logo

Curl of a Vector Function in Curvilinear Coordinates 📂Mathematical Physics

Curl of a Vector Function in Curvilinear Coordinates

Theorem

In the curvilinear coordinate system, the curl of the vector function F=F(q1,q2,q3)=F1q^1+F2q^2+F3q^3\mathbf{F}=\mathbf{F}(q_{1},q_{2},q_{3})=F_{1}\hat{\mathbf{q}}_{1}+F_{2}\hat{\mathbf{q}}_{2}+F_{3}\hat{\mathbf{q}}_{3} is as follows.

∇×F=q^1h2h3(∂(F3h3)∂q2−∂(F2h2)∂q3)+q^2h1h3(∂(F1h1)∂q3−∂(F3h3)∂q1)+q^3h1h2(∂(F2h2)∂q1−∂(F1h1)∂q2)=1h1h2h3∣h1q^1h2q^2h3q^3∂∂q1∂∂q2∂∂q3F1h1F2h2F3h3∣ \begin{align*} \nabla \times \mathbf{F} &= \frac{\hat{\mathbf{q}}_{1}}{h_{2}h_{3}}\left( \dfrac{\partial (F_{3}h_{3})}{\partial q_{2}} - \dfrac{\partial (F_{2}h_{2})}{\partial q_{3}} \right) + \frac{\hat{\mathbf{q}}_{2}}{h_{1}h_{3}}\left( \dfrac{\partial (F_{1}h_{1})}{\partial q_{3}} - \dfrac{\partial (F_{3}h_{3})}{\partial q_{1}} \right) \\ &\quad+ \frac{\hat{\mathbf{q}}_{3}}{h_{1}h_{2}}\left( \dfrac{\partial (F_{2}h_{2})}{\partial q_{1}} - \dfrac{\partial (F_{1}h_{1})}{\partial q_{2}} \right) \\ &= \frac{1}{h_{1}h_{2}h_{3}} \begin{vmatrix} h_{1}\hat{\mathbf{q}}_{1} & h_{2}\hat{\mathbf{q}}_{2} & h_{3}\hat{\mathbf{q}}_{3} \\[0.5em] \dfrac{\partial }{\partial q_{1}} & \dfrac{\partial }{\partial q_{2}} & \dfrac{\partial }{\partial q_{3}} \\[1em] F_{1}h_{1} & F_{2}h_{2} & F_{3}h_{3} \end{vmatrix} \end{align*}

hih_{i} is the scale factor.

Formula

  • Cartesian coordinates:

    h1=h2=h3=1 h_{1}=h_{2}=h_{3}=1

    ∇×F=(∂Fz∂y−∂Fy∂z)x^+(∂Fx∂z−∂Fz∂x)y^+(∂Fy∂x−∂Fx∂y)z^ \nabla \times \mathbf{F} = \left( \dfrac{\partial F_{z}}{\partial y}-\dfrac{\partial F_{y}}{\partial z} \right)\hat{\mathbf{x}}+ \left( \dfrac{\partial F_{x}}{\partial z}-\dfrac{\partial F_{z}}{\partial x} \right)\hat{\mathbf{y}}+ \left( \dfrac{\partial F_{y}}{\partial x}-\dfrac{\partial F_{x}}{\partial y} \right)\hat{\mathbf{z}}

  • Cylindrical coordinates:

    h1=1,h2=ρ,h3=1 h_{1}=1,\quad h_{2}=\rho,\quad h_{3}=1

    ∇×F=(1ρ∂Fz∂ϕ−∂Fϕ∂z)ρ^+(∂Fρ∂z−∂Fz∂ρ)ϕ^+1ρ(∂(ρFϕ)∂ρ−∂Fρ∂ϕ)z^ \nabla \times \mathbf{F} = \left(\frac{1}{\rho}\dfrac{\partial F_{z}}{\partial \phi} - \dfrac{\partial F_{\phi}}{\partial z} \right)\boldsymbol{\hat \rho} + \left(\dfrac{\partial F_{\rho}}{\partial z} - \dfrac{\partial F_{z}}{\partial \rho} \right)\boldsymbol{\hat \phi} + \frac{1}{\rho}\left(\dfrac{\partial (\rho F_{\phi})}{\partial \rho} - \dfrac{\partial F_{\rho}}{\partial \phi} \right)\mathbf{\hat{\mathbf{z}}}

