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Curl of Vector Functions in 3D Cartesian Coordinates 📂Mathematical Physics

Curl of Vector Functions in 3D Cartesian Coordinates

Definition

For a vector function F(x,y,z)=(Fx,Fy,Fz)=Fxx^+Fyy^+Fzz^\mathbf{F}(x,y,z)=(F_{x},F_{y},F_{z})=F_{x}\hat{\mathbf{x}} + F_{y}\hat{\mathbf{y}} + F_{z}\hat{\mathbf{z}}, the following vector is defined as the curl of F\mathbf{F}, denoted as ∇×F\nabla \times \mathbf{F}.

∇×F=(∂Fz∂y−∂Fy∂z)x^+(∂Fx∂z−∂Fz∂x)y^+(∂Fy∂x−∂Fx∂y)z^=∣x^y^z^∂∂x∂∂y∂∂zFxFyFz∣ \begin{align} \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}} \label{def1} \\ &=\begin{vmatrix} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ \dfrac{ \partial }{ \partial x} & \dfrac{ \partial }{ \partial y } & \dfrac{ \partial }{ \partial z} \\ F_{x} & F_{y} &F_{z}\end{vmatrix} \label{def2} \end{align}

  • (2)(2) serves as an easy-to-remember formula for the curl of F\mathbf{F}. Think of it as a determinant and expand accordingly.

Explanation

Curl translates to rotation. However, since the term ‘rotation’ is too common and may lead to confusion with rotation instead of curl, the term ‘curl’ is preferred at the Fresh Shrimp Sushi Restaurant.

∇×F\nabla \times \mathbf{F} is a vector that indicates in which direction the physical quantity F\mathbf{F} is rotating. If you place the direction of ∇×F\nabla \times \mathbf{F} as the axis (thumb) and apply the right-hand rule, the direction in which your right hand wraps around corresponds to the direction of rotation of F\mathbf{F}. The magnitude of the vector ∇×F\nabla \times \mathbf{F} represents the extent of the rotation.

Using Einstein notation and Levi-Civita symbol, it can be expressed as follows. If ∇j=∂∂xj\nabla_{j} = \dfrac{\partial }{\partial x_{j}} is denoted,

∇×F=ϵijke^i∇jFk \nabla \times \mathbf{F} = \epsilon_{ijk}\hat{\mathbf{e}}_{i}\nabla_{j}F_{k}

Meanwhile, note that the value denoted as (1)(1) in the definition is expressed as ∇×F\nabla \times \mathbf{F}. Although ∇\nabla is referred to as the del operator, thinking of it as having a meaning on its own could easily lead to confusion, mistaking ∇⋅F\nabla \cdot \mathbf{F} or ∇×F\nabla \times \mathbf{F} as the dot product and cross product, respectively. Hence, it’s best to understand ∇\nabla merely as a convenient notation, and it might be even better to think of the del operator as synonymous with the gradient. The del operators, encompassing the gradient, divergence, and curl, will be discussed in more detail below.

Points to Note

∇×F\nabla \times \mathbf{F} is not the cross product of ∇\nabla and F\mathbf{F}

∇×F\nabla \times \mathbf{F} is definitely not the cross product of ∇\nabla and F\mathbf{F}.

∇×F\nabla \times \mathbf{F} is merely a vector containing some information about F\mathbf{F}. We think of ∇\nabla as a vector like ∇=∂∂xx^+∂∂yy^+∂∂zz^\nabla = \dfrac{ \partial }{ \partial x}\hat{\mathbf{x}} + \dfrac{ \partial }{ \partial y}\hat{\mathbf{y}} + \dfrac{ \partial }{ \partial z}\hat{\mathbf{z}}, and the result matches perfectly with (1)(1), so it’s denoted as ∇×F\nabla \times \mathbf{F} for convenience. Assuming ∇\nabla is an actual vector would lead to strange results.

The following equation holds for two vectors A,B\mathbf{A}, \mathbf{B}:

∇×(A×B)=(B⋅∇)A−(A⋅∇)B+A(∇⋅B)−B(∇⋅A) \nabla \times (\mathbf{A} \times \mathbf{B}) = (\mathbf{B} \cdot \nabla)\mathbf{A} - (\mathbf{A} \cdot \nabla)\mathbf{B} + \mathbf{A} (\nabla \cdot \mathbf{B}) - \mathbf{B} (\nabla \cdot \mathbf{A})

If ∇\nabla were truly a vector, we could substitute it into the above formula and obtain the following results.

∇×(∇×F)=(F⋅∇)∇−(∇⋅∇)F+∇(∇⋅F)−F(∇⋅∇) \nabla \times (\nabla \times \mathbf{F})=(\mathbf{F} \cdot \nabla)\nabla - (\nabla \cdot \nabla)\mathbf{F} + \nabla (\nabla \cdot \mathbf{F}) - \mathbf{F} (\nabla \cdot \nabla)

However, the correct result is as follows.

