Prof.Ghrist at his best!

To understand why this is true, we must start with the Fundamental Theorem of Vector calculus. If $F$ is a conservative field ( i.e $F = \nabla \phi$ ), then

$\int\limits_{A}^{B} F.dr = \int\limits_{A}^{B} \nabla\phi .dr = \phi_{A} - \phi_{B}$

What this means is that the value is dependent only on the initial and final positions. The path that you take to get from A to B is not important.

Now if the path of integration is a closed loop, then points A and B are the same, and therefore:

$\int\limits_{A}^{A} F.dr = \int\limits_{A}^{A} \nabla\phi .dr = \phi_{1} - \phi_{1} = 0$

Now that we are clear about this, according to Stokes theorem the same integral for a closed region can be represented in another form:

$\int_{C} F.dr = \int\int_{A} (\nabla X F) .\vec{n} dA = 0$

From this we get that Curl = $\nabla X F = 0$ for a conservative field (i.e $F = \nabla \phi$). Therefore when a conservative field is operated on by a curl operator ($\nabla X$), it yields 0.

Bravo Prof.Ghrist! Beautifully said 😀

How to visualize Flux ?

Sometime ago I was asked how to visualize Flux in the context of Gauss law.

$\int\int_{S} {\bf E. \vec{n}} dS = \frac{q}{\epsilon}$

I believe one of the primary reasons why people get thrown away by this idea of flux is due to that double integral sign. And when explained what that integral meant, a lot of people felt at ease.

What is Flux ?

Flux is a measure of how much stuff is entering or leaving a surface.

What does the Integral mean ?

$\int\int_{S} {\bf E. \vec{n}} dS$

Why is the above integral a representation of Flux?

To understand why let’s take the example where you know the electric field and want to find the flux across a sphere. How would you go about finding that ?

Well lets start with a cube and wrap it around the charge and calculate the stuff coming in and out of each surface of this cube. This won’t give the actual value but an approximate.

$Flux \approx Flux_{face-1} + Flux_{face-2} + \hdots + Flux_{face-6}$

$Flux \approx E_{1} \Delta S + E_{2} \Delta S + \hdots + E_{6} \Delta S$

$Flux \approx \sum\limits_{i=1}^{6} {\bf E.\vec{n}} \Delta S$

where $\Delta S$ is the area of the surface.

Now to find a better approximate, you can move from a cube to higher dimensions. And as a result we will get better and better approximates for the Flux.

$Flux \approx \sum\limits_{i=1}^{N} {\bf E.\vec{n}} \Delta S$

where N is the number of surface elements.

But is imperial to note that as we increase the number of surface elements, the surface area must also decrease for better approximation.

And this approximation for the flux becomes the actual value when the area of the surface elements tends to $0$ i.e

$Flux = \sum\limits_{i=1}^{N} {\bf E.\vec{n}} \Delta S$ as $\Delta S \to 0$ $N \to \infty$

This is what is written out as an Integral as :

$Flux = \int\int_{S} {\bf E. \vec{n}} dS$

Now although in this post we have laid emphasis on the surface being a sphere, in theory it can be closed or even open. This analysis would be valid at all times.