  • Spherical coordinates:

    h1=1,h2=r,h3=rsin⁡θ h_{1}=1,\quad h_{2}=r\quad, h_{3}=r\sin\theta

    ∇×F=1rsin⁡θ(∂(Fϕsin⁡θ)∂θ−∂Fθ∂ϕ)r^+1r(1sin⁡θ∂Fr∂ϕ−∂(rFϕ)∂r)θ^+1r(∂(rFθ)∂r−∂Fr∂θ)ϕ^ \nabla \times \mathbf{F} = \frac{1}{r\sin\theta}\left(\dfrac{\partial (F_{\phi} \sin\theta)}{\partial \theta} - \dfrac{\partial F_{\theta}}{\partial \phi} \right)\mathbf{\hat r} + \frac{1}{r}\left(\frac{1}{\sin\theta}\dfrac{\partial F_{r}}{\partial \phi} - \dfrac{\partial (r F_{\phi})}{\partial r} \right)\boldsymbol{\hat \theta} + \frac{1}{r}\left(\dfrac{\partial (r F_{\theta})}{\partial r} - \dfrac{\partial F_{r}}{\partial \theta} \right)\boldsymbol{\hat \phi}

Derivation

Method 1 1

Let’s denote the curvilinear coordinates by (q1,q2,q3)(q_{1}, q_{2}, q_{3}).

F=F1q^1+F2q^2+F3q^3 \mathbf{F} = F_{1}\hat{\mathbf{q}}_{1} + F_{2}\hat{\mathbf{q}}_{2} + F_{3}\hat{\mathbf{q}}_{3}

Since curl has linearity,

∇×F=∇×(F1q^1+F2q^2+F3q^3)=∑i=13∇×(Fiq^i) \begin{equation} \nabla \times \mathbf{F} = \nabla \times \left( F_{1}\hat{\mathbf{q}}_{1} + F_{2}\hat{\mathbf{q}}_{2} + F_{3}\hat{\mathbf{q}}_{3} \right) = \sum_{i=1}^{3} \nabla \times (F_{i}\hat{\mathbf{q}}_{i}) \end{equation}

Gradient in curvilinear coordinates

∇f=1h1∂f∂q1q^1+1h2∂f∂q2q^2+1h3∂f∂q3q^3=∑i=131hi∂f∂qiq^i \nabla f= \frac{1}{h_{1}}\frac{ \partial f }{ \partial q_{1} } \hat{\mathbf{q}}_{1} + \frac{1}{h_{2}}\frac{ \partial f }{ \partial q _{2}}\hat{\mathbf{q}}_{2}+\frac{1}{h_{3}}\frac{ \partial f }{ \partial q_{3} } \hat{\mathbf{q}}_{3}=\sum \limits _{i=1} ^{3}\frac{1}{h_{i}}\frac{ \partial f}{ \partial q_{i}}\hat{\mathbf{q}}_{i}

By the gradient formula, we get the following.

∇qi=∑m=131hm∂qi∂qmq^m=1hiq^i \nabla q_{i} = \sum \limits _{m=1} ^{3}\frac{1}{h_{m}}\frac{ \partial q_{i}}{ \partial q_{m}}\hat{\mathbf{q}}_{m} = \frac{1}{h_{i}}\hat{\mathbf{q}}_{i}

  ⟹  hi∇qi=q^i \begin{equation} \implies h_{i}\nabla q_{i} = \hat{\mathbf{q}}_{i} \end{equation}

Multiplication formula with the del operator

∇×(fA)=(∇f)×A+f(∇×A) \nabla \times (f \mathbf{A}) = (\nabla f) \times \mathbf{A} + f (\nabla \times \mathbf{A})

Substituting (2)(2) into (1)(1) and applying the multiplication formula yields the following.

