I Coordinate systems on the 2-sphere

[kent davidge]

If I define the two dimensional sphere in the usual way, this gives me a metric $ds^2 = r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2$. Can I just define a new coordinate system giving a point coordinates $(\theta', \phi') = (\theta r^2, \phi r^2 \sin^2 \theta)$?. This gives me the metric $ds^2 = d\theta'^2 + d\phi'^2$.

 

[Orodruin]

You can do any coordinate change you want, it will not give you a metric of that form.

 

[kent davidge]

I thought I could show that the 2-sphere is locally bijective to R² by showing that there can be local coordinate transformations that put the metric on the euclidean form at any point on the 2-sphere.

 

[Orodruin]

You cannot choose coordinates such that the coordinate basis is Euclidean at one point. You cannot choose global such coordinates. The sphere is not flat!

 

[kent davidge]

can't believe that the sphere is "locally like R²" but its metric can't locally be equal to that of R².

 

[PeroK]

I thought I could show that the 2-sphere is locally bijective to R² by showing that there can be local coordinate transformations that put the metric on the euclidean form at any point on the 2-sphere.

Compare there two statements:

Given any point, you can find a coordinate transformation that makes the metric approximately Euclidean at that point.

You can find a coordinate transformation that makes the metric Euclidean at every point.

 

[kent davidge]

Compare there two statements:

Given any point, you can find a coordinate transformation that makes the metric approximately Euclidean at that point.

You can find a coordinate transformation that makes the metric Euclidean at every point.

oh excuse me if my post was misleading. It should read: I thought we could, at any point, find a coordinate transformation that would give us the euclidean metric. These would be different transformations for different points, not only one global coordinate transformation that would hold at any point.

 

[PeroK]

oh excuse me if my post was misleading. It should read: I thought we could, at any point, find a coordinate transformation that would give us the euclidean metric. These would be different transformations for different points, not only one global coordinate transformation that would hold at any point.

The sphere is not flat anywhere.

I'm not sure where in your previous post, the local point was identified. It looked like a general, global transformation.

 

[kent davidge]

I'm not sure where in your previous post, the local point was identified. It looked like a general, global transformation.

yes, I would ask you to forget about that post, and look at my previous one, because that's what I mean.

 

[PeroK]

yes, I would ask you to forget about that post, and look at my previous one, because that's what I mean.

You can't get the Euclidean metric anywhere on sphere. But, on a small region it is approximately Euclidean.

 

[kent davidge]

You can't get the Euclidean metric anywhere on sphere. But, on a small region it is approximately Euclidean.

I not get it. Any region of the sphere looks exactly the same physically. So why can't we conclude that it's possible to cover every region of the sphere with an Euclidean metric (but not to extend that metric on to another region)?

 

[PeroK]

I not get it. Any region of the sphere looks exactly the same physically. So why can't we conclude that it's possible to cover every region of the sphere with an Euclidean metric (but not to extend that metric on to another region)?

Because the sphere is not flat anywhere.

 

[suremarc]

I’ve never studied differential geometry seriously before, but here’s my thoughts:

Suppose we have a chart (U, φ) on a Riemannian manifold (M, g). This gives us a coordinate expression for the Riemann curvature tensor R^a_bcd on U. Now, a smooth change of variables amounts to a separate chart (U, ψ) such that the transition map ψ(φ^-1) is a diffeomorphism. This induces a transformation of our coordinate expression of R^a_bcd via the chain rule.

However, it’s not hard to see that a diffeomorphism can’t transform a nonzero tensor into zero at any point, since it is bijective at every point. Moreover, a diffeomorphism leaves invariants of a tensor unchanged. So this gives us a set of necessary conditions for two metrics to be equivalent by a change of variables. In 2 dimensions this reduces to the condition that their scalar curvatures have the same sign (or are both zero), I think.

Interestingly, the quadratic forms given by the metrics at a single point are equivalent for the sphere and plane. However, the metrics themselves are not equivalent on any open subset due to the smooth structure.

 

[stevendaryl]

If I define the two dimensional sphere in the usual way, this gives me a metric $ds^2 = r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2$. Can I just define a new coordinate system giving a point coordinates $(\theta', \phi') = (\theta r^2, \phi r^2 \sin^2 \theta)$?. This gives me the metric $ds^2 = d\theta'^2 + d\phi'^2$.

