Derivation of the Curl formula in cartesian coordinates.

Question

By calculating the circulation per area of a vector field

$$F(x,y,z) = F_x(x,y,z)\vec{x} + F_y(x,y,z)\vec{y} + F_z(x,y,z)\vec{z}$$

in a small rectangle around $(x_0, y_0, z_0)$ on the $xy$ plane, it can be shown the limit as the sides of the rectangle approach zero is

$$\left(\frac{\partial F_y(x_0, y_0, z_0)}{\partial x} - \frac{\partial F_x(x_0, y_0, z_0)}{\partial y}\right)$$

The same calculation however is not that straightforward if the rectangle does not lie in the $xy$, $yz$, or $xz$ planes. Now if $\vec{n}$ is the normal of the plane, I thought that by performing a change of basis such that $\vec{n} \rightarrow \vec{z'} $ and by following the previous calculations we could show that the limit of the circulation per area is

$$ \left(\frac{\partial F_{y'}(x'_0, y'_0, z'_0)}{\partial x'} - \frac{\partial F_{x'}(x_0, y_0, z_0)}{\partial y'}\right) $$

This is also the inner product of the curl of the vector field and the normal $\vec{n}$

As such the two should be equal:

$$\left(\frac{\partial F_{y'}(x'_0, y'_0, z'_0)}{\partial x'} - \frac{\partial F_{x'}(x'_0, y'_0, z'_0)}{\partial y'}\right) = \\ \left[\left(\frac{\partial F_z(x_0, y_0, z_0)}{\partial y} - \frac{\partial F_y(x_0, y_0, z_0)}{\partial z} \right)\vec{x} + \left(\frac{\partial F_z(x_0, y_0, z_0)}{\partial x} - \frac{\partial F_x(x_0, y_0, z_0)}{\partial z} \right)\vec{y} + \left(\frac{\partial F_y(x_0, y_0, z_0)}{\partial x} - \frac{\partial F_x(x_0, y_0, z_0)}{\partial y} \right)\vec{z}\right] \cdot \vec{n} $$ I've been trying to prove the above equality for some time without success, specifically I am not sure how to handle the transformations correctly. Any help with this is much appreciated!

Answer 1

A simpler approach is via integral theorems. As stated in the question, the special cases for a rectangle in the $xy$ , $yz$ , $zx$ planes are well understood. According to Green's theorem : $$ \begin{cases} \iint_{xy} \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) dx\, dy = \oint_{xy} \left( F_x\, dx + F_y\, dy \right) \\ \iint_{yz} \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) dy\, dz = \oint_{xy} \left( F_y\, dy + F_z\, dz \right) \\ \iint_{zx} \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) dz\, dx = \oint_{xy} \left( F_z\, dz + F_x\, dx \right) \end{cases} $$ But instead of rectangles, we take half rectangles, or better: the triangles $OAB$ , $OBC$ , $OAC$ respectively:

Thanks to Green's theorem we can replace area integrals by line-integrals; mind that they are counter-clockwise. Then it is clear that, irrespective of any further content: $$ \oint_{OAB} + \oint_{OBC} + \oint_{OAC} + \oint_{ABC} = 0 $$ Assuming that the operator rot(ation) is not defined yet in general, this means that we now have an expression for it: $$ 2 \iint_{ABC} \vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n}\, dA = \\ - \iint_{xy} \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) dx\, dy - \iint_{yz} \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) dy\, dz - \iint_{zx} \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) dz\, dx $$ Continuing with infinitesimal volumes / areas and flipping normals on the right hand side, so that they become the components of the normal at the left hand side: $$ \vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n}\, \Delta A = \\ \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right)\cdot n_x\, \Delta A + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right)\cdot n_y\, \Delta A + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\cdot n_z\, \Delta A $$ Leaving out the infinitesimal area $\,\Delta A\,$ gives us the same answer as found by the OP themselves.
A somewhat neater approach is to calculate mean values and let the area of the (red) triangle go to zero: $$ \vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n} = \lim_{ABC \to 0} \frac{\iint_{ABC} \vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n}\, dA}{\iint_{ABC} dA} $$ Note. I've encountered essentially the same method at several places elsewhere in physics (I think it's with stress and strain). Aanyway, a related subject is : What does shear mean?

Answer 2

It is possible to prove this by taking Green's theorem and applying rotations. So what we do is rotate the whole space so that the normal is in the z direction (or whatever Green's Theorem configuration we like) and then use the rotation information to get an expression for the original space.

