# About coordinate transformations in general

1. Sep 17, 2010

### friend

I want to make sure of my understanding of coordinate transformations.

First of all, is it true that if $$${x_i}$$$ is a coordinate system on a manifold, then

$$${q_j} = {q_j}({x_i})$$$

is a coordinate transform from "x" space to "q" space? If so can "x" be a flat space and "q" a curved space? Thanks.

2. Sep 18, 2010

### Eynstone

q must be congruous & compatible for the transform to be valid & not just local. Whether to call a manifold 'flat' is by one's personal choice.

3. Sep 18, 2010

Usually whether the manifold is flat or not - that depends on the curvature tensor. Curvature tensor is a tensor - it transforms in a linear way. If it is zero in one coordinate system, it is zero in any other coordinate system.

But you can have your own definition of being flat or not-flat. It needs to be given first in order to answer your question. Give a simple example of what is on your mind - and then you will certainly get some help.

4. Sep 18, 2010

### lavinia

The $$${q_j}$$$ can not be arbitrary. The mapping from x's to q's must at least be a homeomorphism but often it is required that it have more structure e.g. it must be a diffeomorphism.

The geometry of a space does not depend upon coordinates but rather upon measurement of distance and angle. This requires further structure such as a Riemannian metric.

Often when one parameterizes a region on a manifold, the parameter domain is just a subset of Euclidean space and so is flat while the region on the surface may curve around in a higher dimensional space and thus is curved. But this has nothing to do with the coordinates transformations themselves but rather with the geometry of the regions.

5. Sep 19, 2010

### friend

I'm trying to understand how the curvature (derived from the metric, and thus the metric) can be inherant in the manifold and also how you can have your own definition of "flat".

Can the metric be defined apart from a coordinate systems, say, d(p1,p2), where p1 and p2 are points on the manifold somehow specified without coordinates and d(p1,p2) is the distance between p1 and p2? But then again a distance is a number, which can be arbitrarily scaled, right? Or is the curvature scale invariant?

6. Sep 19, 2010

The curvature tensor is scale invariant. But not the so called "scalar curvature".

Moreover, "flatness" depends on the "parallel transport". Two metrics differing by a constant scale factor define the same parallel transport. Parallel transport has a simple geometrical interpretation and can be defined without referring to coordinates. The only thing you need are "differentiable curves" and "tangent vectors". You do not even need a "metric".

7. Sep 19, 2010

### lavinia

A Riemannian metric is independent of the coordinate system. The distance between points is derived from the metric and for a complete manifold is equal to the length of a distance minimizing geodesic.

The curvature 2 forms are intrinsic because they are derived from the metric and not from the particualr realization of the manifold in another manifold e.g. in Euclidean space.

8. Sep 23, 2010

### friend

For example, consider the projection of the hemisphere onto a plane. Here we have

$$$x = r \cdot \sin \theta \cdot \cos \varphi$$$

$$$y = r \cdot \sin \theta \cdot \sin \varphi$$$

for fixed radius r, and the inverse,

$$$\theta = {\cos ^{ - 1}}(\frac{{\sqrt {{r^2} - {x^2} - {y^2}} }}{r})$$$

$$$\varphi = atan 2(y,x)$$$,

where atan2 is a variant of arctan, only it covers all quadrants.

The point is, here

$$$\theta = \theta (x,y)$$$ and $$$\varphi = \varphi (x,y)$$$

This would appear to be a coordinate transformation from flat space of $$$(x,y)$$$, to a curved space of $$$(\theta ,\varphi )$$$. Am I missing something?

Last edited: Sep 23, 2010
9. Sep 24, 2010

It can be viewed in three ways:

1) you have a flat space, with flat parallel transport in coordinates (x,y), you have a curved hemisphere with the natural parallel transport of the sphere and in coordinates $$(\theta,\varphi)$$ , you have a map from a flat Riemannian manifold to a non-flat one.
2) Both $$(x,y)$$ and $$(\theta,\varphi)$$ are just different coordinates on the open unit disc of the flat plane
3) Both $$(x,y)$$ and $$(\theta,\varphi)$$ are just different coordinates on the curved hemisphere

10. Sep 28, 2010

### friend

I'm trying to reconcile these statements with others that say that the curvature cannot change with a change of coordinates, which would preclude a coordinate transform from flat to curved space.

