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The divergence operator in a rotated reference frame

  1. Jul 12, 2012 #1
    One can easily prove that [itex]\nabla \cdot f[/itex] 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 [itex]\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'}[/itex] where [itex]\mathbf r' = U \mathbf r[/itex] is a coordinate transformation with U orthogonal.

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

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

    This gives that [itex]\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'[/itex]

    Where did I err?
  2. jcsd
  3. Jul 13, 2012 #2
    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 [itex]\mathbf e = U^T \mathbf e'[/itex] and things work out, i.e. [itex]\nabla = \mathbf e^T \cdot \partial = \mathbf e'^T \cdot \partial '[/itex]
  4. Jul 15, 2012 #3
    Just bust out some sines and cosines and you should find it all falls out after a bit of tedious algebra.
  5. Jul 15, 2012 #4
    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.
  6. Jul 15, 2012 #5
    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 :)
  7. Jul 15, 2012 #6
    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.
  8. Jul 15, 2012 #7
    This might be easier to do in polar coordinates. Just a wild guess off the top of my head.
  9. Jul 15, 2012 #8
    To be fair, your method is actually more appropriate and does generalise nicely. I just put mine forward as a nice explicit way to show that it's definitely true.

    I usually go explicit first in the simplest case, then use nicer and more general notation and then finally generalise a result. I just find it helps build the intuition along the way as nice notation often distances you from that physical interpretation of what's going on.
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