# 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!

# Beautiful proofs(#2): Euler’s Sum

$1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \hdots = \frac{\pi^2}{6}$

Say what? This one blew my mind when I first encountered it. But it turns out Euler was the one who came up with it and it’s proof is just beautiful!

Prerequisite
Say you have a quadratic equation $f(x)$ whose roots are $r_1,r_2$, then you can write $f(x)$ as follows:

$f(x) = (x-r_1)(x-r_2) = 0$  (or)

$f(x) = (r_1-x)(r_2-x) = 0$  (or)

$f(x) = (1- \frac{x}{r_1})(1- \frac{x}{r_2}) = 0$

$f(x) = 1 - (\frac{1}{r_1} + \frac{1}{r_2}) + \frac{x^2}{r_1 r_2} = 0$

As for as this proof is concerned we are only worried about the coefficient of $x$, which you can prove that for a n-degree polynomial is:

$a_1 = - (\frac{1}{r_1} + \frac{1}{r_2} + \hdots + + \frac{1}{r_n})$

where $r_1,r_2 \hdots r_n$ are the n-roots of the polynomial.

Now begins the proof

It was known to Euler that

$f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - \frac{1}{3!}y + \hdots$

But this could also be written in terms of the roots of the equation as:

$f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - (\frac{1}{r_1} + \frac{1}{r_2} + \hdots + + \frac{1}{r_n})y + \hdots$

Now what are the roots of $f(y)$ ?. Well, $f(y) = 0$ when $\sqrt{y} = n \pi$ i.e $y = n^2 \pi^2$ *

The roots of the equation are $y = \pi^2, 4 \pi^2, 9 \pi^2, \hdots$

Therefore,

$f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - \frac{1}{3!}y + \hdots = 1 -( \frac{1}{\pi^2} + \frac{1}{4 \pi^2} + \hdots )y + \hdots$

Comparing the coefficient of y on both sides of the equation we get that:

$\frac{1}{6} = \frac{1}{\pi^2} + \frac{1}{4 \pi^2} + \frac{1}{ 9 \pi^2} + \hdots$

$\zeta(2) = \frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \hdots$

Q.E.D

* n=0 is not a root since
$\frac{sin(\sqrt{y})}{\sqrt{y}} = 1$ at y = 0

** Now if all that made sense but you are still thinking : Why on earth did Euler use this particular form of the polynomial for this problem, read the first three pages of this article. (It has to do with convergence)

# Why on earth is matrix multiplication NOT commutative ? – Intuition

One is commonly asked to prove in college as part of a linear algebra problem set that matrix multiplication is not commutative. i.e If A and B are two matrices then :

$AB \neq BA$

But without getting into the Algebra part of it, why should this even be true ? Let’s use linear transformations to get a feel for it.

If A and B are two Linear Transformations namely Rotation and Shear. Then it means that.

$(Rotation)(Shearing) \neq (Shearing)(Rotation)$

Is that true? Well, lets perform these linear operations on a unit square and find out:

(Rotation)(Shearing)

(Shearing)(Rotation)

You can clearly see that the resultant shape is not the same upon the two transformations. This means that the order of matrix multiplication matters a lot ! ( or matrix multiplication is not commutative.)

# Basis Vectors are instructions !

Basis vectors are best thought of in the context of roads.

Imagine you are in a city – X which has only roads that are perpendicular to one another.

You can reach any part of the city but the only constraint is that you need to move along these perpendicular roads to get there.

Now lets say you go to another city-Y which has a different structure of roads.

In this case as well you can get from one part of the city to any other, but you have to travel these ‘Sheared cubic’ pathways to get there.

Just like these roads determine how you move about in the city, Basis Vectors encode information on how you move about on a plane. What do I mean by that ?

The basis vector of City-X is given as:

$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

This to be read as – ” If you would like to move in City-X you can only do so by taking 1 step in the x-direction or 1 step in the y-direction ”

The basis vector of City-Y is given as:

$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$

This to be read as – ” If you would like to move in City-Y you can only do so by taking 1 step in the x-direction or  1 step along the diagonal OB ”

Conclusion:

By having the knowledge about the Basis Vectors of any city, you can travel to any destination by merely scaling these basis vectors.

