Non-local velocities should not be a problem since we have a flat spacetime in the [I]physical context. Parallel-Transport is not so serious an issue in flat spacetime.

[Bill_K]

Nice try, but you forgot what happens when for example g00 is a function of the X's. You can define T = ∫√g00 dt holding the X's constant, that part's Ok. But when you differentiate it, you'll get more terms: dT = √g00 dt + .... In other words, dT = √g00 dt is not an exact differential.

[Anamitra]

The integrals are actually definite integrals.[I have included this point separately now to avoid any misininterpretation]

[Dickfore]

First of all, your 'GR metric' is not the most general there is, because there are no off-diagonal terms, like:

2 g_{0 1} \, dx^{0} \, dx^{1}

Second, you forget that g_{0 0} is a function of x^{0}, x^{1}, x^{2}, x^{3}, so, when you do the integral:

T - T_{0} = \int{\sqrt{g_{0 0}} \, dx^{0}}

T is still a function of x^{1}, x^{2}, x^{3} parametrically. Therefore:

dT = \sqrt{g_{0 0}} \, dx^{0} + \int{\frac{1}{2 \, \sqrt{g_{0 0}}} \, \sum_{k = 1}^{3}{\frac{\partial g_{0 0}}{\partial x^{k}} \, dx^{k}} \, dx^{0}}

Good luck trying to express the metric through this mess.

[Anamitra]

Well, for orthogonal systems
{g}_{ij}{=}{0}
if {i}{\ne}{j}

We consider two tangential four vectors at the intersection of a pair of grid-lines[say x1 and x2] in an orthonormal system.

a=(0,e1,0,0)
b=(0,0,e2,0)

a.b=0 if the vectors, a and b are orthogonal.

Again,

{a}{.}{b}{=}{g}_{12}
Therefore in an orthogonal system quantities like g(1,2) are zero.

[My calculations relate to orthogonal systems or systems which might be reduced to an orthogonal one(after reducing it to the orthogonal form)]

[Dickfore]

[My calculations relate to orthogonal systems or systems which might be reduced to an orthogonal one(after reducing it to the orthogonal form)]


Although this might be done at any point in space-time, it cannot be done at all points simultaneously for a general metric by an arbitrary coordinate transformation.

[Anamitra]

Although this might be done at any point in space-time, it cannot be done at all points simultaneously for a general metric by an arbitrary coordinate transformation.

If the coordinates in the general metric are of a global nature we may carry out a transformation as indicated in the first post to get a global flat spacetime

[Dickfore]

global nature

Please clarify this term.

[Anamitra]

Quantities like t,z,y z or t,r theta , phi used to describe the General metric should cover a large region of spacetime. Just think of the t,r,theta,phi we use in the description of Schwarzschild Geometry.Distant points have these labels[having different values].

We are not using different coordinate systems for using these labels[one can stay comfortably in the same t,r,theta,phi system if he wants to]

[Dickfore]

Quantities like t,z,y z or t,r theta , phi used to describe the metric should cover a large region of spacetime. Just think of the t,r,theta,phi we use in the description of Schwarzchild Geometry.Distant points have these labels[having different values].

We are not using separate coordinate systems for using these labels[one can stay comfortably in the same t,r,theta,phi system if he wants to]

Uhhh, this is what curvilinear coordinates do. However, it has nothing to do with your deductions in this thread.

[TrickyDicky]

Although this might be done at any point in space-time, it cannot be done at all points simultaneously for a general metric by an arbitrary coordinate transformation.

The EFE give a local tensorial relation between physical quantities holding at each point in spacetime, and so do their metric solutions.
I'm not sure what you mean by "it can not be done at all points simultaneously".

[Anamitra]

Uhhh, this is what curvilinear coordinates do. However, it has nothing to do with your deductions in this thread.

In post #1 I have integrated along lines like:t=const,x1=const,x3=constant--that is along lines where only x1 is changing. Integrations of this type extend over a finite path-----the global nature of the coordinates is supportive to this type of work/treatment.

[Ben Niehoff]

Let us consider the General Relativity metric:
{ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{{dx}_{1}}^{2}{-}{g}_{22}{{dx}_{2}}^{2}{-}{{g}_{33}}{{dx}_{3}}^{2} ---------------- (1)
Using the substitutions:
{dT}{=}\sqrt{{g}_{00}}{dt}

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 rest of your claims have a similar problem:

{dX}_{1}{=}\sqrt{{g}_{11}}{dx}_{1}
{dX}_{2}{=}\sqrt{{g}_{22}}{dx}_{2}
{dX}_{3}{=}\sqrt{{g}_{33}}{dx}_{3}

