Deriving the curl of a vector field from the definition of torque.

Question

I just learned about the definition of $\text{curl}\ F$ for some vector field $F(x, y)=M(x, y)\mathbf{e}_1+N(x, y)\mathbf{e}_2$ in $\Bbb{R}^2$ and was wondering how that could be derived from the definition of torque:

$$ \tau=F\times r $$

My interpretation of curl is that it represents an infinitesimal rate of torque at a point, so my intuition was to set $r = \langle \text{d}x,\ \text{d}y \rangle$, since then $\tau$ would represent the torque infinitesimally close to a point $\langle x,\ y\rangle$.

Now of course, for the cross product to make sense $F$ and $r$ have to be in $\Bbb{R}^3$, so we can just add on a third component of zero to both of them. Now we have $$ F = \langle M(x, y),\ N(x, y),\ 0\rangle \\ r = \langle \text{d}x,\ \text{d}y,\ 0\rangle $$ Computing the cross product gives $$ \tau = \langle 0,\ 0,\ M(x, y)\text{d}y-N(x, y)\text{d}x\rangle $$ Then I assumed that since $\tau$ is a vector and curl is a scalar, we can obtain the curl from $\tau$ by computing it's magnitude, which gives: $$ \text{curl}\ F = ||\tau|| = M(x, y)\text{d}y-N(x, y)\text{d}x $$ Obviously that's wrong, but it seems very close (at least in my opinion) to the actual definition of curl: $$ \text{curl}\ F = N_x-M_y $$ My "curl" has $M\text{d}y$ instead of $M_y$ and $N\text{d}x$ instead of $N_x$, and is off by a factor of $-1$.

Am I on the right track? I feel really close, but I just can't figure out where to go from here to get the right definition of curl. Is curl actually even equivalent to torque, or is it equivalent to something else?

Answer 1

Imagine a small disk floating in a river, perhaps it's a cork or something.

As the water brushes against it a torque is induced. Let's see if we can figure out what the net torque is. First we need to know the force at a given point on the circumference. What we need to look at is the relative velocity of the point with respect to the center. If the center of the cork is $(x, y)$, our point is $(x + \Delta x, y + \Delta y)$, and the velocity field is denoted by $F$, then the relative velocity is: $$F(x + \Delta x, y + \Delta y) - F(x, y)$$

Now for small $\Delta x$ and $\Delta y$ this is approximately equal to the jacobian matrix of $F$ times the vector $(\Delta x, \Delta y)$: $$\begin{pmatrix}\partial_xF_x & \partial_yF_x \\ \partial_xF_y & \partial_yF_y\end{pmatrix}\begin{pmatrix}\Delta x \\ \Delta y\end{pmatrix} = \begin{pmatrix}\partial_xF_x\Delta x + \partial_yF_x\Delta y \\ \partial_xF_y\Delta x + \partial_yF_y\Delta y\end{pmatrix}$$

Now divide the circumference of the cork into a fixed number of arcs.

We will approximate the force on an arc as the relative velocity of the water at the center point of the arc times the length of the arc, which we'll call $\ell$, times some proportionality constant. By changing units we can always make the proportionality constant 1 so we will ignore it. Now we need the component of the force which is perpendicular to the radius connecting the center of our cork to the center of our arc. To get this we just need to project our force onto $(-\Delta y, \Delta x)$. Then multiplying by the length of the level arm, ie the radius of our cork, which we'll call r, we get the torque contributed by that arc:

$$\begin{align}\tau &= \left(\begin{pmatrix}-\Delta y \\ \Delta x\end{pmatrix} \cdot \begin{pmatrix}\partial_xF_x\Delta x + \partial_yF_x\Delta y \\ \partial_xF_y\Delta x + \partial_yF_y\Delta y\end{pmatrix}\right)\ell r \\ &= (\partial_xF_y\Delta x^2 + \partial_yF_y\Delta x\Delta y - \partial_xF_x\Delta x\Delta y - \partial_yF_x\Delta y^2)\ell r\end{align}$$

Then just sum up the torque contribution of each arc: $$\tau = \sum_i \tau_i$$

The next step is to make our approximation better by increasing the number of arcs. First turn $\Delta x$ and $\Delta y$ into $r\cos(\theta)$ and $r\sin(\theta)$ respectively. Turn that sum into an integral, and turn the length of each arc $\ell$ into $rd\theta$ and integrate from $0$ to $2\pi$.

$$\tau = \int_0^{2\pi}\left[\partial_yF_x(r\cos(\theta))^2 + \partial_yF_y(r\cos(\theta))(r\sin(\theta)) - \partial_xF_x(r\cos(\theta))(r\sin(\theta)) - \partial_xF_y(r\sin(\theta))^2\right]r^2d\theta$$

Now that looks like quite the monster, but it's a lot easier to integrate than you'd think. the two terms in the middle just come out to $0$. and the $\cos^2(\theta)$ and $\sin^2(\theta)$ just turn into $\pi$. So we're left with: $$\tau = \pi r^2(\partial_yF_x - \partial_xF_y)$$

Now obviously $\pi r^2$ is just the area of our cork and the bit on the right is the z-component of the curl of $F$. So the relationship you were looking for is that the curl is the torque per unit area that a small cork will feel in a velocity field $F$.