∇×(Fiq^i)=∇×(Fihi∇qi)=[∇(Fihi)]×∇qi+Fihi∇×(∇qi)=[∇(Fihi)]×∇qi \begin{align*} \nabla \times (F_{i}\hat{\mathbf{q}}_{i}) &= \nabla \times (F_{i}h_{i}\nabla q_{i}) \\ &= [\nabla (F_{i}h_{i})] \times \nabla q_{i} + F_{i}h_{i} \nabla \times (\nabla q_{i}) \\ &= [\nabla (F_{i}h_{i})] \times \nabla q_{i} \end{align*}

The last equality holds because the curl of the gradient is 0\mathbf{0}. Substituting (2)(2) back into the formula and expanding the gradient, we get

[∇(Fihi)]×∇qi=[∑m=131hm∂(Fihi)∂qmq^m]×1hiq^i=1hj∂(Fihi)∂qjq^j×1hiq^i+1hk∂(Fihi)∂qkq^k×1hiq^i=1hihk∂(Fihi)∂qkq^j−1hihj∂(Fihi)∂qjq^k \begin{align*} [\nabla (F_{i}h_{i})] \times \nabla q_{i} &= \left[ \sum_{m=1}^{3} \dfrac{1}{h_{m}}\dfrac{\partial (F_{i}h_{i})}{\partial q_{m}} \hat{\mathbf{q}}_{m} \right] \times \frac{1}{h_{i}}\hat{\mathbf{q}}_{i} \\ &= \dfrac{1}{h_{j}}\dfrac{\partial (F_{i}h_{i})}{\partial q_{j}} \hat{\mathbf{q}}_{j} \times \frac{1}{h_{i}}\hat{\mathbf{q}}_{i} + \dfrac{1}{h_{k}}\dfrac{\partial (F_{i}h_{i})}{\partial q_{k}} \hat{\mathbf{q}}_{k} \times \frac{1}{h_{i}}\hat{\mathbf{q}}_{i} \\ &= \dfrac{1}{h_{i}h_{k}}\dfrac{\partial (F_{i}h_{i})}{\partial q_{k}} \hat{\mathbf{q}}_{j} - \dfrac{1}{h_{i}h_{j}}\dfrac{\partial (F_{i}h_{i})}{\partial q_{j}} \hat{\mathbf{q}}_{k} \end{align*}

Here, we set the index as q^i×q^j=q^k\hat{\mathbf{q}}_{i} \times \hat{\mathbf{q}}_{j} = \hat{\mathbf{q}}_{k}. By substituting this into (1)(1), we get

∇×F=(1h1h3∂(F1h1)∂q3q^2−1h1h2∂(F1h1)∂q2q^3)+(1h2h1∂(F2h2)∂q1q^3−1h2h3∂(F2h2)∂q3q^1)+(1h3h2∂(F3h3)∂q2q^1−1h3h1∂(F3h3)∂q1q^2)=h1q^1h1h2h3(∂(F3h3)∂q2−∂(F2h2)∂q3)+h2q^2h1h2h3(∂(F1h1)∂q3−∂(F3h3)∂q1)+h3q^3h1h2h3(∂(F2h2)∂q1−∂(F1h1)∂q2)=1h1h2h3∣h1q^1h2q^2h3q^3∂∂q1∂∂q2∂∂q3F1h1F2h2F3h3∣ \begin{align*} \nabla \times \mathbf{F} &= \left( \dfrac{1}{h_{1}h_{3}}\dfrac{\partial (F_{1}h_{1})}{\partial q_{3}} \hat{\mathbf{q}}_{2} - \dfrac{1}{h_{1}h_{2}}\dfrac{\partial (F_{1}h_{1})}{\partial q_{2}} \hat{\mathbf{q}}_{3} \right) + \left( \dfrac{1}{h_{2}h_{1}}\dfrac{\partial (F_{2}h_{2})}{\partial q_{1}} \hat{\mathbf{q}}_{3} - \dfrac{1}{h_{2}h_{3}}\dfrac{\partial (F_{2}h_{2})}{\partial q_{3}} \hat{\mathbf{q}}_{1} \right) \\ &\quad+ \left( \dfrac{1}{h_{3}h_{2}}\dfrac{\partial (F_{3}h_{3})}{\partial q_{2}} \hat{\mathbf{q}}_{1} - \dfrac{1}{h_{3}h_{1}}\dfrac{\partial (F_{3}h_{3})}{\partial q_{1}} \hat{\mathbf{q}}_{2} \right) \\ &= \frac{h_{1}\hat{\mathbf{q}}_{1}}{h_{1}h_{2}h_{3}}\left( \dfrac{\partial (F_{3}h_{3})}{\partial q_{2}} - \dfrac{\partial (F_{2}h_{2})}{\partial q_{3}} \right) + \frac{h_{2}\hat{\mathbf{q}}_{2}}{h_{1}h_{2}h_{3}}\left( \dfrac{\partial (F_{1}h_{1})}{\partial q_{3}} - \dfrac{\partial (F_{3}h_{3})}{\partial q_{1}} \right) \\ &\quad+ \frac{h_{3}\hat{\mathbf{q}}_{3}}{h_{1}h_{2}h_{3}}\left( \dfrac{\partial (F_{2}h_{2})}{\partial q_{1}} - \dfrac{\partial (F_{1}h_{1})}{\partial q_{2}} \right) \\ &= \frac{1}{h_{1}h_{2}h_{3}} \begin{vmatrix} h_{1}\hat{\mathbf{q}}_{1} & h_{2}\hat{\mathbf{q}}_{2} & h_{3}\hat{\mathbf{q}}_{3} \\[0.5em] \dfrac{\partial }{\partial q_{1}} & \dfrac{\partial }{\partial q_{2}} & \dfrac{\partial }{\partial q_{3}} \\[1em] F_{1}h_{1} & F_{2}h_{2} & F_{3}h_{3} \end{vmatrix} \end{align*}

