# Solving Curl A in Spherical Coordinates: Tips & Hints

• phos19
In summary, the conversation discusses a problem involving curl and spherical coordinates. The equations are given and the speaker has tried to solve them using different methods. However, they are looking for a simpler solution or a "trick" to solve the problem. They mention a vector identity and provide some hints for solving the problem. Ultimately, the conversation suggests substituting a specific equation to solve the problem.
phos19
Homework Statement
Let $$\vec{B} =\dfrac{1}{4 \pi} \dfrac{-3}{r^4} ( 3\cos^2{\theta} - 1) \hat{u_r} + \dfrac{1}{4 \pi} \dfrac{1}{r^4} ( - 6 \cos{\theta} \sin{\theta} ) \hat{u_{\theta}}$$ (spherical unit vectors)

Find ##\vec{A}## such that ## \vec{B} = \nabla \times \vec{A}##
Relevant Equations
(The ##\vec{B}## is divergenceless !)
I've tried writing the curl A (in spherical coord.) and equating the components, but I end up with something that is beyond me:

{\displaystyle {\begin{aligned}{B_r = \dfrac{1}{4 \pi} \dfrac{-3}{r^4} ( 3\cos^2{\theta} - 1) =\frac {1}{r\sin \theta }}\left({\frac {\partial }{\partial \theta }}\left(A_{\varphi }\sin \theta \right)-{\frac {\partial A_{\theta }}{\partial \varphi }}\right)&\\B_{\theta}= \dfrac{1}{4 \pi} \dfrac{1}{r^4} ( - 6 \cos{\theta} \sin{\theta} ) ={}+{\frac {1}{r}}\left({\frac {1}{\sin \theta }}{\frac {\partial A_{r}}{\partial \varphi }}-{\frac {\partial }{\partial r}}\left(rA_{\varphi }\right)\right)&\\B_{\varphi}= 0={}+{\frac {1}{r}}\left({\frac {\partial }{\partial r}}\left(rA_{\theta }\right)-{\frac {\partial A_{r}}{\partial \theta }}\right)&\end{aligned}}}

Is there a "trick" to solve this , or maybe some vector identity to simplify the problem ?
Any hints are greatly appreciated , thanks!

Last edited:
You can assume that nothing depends on $\phi$, and the third equation is satisfied by $A_r = A_\theta = 0$. That leaves $$\begin{split} B_r &= \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} (A_\phi \sin \theta) \\ B_\theta &= -\frac{1}{r} \frac{\partial}{\partial r} (r A_\phi) \end{split}$$ and now substituting $A_\phi = r^\alpha f(\theta)$ will solve the problem.

vanhees71, PhDeezNutz and topsquark

## 1. How do I convert Cartesian coordinates to spherical coordinates?

To convert Cartesian coordinates (x, y, z) to spherical coordinates (r, θ, φ), use the following formulas:
r = √(x² + y² + z²)
θ = arccos(z / r)
φ = arctan(y / x)

## 2. What is the formula for calculating the curl in spherical coordinates?

The formula for calculating the curl (A) in spherical coordinates is:
∇ × A = 1 / (r sin θ) * [∂(Aₓ sin θ) / ∂θ - ∂Aᵧ / ∂φ + ∂(Aᵧ cos θ) / ∂r]ᵢₓ ᵥᵢₓ + [∂Aᵧ / ∂θ + ∂(Aₓ cos θ) / ∂r]ʲᵧ ᵥᵢₓ + 1 / r * [∂(rAᵧ) / ∂θ - ∂(rAₓ) / ∂φ]ᵏᵧ ᵥᵢₓ

## 3. What are some common mistakes when solving for curl A in spherical coordinates?

Some common mistakes when solving for curl A in spherical coordinates include:
- Forgetting to convert the Cartesian coordinates to spherical coordinates
- Incorrectly applying the conversion formulas
- Forgetting to take into account the unit vectors (ᵢ, ʲ, ᵏ)
- Making errors in differentiating the components of A
- Not simplifying the final expression properly

## 4. How can I simplify the expression for curl A in spherical coordinates?

To simplify the expression for curl A in spherical coordinates, you can use the following identities:
- sin θ / r = ∂(1 / r) / ∂θ
- cos θ / r = ∂(1 / r) / ∂r
- sin θ / r² = -∂(cos θ) / ∂θ
- cos θ / r² = ∂(sin θ) / ∂θ

## 5. Are there any helpful tips for solving curl A in spherical coordinates?

Some helpful tips for solving curl A in spherical coordinates include:
- Double check your conversion from Cartesian to spherical coordinates
- Use the identities mentioned above to simplify the expression
- Pay attention to the unit vectors (ᵢ, ʲ, ᵏ) and make sure they are included in the final expression
- Practice differentiating spherical coordinates to become more familiar with the process
- Check your final answer by converting it back to Cartesian coordinates and comparing it to the original expression for curl A

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