∇×(∇×F)=∇(∇⋅F)−∇2F \nabla \times (\nabla \times \mathbf{F})=\nabla(\nabla \cdot \mathbf{F})-\nabla ^{2} \mathbf{F}

Another example exists. Since the cross product of vectors has the property of anticommutativity, if ∇×F\nabla \times \mathbf{F} were a cross product, the following equation should hold:

∇×F=?−F×∇ \nabla \times \mathbf{F} \overset{?}{=} - \mathbf{F} \times \nabla

Therefore, ∇\nabla is not a vector, and it can be understood that ∇×F\nabla \times \mathbf{F} is not the cross product of ∇\nabla and F\mathbf{F}. Instead of being a vector, ∇×\nabla \times should be considered as a function in itself. In physics, functions that take other functions as variables are called operators.

So, what’s the difference between ∇×F\nabla \times \mathbf{F} and F×∇\mathbf{F} \times \nabla?

∇×\nabla \times is an operator defined as follows, taking a vector function as its variable:

∇×(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}}

In other words, ∇×F\nabla \times \mathbf{F} is the function value when the variable F\mathbf{F} is substituted into the operator (function) ∇×\nabla \times. Of course, this in turn is a vector function of the variables (x,y,z)(x,y,z). While ∇×F\nabla \times \mathbf{F} is the function value of ∇×\nabla \times, F×∇\mathbf{F} \times \nabla is an operator in itself. Although it’s not a commonly used expression, it can be defined as the following differential operator if we were to define it.

F×∇=∣x^y^z^FxFyFz∂∂x∂∂y∂∂z∣=(Fy∂∂z−Fz∂∂y)x^+(Fz∂∂x−Fx∂∂z)y^+(Fx∂∂y−Fy∂∂x)z^ \begin{align*} \mathbf{F} \times \nabla &= \begin{vmatrix} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ F_{x} & F_{y} &F_{z} \\ \dfrac{ \partial }{ \partial x} & \dfrac{ \partial }{ \partial y } & \dfrac{ \partial }{ \partial z} \end{vmatrix} \\ &= \left( F_{y}\dfrac{ \partial }{ \partial z} - F_{z}\dfrac{ \partial }{ \partial y} \right)\hat{\mathbf{x}} + \left( F_{z}\dfrac{ \partial }{ \partial x} - F_{x}\dfrac{ \partial }{ \partial z} \right)\hat{\mathbf{y}} + \left( F_{x}\dfrac{ \partial }{ \partial y} - F_{y}\dfrac{ \partial }{ \partial x} \right)\hat{\mathbf{z}} \end{align*}

Derivation

Now, let’s consider a function that indicates the direction of rotation (clockwise or counterclockwise) of a rotating vector function. It’s important to note that no direction within the plane of rotation can specify the direction of rotation. Look at the diagram below.

  • Vector −x^-\hat{\mathbf{x}} can explain the movement at point AA, but not at point BB.
  • Vector y^\hat{\mathbf{y}} can explain the movement at point CC, but not at point DD.
  • Vector x^+y^\hat{\mathbf{x}} + \hat{\mathbf{y}} can explain the path FF, but not GG.

This is also true for clockwise rotation. Now, you should sense the need to move out of the plane of rotation to specify the direction of rotation. In fact, there’s already a good method to determine this: using the right-hand rule, which determines the axis of rotation in the direction of the thumb when the right hand wraps around. Therefore, in the xyxy-plane, the axis (direction) of counterclockwise rotation is z^\hat{\mathbf{z}}, and the axis (direction) of clockwise rotation is −z^-\hat{\mathbf{z}}.

Now, let’s find a value that indicates the z^\hat{\mathbf{z}} direction when F\mathbf{F} is rotating counterclockwise in the xyxy-plane, in other words, a positive value. Let’s represent the rotation simply with a rectangle as below.

Path ① moves from point aa to point bb, and let’s say F(a)=(1,0,0)\mathbf{F}(a) = (1,0,0) and F(b)=(0,1,0)\mathbf{F}(b) = (0,1,0). Then, as xx changes by +1 from point aa to bb, and FyF_{y} also changes by +1, we obtain the following.

∂Fy∂x>0 \dfrac{\partial F_{y}}{\partial x} \gt 0

In the same manner, on the path where the point moves from bb to cc, yy changes by +1, and FxF_{x} changes by -1. Checking all four paths, we find that:

∂Fy∂x>0in path 1◯, 3◯ \dfrac{\partial F_{y}}{\partial x} \gt 0 \quad \text{in path 1◯\textcircled{1}, 3◯\textcircled{3}}

∂Fx∂y<0in path 2◯, 4◯ \dfrac{\partial F_{x}}{\partial y} \lt 0 \quad \text{in path 2◯\textcircled{2}, 4◯\textcircled{4}}

Therefore, for a vector F\mathbf{F} that rotates counterclockwise as above, the value below is always positive:

∂Fy∂x−∂Fx∂y>0 \dfrac{\partial F_{y}}{\partial x} - \dfrac{\partial F_{x}}{\partial y} \gt 0