■

Method 2 2

To find the first component of ∇×F\nabla \times \mathbf{F}, consider the closed curved surface where the q1q_{1} coordinates are constant.

Here, ⊙\odot means the direction emerging through the plane. Therefore, since dσ=h2h3dq2dq3q^1d\boldsymbol{\sigma} = h_{2}h_{3} dq_{2}dq_{3}\hat{\mathbf{q}}_{1},

∫∇×F⋅dσ≈(∇×F)h2h3dq2dq3⋅q^1 \int \nabla \times \mathbf{F} \cdot d\boldsymbol{\sigma} \approx (\nabla \times \mathbf{F})h_{2}h_{3}dq_{2}dq_{3} \cdot \hat{\mathbf{q}}_{1}

Stokes’ theorem

∫S(∇×v)⋅da=∮Pv⋅dl \int_{\mathcal{S}} (\nabla \times \mathbf{v} )\cdot d\mathbf{a} = \oint_{\mathcal{P}} \mathbf{v} \cdot d\mathbf{l}

By Stokes’ theorem, we get the following.

(∇×F)h2h3dq2dq3⋅q^1=∮F⋅dr (\nabla \times \mathbf{F})h_{2}h_{3}dq_{2}dq_{3} \cdot \hat{\mathbf{q}}_{1} = \oint \mathbf{F} \cdot d \mathbf{r}

Therefore, the first component of ∇×F\nabla \times \mathbf{F} can be obtained by calculating the following.

(∇×F)1=(∇×F)⋅q^1=1h2h3dq2dq3∮F⋅dr (\nabla \times \mathbf{F})_{1} = (\nabla \times \mathbf{F}) \cdot \hat{\mathbf{q}}_{1} = \frac{1}{h_{2}h_{3}dq_{2}dq_{3}}\oint \mathbf{F} \cdot d \mathbf{r}

Let’s think about the closed curve integral divided into 1◯\textcircled{1}~4◯\textcircled{4}.

∮F⋅dr=∫1◯F⋅dr+∫2◯F⋅dr+∫3◯F⋅dr+∫4◯F⋅dr \oint \mathbf{F} \cdot d \mathbf{r} = \int_{\textcircled{1}} \mathbf{F} \cdot d \mathbf{r} + \int_{\textcircled{2}} \mathbf{F} \cdot d \mathbf{r} + \int_{\textcircled{3}} \mathbf{F} \cdot d \mathbf{r} + \int_{\textcircled{4}} \mathbf{F} \cdot d \mathbf{r}

Calculating the integral over the path 1◯\textcircled{1}, we get

∫1◯F⋅dr≈F(q2,q3)⋅h2(q2,q3)dq2q^2=F2h2dq2 \int_{\textcircled{1}} \mathbf{F} \cdot d \mathbf{r} \approx \mathbf{F}(q_{2},q_{3}) \cdot h_{2}(q_{2},q_{3}) dq_{2}\hat{\mathbf{q}}_{2} = F_{2}h_{2}dq_{2}

