Apparent size

October 26, 2019May 5, 2020 | Cosmoshivani

Sitting at the dining table, once I was calculating how planets appear to the human eye from our planet. I was trying to calculate the angular diameter or apparent size of planets using trigonometry. I have no idea where that notebook is, but today, I found an awesome blog which gave me enough background to start doing some calculations again.

This time I am going to calculate the apparent size of ISS. I have seen it floating across the sky a few times.

Angular diameter, $\delta = 2 sin^{-1} \frac{D}{2d}$
$d$ : Distance of ISS from Earth = $400 km$
$D$ : Length of ISS = $109 m$
$\delta$ : angular diameter of ISS = ?
$\delta = 2 sin^{-1} \frac{109}{2 \times 400 \times 10^3}$
$= 2 sin^{-1} (0.000136)$
$= 2 \times 0.0079 = 0.0156 \deg$
$0.0156 \deg = 0.9 arcminute$
Its almost equal to the apparent diameter given on Wikipedia, which is 1 arcminute.

Black Holes

October 24, 2019February 1, 2021 | Cosmoshivani

As I have experienced so far, the origin of all cosmological wonders is Einstein’s field equations. Their solutions have predicted and explained so many phenomena. Like Gravitational waves, Black holes, the precession of Mercury’s orbit, the curvature of space-time etc.

And in this blogpost, I will try to find the Schwarzschild radius of some familiar stars.

So, if you squeeze matter inside a certain radius called Schwarzschild radius, you get a black hole. Its also sometimes called the event horizon, from where nothing can escape.

Lets start doing some calculations now,

We already have the escape velocity formula for a classical body, for example Earth, written as

$v = \sqrt {\frac{2 G M}{R}}$

If escape velocity becomes equal to the speed of light (the speed limit of the universe), we have $v = c$

$c = \sqrt{\frac{2 G M}{R}}$

$R = \frac{2 G M}{c^2} \thinspace ... \thinspace \left(1\right)$

Let’s call this radius $R_s$

Using this formula we can evaluate the Schwarzschild Radius of any object. i.e. any object if stuffed inside this radius will theoretically become a black hole. To take a few examples, I will be evaluating $R_s$ of some familiar stars.

But before that lets just check whether we are on the right track. Lets check whether $R_s$ of earth actually turns out to be $8.87 \times 10^{-3}m$ as per Wikipedia.

In $eq. 1$ , $\frac{2 G}{c^2}$ is constant and its value is nearly $1.48 \times 10^{-27}$ so plugging in the mass of earth which is nearly $5.97 \times 10^{24} kg$ we get, $8.85 \times 10^{-3}m$ . And it is close to the actual value which is $8.87 \times 10^{-3}m$ , so we are good to go now.

Lets start with Sirius, brightest star in night sky. There are actually two stars in this system. Sirius A and Sirius B. Sirius A is the star we can actually see. But Sirius B is a white dwarf and can only be seen using a powerful telescope.

It has $2.063$ solar masses which, when converted is nearly $4.1 \times 10^{30} kg$

Again, using $eq.1$ , we now have

$R_s = 1.48 \times 10^{-27} \times 4.1 \times 10^{30} \approx 6 km$

The next star I am taking belongs to a collection of stars called summer triangle. Deneb, which is also a part of the beautiful Cygnus constellation, with 19 solar masses or $3.78 \times 10^{31} kg$

$R_s = 1.48 \times 10^{-27} \times 3.78 \times 10^{31} \approx 56 km$

Which is quite large! Because for sun $R_s$ is nearly $3 km$ . But its reasonable I think since Deneb is 200 times bigger than the sun!

Dust of Butterfly Wings

October 19, 2019February 1, 2021 | Cosmoshivani

If you ever accidentally touched the wings of a butterfly, you must have seen fine dust on your fingers. But its not just an ordinary dust.

Instrument used: Gemko Labwell Microscope

Magnification: 100x

Camera: 8MP

These are the scales of the butterfly wing!

Thanks to Destin of Smarter Every Day channel, who introduced me to this beauty of nature. I wanted to see the butterfly wings since the day I watched his video.

Scales of a moth wing (Found the wing somewhere in the house)

Close your eyes

October 19, 2019August 4, 2020 | Cosmoshivani

Close your eyes and imagine yourself flying above the ground, up above the clouds, passing the satellites you see the blueness of the Earth.

Another blink and you are out of the solar system. You see millions and millions of rocks circling around and dancing under the force of Sun’s gravity.

Blink again and you are out of the galaxy. You see the shimmering stars spiraling around a singularity, which can only be seen if you somehow found a way to step into the fourth dimension. The so called black hole.

