## During a astrophysics lecture…

Prof: You can attend a colloquium ; sleep through the entire session but if you wanted to ask a really intelligent sounding question at the end go with ‘Excuse me, what about the magnetic fields?’ … Works like a charm ! ðŸ˜€ ðŸ˜€

## 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:

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

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 ðŸ˜€

## An infinite joke

$\frac{1}{0} = \infty$

Now flip this over by 90 degree counter clockwise :

$- 10 = 8$

$- 18 = 0$

Now flip this over again by 90 degree clockwise :

$\frac{1}{\infty} = 0$