How would one arrive at the formulas for divergence and curl?

It has been some years since I've taken multivariable calculus now, but there's something I really never understood: how people would discover the expressions for divergence and curl. I mean, the books usually say the formulas and then show that with that it's possible to view divergence as a measure of how much a vector field diverges locally and curl the analog for rotation locally.

Now, it's not clear that if you pick those expressions it will give this interpretation. Books usually say: "we take those formulas because they work" and well, I know that. What I want to know is: imagining we want to find two operators $\operatorname{div}$ and $\operatorname{curl}$ on vector fields such that $\operatorname{div}$ gives local divergence and $\operatorname{curl}$ gives local rotation, how could we deduce the definitions that would work?

I'm questioning this because currently I'm studying differential forms on manifolds, and to appreciate the definition of exterior derivative I thought it would be good to go back and see where the definitions of divergence and curl come from.

Based then on the exterior derivative, I've found out that if $v\in \mathfrak{X}(\mathbb{R}^3)$ is a vector field and we consider the usual cartesian coordinates in $\mathbb{R}^3$ then

$$\nabla \times v = \sum_{i=1}^3 \nabla v^i \times \dfrac{\partial}{\partial x^i} \qquad \nabla\cdot v = \sum_{i=1}^3 \nabla v^i \cdot \dfrac{\partial}{\partial x^i}$$

I then started to try seeing if these formulas were any easier to find out, but I couldn't get anythin from it.

Thanks very much in advance.

Solution 1:

The complete answer is given on pages 22-27 of my 2011 vector calculus notes. I think many good calculus text include these heuristic arguments, I found them in Thomas' calculus a few editions back. Long story short, what you should really do to understand is to prove Greene's and Stokes' Theorems, this will give you deeper insight into the nature of your question. Let me summarize the method here:

1. The flux of $\langle P,Q \rangle$ through a little rectangle with corners $(x,y), (x+\triangle x,y), (x,y+\triangle y), (x+\triangle x,y+\triangle y)$ is easily calculated by multiplying components of $\langle P,Q \rangle$ with width ($\triangle x$) and height ($\triangle y$) with the values of the components at the corners. Then divide by the area $\triangle x \triangle y$ of the rectangle and pass to the limits $\triangle x,\triangle y \rightarrow 0$ to obtains partial derivatives which we recognize as the divergence of $\langle P,Q \rangle$. It follows that, $\nabla \cdot \langle P,Q \rangle$ measures the flux area density of the vector field $\langle P,Q \rangle$. The extension to three variables is including in my notes where I discuss a sketch of the proof of Gauss' Theorem.

2. The curl is seen from doing a similar argument to calculate the circulation of $\langle P,Q \rangle$ around a little rectangle with corners $(x,y), (x+\triangle x,y), (x,y+\triangle y), (x+\triangle x,y+\triangle y)$. To find the circulation from $(x,y)$ to $(x+\triangle x,y)$ we multiply $P(x,y) \triangle x$ as the path is horizontal so only the $x$-component lines up with the segment to contribute some circulation. If you take into account the directions of each segment and divide by the area $\triangle x \triangle y$ of the rectangle and then pass to the limit $\triangle x,\triangle y \rightarrow 0$ and the $z$-component of the curl appears. It follows that, $\nabla \times \langle P,Q,0 \rangle$ measures the circulation area density of the vector field $\langle P,Q,R \rangle$ in the $z$-direction. To extend to a three component vector field we need a vector to describe the possible circulations along a space curve in all three directions.

All of this said, I really would rather give the less helpful answer that $d$ is the natural exterior derivative operation on the exterior algebra of $\mathbb{R}^3$ and it is just the case that: 1. $df = \omega_{\nabla f}$, 2. $d\omega_{\vec{F}} = \Phi_{\nabla \times \vec{F}}$ and 3. $d \Phi_{\vec{G}} = \nabla \cdot G dx \wedge dy \wedge dz$

where $\omega_{\langle a,b,c \rangle} = adx+bdy+cdz$ and $\Phi_{\langle a,b,c \rangle} = ady \wedge dz+bdz \wedge dz+cdx \wedge dy$. Therefore, gradient, curl and divergence are just different levels of the cohomological operator which ultimately reveals the deeper shape of space.

Solution 2:

I'm not sure if this is the type of thing you are looking for, and it is only a partial answer. I don't know about $\operatorname{curl}$, but for $\operatorname{div}$ I've always thought that this came from expectations of what we are looking for. In a physical sense it seems sensible that, given a domain $\Omega$ to study, we may want to know how much 'stuff' (could be water, oil, air, etc.) is coming into and out of the domain at some point in time. If we call this quantity the divergence it is not hard to see that this is the net flux (sum of pointwise velocities) across the boundary of $\Omega$ (I'll call the boundary $\Gamma$). Hence what I am calling the divergence will be the integral of the normal component of the boundary times the velocity vector of the substance I am interested in. I.e. for $\mathbf{v}$ some vector in $d$-dimensional function space $V^{d}$ this would be written as

$$\int_{\Gamma} \mathbf{v}\cdot \nu\ \text{d}s,$$

where $\nu$ is the outward facing normal on $\Gamma$.

Using integration by parts (I don't want to say the divergence theorem here) we can see that the integral of the normal component over the boundary is equal to the 'divergence' integrated over the interior of the domain.

$$\int_{\Gamma} \mathbf{v}\cdot \nu\ \text{d}s = \int_{\Omega} \frac{\partial}{\partial x_{1}}\mathbf{v}\ \text{d}\mathbf{x} + \int_{\Omega} \frac{\partial}{\partial x_{2}}\mathbf{v}\ \text{d}\mathbf{x} + \ldots + \int_{\Omega} \frac{\partial}{\partial x_{d}}\mathbf{v}\ \text{d}\mathbf{x} \\ = \int_{\Omega}\nabla\cdot\mathbf{v}\ \text{d}\mathbf{x} \hspace{12em}$$

The term 'divergence' is just a name given to the operator $\nabla\cdot$ which was found to have the desired property. Given integration by parts this falls into place for me.