Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Curl vs exterior derivative in spherical coords

  1. Dec 15, 2011 #1
    I am trying to get a good grasp of the relation between the curl of a vector field and the exterior derivative of a 1-form field. In cartesian coordinates for flat R^3 the relationship is misleadingly simple. However, it still requires us to make an identification of the 2-form basis [tex]dx \wedge dy[/tex] with the basis vector [tex]\mathbf{\hat{e}}_z[/tex], the justification of which in general is not clear to me.

    Consider instead spherical coordinates on R^3. Given a vector field

    [tex]\mathbf{A}=A'^r\mathbf{\hat{r}} +A'^{\theta}\mathbf{\hat{\theta}} +A'^{\phi}\mathbf{\hat{\phi}} [/tex]

    The basis vectors here are unit vectors. We have put a prime on the components to distinguish them from the components with respect to the coordinate basis of the tangent space. In the coordinate basis we have

    [tex]\mathbf{A}=A^r\mathbf{\hat{e}}_r +A^{\theta}\mathbf{\hat{e}}_\theta +A^{\phi}\mathbf{\hat{e}}_\phi [/tex]

    where [tex]\mathbf{\hat{e}}_r = \mathbf{\hat{r}}[/tex] [tex]\mathbf{\hat{e}}_\theta=r\mathbf{\hat{\theta}}[/tex] [tex]\mathbf{\hat{e}}_\phi=r\sin\theta\mathbf{\hat{\phi}}[/tex]

    The [tex]\mathbf{\hat{e}}_j[/tex] basis vectors are essentially identical to [tex]\partial_j[/tex]. (The relationship between [tex]\mathbf{\hat{e}}_r,\mathbf{\hat{e}}_\theta,\mathbf{\hat{e}}_\phi[/tex] and [tex]\mathbf{\hat{e}}_x=\mathbf{\hat{x}},\mathbf{\hat{e}}_y=\mathbf{\hat{y}}[/tex][tex]\mathbf{\hat{e}}_z=\mathbf{\hat{z}}[/tex] is identical to the relationship between [tex]\partial_r,\partial_\theta,\partial_\phi[/tex] and [tex]\partial_x,\partial_y,\partial_z[/tex]) Now the curl of A is usually given with respect to the unit-vector (orthonormal) basis:

    [tex]\nabla\times\mathbf{A}=\frac{1}{r\sin\theta}\left(\frac{\partial}{\partial\theta}(A'^\phi\sin\theta)-\frac{\partial A'^\theta}{\partial\phi}\right)\mathbf{\hat{r}} + \frac{1}{r} \left( \frac{1}{\sin\theta} \frac{\partial A'^r}{\partial \phi}- \frac{\partial}{\partial r}(rA'^{\phi}) \right)\mathbf{\hat{\theta}} + \frac{1}{r} \left( \frac{\partial}{\partial r}(rA'^{\theta}) - \frac{\partial A'^r}{\partial \theta} \right)\mathbf{\hat{\phi}}[/tex]

    We can form a basis [tex]dr,d\theta,d\phi[/tex] dual to the coordinate basis and consider the 1-form [tex]A=A_r dr + A_\theta d\theta + A_\phi d\phi[/tex] with exterior derivative [tex]dA = \left( \frac{\partial A_\phi}{\partial\theta} - \frac{\partial A_\theta}{\partial\phi} \right)d\theta \wedge d\phi + \left( \frac{\partial A_r}{\partial\phi} - \frac{\partial A_\phi}{\partial r} \right)d\phi \wedge dr + \left( \frac{\partial A_\theta}{\partial r} - \frac{\partial A_r}{\partial\theta} \right)dr \wedge d\theta[/tex] the components of which are not clearly related to those of the curl.

    It can be shown that the 1-form components (coordinate basis) are related to the vector components (normalized basis) by [tex]A'^r = A_r[/tex][tex]A'^\theta = \frac{1}{r} A_\theta[/tex][tex]A'^\phi = \frac{1}{r\sin\theta} A_\phi[/tex]

    If we write the expression for the curl in terms of the 1-form components and the coordinate basis vectors we get fairly close to the exterior derivative expression:

    [tex]\nabla\times\mathbf{A}= \left( \frac{\partial A_\phi}{\partial\theta} - \frac{\partial A_\theta}{\partial\phi} \right) \frac{1}{r^2 \sin\theta} \mathbf{\hat{e}}_r + \left( \frac{\partial A_r}{\partial\phi} - \frac{\partial A_\phi}{\partial r} \right)\frac{1}{r^2 \sin\theta} \mathbf{\hat{e}}_\theta +\left( \frac{\partial A_\theta}{\partial r} - \frac{\partial A_r}{\partial\theta} \right) \frac{1}{r^2 \sin\theta} \mathbf{\hat{e}}_\phi [/tex]
    [tex]dA = \left( \frac{\partial A_\phi}{\partial\theta} - \frac{\partial A_\theta}{\partial\phi} \right)d\theta \wedge d\phi + \left( \frac{\partial A_r}{\partial\phi} - \frac{\partial A_\phi}{\partial r} \right)d\phi \wedge dr + \left( \frac{\partial A_\theta}{\partial r} - \frac{\partial A_r}{\partial\theta} \right)dr \wedge d\theta[/tex]

    So what is the relationship between these two expressions. How do I understand that they are equivalent?

    [tex]\frac{1}{r^2 \sin\theta} [/tex] is equal to the Jacobian determinant [tex]\left| \frac{\partial (r,\theta,\phi)}{\partial (x,y,z)} \right|[/tex] but I am not sure if that is coincidental or relevant here.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted
Similar Discussions: Curl vs exterior derivative in spherical coords
  1. Exterior Derivative (Replies: 2)

  2. Exterior derivative (Replies: 18)

  3. Exterior derivative (Replies: 6)

  4. Exterior derivative (Replies: 5)

Loading...