# The General Relativity Metric and Flat Spacetime

• Anamitra
Can you please elaborate?In summary, the General Relativity metric corresponds to flat spacetime. The integrals are definite integrals and the metric equation can be described globally with the variables T,X1,X2,X3. Parallel-Transport is not so serious an issue in 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[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.

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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.

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

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.

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)]

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Anamitra said:
[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.

Dickfore said:
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

Anamitra said:
global nature

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]

Anamitra said:
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 Schwarzschild 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.

Dickfore said:
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".

Dickfore said:
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.

Anamitra said:
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.

Anamitra said:
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 said:
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.

Ben Niehoff said:
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 don't have a bunch of $${dx}^{\alpha}$$ to trouble me.

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

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

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

Anamitra said:
$${ds}^{2}{=}{dx}^{2}{+}{dy}^{2}{+}{dz}^{2}$$

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.

Effectively we have two coordinates due to the constraint.

You don't have three independent coordinates.

Anamitra said:
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?

The variables in the metric are not independent--we don't 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.]

Anamitra said:
The variables in the metric are not independent--we don't 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:

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

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

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

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.

Anamitra said:
$${ds}^{2}{=}{dx}^{2}{+}{dy}^{2}{+}{dz}^{2}$$

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.

Ben Niehoff said:
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?

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:

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

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

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

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 $ddv = 0$).

$$x^{1} = \theta$$
$$x^{2} = \sin{(\theta)} \, \phi$$
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.

Ben Niehoff said:
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]

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

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

Consider the metric:

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

Transformations:

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

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

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

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

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