- #1
pellman
- 683
- 6
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 dx \wedge dy with the basis vector \mathbf{\hat{e}}_z, the justification of which in general is not clear to me.
Consider instead spherical coordinates on R^3. Given a vector field
\mathbf{A}=A'^r\mathbf{\hat{r}} +A'^{\theta}\mathbf{\hat{\theta}} +A'^{\phi}\mathbf{\hat{\phi}}
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
\mathbf{A}=A^r\mathbf{\hat{e}}_r +A^{\theta}\mathbf{\hat{e}}_\theta +A^{\phi}\mathbf{\hat{e}}_\phi
where \mathbf{\hat{e}}_r = \mathbf{\hat{r}} \mathbf{\hat{e}}_\theta=r\mathbf{\hat{\theta}} \mathbf{\hat{e}}_\phi=r\sin\theta\mathbf{\hat{\phi}}
The \mathbf{\hat{e}}_j basis vectors are essentially identical to \partial_j. (The relationship between \mathbf{\hat{e}}_r,\mathbf{\hat{e}}_\theta,\mathbf{\hat{e}}_\phi and \mathbf{\hat{e}}_x=\mathbf{\hat{x}},\mathbf{\hat{e}}_y=\mathbf{\hat{y}}\mathbf{\hat{e}}_z=\mathbf{\hat{z}} is identical to the relationship between \partial_r,\partial_\theta,\partial_\phi and \partial_x,\partial_y,\partial_z) Now the curl of A is usually given with respect to the unit-vector (orthonormal) basis:
\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}}
We can form a basis dr,d\theta,d\phi dual to the coordinate basis and consider the 1-form A=A_r dr + A_\theta d\theta + A_\phi d\phi with exterior derivative 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 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 A'^r = A_rA'^\theta = \frac{1}{r} A_\thetaA'^\phi = \frac{1}{r\sin\theta} A_\phi
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:
\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
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
So what is the relationship between these two expressions. How do I understand that they are equivalent?
\frac{1}{r^2 \sin\theta} is equal to the Jacobian determinant \left| \frac{\partial (r,\theta,\phi)}{\partial (x,y,z)} \right| but I am not sure if that is coincidental or relevant here.
Consider instead spherical coordinates on R^3. Given a vector field
\mathbf{A}=A'^r\mathbf{\hat{r}} +A'^{\theta}\mathbf{\hat{\theta}} +A'^{\phi}\mathbf{\hat{\phi}}
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
\mathbf{A}=A^r\mathbf{\hat{e}}_r +A^{\theta}\mathbf{\hat{e}}_\theta +A^{\phi}\mathbf{\hat{e}}_\phi
where \mathbf{\hat{e}}_r = \mathbf{\hat{r}} \mathbf{\hat{e}}_\theta=r\mathbf{\hat{\theta}} \mathbf{\hat{e}}_\phi=r\sin\theta\mathbf{\hat{\phi}}
The \mathbf{\hat{e}}_j basis vectors are essentially identical to \partial_j. (The relationship between \mathbf{\hat{e}}_r,\mathbf{\hat{e}}_\theta,\mathbf{\hat{e}}_\phi and \mathbf{\hat{e}}_x=\mathbf{\hat{x}},\mathbf{\hat{e}}_y=\mathbf{\hat{y}}\mathbf{\hat{e}}_z=\mathbf{\hat{z}} is identical to the relationship between \partial_r,\partial_\theta,\partial_\phi and \partial_x,\partial_y,\partial_z) Now the curl of A is usually given with respect to the unit-vector (orthonormal) basis:
\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}}
We can form a basis dr,d\theta,d\phi dual to the coordinate basis and consider the 1-form A=A_r dr + A_\theta d\theta + A_\phi d\phi with exterior derivative 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 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 A'^r = A_rA'^\theta = \frac{1}{r} A_\thetaA'^\phi = \frac{1}{r\sin\theta} A_\phi
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:
\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
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
So what is the relationship between these two expressions. How do I understand that they are equivalent?
\frac{1}{r^2 \sin\theta} is equal to the Jacobian determinant \left| \frac{\partial (r,\theta,\phi)}{\partial (x,y,z)} \right| but I am not sure if that is coincidental or relevant here.
