# Rigid motion in curvilinear coordinates

I would like someone to clarify this since it has bedazzled me and can't seem to get a grip on it. Consider a 3D real space and Euclidean coordinates ($x_1,x_2,x_3$), with an associated standard basis E={$e_1,e_2,e_3$}. Now also consider new curvilinear coordinates ($y_1,y_2,y_3$) such that $y_i=f_i(x_1,x_2,x_3)$. To each point y there is a local covariant basis $g_i=\frac{\partial x_j}{\partial y_i}e_j$ (summation convention in use. In matrix notation just consider G=J E, with J a $3\times 3$ matrix). The equation relates the Euclidean frame and the local frame.

First question: It is my understanding that vectors live on the tangent space of the base space at point y. Thus a vector at y (or the respective x) lives in a different space than a vector at the origin. In other words, the local basis and the Euclidean basis are defined at different points. How can then, they be associated? i.e. how can G=J E hold since the two frames are defined at different points?

(In a flat space space, one could transport parallely the E frame, in a very trivial way, from the origin to the point x and then associated the two frames at the same point. In general manifolds one should define a connection.)

Second question: Why does this matter? Because in order to describe a rigid motion, you also need the coordinates of the origin i.e. you need an affine space. Thus, should the relation between the E frame (at the origin) and the local frame G (at x) be something like:

$$g_i=\frac{\partial x_j}{\partial y_i}e_j'$$

where,

$$e_j'=e_j+T$$

that is, the transported E at the point x?

EDIT:

Regarding my second question, should the transformation from global to local coordinates be defined as, $$\begin{bmatrix} u_1\\ u_2\\ u_3\\ 0 \end{bmatrix}=\begin{bmatrix} J & \mathbf{x}\\ \mathbf{0}& 1 \end{bmatrix}\begin{bmatrix} v_1\\ v_2\\ v_3\\ 0 \end{bmatrix}$$

where $v_i$ are the components of a vector located at x in the global frame , and $u$ the components in the local frame at x, and $J=[\frac{\partial x_j}{\partial y_i}]$ (i.e. G=J E), using of course homogenous coordinates.

-

## 1 Answer

What you are looking for is the notion of parallel transport or connexion. Like you observed, in general the tangent spaces of two distinct points on a smooth manifold are not comparable. As it turns out, in general, to relate the tangent spaces between two points on a smooth manifold, it is not sufficient to just specify which two points you want to compare: it is also necessary to specify which path you use to get from one point to the other. This is because on a general smooth manifold it is not guaranteed that you can pick a consistent set of transformations to relate the tangent spaces between all possible pairs of points. (A manifold for which such consistent choice is possible is very special, and is called a parallelizable manifold.)

A particular case to consider the the two dimensional sphere. If there is a consistent choice of transformations that relate the tangent spaces between all possible pairs of points, you can fix a set of basis vectors at one point, and using the transformation, you can extend this to a set of basis vectors in the tangent space of every point. For the two dimensional sphere, however, the hairy ball theorem says that this is not possible, if you additionally require that the parallel transports of a vector be continuous.

-
Thanks Willie, however i was aware of the notion of "connection/ parallel transport". There is no need to go to general manifold theory in my question. Just consider that we have the plain old 3D Euclidean space. Please see the edit on the question which clarifies my question further. The problem is to specifically derive the transformation from local to global. –  Jorge Jan 24 '12 at 17:10