Help me with Cylinder -coordinates problem, back to Cartesian or not? How to do it fast?

Question

Source of the problem, 3b here.

Problem Question

Electricity density in cylinder coordinates is $\bar{J}=e^{-r^2}\bar{e}_z$. Current creates magnetic field of the form $\bar{H}=H(r)\bar{e}_{\phi}$ where $H(0)=0$. Define $H(r)$ from the maxwell equations

$$\nabla\times\bar{H}=\frac{1}{r}\begin{vmatrix}\bar{e}_r & r\bar{e}_\phi & \bar{e}_z \\ \partial_r & \partial_\phi & \partial_z \\ H_r & r H_\phi & H_z \\ \end{vmatrix} = \bar{J}.$$

So

$$\begin{align} \nabla\times\bar{H} &= \frac{1}{r} \bar{e}_r \begin{vmatrix} \partial_\phi & \partial_z \\ r H_\phi & H_z \\ \end{vmatrix} - r\bar{e}_\phi \begin{vmatrix} \partial_r & \partial_z \\ H_r & H_z \\ \end{vmatrix} + \bar{e}_z \begin{vmatrix} \partial_r & \partial_\phi \\ H_r & r H_\phi \\ \end{vmatrix} \\ &= \frac{1}{r} \left( \bar{e}_r (\partial_\phi H_z-\partial_z H_\phi) - r\bar{e}_\phi (\partial_r H_z-\partial_z H_r) + \bar{e}_z (\partial_r rH_\phi-\partial_\phi H_r) \right). \end{align}$$

I messed the calculations here up when I tried to go back to Cartesian coordinates because it is otherwise hard for me to see the math. So I tried to think things with them

$$\begin{cases} x=r\cos(\phi) \\ y=r\sin(\phi) \\ z=h \\ \end{cases}$$

but it took me many pages of erroneous calculations and I could not finish on time. Now my friend suggested the below.

My friend's approach which I could not understand yet or his purpose, something to do with independence

$$\begin{align} H &= H(r)e_\phi\\ &=0+H_\phi e_\phi+0 \end{align}$$

where

$$\begin{cases} H_\phi = H(r) <---\text{ independent of R}\\ H_r = 0 \\ H_2 =0 \\ \end{cases}.$$

Could someone explain what I am doing wrong in going back to the Cartesian? I know it is not wrong but it is extremely slow way of doing things. I am not sure whether I was meant to remember the page 817 here or what is really essential to solve this problem?

Answer 1

This service blocked my original answer with stupid two days' ban so shortly

$$\nabla\times\bar{H}= \begin{pmatrix}\partial_x \\ \partial_y \\ \partial_y\end{pmatrix}\times\bar{H}\not = \begin{pmatrix}\partial_r \\ \partial_\alpha \\ \partial_\phi\end{pmatrix}\times\bar{H}$$

where I want to stress

$$\nabla \not = \begin{pmatrix}\partial_r \\ \partial_\alpha \\ \partial_\phi\end{pmatrix}.$$

Then with the WW, you will get first degree-differential equation. Sorry I am now missing all references but this was the crux point to realize, not to mix the $\nabla$ from cartesian coordinates to polar coordinates.

I will update this if I can find the original answer, stupid censorship, well perhaps this is just gamification -- I lost my answer, stupid. Now I am too angry to concentrate on this junk, $\nabla$ is defined in cartesian -- to calculate mock $\nabla$ in polar coordinates you need to do some weekend calculations... (I mean to verify the conversion formula).

Answer 2

What you are meant to remember is simply this:

Let $\vec{v}$ and $\vec{w}$ be two vectors in some vector space $V$. Let $\vec{e}_1, \ldots, \vec{e}_n$ be a basis of $V$, and let $\vec{v} = \sum v_i \vec{e}_i$ and $\vec{w} = \sum w_i\vec{e}_i$. Then $\vec{v} = \vec{w}$ if and only if for every $1 \leq i \leq n$ we have $v_i = w_i$.

In other words, the basis decomposition of a vector is unique. Hence the vector equation

$$ \nabla\times\vec{H} = \vec{J} $$

is exactly equivalent to the equation on their components expressed relative to a basis (and in this case we take the basis vectors of the cylindrical coordinates system)

$$ \begin{cases} 0 = J_r &= \left(\nabla\times H\right)_r = \frac{1}{r}(\partial_\phi H_z - r\partial_z H_\phi) \\ 0 = J_\phi &= \left(\nabla\times H\right)_\phi = \partial_rH_z - \partial_z H_r \\ e^{-r^2} = J_z &= \left(\nabla\times H\right)_z = \frac{1}{r}\left(\partial_r(rH_\phi) - \partial_\phi H_r\right) \end{cases}$$

Now by assumption $H_r = H_z = 0$, since $\vec{H}$ only has the $\phi$ component, which we write as $H_\phi$. By assumption $H_\phi$ depends only on $r$. This shows that the first two equations in the system are automatically satisfied. To solve for $H_\phi$, you examine the third equation, which now reduces to

$$ r e^{-r^2} = \partial_r (r H_\phi) $$

Noticing that

$$ \partial_r e^{-r^2} = -2r e^{-r^2} $$

you get

$$ \partial_r \left( -\frac12 e^{-r^2} + C\right) = \partial_r (r H_\phi) $$

Taking the antiderivative on both sides, you have the solution for $H_\phi$ in terms of $r$, with a free parameter $C$, which you can then set using the boundary condition $H_\phi(r=0) = 0$.