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There is the well known expression for the divergence of a vector field $V$ as the limit of smaller and smaller surfaces of the flux of a surface.

However it occurred to me that there is another way to describe the divergence which I could not figure out how to search to read more about or to even find out if it is correct.

If $V = \sum V^i e_i$ is a vector field described in the standard basis then the derivative is $$\sum_{i,j}\partial_jV^i e^j\otimes e_i$$ The trace of this is the divergence. Now from what I understand of the trace $\sum\partial_iV^i$ it is proportional to averaging the tensor over all all unit vectors $v$ and their corresponding forms $w$ in the following sense. $$\text{div}(V) = \text{tr}(\sum_{i,j}\partial_jV^i e^j\otimes e_i) \propto \int_C (\sum_{i,j}\partial_jV^i e^j\otimes e_i)(v,w)$$ where $C$ is the unit circle and $w$ is $v$ as a form. Concretely with $v=(\cos\theta,\sin\theta)$ then this becomes $$\int_0^{2\pi} \partial_1V^1\cos^2\theta + \partial_1V^2\cos\theta\sin\theta +\partial_2V^1\cos\theta\sin\theta +\partial_2V^2\sin^2\theta d\theta$$ The terms with $\sin\theta\cos\theta$ evaluate to 0. So this gives $$c(\partial_1V^1 +\partial_2V^2)$$ for a constant $c$.

The initial integral can be interpreted as saying for each direction we look at the derivative of $V$ in that direction $v$ and then project that onto $v$ (this is the role $w$ plays). Said another way this is how much the vector field changes in direction $v$ but only the part along $v$ and only the magnitude change. And then we kind of average over all directions. This is different from the usual flux over surface description in that first off it deals with the derivative of the vector field whereas the the integral of the flux does not deal with the derivative of the vector field.

Is there a description like this for the curl?

user782220
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    Sure, your approach to divergence is like taking the directional derivative $\partial_v V$ and averaging its dot product with $v$. The curl is like doing the same thing but taking the cross product instead (hence the notation $\nabla\cdot V$ and $\nabla\times V$). In tensor notation, you will invoke the Levi-Civita symbol. –  Apr 26 '19 at 08:08
  • And what is the meaningful intuitive description of that? – user782220 Apr 26 '19 at 22:37
  • Specifically I mean how $\varepsilon^{ijk}\partial_m V ^n$ with trace applied to $j,m$ and $k,n$ has what meaning similar to the trace applied giving the diviergence is a averaging as mentioned above. – user782220 Apr 28 '19 at 06:11
  • Maybe helpful : What does shear mean? . – Han de Bruijn Apr 28 '19 at 12:16
  • There's a nice video on divergence and curl by 3Blue1Brown that gives that gives some great visual intuition: https://www.youtube.com/watch?v=rB83DpBJQsE – Emory Sun May 03 '19 at 16:06

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The curl of a vector field is only defined in three dimensions.

Imagine a small circle perpendicular to the vector $e_i$. The $e_i$ component of the curl, $(\nabla \times V)\cdot e_i$, can be described as the average over all points $v$ on that circle of the derivative of $V$ in the direction counterclockwise tangent to the circle at $v$.

For example, suppose that we want to find the z-component of curl and we have a unit circle in the xy-plane parameterised as $v(\theta)=(\cos\theta, \sin \theta)$. The counterclockwise tangent vectors are $w(\theta)=(-\sin\theta, \cos\theta)$. This z-component is proportional to

$\displaystyle \int_0^{2\pi} \partial_iV^j v_iw_j \mathrm{d}\theta=\int_0^{2\pi} (-\partial_1V^1 \cos\theta\sin\theta +\partial_1V^2\cos^2\theta-\partial_2V^1\sin^2\theta+\partial_2V^2\sin\theta\cos\theta)\ \mathrm{d}\theta=c(\partial_1V^2-\partial_2V^1)$

Angela Pretorius
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