Covariant Basis

One lecturer presented it by doing the work for a specific basis, polar coordinates, and then translating that work to a generic template.

Once the axis for zero degrees is set (to be used for theta) and unit for length is set (to be used for r) we have a function \overrightarrow R(r,\theta).

\overrightarrow e_r = \dfrac {\partial \overrightarrow R (r, \theta)}{\partial r}

\overrightarrow e_\theta = \dfrac {\partial \overrightarrow R (r, \theta)}{\partial \theta}

For the general case:

\overrightarrow R (Z^1, Z^2)

\overrightarrow e_{Z^1} = \dfrac {\partial \overrightarrow R (Z^1, Z^2)} {\partial Z^1}

\overrightarrow e_{Z^2} = \dfrac {\partial \overrightarrow R (Z^1, Z^2)} {\partial Z^2}

One last thing, functions like \overrightarrow R(r,\theta) and \overrightarrow R (Z^1, Z^2) are Position Vectors.

Scratch Work Below

The work below deserves its own place elsewhere, we are putting it here for now so we won’t lose it.

What we have below we might jokingly say takes the form “mess = single symbol”.

Z^{i'} (Z (Z')) = Z^{i'}

The above might refer to the example shown below. For this story that ‘i’ goes from 1 to 3:

r(x(r,\theta,\phi), y( r,\theta,\phi), z( r,\theta,\phi) ) \equiv r

\theta(x(r,\theta,\phi), y( r,\theta,\phi), z( r,\theta,\phi) ) \equiv \theta

\phi(x(r,\theta,\phi), y( r,\theta,\phi), z( r,\theta,\phi) ) \equiv \phi

we are 34 minutes into 4a of MathTheBeautiful by Pavel Grinfeld

We can dive a little deeper into the above idea if we do some work in two dimensions with Cartesian and Polar Coordinates.

x(r,\theta) = r cos(\theta)

y(r,\theta) = r sin(\theta)

r(x,y) = \sqrt{x^2 + y^2}

\theta(x,y) = arctan \dfrac {y} {x}

Let’s take r(x,y) = \sqrt{x^2 + y^2} and see what happens when we replace x and y with their functions:

  • x = r cos(\theta)
  • y = r sin(\theta)

\sqrt{x^2 + y^2}

\sqrt{r^2 cos^2(\theta) + r^2 sin^2(\theta)}

\sqrt{r^2[(cos^2(\theta) + sin^2(\theta)]}

r \sqrt{cos^2(\theta) + sin^2(\theta)}

r \sqrt{1}

r (1)

r