Now it's easy to extend this to 3d. Imagine a ball instead of a disk, totally submerged in a river. The same basic argument can be applied in the 3 coordinate planes and you will get the 3 components of the curl. This also gives you a justification for why the components of the curl actually constitute a real vector. We know already that torques are vectors and that an arbitrary torque in 3d can be decomposed into torques along the 3 coordinate axes. And so we just go in reverse, we have 3 torques along the coordinate axes acting on this ball, so they must be the components of an arbitrary torque.

Answer 2

The curl is a vector operator, the result is a vector and you end up with a vector field in 3D. The field $F=\langle M(x,y,z), N(x,y,z), P(x,y,z)\rangle$ is conservative, so each of $M$, $N$, $P$ is a partial derivative with respect to a direction of the potential function of the field, $f(x,y,z)$. The curl cross product is then ($\mathbf F$ here is field): $$\mathbf{curl}\text{ }\mathbf F(x,y,z)=\begin{vmatrix} \hat i & \hat j & \hat k \\ dx & dy & dz \\ \mathbf M & \mathbf N & \mathbf P \end{vmatrix} = \hat i (\mathbf P dy - \mathbf N dz) + \ldots$$

where $\mathbf M=\frac{\partial f}{\partial x}$, $\mathbf N=\frac{\partial f}{\partial y}$, and $\mathbf P=\frac{\partial f}{\partial z}$.

The torque cross product is ($\mathbf F$ here is force):

$$\tau=\mathbf r \times \mathbf F = \begin{vmatrix} \hat i & \hat j & \hat k \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{vmatrix} = \hat i (F_z r_y - F_y r_z) + \ldots$$

The curl definition is infinitesimal rotation of a vector field and in that respect I see a similarity, i.e., curl of a field looks like torque field for infinitesimally small position vectors at each point in the field.

Answer 3

Let $\mathbf p\in \Bbb R^3$ on $xy$-plane. We write the circle $C_{\mathbf p}(r)$, whose interior is the open disc $D_{\mathbf p}(r)$. We define a sufficiently differentiable $xy$-planar vector field $\mathbf F$. The torque of our field has measure $τ_{\mathbf F}$, and we interpret $τ_{\mathbf F}dθ$ as the infinitesimal amount of field 's torque over a circle 's angle $dθ$ or an infinitesimal arc $ds=rdθ$. We finally write as usual $\mathbf F= \mathbf F_{\bot}+\mathbf F_{||}$. Then we think as follows:

$$\frac {1}{πr^2} \oint_{C_{\mathbf p}(r)} τ_{\mathbf F}dθ=\frac {1}{πr^2}\oint_{C_{\mathbf p}(r)} F_{\bot}rdθ=\frac {1}{πr^2} \oint_{C_{\mathbf p}(r)} \mathbf F_{\bot}\cdot d\mathbf s=\frac {1}{πr^2}\int_{D_{\mathbf p}(r)} (\nabla \times \mathbf F) \cdot \mathbf k dA\approx \frac {1}{πr^2} [(\nabla \times \mathbf F) \cdot \mathbf k](\mathbf p)πr^2= [(\nabla \times \mathbf F) \cdot \mathbf k](\mathbf p),$$

where we used in order the equalities $τ_{\mathbf F}dθ=τ_{\mathbf F_{\bot}}dθ=F_{\bot}rdθ=F_{\bot}ds=\mathbf F_{\bot}\cdot d\mathbf s$, for $\mathbf F_{\bot}||d\mathbf s$; the Stokes Theorem; the equality $\nabla \times \mathbf F= \nabla \times (\mathbf F_{\bot}+\mathbf F_{||})=\nabla \times \mathbf F_{\bot}+\nabla \times \mathbf F_{||}=\nabla \times \mathbf F_{\bot}$, for $\nabla \times \mathbf F_{||}=\mathbf 0$; and the approximation as $r\to 0$.

As a result we have shown that

the overall torque 's density of a vector field round a circle with centre p and radius r, as r vanishes, equals the measure of rotor 's normal component of vector field at p,

precisely

$$\lim_{r\to 0}\frac {1}{πr^2} \oint_{C_{\mathbf p}(r)} τ_{\mathbf F}dθ=[(\nabla \times \mathbf F) \cdot \mathbf k](\mathbf p).$$