All the mysteries of the universe live either in the quantum world or inside a singularity. Let’s get out of here. Close your eyes, you have seen enough suns.

What you see now are the biggest structures discovered by humans so far. These are dancing and swirling threads, threads of light and matter. Cluster and superclusters of galaxies, forming filaments.

Blink again, and you will be in empty space, which ofcourse is only empty to your eyes. You wonder what’s beyond the universe. No matter how many times you squeeze your eyes you are still in the blackness of space.

Why, you ask…

Well, the superpowers that you have of imagining the whole universe inside your mind is a collection of decades of work. We have different theories regarding what lies beyond; none of which have yet been proven.

Now its up to you which reality you want to choose. You may end up in a bubble of universes, each with its own set of laws. Or, in a parallel universe, similar to ours but just a different set of events, where you might never have been born.

So, have you chosen one?

Now, close you eyes…

Day 13: Infinite potential well (1D)

October 15, 2019February 10, 2021 | Cosmoshivani

Solving Schrödinger equation for infinite potential well:

The general method to solve the SE is by solving the TISE for $\psi$ and then multiplying it by $\phi$ .

The potential function is defined as:

$V = 0$ for $\leq x \leq a$ and $V =\infty$ for $x < 0, x > a$

The idea behind taking this non-realistic potential is described quite in detail in this Wikipedia Entry. It’s clearly written that if you have a particle in a box that’s very large, the probability to find it at a certain point is fixed. But, as the box gets smaller, in the nanometer range (think about the zoom in scenes from Ant Man), all you can get is the probability distribution. The particle cannot be located exactly inside the well, but only the probability to find it somewhere inside the box can be calculated. And that’s done by Schrödinger equation.

The TISE (Time Independent Schrödinger Equation) after plugging the potential looks like this,

$\frac{-\hbar^2}{2m} \frac{d^2\psi}{d x^2} = E\psi$

$\frac{d^2\psi}{d x^2} = \frac{-2m}{\hbar^2} E \psi$

It’s a $2^{nd}$ order linear differential equation with constant coefficients.

Let’s simplify the equation by replacing all these constants with a single constant.

Let, $k^2 = \frac{2m}{\hbar^2} E$

Or, $k = \frac{\sqrt{2mE}}{\hbar}$

After plugging it back in the equation, we get a nice looking differential equation.

$\frac{d^2\psi}{dx^2} = -k^2 \psi$

$\frac{d^2\psi}{dx^2} + k^2 \psi = 0$

Let’s solve this equation by a general method,

First we need to make the characteristic equation. For that, we assume $\frac{d}{dx} = m$ and $\frac{d^2}{dx^2} = m^2$

Now, our equation looks like quadratic equation, which can easily be solved.

$\left(m^2 + k^2 \right)\psi = 0$

$m = \pm ik$

We have two complex roots. Let’s see how we can write the general solution for such roots.

If you have complex roots of form $a \pm ib$ , then the general solution looks like this,

$y = c_1 e^{ax} sin\left(bx\right) + c_2 e^{ax} cos\left(bx\right)$

In our case, we have complex roots of the form $0 \pm ik$ , so the solution will be,

$\psi\left(x\right) = A e^{0x} sin\left(kx\right) + B e^{0x} cos\left(kx\right)$

Or, $\psi\left(x\right) = A sin\left(kx\right) + B cos\left(kx\right)$

We now have the wave function for infinite potential well. But, there are still some unknown constants. So we need to evaluate the values of $A$ , $B$ , and $k$ .

For that, we have to apply the boundary conditions. Only those wave functions are valid who follow these conditions. And one of these conditions is, the wave function $\psi$ and its derivative should be continuous. And, for the wave function to be continuous, $\psi\left(x=0\right) = 0$ and $\psi\left(x=a\right) = 0$ . Which basically means, the wave function vanishes at the boundaries, i.e. particle cannot go out of the potential well.

Let’s apply these conditions to our solution,

At $x = 0$ , $\psi\left(0\right) = A sin 0 + B cos 0 = 0$

Or, $\psi\left(0\right) = B = 0$

At $x = a$ , $\psi\left(a\right) = A sin ka + B cos ka = 0$

Or, $\psi\left(a\right) = A sinka = 0$

From here we have either $A = 0$ , (which doesn’t tell us anything) or $sinka = 0$ . To satisfy this last case, the value of $ka$ should be an integer multiple of $\pi$ .

So we have $ka = \pm \pi$ , where $n$ is a Natural number.

We can now write our wave function again, and this time we plug in the values of two constants we have evaluated so far, $k =\frac{n\pi}{a}$ , and $B = 0$

$\psi\left(x\right) = A sin\left(\frac{n \pi}{a} x\right)$

Or, $\psi\left(x\right) = A sin\left(kx\right)$

We have now come one step closer to our final solution. We just need $A$ . And it can be done by normalization.

