# Statistical Mechanics

I have planned to bring my class notes on internet now. In this very first article, I’d like to put some light on Ensembles which are the most essential parts of Statistical Mechanics. After the introduction of Ensembles, we shall proceed to other important topics like μ and γ spaces, postulates of Statistical Mechanics, Liouville’s theorem, partition function etc. and much more. As always, I have tried my best to keep the language simple and graspable.

# Nanostory of Nanotechnology

Fullerene

Well, this is not going to be a nano [very short] story either of fairies or aliens. This is a big story of Nanotechnology, one of the most advanced topics in physics. Wait. It’s not going to be so hard or advanced to read. It is really going to be a good story because I’m not going to teach you about this stuff. I am trying to say and save it’s history on MY DIGITAL NOTEBOOK. I think you all should also read this.
Nanotechnology has become a widely discussed topic today in newspapers, magazines, journals, blogs and even in television ads. It’s very common that some organization announcing ‘yet another’ “nano-conference”. Nanotechnology or nano tech in short, refers to the technology of creating materials, devices and functions using atomically manipulated matter. If you didn’t understand it clearly yet read further lines. Continue reading

# Fox – Rabbit Chase Problems

##### Part I:

A fox chases a rabbit. Both run at the same speed $v$. At all times, the fox runs directly toward the instantaneous position of the rabbit , and the rabbit runs at an angle $\alpha$ relative to the direction directly away from the fox. The initial separation between the fox and the rabbit is $l$.

When and where does the fox catch the rabbit (if it does)? If it never does, what is their eventual separation?

##### Part II:

Similarly think about the same situation, except now let the rabbit always move in the straight line of its initial direction in above part of the question.

When and where does the fox catch the rabbit (if it does)? If it never does, what is their eventual separation?

## Solutions

#### Part I

The relative speed of the fox and the rabbit, along the line connecting them, is always $v_{\text{rel}}= v- v \cos \alpha$. Therefore, the total time needed to decrease their separation from $l$ to zero is $T=\dfrac{l}{v-v \cos \alpha} =\dfrac{l}{v(1-\cos \alpha)} \ \ldots (1)$ which is valid unless $\alpha=0$, in which case the fox never catches the rabbit.
The location of their meeting is a little trickier to obtain. We have two methods to do :

##### SLICK METHOD

Imagine that the rabbit chases another rabbit, which chases another rabbit, etc. Each animal runs at an angle $\alpha$ relative to the direction directly away from the animal chasing it. The initial positions of all the animals lie on a circle, which is easily seen to have radius $R=\dfrac{l/2}{\sin (\alpha/2)} \ \ldots (2)$.
The center of the circle is the point, O, which is the vertex of the isosceles triangle with vertex angle $\alpha$, and with the initial fox and rabbit positions as the othe two vertices. By symmetry, the positions of the animals at all times must lie on a circle with center O. Therefore, O, is the desired point where they meet. The animals simply spiral into O.

#### Remark

An equivalent solution is the the following:

At all times, the rabbit’s velocity vector is obtained by rotating the fox’s velocity vector by angle $\alpha$. The meeting point O, is therefore the vertex of the above mentioned isosceles triangle,

##### MESSIER METHOD

The speed of the rabbit in the direction orthogonal to the line connecting the two animals in $v \sin \alpha$. Therefore, during a time $dt$, the direction of the fox’s motion changes by an angle $d\theta =\dfrac {v \sin \alpha}{l_t} dt$ , where $l_t$ is the separation at time $t$. Hence the change in the fox’s velocity has magnitude $|d\overrightarrow{v}|=v d\theta =v (v \sin \alpha dt/l_t)$. The vector $d\overrightarrow{v}$ is orthogonal to $\overrightarrow{v}$, therefore, to get the $x$-component of $d\overrightarrow{v}$, we need to multiply $|d\overrightarrow{v}|$ by $v_y/v$. Similar reasoning holds for $y$-component of $d\overrightarrow{v}$, so we arrive at the two equations $\dot{v_x}= \frac{vv_y \sin \alpha}{l_t} \ \ldots (3)$ $\dot{v_y}=- \frac{vv_x \sin \alpha}{l_t} \ \ldots (4)$
Now, we know that $l_t =\{ l-v(1-\cos \alpha) t \}$. Multiplying the above equations (3) and (4) by $l_t$ , and integrating from the initial to final times, yields $v_{x,0}l+v(1-\cos \alpha)X=v \sin \alpha \, Y \ \ldots (5)$ $v_{y,0}l+v(1-\cos \alpha)Y=-v \sin \alpha \, X \ \ldots (6)$
where (X,Y) is the total displacement vector and $(v_{x,0},v_{y,0})$ is the initial velocity vector. Putting all the X and Y terms on the right sides, and squaring and adding the equations, we get $l^2v^2=(X^2+Y^2)(v^2 \sin^2 \alpha +v^2{(1-\cos \alpha)}^2). \ \ldots (7)$ Therefore , the net displacement is
$R=\sqrt{X^2+Y^2}=\dfrac{l}{\sqrt{2(1-\cos \alpha)}}=\dfrac{l/2}{\sin (\alpha/2)} \ \ldots (8)$
To find the exact location, we can, without loss of generality, set $v_{x,0} =0$, in which case we find $Y/X=(1-\cos \alpha)/\sin \alpha =\tan \alpha/2$. This agrees with the result of the first solution. $\Box$