If you transform from $\theta, \phi$ to $\theta', \phi'$, then the metric tensor (actually, the "line element", but you can figure out the metric tensor from that) transforms as follows:

$r^2 d\theta^2 + r^2 sin^2(\theta) \Rightarrow r^2 (\frac{\partial \theta}{\partial \theta'} d\theta' + \frac{\partial \theta}{\partial \phi'} d\phi')^2 + r^2 sin^2(\theta) (\frac{partial \phi}{\partial \theta'} d\theta' + \frac{\partial \phi}{\partial \phi'} d\phi')^2$

In your particular case, you want $\theta' = r \theta$ and $\phi' = r sin(\theta) \phi$. You have to get $\theta, \phi$ in terms of $\theta', \phi'$ first:

$\theta = \frac{1}{r} \theta'$
$\phi = \frac{1}{r sin(\frac{\theta'}{r})} \phi'$

So

$\frac{\partial \theta}{\partial \theta'} = \frac{1}{r}$
$\frac{\partial \theta}{\partial \phi'} = 0$
$\frac{\partial \phi}{\partial \theta'} = - \frac{cos(\frac{\theta'}{r})}{r^2 sin^2(\frac{\theta'}{r})} \phi'$
$\frac{\partial \phi}{\partial \phi'} = \frac{1}{r sin(\frac{\theta'}{r})}$

So in terms of $\phi', \theta'$ you have:

$r^2 (\frac{1}{r} d \theta' + 0)^2 + r^2 sin^2(\frac{\theta'}{r}) (- \frac{cos(\frac{\theta'}{r})}{r^2 sin^2(\frac{\theta'}{r})} \phi' d \theta' + \frac{1}{r sin(\frac{\theta'}{r})} d\phi')^2$

$= (d \theta')^2 + (d \phi')^2 +$ a bunch of other messy terms involving $(d \theta')^2$ and $d \theta' d\phi'$.

 

[kent davidge]

You are clearly not understanding what I am asking. That's because I have been unable to describe what I mean properly. But after thinking more carefully for some time I now can tell you what I mean in symbols/equations:

Given a manifold whose metric is $g_{\mu \nu} (x) dx^\mu dx^\nu$ (1) in a given coordinate system $\{ x \}$. For sack of discussion, suppose this manifold has curvature. The metric around any point $x = X$ is $g_{\mu \nu} (X) dx^\mu dx^\nu$ (2).

Now the coefficients $g_{\mu \nu} (X)$ are just constants, because they are function being evaluated at a specific point $X$. If I use (2) to evaluate the curvature tensor, it will vanish. (And so will any other quantity depending on the metric coefficients, like the connection.) In this way we see that around any point $x = X$ the metric is that of $\mathbb{R}^n$, and perhaps I can even make a coordinate transformation $$\delta_{\sigma \rho} = g_{\mu \nu} (X) \frac{\partial x^\mu}{\partial u^\sigma} \frac{\partial x^\nu}{\partial u^\rho}$$ to put the metric into the Euclidean form (Or Lorentz form if the manifold is Lorentzian). But I'm not sure. Anyways, that's what I was trying to say since my post #1. But unfortunately I used the wrong description there.

Now it's clear that if I want to have a complete description of my manifold, I should use $x = X$ in the curvature tensor computed from the metric (1), in which case I will get nonvanishing components, if the manifold has curvature.

 

[stevendaryl]

Given a manifold whose metric is $g_{\mu \nu} (x) dx^\mu dx^\nu$ (1) in a given coordinate system $\{ x \}$. For sack of discussion, suppose this manifold has curvature. The metric around any point $x = X$ is $g_{\mu \nu} (X) dx^\mu dx^\nu$ (2).

Now the coefficients $g_{\mu \nu} (X)$ are just constants, because they are function being evaluated at a specific point $X$.

I don't understand what you mean. They aren't constants, they are functions of $X$.

If I use (2) to evaluate the curvature tensor, it will vanish.

The curvature tensor depends not just on the metric tensor at a point, but also its first and second derivatives. Those aren't zero.

perhaps I can even make a coordinate transformation to put the metric into the Euclidean form (Or Lorentz form if the manifold is Lorentzian).

