Understanding math – Ecstasy Shots

Using Complex numbers in Classical Mechanics

When one is solving problems on the two dimensional plane and you are using polar coordinates, it is always a challenge to remember what the velocity/acceleration in the radial and angular directions ( $v_r , v_{\theta}, a_r, a_{\theta}$ ) are. Here’s one failsafe way using complex numbers that made things really easy :

$z = re^{i \theta}$

$\dot{z} = \dot{r}e^{i \theta} + ir\dot{\theta}e^{i \theta} = (\dot{r} + ir\dot{\theta} ) e^{i \theta}$

From the above expression, we can obtain $v_r = \dot{r}$ and $v_{\theta} = r\dot{\theta}$

$\ddot{z} = (\ddot{r} + ir\ddot{\theta} + i\dot{r}\dot{\theta} ) e^{i \theta} + (\dot{r} + ir\dot{\theta} )i \dot{\theta} e^{i \theta}$

$\ddot{z} = (\ddot{r} + ir\ddot{\theta} + i\dot{r}\dot{\theta} + i \dot{r} \dot{\theta} - r\dot{\theta}\dot{\theta} )e^{i \theta}$

$\ddot{z} = (\ddot{r} - r(\dot{\theta})^2+ i(r\ddot{\theta} + 2\dot{r}\dot{\theta} ) )e^{i \theta}$

From this we can obtain $a_r = \ddot{r} - r(\dot{\theta})^2$ and $a_{\theta} = (r\ddot{\theta} + 2\dot{r}\dot{\theta})$ with absolute ease.

Something that I realized only after a mechanics course in college was done and dusted but nevertheless a really cool and interesting place where complex numbers come in handy!

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.

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.

Beautiful Proofs(#3): Area under a sine curve !

So, I read this post on the the area of the sine curve some time ago and in the bottom was this equally amazing comment :

Screenshot from 2017-06-07 00:19:11

$\sum sin(\theta)d\theta =$ Diameter of the circle/ The distance covered along the x axis starting from $0$ and ending up at $\pi$ .

And therefore by the same logic, it is extremely intuitive to see why:

$\int\limits_{0}^{2\pi} sin/cos(x) dx = 0$

Because if a dude starts at $0$ and ends at $0/ 2\pi/ 4\pi \hdots$ , the effective distance that he covers is 0.

If you still have trouble understanding, follow the blue point in the above gif and hopefully things become clearer.

nth roots of unity : A geometric approach

When one is dealing with complex numbers, it is many a times useful to think of them as transformations. The problem at hand is to find the nth roots of unity. i.e

$z^n = 1$

Multiplication as a Transformation

Multiplication in the complex plane is mere rotation and scaling. i.e

$z_{1} = r_{1}e^{i\theta_{1}}, z_{2} = r_{2}e^{i\theta_{2}}$

$z_{1}z_{2} = \underbrace{r_{1} r_{2}}_{scaling} \underbrace{e^{i(\theta_{1} + \theta_{2})}}_{rotation}$

Now what does finding the n roots of unity mean?

If you start at 1 and perform n equal rotations( because multiplication is nothing but rotation + scaling ), you should again end up at 1.

We just need to find the complex numbers that do this.i.e

$z^n = 1$

$\underbrace{zz \hdots z}_{n} = 1$

$z = re^{i\theta}$

$r^{n}e^{i(\theta + \theta + \hdots \theta)} = 1e^{2\pi k i}$

$r^{n}e^{in\theta} =1e^{2\pi k i}$

This implies that :

$\theta = \frac{2\pi k}{n}, r = 1$

And therefore :

$z = e^{\frac{2\pi k i}{n}}$

Take a circle, slice it into n equal parts and voila you have your n roots of unity.

Okay, but what does this imply ?

Multiplication by 1 is a $360^o/0^o$ rotation.

drawing

When you say that you are multiplying a positive real number(say 1) with 1 , we get a number(1) that is on the same positive real axis.

Multiplication by (-1) is a $180^o$ rotation.

drawing-1

When you multiply a positive real number (say 1) with -1, then we get a number (-1) that is on the negative real axis

The act of multiplying 1 by (-1) has resulted in a 180^o transformation. And doing it again gets us back to 1.

Multiplication by $i$ is a $90^o$ rotation.

drawing-2

Similarly multiplying by i takes 1 from real axis to the imaginary axis, which is a 90^o rotation.

This applies to -i as well.

That’s about it! – That’s what the nth roots of unity mean geometrically. Have a good one!

Why is the area under one hump of a sine curve exactly 2?

Girls' Angle

I was talking with a student recently who told me that he always found the fact that $latex int_0^{pi} sin x , dx = 2$ amazing. “How is it that the area under one hump of the sine curve comes out exactly 2?” He asked me if there is an easy way to see that, or is it something you just have to discover by doing the computation.

If you’ve wondered about this too, perhaps you’ll find the following of interest.

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