Is it possible that there may be contexts in which these statements apply? For example, I'm wondering about the difference between a "re-parameterization" and a "diffeomorphism". I can certainly imagine that a re-parameterization of a surface is not going to change its curvature. But then General Relativity talks about diffeomorphism invariance, where the laws of physics do not change with a diffeomorphism, the entire point of which is that a change in curvature does not change physics. Is this diffeomorphism a coordinate transformation as described above? Thanks.

11. Sep 29, 2010

The entire point is that a change of coordinates does not change the curvature. Changing the curvature changes physics. That's the entire point.

Curvature is coded in the curvature tensor, not in coordinates.

12. Oct 10, 2010

### friend

It seems to me that it would be a straightforward procedure to prove that curvature is invariant wrt coordintate changes since we have:

The scalar curvature is

$$$R = {g^{\sigma \nu }}{R_{\sigma \nu }}$$$

where $$${R_{\sigma \nu }}$$$ is the Ricci tensor

$$${R_{\sigma \nu }} = {R^\rho }_{\sigma \rho \nu }$$$

where $$${R^\rho }_{\sigma \mu \nu }$$$ is the Rieman curvature tensor

$$${R^\rho }_{\sigma \mu \nu } = {\partial _\mu }{\Gamma ^\rho }_{\nu \sigma } - {\partial _\nu }{\Gamma ^\rho }_{\mu \sigma } + ({\Gamma ^\rho }{_\mu _\lambda }{\Gamma ^\lambda }_{\nu \sigma } - {\Gamma ^\rho }{_\nu _\lambda }{\Gamma ^\lambda }_{\mu \sigma })$$$

where $$${\Gamma ^\rho }_{\nu \sigma }$$$ is the Christoffel symbol

$$${\Gamma ^\rho }_{\nu \sigma } = \frac{1}{2}{g^{\rho \alpha }}(\frac{\partial }{{\partial {q^\sigma }}}{g_{\alpha \nu }} + \frac{\partial }{{\partial {q^\nu }}}{g_{\alpha \sigma }} - \frac{\partial }{{\partial {q^\alpha }}}{g_{\nu \sigma }})$$$

where g is the metric tensor such that

$$${\rm{g = }}{{\rm{J}}^T}{\rm{J}}$$$

where J is the jacobian

$$${\rm{J}} = \left[ {\begin{array}{*{20}{c}} {\frac{{\partial {q^1}}}{{\partial {x^1}}}}&{\frac{{\partial {q^1}}}{{\partial {x^2}}}}& \cdots &{\frac{{\partial {q^1}}}{{\partial {x^n}}}}\\ {\frac{{\partial {q^2}}}{{\partial {x^1}}}}&{\frac{{\partial {q^2}}}{{\partial {x^2}}}}& \cdots &{\frac{{\partial {q^2}}}{{\partial {x^n}}}}\\ \vdots & \vdots & \ddots & \vdots \\ {\frac{{\partial {q^m}}}{{\partial {x^1}}}}&{\frac{{\partial {q^m}}}{{\partial {x^2}}}}& \cdots &{\frac{{\partial {q^m}}}{{\partial {x^n}}}} \end{array}} \right]$$$

where $$${q^\mu } = {q^\mu }({x^i})$$$ is the coordinate transformation.

I'm thinking that if someone went through all the tedious partial derivatives they might prove that the scalar curvature is invariant, or at least under what conditions of the partials that it remains invariant. Has anyone seen such a proof ever worked out? Or would someone like to work it out? I'm not sure I have that much skill or patience. Thanks.