In general, if you're evaluating a tensor $g_{\mu \nu}$ at a point $X$, then you can, in the neighborhood of $X$, choose a coordinate system and approximate $g_{\mu \nu}$ by a power series.

$g_{\mu \nu}(X') \approx g_{\mu \nu} + (\frac{\partial}{\partial x^\lambda} g_{\mu \nu} ) (X^\lambda -X'^\lambda) + \frac{1}{2} (\frac{\partial^2}{\partial x^\lambda \partial x^\tau} g_{\mu \nu}) (X^\lambda - X'^\lambda) (X^\tau - X'^\tau) + ...$

(where on the right-side of $\approx$, $g_{\mu \nu}$ and its derivatives are evaluated at the point $X$)

You can choose a coordinate system where $\frac{\partial}{\partial x^\lambda} g_{\mu \nu} = 0$ at the point $X$, and where $g_{\mu \nu}$ is diagonal with entries $\pm 1$, but in general, you can't make the second derivatives equal to 0. The curvature tensor depends on the first and second derivatives of the metric, as well as the metric itself.

 

[kent davidge]

OK. I was reading a site called Relativity Made Easy and I saw some equations that gave me a hint into how to approach my question from another angle so that hoppefully you can understand me.

Consider the expression for the levi-civita connection coefficients $$\Gamma^\lambda_{\mu \nu} = \frac{dx^\lambda}{d yi^\alpha} \frac{\partial^2 y^\alpha}{\partial x^\mu x^\nu}$$ where the $\xi$ coordinates satisfy the equations $d^2 y^\alpha / d \tau^2 = 0$ with $\tau$ being the proper-time.

The above equation can be solved for $y$, right? If so, we are extending the "flat" coordinates to a small region of our "non-flat" manifold. However, I admit I've not seen this being done in the literature, so I'm not sure whether or not this works.

 

[stevendaryl]

OK. I was reading a site called Relativity Made Easy and I saw some equations that gave me a hint into how to approach my question from another angle so that hoppefully you can understand me.

Consider the expression for the levi-civita connection coefficients $$\Gamma^\lambda_{\mu \nu} = \frac{dx^\lambda}{d yi^\alpha} \frac{\partial^2 y^\alpha}{\partial x^\mu x^\nu}$$ where the $\xi$ coordinates satisfy the equations $d^2 y^\alpha / d \tau^2 = 0$ with $\tau$ being the proper-time.

The above equation can be solved for $y$, right? If so, we are extending the "flat" coordinates to a small region of our "non-flat" manifold. However, I admit I've not seen this being done in the literature, so I'm not sure whether or not this works.

Yes, you can find some coordinate system $y^\alpha$ that works. What that amounts to is finding a coordinate system $y^\alpha$ so that $\frac{\partial g_{\alpha \beta}}{\partial y^\gamma} = 0$ at a particular point. That doesn't mean that it's a "flat" coordinate system, because flatness is independent of which coordinate system you're using. As I said, curvature depends on the second derivatives of the metric (or alternatively, the first derivatives of $\Gamma^\alpha_{\beta \gamma}$). You can make the $\Gamma$ matrices equal to zero at a single point, but their derivatives won't be zero.

You can, more generally, come up with a coordinate system so that $\Gamma$ is zero everywhere along a particular geodesic. That's Fermi normal coordinates for the geodesic.

https://en.wikipedia.org/wiki/Fermi_coordinates

 

[lavinia]

One thing that can be done for any Riemannian surface not only the standard sphere is to define isothermal coordinates in a neighborhood of any point. In isothermal coordinates the metric has the form $ds^2 = ρ(dx^2+dy^2)$ where $ρ$ is a positive nowhere zero function. So at any point the metric differs from the standard flat metric by a scale factor $ρ$. In general, $ρ$ is not constant and varies from point to point so the metric is only conformal to the flat metric. Conformal means that angles are preserved but not necessarily lengths. In general a manifold that admits isothermal coordinates $ds^2 = ρΣ_{i}dx_{i}^2$ is said to be "conformally flat".

It is a different question to ask whether a topological sphere can be given a metric that is flat.