As an example, lets say need to get to the point (3,2), then in City-X,  you would take 2 steps in the x-direction and 3 steps in the y-direction

$\begin{bmatrix} 3 \\ 2 \end{bmatrix} = 3* \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 2 * \begin{bmatrix} 0 \\ 1 \end{bmatrix}$

And similarly in City-Y, you would take 1 step along the x -direction and 2 steps along the diagonal OB.

$\begin{bmatrix} 3 \\ 2 \end{bmatrix} = 1* \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 2 * \begin{bmatrix} 1 \\ 1 \end{bmatrix}$

Destination Arrived 😀

# On the travelling wave: An intuition

The aim of this post is to understand the travelling wave solution. It is sometimes not explained in textbook as to why the solution “travels”.

We all know about our friend – ‘The sinusoid’.

$y = sin(x)$

y becomes 0 whenever sin(x) = 0 i.e $x = n \pi$

Now the form of the travelling sine wave is as follows:

$y= sin(x - \omega t)$

When does the value for y become 0 ? Well, it is when

$x - \omega t = n \pi$

$x = n \pi + \omega t$

As you can see this value of x is dependent on the value of time ‘t’, which means as time ticks, the value of x is pushed forward/backward by a $\omega$.

When the value of $\omega > 0$, the wave moves forward and when $\omega < 0$, the wave moves backward.

Here is a slowly moving forward sine wave.

By virtue of everyday usage, the fact
that (-1) x (-1) = 1 has been engraved onto our heads. But, only
recently did I actually sit down to explore why, in general negative
times negative yields a positive number !

## The Intuitive Argument

Let’s play a game called “continue the pattern”. You would be surprised, how intuitive the results are:

2 x 3 = 6

2 x 2 = 4

2 x 1 = 2

2 x 0 = 0

2 x (-1) = ??  (Answer : -2 )

2 x (-2 ) = ?? (Answer : -4 )

2 x ( -3) = ?? (Answer : -6 )

The number on the right-hand side keeps decreasing by 2 !

Therefore positive x negative = negative.

2 x -3 = -6

1 x -3 = -3

0 x -3 = 0

-1 x -3 = ?? (Answer : 3)

-2 x -3 = ?? (Answer : 6)

The number on the right-hand side keeps increasing by 3.

Therefore negative x negative = positive.

Pretty Awesome, right? But, let’s up the ante and compliment our intuition.

## The Banker’s Interpretation

We can think of negative number as a debt.

If my bank account contains $-3 then I owe the bank$3.

Suppose that my debt is multiplied by 2, then surely the debt becomes $6. So it makes perfect sense to insist that 2 x -3= -6. What then should -2 x -3 be? Well if the bank kindly writes off ( takes away ) two debts of$3 each. My account balance is now 0. It is exactly as though i had deposited \$6 to my account.

So, in banking terms -2 x -3 = +6

## The Number Line Approach.

Imagine a number line on which you walk. Multiplying x*y is taking x steps, each of size y.

Negative
steps require you to face the negative end of the line before you start
walking and negative step sizes are backward (i.e., heel first) steps.

So, -x*-y means to stand on zero, face in the negative direction, and then take x backward steps, each of size y.

Ergo,
-1 x -1 means to stand on 0, face in the negative direction, and then
take 1 backward step.

This lands us smack right on +1 !

I am saving the best for the last.

Obviously I cannot call -1x -1 as -1 since if that were the case then :

-1 x 1 = -1 and also -1 x -1 = -1. We get,

-1 x 1 = -1 x -1

i.e then -1 = 1 by virtue of cancellation.Which is absurd.

Hence it is +1

Powerful technique!

## Concluding remarks.

Hope you enjoyed the post and Pardon me if you found this to be
rudimentary for your taste. This post was inspired by Joseph H.
Silverman’s Book – A friendly Introduction to Number Theory. If you are passionate about numbers or math, in general it is a must read.

There are several other arithmetic methods that prove the same, if you are interested feel free to explore.

Have a Good Day!