We start by applying a rotation around the x and y axis

$$ \left(\begin{array}{c} x' \\ y' \\ z' \end{array}\right) = \left(\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i\end{array}\right) \cdot \left(\begin{array}{c} x \\ y \\ z\end{array}\right) $$

This rotates the surface so that its normal at the required point, points upwards. This means,

$$\left(\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right) = \left(\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i\end{array}\right) \cdot \vec{n} $$

and by using the inversion property of rotation matrices,

$$ \vec{n} = \left(\begin{array}{ccc} a & d & g \\ b & e & h \\ c & f & i\end{array}\right) \cdot \left(\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right) = \left(\begin{array}{c} g \\ h \\ i \end{array}\right) $$

Notice that the normal $\vec{n}$ is the last row of our rotation matrix. Since the first and second row are also unit vectors orthogonal to $\vec{n}$ we can write $\vec{n}$ as their cross product,

$$ \vec{n} = \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ a & b & c \\ d & e & f \end{array}\right| $$

and hence,

$$ n_x = \left|\begin{array}{cc}b & c \\ e & f \end{array}\right| n_y = \left|\begin{array}{cc}d & f \\ a & c \end{array}\right|, n_z = \left|\begin{array}{cc}a & b \\ d & e \end{array}\right|$$ We also want to rotate our vector field appropriately: $$ \left(\begin{array}{c} F_{x'}(P') \\ F_{y'}(P') \\ F_{z'}(P') \end{array}\right) = \left(\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i\end{array}\right) \cdot \left(\begin{array}{c} F_x(P) \\ F_y(P) \\ F_z(P)\end{array}\right) $$

Finally we can derive our expression using simple determinant properties and the linear relationships that we wrote above,

$$\begin{gather}\frac{\partial F_{y'}}{\partial x'} - \frac{\partial F_{x'}}{\partial y'} \end{gather} = \\ \left|\begin{array}{cc} \frac{\partial}{\partial x'} & \frac{\partial}{\partial y'} \\ F_{x'} & F_{y'} \end{array}\right| = \\ \left|\begin{array}{ccc} a\frac{\partial}{\partial x} + b\frac{\partial}{\partial y} + c\frac{\partial}{\partial z} & d\frac{\partial}{\partial x} + e\frac{\partial}{\partial y} + f\frac{\partial}{\partial z} \\ aF_{x} + bF_{y} + cF_{z} & dF_{x} + eF_{y} + fF_{z} \end{array}\right| = \\ \left| \begin{array}{cc} a\frac{\partial}{\partial x} & e\frac{\partial}{\partial y} \\ aF_x & eF_y\end{array}\right| + \left| \begin{array}{cc} a\frac{\partial}{\partial x} & f\frac{\partial}{\partial z} \\ aF_x & fF_z\end{array}\right| + \left| \begin{array}{cc} b\frac{\partial}{\partial y} & d\frac{\partial}{\partial x} \\ bF_y & dF_x\end{array}\right| + \\ \left| \begin{array}{cc} b\frac{\partial}{\partial y} & f\frac{\partial}{\partial z} \\ bF_y & fF_z\end{array}\right| + \left| \begin{array}{cc} c\frac{\partial}{\partial z} & d\frac{\partial}{\partial x} \\ cF_z & dF_x\end{array}\right| + \left| \begin{array}{cc} c\frac{\partial}{\partial z} & e\frac{\partial}{\partial y} \\ cF_z & eF_y\end{array}\right| = \\ (bf-ce)\left| \begin{array}{cc} \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_y & F_z \end{array}\right| + (af-cd)\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\ F_x & F_z \end{array}\right| + (ae-db)\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ F_x & F_y \end{array}\right| = \\ \left|\begin{array}{cc}b & c \\ e & f \end{array}\right|\left| \begin{array}{cc} \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_y & F_z \end{array}\right| - \left|\begin{array}{cc}d & f \\ a & c \end{array}\right|\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\ F_x & F_z \end{array}\right| + \left|\begin{array}{cc}a & b \\ d & e \end{array}\right|\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ F_x & F_y \end{array}\right| = \\ n_x\left| \begin{array}{cc} \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_y & F_z \end{array}\right| - n_y\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\ F_x & F_z \end{array}\right| + n_z\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ F_x & F_y \end{array}\right| = \\ \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{array}\right| \cdot \vec{n}$$