The process of dividing a wave function with normalization constant is called normalization. And the normalization constant can be evaluated by taking the modulus square of the wave function and integrating it in the range $0$ to $a$ and then equating it to 1. That was mathematical formula. Physically it would mean that we are trying to refine the wave functions so that the probability density remains equal to 1. Probability density is equated to one because we are assuming that the particle is got to be somewhere in a defined region or in all of space if we are taking the limits from $-\infty$ to $+\infty$ .

$\int_0^a |\psi|^2 dx = \int_0^a |A|^2 sin^2\left(kx\right) = 1$

$|A|^2 \int_0^a \frac{\left(1-cos2x\right)}{2} dx = 1$

$|A|^2 \left(\frac{a}{2}\right) = 1$

$A = \sqrt{\frac{2}{a}}$

This is the final piece of the puzzle, let’s plug it back in.

$\psi = \sqrt{\frac{2}{a}} sin \frac{n\pi x}{a}$

The TISE has given us infinite number of solutions for the infinite potential well that we started with. Each of these wave functions have corresponding energy eigenvalues. And we can now evaluate the general expression of the nth energy level.

Here is the expression that we used previously in our derivation,

$k^2 = \frac{2 m E}{\hbar^2}$

Now that we know the possible values of $k$ , we can find the expression for $E$ ,

$\frac{n^2 \pi ^2}{a^2} = \frac{2 m E}{\hbar^2}$

$E = \frac {n^2\pi ^2 \hbar ^2 }{2 m a^2}$

It is the expression for $n^{th}$ energy level!

The wave function $\Psi = \psi\left(x,t\right)$ can now be written.

$\Psi = \psi\left(x\right) \phi\left(t\right)$

$\Psi = \sqrt{\frac{2}{a}} sin \frac{n\pi x}{a} e^{\frac{-iEt}{\hbar}}$

Reference: Introduction to Quantum Mechanics – David J. Griffiths Available to download on Archive.org

Excursion: lMrCoffey

A question related to Uncertainty Principle

October 9, 2019February 1, 2021 | Cosmoshivani

Show that for a particle of mass m which moves in a one-dimensional infinite potential well of length a, the uncertainties product $\Delta x_n \Delta p_n$ is given by $\Delta x_n \Delta p_n \approx \frac{n \pi \hbar}{\sqrt 12}$

We have to calculate the standard deviations $\Delta x_n$ and $\Delta p_n$ and then multiply them to see if its equal to the given value.

$\Delta x_n = \sqrt {\left<x^2\right> - \left<x\right>^2}$

$\Delta x_n = \sqrt {\left<p_x^2\right> - \left<p_x\right>^2}$

Expectation values of $x$ and $p_x$ ,

$\left<x\right> = \frac{\left<\psi|x\psi\right>}{\left<\psi|\psi\right>}$

$\left<p_x\right> = \frac{\left<\psi|p_x\psi\right>}{\left<\psi|\psi\right>}$

The denominator or the normalization factor is common to both expressions so let’s evaluate it first,

$\left<\psi|\psi\right> = \int_{0}^{a} (\sqrt{\frac{2}{a}} sin\frac{n\pi x}{a})^2 dx = 2$

$\left<\psi|x\psi\right> = \int_{0}^{a} \frac{2}{a} sin^2\frac{n \pi x}{a} x dx = a$

$\left<x\right> = \frac{a}{2}$

$\left<\psi|x^2\psi\right> = \int_{0}^{a} \frac{2}{a} sin^2\frac{n \pi x}{a} x^2 dx = \frac{2a^2}{3} - \frac{a^2}{2 n^2 \pi^2}$

$\left<x^2\right> = \frac{a^2}{3} - \frac{a^2}{4 n^2 \pi^2}$

$\Delta x = \sqrt{\frac{a^2}{12} - \frac{a^2}{4 n^2 \pi^2}}$

$\left<\psi |p_x \psi\right> = \int_{0}^{a} \frac{2}{a} sin\frac{n \pi x}{a} -i \hbar \frac{\partial}{\partial x} sin\frac{n \pi x}{a} dx = 0$

$\left<p_x\right> = 0$

$\left<\psi|p_x^2\psi\right> = \int_{0}^{a} \frac{2}{a} sin^2\frac{n \pi x}{a} p_x^2 dx = \int_{0}^{a} \frac{2}{a} sin^2\frac{n \pi x}{a} (-i\hbar \frac{\partial}{\partial x})^2 dx = \frac{2 n^2 \pi^2 \hbar^2}{a^2}$