# The divergence operator in a rotated reference frame

1. Jul 12, 2012

### nonequilibrium

One can easily prove that $\nabla \cdot f$ is invariant under a rotation of the reference frame, however I would like to prove that the divergence operator itself is invariant (same principle, different approach). In other words I want to prove that $\mathbf \nabla = \mathbf e_x \frac{\partial}{\partial x} + \mathbf e_y \frac{\partial}{\partial y} = \mathbf e_{x'} \frac{\partial}{\partial x'} + \mathbf e_{y'} \frac{\partial}{\partial y'}$ where $\mathbf r' = U \mathbf r$ is a coordinate transformation with U orthogonal.

I think matrix notation will simplify things. Rewrite $\nabla = \mathbf e^T \cdot \partial$ where we define \mathbf e = \left( \begin{align} \mathbf e_x \\ \mathbf e_y \end{align}\right) and \partial = \left( \begin{align} \partial_x \\ \partial_y \end{align} \right).

If I remember correctly, the basis transformation is the inverse of the coordinate transformation, i.e. $\mathbf e = U \mathbf e'$. Also, one can easily check that $\partial = U^T \partial'$ (e.g. $\partial_{x} = \frac{\partial x'}{\partial x} \partial_{x'} + \frac{\partial y'}{\partial x} \partial_{y'} = U_{11} \partial_{x'} + U_{21} \partial_{y'}$)

This gives that $\nabla = \mathbf e^T \cdot \partial = \left( U \mathbf e' \right)^T \cdot \left( U^T \partial' \right) = \mathbf e'^T \; U^T U^T \; \partial' \neq \mathbf e'^T \cdot \partial'$

Where did I err?

2. Jul 13, 2012

### nonequilibrium

Actually, googling it a bit, it seems derivatives and basis vectors transform the same way (apparently called "covariantly", although I'm not yet familiar with such language).

So in that case $\mathbf e = U^T \mathbf e'$ and things work out, i.e. $\nabla = \mathbf e^T \cdot \partial = \mathbf e'^T \cdot \partial '$

3. Jul 15, 2012

### Marioeden

Just bust out some sines and cosines and you should find it all falls out after a bit of tedious algebra.

4. Jul 15, 2012

### nonequilibrium

Seems like bad advice, the whole notion of vector calculus was invented to avoid the tedious algebra. Anyway it seems I was able to work it out as shown in my previous post.

5. Jul 15, 2012

### Marioeden

I say tedious, it's like four lines:
x' = xcos(t) + ysin(t)
y' = -xsin(t) + ycos(t)

(same for the vectors only you have e1 for x and e2 for y)

then just plug in and substitute d/dx = cos(t)d/dx' + sin(t)d/dy' and also for d/dy and you're done :)

6. Jul 15, 2012

### nonequilibrium

Oh first off, my previous reply was a bit "harsh" but I thought I was replying to another thread, in a different context (but also happened to be about vector calculus), so anyway sorry about that.

As for your suggestion: I don't see how that requires less work? You seem to be taking the same steps I did? Also my notation makes it valid for all dimensions, which is a nice perk, and also I don't have to explicitly calculate any products, so I'd say my calculations are neater.

7. Jul 15, 2012