# Curl of a vector in spherical coordinates

The curl of a Vector function in curvilinear coordinate system is given by $$\nabla \times A = \frac 1 {h_1 h_2 h_3} \begin{vmatrix} h_1 \hat e_1 & h_2 \hat e_2 & h_3 \hat e_3\\ \partial \over \partial x_1 & \partial \over \partial x_2 & \partial \over \partial x_3\\ h_1 A_1 & h_2 A_2 & h_3 A_3 \end{vmatrix} \hspace{20 mm} \mathbf{(1)}$$ where $h_1, h_2, h_3$ are scale factors. For spherical coordinates $$h_1 = 1, h_2 = r, h_3 = r\sin\theta$$

However I don't understand (1), which is also not explained in my book. How is it derived?? Can anyone explain me? Even links would be helpful. Thank you!!

• I'm sorry. I thought you were having trouble understanding the matrix expression. Maybe you can derive the formula in $(1)$ by using the spherical change of coordinates, apply the curl in Cartesian coordinates and the chain rule a couple of times. – talmid Jun 21 '12 at 17:46
• thank you for suggestion :) I'll try that and see – Santosh Linkha Jun 21 '12 at 17:47

Before doing the derivation, I'd like to explain the origin of the scale factors $h_i$. We will assume throughout that our curvilinear coordinates $x_1$, $x_2$, and $x_3$ are orthogonal, i.e. that the gradients $\nabla x_1$, $\nabla x_2$, $\nabla x_3$ are orthogonal vectors. We will also assume that they are right-handed, in the sense that $\widehat{e}_1\times\widehat{e}_2=\widehat{e}_3$.

The Origin of the Scale Factors

One important difference between curvilinear coordinates $x_1,x_2,x_3$ and standard $x,y,z$ coordinates is that curvilinear coordinates do not change at unit speed. That is, if we start at a point and move in the direction of $\widehat{e}_i$, we should not expect $x_i$ to increase at unit rate.

One consequence of this is that the gradients $\nabla x_i$ of the curvilinear coordinates are not unit vectors. For $x,y,z$ coordinates, we know that $$\nabla x \;=\; \widehat{\imath},\qquad \nabla y \;=\; \widehat{\jmath},\qquad\text{and}\qquad \nabla z\;=\; \widehat{k}.$$ However, for curvilinear coordinates, we get something like $$\nabla x_1 \;=\; \frac{1}{h_1}\widehat{e}_1,\qquad \nabla x_2 \;=\; \frac{1}{h_2}\widehat{e}_2,\qquad\text{and}\qquad \nabla x_3 \;=\; \frac{1}{h_3}\widehat{e}_3, \tag*{(1)}$$ where $h_1$, $h_2$, and $h_3$ are scalars.

The reciprocal $1/h_i$ of each scale factor represents the rate at which $x_i$ will change if we move in the direction of $\widehat{e}_i$ at unit speed. Equivalently, you can think of $h_i$ as the speed that you have to move if you want to increase $x_i$ at unit rate. For spherical coordinates, it should be geometrically obvious that $h_1 = 1$, $h_2 = r$, and $h_3 = r\sin\theta$.

We can use the scale factors to give a formula for the gradient in curvilinear coordinates. If $u$ is a scalar, we know from the chain rule that $$\nabla u \;=\; \frac{\partial u}{\partial x_1}\nabla x_1 \,+\, \frac{\partial u}{\partial x_2}\nabla x_2 \,+\, \frac{\partial u}{\partial x_3}\nabla x_3$$ Substituting in the formulas from (1) gives us $$\nabla u \;=\; \frac{1}{h_1}\frac{\partial u}{\partial x_1}\widehat{e}_1 \,+\, \frac{1}{h_2}\frac{\partial u}{\partial x_2}\widehat{e}_2 \,+\, \frac{1}{h_3}\frac{\partial u}{\partial x_3}\widehat{e}_3\tag*{(2)}$$ This is the formula for the gradient in curvilinear coordinates.
First, observe that the determinant formula you have given for the curl is equivalent to the following three formulas: $$\begin{gather*} (\nabla\times A)\cdot\widehat{e}_1 \;=\; \frac{1}{h_2h_3}\left|\begin{matrix}\frac{\partial}{\partial x_2} & \frac{\partial}{\partial x_3} \\[8pt] h_2A_2 & h_3A_3\end{matrix}\right| \\[12pt] (\nabla\times A)\cdot\widehat{e}_2 \;=\; \frac{1}{h_3h_1}\left|\begin{matrix}\frac{\partial}{\partial x_3} & \frac{\partial}{\partial x_1} \\[8pt] h_3A_3 & h_1A_1\end{matrix}\right| \\[12pt] (\nabla\times A)\cdot\widehat{e}_3 \;=\; \frac{1}{h_1h_2}\left|\begin{matrix}\frac{\partial}{\partial x_1} & \frac{\partial}{\partial x_2} \\[8pt] h_1A_1 & h_2A_2\end{matrix}\right| \end{gather*}$$ We will prove the first of these formulas. Given any vector field $A$, we can write \begin{align*} A \;&=\; A_1 \widehat{e}_1 \,+\, A_2 \widehat{e}_2 \,+\, A_3 \widehat{e}_3 \\[6pt] &=\; h_1A_1\,\nabla x_1 \,+\, h_2A_2\,\nabla x_2 \,+\, h_3A_3\,\nabla x_3 \end{align*} Taking the curl gives $$\nabla \times A \;=\; \nabla(h_1A_1)\times (\nabla x_1) \,+\, \nabla(h_2A_2)\times(\nabla x_2) \,+\, \nabla(h_3A_3)\times(\nabla x_3)$$ Here we have used the identity $\nabla\times(uF) = (\nabla u)\times F + u(\nabla\times F)$, as well as the fact that the curl of a gradient is zero. Applying formula (1), we get $$\nabla \times A \;=\; \frac{1}{h_1}\nabla(h_1A_1)\times \widehat{e}_1 \,+\, \frac{1}{h_2}\nabla(h_2A_2)\times\widehat{e}_2 \,+\, \frac{1}{h_3}\nabla(h_3A_3)\times\widehat{e}_3$$ When we take the cross products, the $\widehat{e}_1$ component will be $$(\nabla \times A)\cdot\widehat{e}_1 \;=\; \frac{1}{h_3}\nabla(h_3A_3)\cdot\widehat{e}_2 \,-\, \frac{1}{h_2}\nabla(h_2A_2)\cdot\widehat{e}_3.$$ But, by formula (2) for the gradient, $$\nabla(h_3A_3)\cdot\widehat{e}_2 \;=\; \frac{1}{h_2}\frac{\partial}{\partial x_2}(h_3 A_3)\qquad\text{and}\qquad\nabla(h_2A_2)\cdot\widehat{e}_3 \;=\; \frac{1}{h_3}\frac{\partial}{\partial x_3}(h_2 A_2)$$ Therefore, \begin{align*} (\nabla \times A)\cdot\widehat{e}_1 \;&=\; \frac{1}{h_2h_3}\frac{\partial}{\partial x_2}(h_3A_3) \,-\, \frac{1}{h_2h_3}\frac{\partial}{\partial x_3}(h_2A_2) \\[12pt] &=\; \frac{1}{h_2h_3}\left|\begin{matrix}\frac{\partial}{\partial x_2} & \frac{\partial}{\partial x_3} \\[8pt] h_2A_2 & h_3A_3\end{matrix}\right| \end{align*} as desired.