Conversely, if F\mathbf{F} is rotating clockwise, the above value is always negative. Now, we can define the operator curl⁡xy\operatorname{curl}_{xy}, which indicates the direction and magnitude of rotation in the xyxy-plane when the vector function F\mathbf{F} is substituted:

curl⁡xy(F)=(∂Fy∂x−∂Fx∂y)z^ \operatorname{curl}_{xy} (\mathbf{F}) = \left( \dfrac{\partial F_{y}}{\partial x} - \dfrac{\partial F_{x}}{\partial y} \right) \hat{\mathbf{z}}

  • The sign of the z^\hat{\mathbf{z}} component of this function indicates the direction of rotation of F\mathbf{F} in the xyxy-plane:
    • If it’s positive (+), F\mathbf{F} rotates counterclockwise in the xyxy-plane.
    • If it’s negative (-), F\mathbf{F} rotates clockwise in the xyxy-plane.
    • If it’s zero (0), there is no rotation.
  • The magnitude of the z^\hat{\mathbf{z}} component of this function indicates how rapidly F\mathbf{F} is rotating in the xyxy-plane.

Applying this discussion similarly to the yzyz-plane and the zxzx-plane, we can define the vector ∇×F\nabla \times \mathbf{F}, which indicates the direction and magnitude of rotation of F\mathbf{F} in 3-dimensional space, as follows.

∇×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}}

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  • Linearity: ∇×(A+B)=∇×A+∇×B \nabla \times (\mathbf{A} + \mathbf{B}) = \nabla \times \mathbf{A} + \nabla \times \mathbf{B}

  • Multiplication rule:

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

    ∇×(A×B)=(B⋅∇)A−(A⋅∇)B+A(∇⋅B)−B(∇⋅A) \nabla \times (\mathbf{A} \times \mathbf{B}) = (\mathbf{B} \cdot \nabla)\mathbf{A} - (\mathbf{A} \cdot \nabla)\mathbf{B} + \mathbf{A} (\nabla \cdot \mathbf{B}) - \mathbf{B} (\nabla \cdot \mathbf{A})

  • 2nd derivative:

    ∇×(∇f)=0 \nabla \times (\nabla f) = \mathbf{0}

    ∇×(∇×F)=∇(∇⋅F)−∇2F \nabla \times (\nabla \times \mathbf{F}) = \nabla (\nabla \cdot \mathbf{F}) - \nabla^{2} \mathbf{F}

  • 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}

  • Integral formulas ∫V(∇×v)dτ=−∮Sv×da \int_{\mathcal{V}} (\nabla \times \mathbf{v}) d \tau = - \oint_{\mathcal{S}} \mathbf{v} \times d \mathbf{a}

    ∫S∇T×da=−∮PTdl \int_{\mathcal{S}} \nabla T \times d \mathbf{a} = - \oint_{\mathcal{P}} T d \mathbf{l}

  • Integration by parts ∫Sf(∇×A)A⋅da=∫S[A×(∇f)]⋅da+∮PfA⋅dl \int_{\mathcal{S}} f \left( \nabla \times \mathbf{A} \right)\mathbf{A} \cdot d \mathbf{a} = \int_{\mathcal{S}} \left[ \mathbf{A} \times \left( \nabla f \right) \right] \cdot d\mathbf{a} + \oint_{\mathcal{P}} f\mathbf{A} \cdot d\mathbf{l}

    ∫VB⋅(∇×A)dτ=∫VA⋅(∇×B)dτ+∮S(A×B)⋅da \int_{\mathcal{V}} \mathbf{B} \cdot \left( \nabla \times \mathbf{A} \right) d\tau = \int_{\mathcal{V}} \mathbf{A} \cdot \left( \nabla \times \mathbf{B} \right) d\tau + \oint_{\mathcal{S}} \left( \mathbf{A} \times \mathbf{B} \right) \cdot d \mathbf{a}

Proof

Linearity

Using Einstein notation and Levi-Civita symbol, if we denote ∇j=∂∂xj\nabla_{j} = \dfrac{\partial }{\partial x_{j}}, then:

[∇×(A+B)]i=ϵijk∇j(A+B)k=ϵijk∇j(Ak+Bk)=ϵijk∇jAk+ϵijk∇jBk=[∇×A]i+[∇×B]i \begin{align*} \left[ \nabla \times (\mathbf{A} + \mathbf{B}) \right]_{i} &= \epsilon_{ijk} \nabla_{j} (\mathbf{A} + \mathbf{B})_{k} \\ &= \epsilon_{ijk} \nabla_{j} (A_{k} + B_{k}) \\ &= \epsilon_{ijk} \nabla_{j}A_{k} + \epsilon_{ijk} \nabla_{j}B_{k} \\ &= [\nabla \times \mathbf{A}]_{i} + [\nabla \times \mathbf{B}]_{i} \\ \end{align*}

The third equality holds because ∂(Ak+Bk)∂xj=∂Ak∂xj+∂Bk∂xj\dfrac{\partial (A_{k} + B_{k})}{\partial x_{j}} = \dfrac{\partial A_{k}}{\partial x_{j}} + \dfrac{\partial B_{k}}{\partial x_{j}}.

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Further reading