For the path 2◯\textcircled{2},

∫2◯F⋅dr≈F(q2+dq2,q3)⋅h3(q2+dq2,q3)dq3q^3=F3(q2+dq2,q3)h3(q2+dq2,q3)dq3≈[F3h3+∂(F3h3)∂q2dq2]dq3 \begin{align*} \int_{\textcircled{2}} \mathbf{F} \cdot d \mathbf{r} &\approx \mathbf{F}(q_{2}+dq_{2},q_{3}) \cdot h_{3}(q_{2}+dq_{2},q_{3}) dq_{3}\hat{\mathbf{q}}_{3} \\ &= F_{3}(q_{2}+dq_{2},q_{3})h_{3}(q_{2}+dq_{2},q_{3})dq_{3} \\ &\approx \left[ F_{3}h_{3} + \dfrac{\partial (F_{3}h_{3})}{\partial q_{2}}dq_{2} \right]dq_{3} \end{align*}

In the last line, Taylor’s approximation was used.

Taylor’s theorem

f(x+dx)≈f(x)+f′(x)dx f(x+dx) \approx f(x) + f^{\prime}(x)dx

Likewise, calculating the remaining integrals, we get

∫3◯F⋅dr=−∫−3◯F⋅dr≈−F(q2,q3+dq3)⋅h2(q2,q3+dq3)dq2q^2=−F2(q2,q3+dq3)h2(q2,q3+dq3)dq2≈−[F2h2+∂(F2h2)∂q3dq3]dq2 \begin{align*} \int_{\textcircled{3}} \mathbf{F} \cdot d \mathbf{r} &= -\int_{-\textcircled{3}} \mathbf{F} \cdot d \mathbf{r} \\ &\approx -\mathbf{F}(q_{2},q_{3}+dq_{3}) \cdot h_{2}(q_{2},q_{3}+dq_{3}) dq_{2}\hat{\mathbf{q}}_{2} \\ &= -F_{2}(q_{2},q_{3}+dq_{3})h_{2}(q_{2},q_{3}+dq_{3})dq_{2} \\ &\approx -\left[ F_{2}h_{2} + \dfrac{\partial (F_{2}h_{2})}{\partial q_{3}}dq_{3} \right]dq_{2} \end{align*}

∫4◯F⋅dr=−∫−4◯F⋅dr≈F(q2,q3)⋅h3(q2,q3)dq3q^3=F3h3dq3 \int_{\textcircled{4}} \mathbf{F} \cdot d \mathbf{r} = -\int_{-\textcircled{4}} \mathbf{F} \cdot d \mathbf{r} \approx \mathbf{F}(q_{2},q_{3}) \cdot h_{3}(q_{2},q_{3}) dq_{3}\hat{\mathbf{q}}_{3} = F_{3}h_{3}dq_{3}

Therefore,

(∇×F)1=1h2h3dq2dq3∮F⋅dr=1h2h3dq2dq3[∂(F3h3)∂q2dq2dq3−∂(F2h2)∂q3dq3dq2]=1h2h3[∂(F3h3)∂q2−∂(F2h2)∂q3] \begin{align*} (\nabla \times \mathbf{F})_{1} &= \dfrac{1}{h_{2}h_{3}dq_{2}dq_{3}}\oint \mathbf{F} \cdot d \mathbf{r} \\ &= \dfrac{1}{h_{2}h_{3}dq_{2}dq_{3}} \left[ \dfrac{\partial (F_{3}h_{3})}{\partial q_{2}}dq_{2}dq_{3} - \dfrac{\partial (F_{2}h_{2})}{\partial q_{3}}dq_{3} dq_{2} \right]\\ &= \dfrac{1}{h_{2}h_{3}} \left[ \dfrac{\partial (F_{3}h_{3})}{\partial q_{2}} - \dfrac{\partial (F_{2}h_{2})}{\partial q_{3}} \right]\\ \end{align*}

The second and third components are obtained in the same way.

■

See also

  1. https://math.libretexts.org/Bookshelves/Differential_Equations/Introduction_to_Partial_Differential_Equations_(Herman)/06%3A_Problems_in_Higher_Dimensions/6.09%3A_Curvilinear_Coordinates ↩︎

  2. George B. Arfken, MATHEMATICAL METHODS FOR PHYSICISTS (6th Edition), p112 ↩︎