# Component definition in curvilinear coordinates

I'm watching this lecture that gives an introduction to tensors. If we have a coordinate system that's an affine transformation of the Cartesian coordinate system, then the projection of a vector $$v$$ (onto a particular axis) is defined as $$v_m = v.e_m$$ or the dot product of the vector with the corresponding basis vector (mentioned at this timestamp).

Here the prof states that if the "distance representing each coordinate separation" were the same, then the projections would correspond to components of the vector, or that $$v_m=v.e_m=v^m$$, where $$v=v^me_m$$ (Einstein summation convention).

First question: what's a precise way of defining "distance representing each coordinate separation"? Does it refer to the distance between $$x^m=k$$ and $$x^m=k+1$$ keeping all other coordinates fixed?

This is all fine, but what do we do if the surfaces $$x^m=k$$ were curved (curvilinear coordinate system)? And even if they weren't curved, what if the "distance representing each coordinate separation" varied? In either case, how would you define a projection and how would you define a component?

A natural intuitive way of defining the projection/component for a curvilinear coordinate system with equally spaced coordinate separations would be something like this. [Can't upload image for some reason].

Would appreciate any help in clarifying my concepts!

• Math mods: Please merge. – Qmechanic Dec 29 '19 at 5:42

## 2 Answers

Briefly, when you are considering curved coordinate systems, you need to work with local differentials $$(\partial \vec x/\partial x^m)$$ rather than $$(\Delta \vec x/\Delta x^m)$$ over distances where the curvature is important, the same as you work with your instantaneous velocity $$(d \vec x/d t)$$ rather than the average velocity $$(\Delta \vec x/\Delta t)$$ when your velocity has been significantly changing over the time $$\Delta t$$.

• Thanks! But could you elaborate a bit? Currently the answer doesn't clarify any of the questions asked in the post. – Shirish Kulhari Dec 18 '19 at 9:52

One application of base vectors is crystal lattices. In the picture below, the distance of atoms is not the same between axis, and each base vector has a magnitude (in Angstroms for example) different from the others and not unitary. That means "distance between coordinate separations".

The point Q can be represented by coordinates (3,2,2) meaning 3a + 2b + 2c. But the projection of the vector PQ to a axis is the dot product $$PQ.a.cos(\theta)$$. That would be 3 only if a = 1.

If the angles between the base vectors were not $$90^0$$ (and normally they aren't), the method using dot products would result in different numbers for the coordinates even if the base vectors were unitary. The first method results in a set of numbers called contravariant, and the second covariant coordinates.

We can not say that PQ = $$n_1$$a + $$n_2$$b + $$n_3$$c for covariant coordinates (when the angles are different). But it can be shown that there is another set of base vectors a', b' and c' (called reciprocal) for which: PQ = $$n_1$$a' + $$n_2$$b' + $$n_3$$c'

Both systems at least work requiring only one set of basis vectors. For curvilinear coordinates things are more complicated. The base vectors change in general when moving from point to point.

Think about flying between cities far north at opposite sides of the pole. If the plane follow the short path over the pole, the direction of the velocity will change suddenly from south-north to north-south. The plane didn't change directions, the base vectors changed.