# The General Relativity Metric and Flat Spacetime

Gold Member
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.

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Gold Member
$${dq}{=}{R}{Sin}{(}{\theta}{)}{d}{\phi}$$
Because we must have$$dq = \frac{\partial q}{\partial \theta} d\theta + \frac{\partial q}{\partial \phi} d\phi$$your equation implies not only$$\frac{\partial q}{\partial \phi} = R \, \sin \, \theta$$but also$$\frac{\partial q}{\partial \theta} = 0$$As others have pointed out in different ways, there is no simultaneous solution to both those equations.

Mentor
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.

Gold Member
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[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)

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Gold Member
I just came across page 64 of the http://www.blatword.co.uk/space-time/Carrol_GR_lectures.pdf" [Broken] 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.)

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Staff Emeritus
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.

Gold Member
Flatlanders living on a sphere:

"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."

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Gold Member
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:
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