It is perfectly fine, however, to break the metric up as you have into four orthonormal 1-forms. Just don't fool yourself into thinking they are closed. For example, you can write

\begin{align*} \theta^0 &= \sqrt{g_{00}} dt \\ \theta^1 &= \sqrt{g_{11}} dx^1 \\ \theta^2 &= \sqrt{g_{11}} dx^2 \\ \theta^3 &= \sqrt{g_{11}} dx^3 \end{align*}

What you have done is discovered the notion of orthonormal frames; i.e., the fact that any metric on a manifold can be locally brought to the form \text{diag}(-1, 1, 1, 1) (with minuses according to the signature of the metric) via an appropriate choice of basis. Your error is assuming that this choice of basis can be integrated to coordinates; i.e., you have assumed that this basis is a coordinate basis (or holonomic basis) when it is not.

You attempt to integrate it as follows:

Now let us consider the following integrals:
{T}{-}{T}{0}{=}\int\sqrt{{g}_{00}}{dt}
[along lines for which coordinate values of x1,x2 and x3 are constant.]

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',x^1,x^2,x^3)} \; dt' \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.

The flat spacetime metric given by relation (2) seems to be globally valid if the above integrals exist.We may describle spacetime globally with the variables T,X1,X2 and X3 having the metric equation(2).

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.

When is this system of equations integrable in some open region? You will find that the obstruction to integrability is precisely the curvature tensor! Only when the curvature tensor vanishes identically in some open region can some flat-space coordinate system be found.

[Dickfore]

In post #1 I have integrated along lines like:t=const,x1=const,x3=constant--that is along lines where only x1 is changing. Integrations of this type extend over a finite path-----the global nature of the coordinates is supportive to this type of work/treatment.

But you failed to recognize my second criticism in post #4.

[TrickyDicky]

The EFE give a local tensorial relation between physical quantities holding at each point in spacetime, and so do their metric solutions.
I'm not sure what you mean by "it can not be done at all points simultaneously".

Nevermind, I see what you mean now. The important part was the one about coordinate transformations.

[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',x^1,x^2,x^3)} \; dt' \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.




If you take a directional derivative along a curve:In this case along one for which x1=const,x2=const and x3=constant you will not have any problem at all.

For calculating X1-X1(0) we have chosen a curve for which t=const,x2= const and x3=const. If you want take the derivative you should take the directional derivative. The problem you have stated will not be there.

I dont have a bunch of {dx}^{\alpha} to trouble me.

[Anamitra]

For an infinitesimal increment of X1 along a direction given by dt ,dx1,dx2 and dx3 we have,
{dX}^{1}{=}\frac{{dX}^{1}}{{dx}^{\alpha}}{dx}^{\al pha}
For an infinitesimal increment along the x1-axis we have:
{dX}^{1}{=}\frac{{dX}^{1}}{{dx}^{1}}{dx}^{1}

[Ben Niehoff]

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.

[Anamitra]

{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}

[Dickfore]

{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}

uhh, the intial metric had 2 coordinates.

[Anamitra]

Effectively we have two coordinates due to the constraint.

You don't have three independent coordinates.

[Dickfore]

Effectively we have two coordinates due to the constraint.

You don't have three independent coordinates.

So, why does your metric have a sum of squares of three differentials?

[Anamitra]

The variables in the metric are not independent--we dont have three independent variables in the description of the metric--only two[effectively] due to the constraint.

[Given the constraint we have a two dimensional space and not three dimensional space.]

[Dickfore]

The variables in the metric are not independent--we dont have three independent variables in the description of the metric--only two due to the constraint.

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}

[Anamitra]

It is absolutely unnecessary to express the metric in the above form

I have a metric which relates to[describes] a two dimensional space--the surface of a sphere--that is sufficient

[Dickfore]

It is clear you do not know what you are talking about. Good day sir.

[Ben Niehoff]

You gave a metric on R^3, which is not the same thing as a metric on S^2. It is true that S^2 can be isometrically embedded in R^3, but that is entirely beside the point...Your original claim was that you could find N flat coordinates for an N-dimensional manifold! You've instead found N+1 coordinates plus a constraint.

If all you want to do is embed curved manifolds into flat ones, be aware that to isometrically embed a general (1,3)-dimensional manifold into a flat spacetime requires an ambient manifold of something like (3,87) dimensions. This neither simplifies the problem of GR nor provides any useful insight.

[DaleSpam]

{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}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.

[DaleSpam]

be aware that to isometrically embed a general (1,3)-dimensional manifold into a flat spacetime requires an ambient manifold of something like (3,87) dimensions.Cool, do you have a ref for this?