#### Part II:

##### SLICK METHOD

Let $A(t)$ and $B(t)$ be the positions of the fox and the rabbit respectively. Let $C(t)$ be the foot of the perpendicular dropped from $A$ to the line of the rabbit’s path. Let $\alpha_t$ be the angle, dependent to the time, at which the rabbit moves relative to the direction directly away from the fox (so at $t=0, \ \alpha_0=\alpha$ and at $t=\infty , \ \alpha_{\infty}=0$).

The speed at which the distance AB decreases is equal to $v-v \cos \alpha_t$. Therefore, the sum of the distances AB and CB doesn’t change. Initially, the sum is $l+l \cos \alpha$ and in the end , it is $2d$ where $d$ is desired eventual separation. Therefore, the desired eventual separation
$d=\dfrac{l(1+\cos \alpha)}{2} \ \ldots (9)$

##### STRAIGHT FORWARD METHOD

Let $\alpha_t$ be defined as in the first solution, and let $l_t$ be the separation at time $t$. The speed of the rabbit in the direction orthogonal to the line connecting the two animals is $v \sin \alpha_t$. The separation is $l_t$ , so the angle $\alpha_t$ changes at a rate $\dot{\alpha_{t}}= - \dfrac{v \sin \alpha_t}{l_t} \ \ldots (10)$. And $l_t$ changes at a rate $\dot{l_t}=-v(1-\cos \alpha_t) \ \ldots (11)$.
Taking the quotient of the above two equations, separating variables, gives a differential equation $\dfrac{\dot{l_t}}{l_t} = \dfrac {\dot{\alpha_t}(1-\cos \alpha_t)}{\sin \alpha_t} \ \ldots (12)$ which on solving gives $\ln (l_t) = -\ln {(1+\cos \alpha_t)} + \ln (k) \ \ldots (13)$. Where $k$ is the constant of integration. Which gives $k=l_t (1+\cos \alpha_t) \ \ldots (14)$. Applying initial conditions $k_0 = l_0 (1+\cos \alpha_0)= l(1+\cos \alpha) \ \ldots (15)$. Therefore from (14), we get $l(1+\cos \alpha) = l_t(1+\cos \alpha_t)$, or $l_t= \dfrac{l(1+\cos \alpha)}{1+\cos \alpha_t}$.
Setting $t= \infty$ and using $\alpha_{\infty} =0$, gives the final result $l_{\infty} = \dfrac{l(1+\cos \alpha)}{2}$. $\Box$.

#### Remark:

The solution of Part II is valid for all $\alpha$ except $\alpha= \pi$. If $\alpha = \pi$, the rabbit run directly towards the fox and they will meet halfway in time $l/2v$.

# Natural Hazards: Impacts of Catastrophic Collisions

“I have grown, looking at sky and stars.” Many people can say this, including me. We loved to watch stars in night. We enjoyed the Sunrise and Sunsets. And offcourse, millions of people loves NASA’s Astronomy Pic Of the Day and they like to add APOD as their wallpapers. There are many more beautiful phenomena in the universe, like Solar Eclipse, Eagle Nebula, Aurora etc. We feel safe, cheer and enjoy our life. However, it does not mean that a smaller strike of any tragedy cannot happen even as you are reading this blog. For a horrible example, who was knowing about the latest Tsunami in Japan, before the date it occured?

The theory of gravity is not enough developed that you can exactly state the motion of a planet or star or any other material in the space. Some bodies, like dwarf planets, sattelites, meteoroids, have random motions. The most fearful motion is of meteoroids, the rock like structures. Usually these are found in between Mars and Jupiter as a belt. But when a meteoroid leaves the belt, the gravitational attraction of the earth, nearest planet to mars, can pull that into it. When a meteoroid enters the atmosphere, friction causes it to heat up and glow. Then we call it a shooting star or a meteor. Small meteors burn up completely as they pass through the atmosphere. Larger ones end up on earth as meteoroid. Impact craters are formed when a large meteoroid or comet crashes into a planet. One such massive natural hazard happened in past, which destroyed all the Dinosaurs.

