Show that w is solenoidal having spherical polar coordinates

[YogiBear]

Homework Statement

The gradient, divergence and curl in spherical polar coordinates r, ∅, Ψ are
nablaΨ = ∂Φ/∂r * e_r + ∂Φ/∂∅ * e_∅ 1/r + ∂Φ/∂Ψ * e_Ψ * 1/(r*sin(∅))
nabla * a = 1/r * ∂/∂r(r²*a_r) + 1/(r*sin(∅)*∂/∂∅[sin(∅)a_∅] + 1/r*sin(∅) * ∂a_Ψ/∂Ψ
nabla x a =

|e_r r*e_∅ r*sin(∅)*e_Ψ |
|∂/∂r ∂/∂∅ ∂/dΨ |
| 1/r²*sin(∅) |a_r r*a_∅ r*sin(∅)*a_Ψ |

Where a = a_r * e_r + a_∅*e_∅ + a_Ψ*e_Ψ

Suppose that the vector field w has the form w = w_Ψ(r, ∅) e_Ψ Show that w is
solenoidal and find w_Ψ(r, ∅), when w is also irrotational. Find a potential for w in this case.

Homework Equations

The Attempt at a Solution

I started by finding vector potential. however it seems too simple. (Or most likely id id it wrong). In addition i am not sure, if it is even the right way to about it. I can't put in working out of the vector potential i have, as i have trouble with coding it on the forum(My handwriting is not legible). So you can assume i have got it wrong. Any help will be greatly appreciated.

 

[marksman95]

First.

To show that ω is solenoidal implies that the divergence of the vector field is 0. Thats easy to show:

$$∇·\omega = \frac{1}{r\sin\theta}\; \frac{\partial\omega_\phi(r,\theta)}{\partial\phi}$$

and since the φ component of ω does not depend on φ, it's partial derivative equals 0.

So the vector field is solenoidal.

Second.

We must impose that ∇×ω=0.
$$∇×\omega=\begin{vmatrix} ê_r & rê_\theta &r\sin\thetaê_\phi \\ \frac{\partial}{\partial r} & \frac{\partial}{\partial \theta}& \frac{\partial}{\partial \phi} \\ 0&0&r\sin\theta\omega_\phi\end{vmatrix} \;=\; ê_r \left[ \frac{\partial\omega_\phi}{\partial \theta}r\cos\theta \right] - ê_\theta \left[ \frac{\partial\omega_\phi}{\partial r}r\sin\theta \right] $$

And this is 0 ∀r,φ if ω_φ is not really dependent on θ and r. So ω_φ=C (is constant on all variables).

And now I am stuck... I really doubt on this second result. I'm sorry, but I'll wait until someone else come check it. Maybe this help you on finding a path.

 

[YogiBear]

First.

To show that ω is solenoidal implies that the divergence of the vector field is 0. Thats easy to show:

$$∇·\omega = \frac{1}{r\sin\theta}\; \frac{\partial\omega_\phi(r,\theta)}{\partial\phi}$$

and since the φ component of ω does not depend on φ, it's partial derivative equals 0.

So the vector field is solenoidal.

Second.

We must impose that ∇×ω=0.
$$∇×\omega=\begin{vmatrix} ê_r & rê_\theta &r\sin\thetaê_\phi \\ \frac{\partial}{\partial r} & \frac{\partial}{\partial \theta}& \frac{\partial}{\partial \phi} \\ 0&0&r\sin\theta\omega_\phi\end{vmatrix} \;=\; ê_r \left[ \frac{\partial\omega_\phi}{\partial \theta}r\cos\theta \right] - ê_\theta \left[ \frac{\partial\omega_\phi}{\partial r}r\sin\theta \right] $$

And this is 0 ∀r,φ if ω_φ is not really dependent on θ and r. So ω_φ=C (is constant on all variables).

And now I am stuck... I really doubt on this second result. I'm sorry, but I'll wait until someone else come check it. Maybe this help you on finding a path.

I have figured out part 2, very similar to part one. Cross multiply the matrix, and we get 0. I think I am getting somewhere on part three as well. Thank you. And your LaTeX skills are too good xD

 

[marksman95]

MMMM... I think you got that rotational wrong. You cannot cancel those partial derivatives inside the determinant.

Those vectors aren't zero, but their associated components.

Also I think you haven't understood the exercice. It doesn't demand you to show that is irrotational, but to assume it is in order to find out what is the function of the phi component.

 

[YogiBear]

Of course. I rushed to conclusion without reading question fully...

thanks for the help thought, i would have had it completely wrong otherwise.