Dark Matter

November 7, 2019August 25, 2020 | Cosmoshivani

One evening, out of curiosity I started watching a video lecture “Dark Matter and Galaxy Rotation” by Dr. Bob Eagle. After watching it, I got motivated to learn about this topic in a little more detail. I found a series of videos by theoretical physicist Sabine Hossenfelder on dark matter (Part 1, 2, 3) which are also really awesome. Another video showing a debate between scientists on similar topics was quite interesting to watch.

From classical mechanics, we know that the balance between sun’s gravitational force and centripetal force keeps the planets from flying away from their orbits. Assuming that all planets move in circular orbits we can equate these two factors, the gravitational force on the left and centripetal force on the right:

$\frac{G M_s m_p}{r^2} = \frac{m_p v^2}{r}$

$v^2 = \frac{G M_s}{r}$

According to this relation, the speed $v$ of planets decreases as their distance $r$ from sun increases. This inverse relation is visible in the graph below. Its also known as the rotation curve.

Now, if we see the rotation curve for a galaxy we would expect it to be similar. But that’s not what was observed. The rotation curve of different galaxies show this similar property that the velocity of stars does not decrease as we move away from the galactic center.

**Rotation curve of spiral galaxy M33: Wikimedia Commons**

As per the relation, if the distance from the center increases, the velocity of stars should decrease but it doesn’t. If the velocity of stars is as high as it has been measured then they should just fly off. Because the gravitational force from the stars is not enough to balance the centripetal force. We can assume that the mass keeps increasing as we move farther from the center so that the term $\frac{M}{r}$ remains constant. But if it is true then where is all the extra mass?

Surprisingly, when the mass of the different galaxies was calculated using gravitational lensing it was found to be greater than what was calculated from the total mass of the stars. This extra mass is now thought to spread evenly throughout the galaxy in halos and is called dark matter.

A very famous theory called MOND or Modified Newtonian Dynamics was used to explain this observation. It modifies Newtonian Gravity by assuming that at low accelerations, the relation $F \propto \frac{1}{r}$ should be used instead of $\frac{1}{r^2}$ but this theory has long been ruled out because of many reasons, one of them is that Newtonian gravity already works for planets and stars very well.

In the end, I would love to quote Sabine Hossenfelder:

Dark matter is a model which fits the data and observations to some extent and that’s how science works.

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,

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

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

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

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

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

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

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

$\left<p_x\right> = 0$

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

$\left<p_x^2\right> = \frac{n^2 \pi^2 \hbar^2}{a^2}$

$\Delta p_x = \sqrt{ \frac{n^2 \pi^2 \hbar^2}{a^2} }$

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

Exercise 4.6 Quantum Mechanics: Concepts and Applications, 2nd Edition

Series Expansion

February 12, 2018February 1, 2021 | Cosmoshivani

A function can be expanded using Taylor series which is given by,

$f(x) = f(a) + \frac{x-a}{1!} f^{'}(a) + \frac{(x-a)^{2}}{2!} f^{''}(a) + \frac{(x-a)^{3}}{3!} f^{'''}(a) + ...$

If $a=0$ , it’s called Maclaurin series.

$f(x) = f(0) + \frac{x}{1!} f^{'}(0) + \frac{(x)^{2}}{2!} f^{''}(0) + \frac{(x)^{3}}{3!} f^{'''}(0) + ...$

Let’s figure out the Maclaurin series for all trigonometric functions,

$f(x) = sinx$ $f(0) = 0$
$f^{'}(x) = cosx$ $f^{'}(0) = 1$
$f^{''}(x) = -sinx$ $f^{''}(0) = 0$
$f^{'''}(x) = -cosx$ $f^{'''}(0) = -1$

$sinx = \frac{x}{1!} - \frac{(x)^{3}}{3!} + ...$

$f(x) = cosx$ $f(0) = 1$
$f^{'}(x) = -sinx$ $f^{'}(0) = 0$
$f^{''}(x) = -cosx$ $f^{''}(0) = -1$
$f^{'''}(x) = sinx$ $f^{'''}(0) = 0$

$cosx = 1 - \frac{(x)^{2}}{2!} + ...$

$f(x) = tanx$
$f(0) = 0$
$f^{'}(x) = sec^{2}x = 1+tan^{2}x = 1+f(x)^{2}$
$f^{'}(0)=1$
$f^{''}(x) = 2f(x) f^{'}(x)$
$f^{''}(0) = 0$
$f^{'''}(x) = 2f^{'}(x)^{2}+2f(x)f^{''}(x)$
$f^{'''}(0) = 2$

$tanx = \frac{x}{1!} + \frac{2(x)^{3}}{3!} + ...$

$f(x) = secx$
$f(0) = 1$
$f^{'}(x) = f(x) tanx$
$f^{'}(0) = 0$
$f^{''}(x) = f^{'}(x) tanx + f(x) f(x)^{2}$
$f^{''}(0) = 1$
$f^{'''}(x) = f{''}(x) tanx + f^{'}(x) f(x)^{2} + f^{'}(x) f(x)^{2} +2 f(x)^{2} f^{'}(x)$

$secx = 1 + \frac{(x)^{2}}{2!} + ...$

$f(x) = cotx$ $f(0) = \infty$
$f(x) = cosecx$ $f(0) = \infty$
How can you write series expansion if the value of function is infinite at $x=0$ ? I am still trying to figure out a satisfactory answer to this question.

Update:

If a function is not analytic at a point then it cannot be expanded using Taylor series. We have another series expansion for such points and its called Laurent’s expansion.

So, lets evaluate Laurent series expansion of $f(z) = cotz$ at $z=0$
$f(z)=\frac{1}{tanz}$
$cotz = \frac{1}{z + \frac{z^{3}}{3!} + \frac{2(z)^{5}}{15} + ...}$
$cotz = \frac{1}{z[1 + \frac{z^{2}}{3!} + \frac{2(z)^{4}}{15} + ...]}$

Now we can use the long division method to divide $\frac{1}{z}$ by $\frac{1}{1 + \frac{z^{2}}{3!} + \frac{2(z)^{4}}{15} + ...}$ .

So, we get

$cotz = \frac{1}{z} - \frac{z}{3} -\frac{z^{3}}{45} - ...$

Similarly we can evaluate Laurent series expansion of $cosecz$ .

$f(z) = cosecz$ at $z=0$
$f(z) = \frac{1}{sinz}$
$cosecz = \frac{1}{z- \frac {z^{3}}{3!} + \frac{z^{5}}{5!} - ...}$
$cosecz = \frac{1}{z[1- \frac {z^{2}}{3!} + \frac{z^{4}}{5!} - ...]}$