Prof.Ghrist at his best!

Screenshot from 2017-07-28 10:55:27

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.

Screenshot from 2017-07-28 11:29:46

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.

cube.GIF

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.

 

Vol 0, No 0

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.

59076-spheres

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.

 

 

On the direction of the cross product of vectors

One of my math professors always told me:

Understand the concept and not the definition

A lot of times I have fallen into this pitfall where I seem to completely understand how to methodically do something without actually comprehending what it means. And only after several years after I first encountered the notion of cross products did I actually understand what they really meant. When I did, it was purely ecstatic!

 

Why on earth is the direction of cross product orthogonal ? Like seriously…

I mean this is one of the burning questions regarding the cross product and yet for some reason, textbooks don’t get to the bottom of this. One way to think about this is :

It is modeling a real life scenario!!

The scenario being :

When you try to twist a screw (clockwise screws being the convention) inside a block in the clockwise direction like so, the nail moves down and vice versa.

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i.e When you move from the screw from u to v, then the direction of the cross product denotes the direction the screw will move.

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That’s why the direction of the cross product is orthogonal. It’s really that simple!

 

Another perspective

Now that you get a physical feel for the direction of the cross product, there is another way of looking at the direction too:

Displacement is a vector. Velocity is a vector. Acceleration is a vector. As you might expect, angular displacement, angular velocity, and angular acceleration are all vectors, too.

But which way do they point ?

crossproduct

Let’s take a rolling tire. The velocity vector of every point in the tire is pointed in every other direction. BUT every point on a rolling tire has to have the same angular velocity – Magnitude and Direction.

How can we possibly assign a direction to the angular velocity ?

cross-product-simple

Well, the only way to ensure that the direction of the angular velocity is the same for every point is to make the direction of the angular velocity perpendicular to the plane of the tire.
Problem solved!