# Gradient as covector: example in 2-d

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 6; Problem 6.2.

One example of a covariant vector is the gradient. As an example, suppose we have a 2-d scalar field given by ${\Phi=bxy=br^{2}\cos\theta\sin\theta}$. In rectangular coordinates

 $\displaystyle \frac{\partial\Phi}{\partial x}$ $\displaystyle =$ $\displaystyle by\ \ \ \ \ (1)$ $\displaystyle \frac{\partial\Phi}{\partial y}$ $\displaystyle =$ $\displaystyle bx \ \ \ \ \ (2)$

In polar coordinates

 $\displaystyle \frac{\partial\Phi}{\partial r}$ $\displaystyle =$ $\displaystyle 2br\cos\theta\sin\theta\ \ \ \ \ (3)$ $\displaystyle \frac{\partial\Phi}{\partial\theta}$ $\displaystyle =$ $\displaystyle br^{2}\left(\cos^{2}\theta-\sin^{2}\theta\right) \ \ \ \ \ (4)$

Note that because we have absorbed the factor of ${r}$ needed for an incremental displacement in the ${\theta}$ direction into the basis vector ${\mathbf{e}_{\theta}}$, there is no extra factor of ${1/r}$ in the ${\frac{\partial\Phi}{\partial\theta}}$ term, as there would be if we had used unit basis vectors.

Now suppose we have a vector ${v^{i}}$ with components given in rectangular coordinates. Then the scalar product is

$\displaystyle v^{i}\partial_{i}\Phi=byv^{x}+bxv^{y} \ \ \ \ \ (5)$

If we convert ${v}$ to polar coords, then

 $\displaystyle v^{r}$ $\displaystyle =$ $\displaystyle v^{x}\cos\theta+v^{y}\sin\theta\ \ \ \ \ (6)$ $\displaystyle v^{\theta}$ $\displaystyle =$ $\displaystyle -v^{x}\frac{\sin\theta}{r}+v^{y}\frac{\cos\theta}{r} \ \ \ \ \ (7)$

The scalar product now is

 $\displaystyle v^{i}\partial_{i}\Phi$ $\displaystyle =$ $\displaystyle \left(v^{x}\cos\theta+v^{y}\sin\theta\right)\left(2br\cos\theta\sin\theta\right)+\left(-v^{x}\frac{\sin\theta}{r}+v^{y}\frac{\cos\theta}{r}\right)br^{2}\left(\cos^{2}\theta-\sin^{2}\theta\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle brv^{x}\sin\theta\left(2\cos^{2}\theta-\cos^{2}\theta+\sin^{2}\theta\right)+brv^{y}\cos\theta\left(2\sin^{2}\theta+\cos^{2}\theta-\sin^{2}\theta\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle brv^{x}\sin\theta+brv^{y}\cos\theta\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle byv^{x}+bxv^{y} \ \ \ \ \ (11)$

Thus the scalar product is invariant.