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.

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