Last edited: Oct 10, 2010
13. Oct 10, 2010

If you calculate the curvature tensor with your metric, you will get

$$${R^\rho }_{\sigma \mu \nu } = 0$$ 14. Oct 15, 2010 ### friend Prof. Kleinert, in his book, Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, pages 769-772, says there are conditions on $$\[{q^\mu } = {q^\mu }({x^i})$$$ which determine whether it is a transformation from flat to flat space or from flat to curved space. He finds that the affine connection can be written:

$$${\Gamma _{\lambda \nu }}^\mu = {\partial _i}{q^\mu } \cdot {\partial _\lambda }{\partial _\nu }{x^i} = \frac{{\partial {q^\mu }}}{{\partial {x^i}}} \cdot \frac{\partial }{{\partial {q^\lambda }}}\left( {\frac{{\partial {x^i}}}{{\partial {q^\nu }}}} \right)$$$

And when this is put into Riemann cuvature tensor,

$$${R_{\mu \nu \lambda }}^\kappa = {\partial _\nu }{\Gamma ^\kappa }_{\lambda \mu } - {\partial _\lambda }{\Gamma ^\kappa }_{\nu \mu } + ({\Gamma ^\kappa }{_\nu _\omega }{\Gamma ^\omega }_{\lambda \mu } - {\Gamma ^\kappa }{_\lambda _\omega }{\Gamma ^\omega }_{\nu \mu })$$$

he gets,

$$${R_{\mu \nu \lambda }}^\kappa = {\partial _a}{q^\kappa } \cdot \left( {{\partial _\mu }{\partial _\nu } - {\partial _\nu }{\partial _\mu }} \right){\partial _\lambda }{x^a}$$$

He then writes, "A transformation for which $$${x^a}\left( q \right)$$$ have commuting derivatives, while the first derivatives $$${\partial _\mu }{x^a}\left( q \right)$$$ do not, carries a flat-space region into a purely curved one."

Does anyone have experience with the proof of this statement? Thanks.

Last edited: Oct 15, 2010
15. Nov 8, 2010

### friend

When dealing with whether we can have a coordinate transformation from flat to curved space, part of the problem may be that we may be used to considering coordinate transformations as transition maps in the overlap of two coordinate charts on the same manifold. So, of course, the curvature is not going to change since that's dealing with different charts on the same manifold. The overlap is still only on one manifold which only has a unique curvature field in that overlap region.

Let's consider a transformation of a disk described with (x,y) coordinates with a euclidean metric to a hemisphere with coordinates $$$(\theta ,\phi )$$$. I'm wondering if such a coordinate transformation can be achieved by use of a larger space, say 3 dimensional Euclidean space, E3. First, transform from the disc to E3 where there is overlap, then transform from E3 to $$$(\theta ,\phi )$$$ coordinates of the hemisphere where again there is overlap. Do the E3 coordinates cancel out?

16. Nov 8, 2010

Is Kleinert giving some examples? Just one would do.

17. Nov 8, 2010

### friend

Actually, it's a 1500 page book, but I have not found such examples. You would think this is a fundamental question that should be studied up front, but if he does address these issues, then his proof is distributed somewhere amoung those 1500 pages. Obviously, the primary example of transformation from flat to curved space is going from a disc to a hemisphere. And the obvious question is does the metric still transform like a tensor in that situation. But I've not seen anywhere on the Web or any book I have where this question is addressed. If it were not possible to do this type of transformation, you'd think they would show it.

18. Nov 8, 2010

This is what makes me suspicious. You can easily find examples with functions that are nondifferentiable a certain number of times here and there, but to have it over an open set? I really would like to see one example - otherwise his whole theory may well have an empty content. And any reasonable map from a disk to a hemisphere will be differentaible as many times as you want except, perhaps along one or several radii - if you really want it.

19. Nov 14, 2010

### friend

Now I'm not so sure a coordinate transformation can go from a flat space to a curved space. For it seems that coordinates are something inscribed on a space, and a coordinate transformation would simply be a change of coordinates on that same space. Perhaps it would be better to a those kind of transformations, a space transformation, not a coordinate transformation. Or it may be that a transformation of the type: $$${q^\mu } = {q^\mu }({x^i})$$$ is always considered to be a "coordinate transformation" since the $$${q^\mu }$$$ are always considered coordinates. Anyone have any experience with this distinction?

20. Nov 14, 2010

As long as it is not said explicitly when you just see $$q^i(x^\mu)$$ it can be interpreted both ways. Most authors are very clear when they write their formulas, some other prefer to be fuzzy - perhaps that is the way they are thinking.