[Anamitra]

Lets write:
{tan}{\phi}{=}\frac{y}{x} -----------(1)

{Cos}{\theta}{=}\frac{z}{R} ----------- (2)
R=const
Frem(1)

{sec}^{2}{\phi}{d}{\phi}{=}\frac{xdy{-}{ydx}}{{x}^{2}}

Or,

{d}{\phi}{=}\frac{xdy-ydx}{{x}^{2}{+}{y}^{2}}

From(2):

{-}{Sin}{d}{\theta}{=}\frac{dz}{R}
{d}{\theta}{=}{-}{dz}\frac{1}{\sqrt{{{R}^{2}{-}{z}^{2}}}}

{ds}^{2}{=}{R}^{2}{[}{d}{\phi}^{2}{+}{Sin}^{2}{\theta}{d}{\theta}^{2}{]}

{ds}^{2}{=}{R}^{2}{[}{(}\frac{xdy-ydx}{{x}^{2}{+}{y}^{2}}{)}^{2}{+}\frac{1}{{R}^{2}} {(}\frac{{xdx}{+}{y}{dy}}{\sqrt{{{R}^{2}{-}{x}^{2}{-}{y}^{2}}}}{)}^{2}{]}

You may use:

{R}^{2}{-}{z}^{2}{=}{x}^{2}{+}{y}^{2}

[DaleSpam]

You still have 3 coordinates. None of what you wrote in the OP involved adding extra coordinates to the manifold.

[Ben Niehoff]

{ds}^{2}{=}{R}^{2}{[}{(}\frac{xdy-ydx}{{x}^{2}{+}{y}^{2}}{)}^{2}{+}\frac{{x}^{2}{+}{ y}^{2}}{{R}^{2}}{(}\frac{dz}{\sqrt{{x}^{2}{+}{y}^{ 2}}}{)}^{2}{]}

As Dale points out, this still has too many coordinates.

In addition, it is not of the form

ds^2 = du^2 + dv^2

for any coordinate functions u, v. (Remember, the one-forms du, \ dv should be closed, i.e. ddu = 0 and ddv = 0).

[Dickfore]

According to your 'prescription':

x^{1} = \theta


x^{2} = \sin{(\theta)} \, \phi

Then:

dx^{1} = d\theta

But:

dx^{2} = \sin{(\theta)} \, d\phi + \phi \, \cos{(\theta)} \, d\theta

it contains the differentials of both angles. Solving from these equations for d\theta and d\phi, we get:

d\theta = dx^{1}


d\phi =\frac{dx^{2} - \phi \, \cos{(\theta)} \, d\theta}{\sin{(\theta)}} = \frac{dx^{2} - \frac{x^{2}}{\sin{(x^{2})}} \, dx^{1}}{\sin{(x^{2})}}

Then, the metric is rewritten as:

ds^{2} = (dx^{1})^{2} + \left(dx^{2} - \frac{x^{2}}{\sin{(x^{2})}} \, dx^{1}\right)^{2}



ds^{2} = \left[1 + \frac{(x^{2})^{2}}{\sin^{2}{(x^{2})}}\right] \, (dx^{1})^{2} - 2 \, \frac{x^{2}}{\sin{(x^{2})}} \, dx^{1} \, dx^{2} + (dx^{2})^{2}

This is not a metric for a flat space. So, the coordinates x^{1} and x^{2} do not describe a flat space.

[Anamitra]

In addition, it is not of the form

ds^2 = du^2 + dv^2

for any coordinate functions u, v. (Remember, the one-forms du, \ dv should be closed, i.e. ddu = 0 and ddv = 0).

The metric I have written is the same as[equivalent to]

{ds}^{2}{=}{R}^{2}{(}{d}{\phi}^{2}{+}{Sin}^{2}{\th eta}{d}{\theta}^{2}{)}
Do you find any problem now?

[Anamitra]

Given Metric:
{ds}^{2}{=}{R}^{2}{(}{d}{\phi}^{2}{+}{sin}^{2}\the ta{d}{\theta}^{2}{)}

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}

{ds}^{2}{=}{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}{)}

[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}{\th eta}{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.

[DaleSpam]

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.

[DaleSpam]

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.

[DaleSpam]

No problem, the way he fights you every step of the way can indeed be very frustrating. See: http://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.

[DaleSpam]

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}{]}

[DaleSpam]

We start from a fixed point theta0Yes, 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.

[DaleSpam]

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.

[DaleSpam]

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 &= R \; d ( \sin \theta \; d\phi) \\ 0 &= 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.

[DaleSpam]

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 &= dr^2 +r^2 d\theta^2+r^2 sin^2(\theta) d\phi^2\\ (ds|_{r=1})^2 &= 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 should not be a problem since we have a flat spacetime in the [I]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 Carroll Lectures on General Relativity (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 Schwarszchild 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',x^1,x^2,x^3)} \; dt' \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:
http://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