I recently completed an introductory course on multivariate calculus, and I'm still trying to come to grips with the concepts taught in the vector calculus segment. Right now, I'm reviewing the concept of divergence.

I understand the lexical definition of divergence, that (in $\mathbb{R}^3$ at least) it's the volumetric density of the outward flux of a vector field. The formulaic definition that Wikipedia and Wolfram offer makes similar sense: $$\mathrm{div}\,\mathbf{F}=\lim_{V \to 0}{\iint_S{\frac{\mathbf{F}\cdot\mathbf{n}}{V}dS}}$$

What I don't understand, however, is how $\mathrm{div}\,\mathbf{F}=\nabla\cdot\mathbf{F}$ follows from the above statement. Something like $\mathrm{div}\,\mathbf{F}=|\nabla\mathbf{F}|$ seems to make more sense to me, because $\mathrm{div}\,\mathbf{F}=\nabla\cdot\mathbf{F}$ just adds up the partials in what are essentially three random directions ($\mathbf{i},\,\mathbf{j},\mathbf{k}$, after all, are the conventional basis vectors for $\mathbb{R}^3$), whereas $\mathrm{div}\,\mathbf{F}=|\nabla\mathbf{F}|$ isn't as... arbitrary? I suppose.

In short: how do you proceed from the formal definition of divergence as the limit of flux over volume as volume goes to zero to getting that divergence is equal to the dot product of nabla and the vector field?

  • $\begingroup$ I don't understand what you mean by "lexical definition." In the second definition, $i, j, k$ can be replaced by any orthonormal basis. You can verify this from the first definition by computing the integral in the first definition in the special case that $V$ is a box. $\endgroup$ – Qiaochu Yuan Feb 11 '13 at 6:29
  • $\begingroup$ If you want to think in terms of the Jacobian $\nabla\mathbf F$, then the divergence is its trace, which is invariant under orthonormal changes of basis, so it's not arbitrary either. $\endgroup$ – Rahul Feb 11 '13 at 7:05
  • $\begingroup$ By the way, it looks like you previously posted as Samuel, is that right? If so, you may want to merge your accounts. $\endgroup$ – Rahul Feb 11 '13 at 7:21

Let $V$ be a cubical volume of edge length $\ell$. As $\ell$ is small, $F$ is approximately constant over each of the faces (note we do not say it is constant over the box).

Let $r$ be the center of the box. The divergence is then

$$\mathrm{div} F = \lim_{\ell \to 0} \frac{1}{\ell^3} [F(r + \ell \hat x/2) \ell^2 - F(r - \ell \hat x/2)\ell^2 + \text{terms like this for $y$, $z$}]$$

A simple rearrangement shows that you get exactly the definition of $\partial F/\partial x$ from the term I've written out, and similarly for the other terms. This is all obtained by evaluating the integral for a particularly chosen geometry of the volume.

As the logic is much the same for the curl, I'd say more this is a way of defining $\nabla$ and how it works, what its expression in various coordinate systems must be. In fact, this construction is used to derive $\nabla$ and its coordinate expression for more general manifolds. It's not that $\nabla$ is defined out of the blue and we show that divergence is related to it. Rather, we found divergence and curl and realized that the geometric arguments used lent themselves to relating them both to a vector differential operator.

  • $\begingroup$ @murphid, "A simple rearrangement shows that you get exactly the definition of ∂F/∂x∂F/∂x". Can you please say whats that rearrangement?? $\endgroup$ – Prasanna Jun 7 '16 at 18:48
  • $\begingroup$ @Prasanna Just multiply the $1/\ell^3$ through, and you get a centered limit reducing to a 1d derivative. $\endgroup$ – Muphrid Jun 7 '16 at 20:19
  • $\begingroup$ Does this mean we can define $f'(x)=\lim_{\ h \to 0} \frac{f(x+h)-f(x-h)}{\ h}$? $\endgroup$ – Prasanna Jun 8 '16 at 5:17
  • $\begingroup$ @Prasanna Almost. Notice the factors of 2 in my expression. Otherwise, yes, that is the basic idea. $\endgroup$ – Muphrid Jun 8 '16 at 14:07

You can use Gauss(Divergence) theorem:

$\mathrm{div}\,\mathbf{F}=\lim_{V \to 0}{\iint_S{\frac{\mathbf{F}\cdot\mathbf{n}}{V}dS}} = \lim_{V \to 0}{\iiint_V{\frac{\nabla\cdot\mathbf{F}}{V}dV}} = \nabla\cdot\mathbf{F}$

Second equality follows from Divergence theorem and third equality is simple exercise(assuming $\nabla\cdot\mathbf{F}$ is continuous).

  • 1
    $\begingroup$ ...Did you just prove that $\operatorname{div}\mathbf F=\operatorname{div}\mathbf F$? $\endgroup$ – Rahul Feb 11 '13 at 7:52
  • $\begingroup$ ohhh sorry I've forgot that we have non standart notation here. Because I have never seen divergence to be defined by integral as in user59093's question. $\endgroup$ – tom Feb 11 '13 at 7:59
  • $\begingroup$ edited to match notation $\endgroup$ – tom Feb 11 '13 at 8:01

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