# A Brief Discussion on Participation of \mu-mesons (muons) based upon Relativistic Dynamics

 The life time of muons { $\mu$-mesons } is $2.2 \times 10^{-6}$ seconds and their speed $0.998c$, so that they can cover only a distance of $0.998c \times 2.2 \times 10^{-6}$ or 658.6 meters in their entire lifetime, and yet they are found in profusion at sea level, i.e., at a depth of 10 kilometers from the upper atmosphere where they are produced. How may this be explained on the basis of (i) Lorentz – Fitzgerald Contraction; (ii) Time Dilation.

# Consequences of Light Absorption – The Jablonski Diagram

Image via Wikipedia

According to the Grotthus – Draper Law of photo-chemistry, also called the principal of photo chemical activation, Only that light which is absorbed by a system can bring about a photochemical change. However it is not essential that the light which is absorbed must bring about a photochemical change. The absorption of light may result in a number of other phenomena as well. For instance, the light absorbed may cause only a decrease in the intensity of the incident radiation. This event is governed by the Beer-Lambert Law. Secondly, the light absorbed may be re-emitted almost instantaneously (within $10^{-8}$ second) in one or more steps. This phenomenon is known as fluorescence. The emission in fluorescence bearer with the removal of the source of light. Sometimes the light absorbed is given out slowly and even long after the removal of the source of light. This phenomenon is known as phosphorescence.
The phenomena of fluorescence and phosphorescence are best explained with the help of the Jablonski Diagram.

In order to understand this diagram, we need to define some terminology. Most molecules have an even number of electrons and thus in the ground state, all the electrons are spin paired. The quantity $\mathbf {2S+1}$, where $S$ is the total electron spin, is known as the spin multiplicity of a state. When the spins are paired $\uparrow \downarrow$ as shown in the figure, the upward orientation of the electron spin is cancelled by the downward orientation so that $\mathbf {S=0}$.

$s_1= + \frac {1}{2}$ ; $s_2= - \frac {1}{2}$ so that $\mathbf{S}=s_1+s_2 =0$.
Hence, $\mathbf {2S+1}=1$

Thus, the spin multiplicity of the molecule is 1. We express it by saying that the molecule is in the singlet ground state.

When by the absorption of a photon of a suitable energy $h \nu$, one of the paired electrons goes to a higher energy level (excited state), the spin orientations of the single electrons may be either parallel or antiparallel. [see image]

•If spins are parallel, $\mathbf {S=1}$ or $\mathbf {2S+1=3}$ i.e., the spin multiplicity is 3. This is expressed by saying that the molecule is in the triplet excited state.
• If the spins are anti-parallel, then
$\mathbf{S=0}$ so that $\mathbf {2S+1=1}$ i.e., the singlet excited state, as already mentioned.

Since the electron can jump to any of the higher electronic states depending upon the energy of the photon absorbed, we get a series of singlet excited states, $\{S_n\}$ where $n \ge 1$ and a series of triplet excited state $\{T_n\}$ where $n \ge 1$. Thus $S_1, \, S_2, \, S_3, ....$ etc are respectively known as first singlet excited state, second singlet excited state and so on. Similarly, in $T_1, \, T_2,\, .....$ are respectively known as first triplet excited state, second triplet excited state and so on.

Make sure, you are not confused in $\mathbf{S}$ & $S_n$

# Classical Theory of Raman Scattering

The classical theory of Raman effect, also called the polarizability theory, was developed by G. Placzek in 1934. I shall discuss it briefly here. It is known from electrostatics that the electric field $E$ associated with the electromagnetic radiation induces a dipole moment $\mu$ in the molecule, given by
$\mu = \alpha E$ …….(1)
where $\alpha$ is the polarizability of the molecule. The electric field vector $E$ itself is given by
$E = E_0 \sin \omega t = E_0 \sin 2\pi \nu t$ ……(2)
where $E_0$ is the amplitude of the vibrating electric field vector and $\nu$ is the frequency of the incident light radiation.

Thus, from Eqs. (1) & (2),
$\mu= \alpha E_0 \sin 2\pi \nu t$ …..(3)
Such an oscillating dipole emits radiation of its own oscillation with a frequency $\nu$, giving the Rayleigh scattered beam. If, however, the polarizability varies slightly with molecular vibration, we can write
$\alpha =\alpha_0 + \frac {d\alpha} {dq} q$ …..(4)
where the coordinate q describes the molecular vibration. We can also write q as:
$q=q_0 \sin 2\pi \nu_m t$ …..(5)
Where $q_0$ is the amplitude of the molecular vibration and $\nu_m$ is its (molecular) frequency. From Eqs. 4 & 5, we have
$\alpha =\alpha_0 + \frac {d\alpha} {dq} q_0 \sin 2\pi \nu_m t$ …..(6)
Substituting for $\alpha$  in (3), we have
$\mu= \alpha_0 E_0 \sin 2\pi \nu t + \frac {d\alpha}{dq} q_0 E_0 \sin 2\pi \nu t \sin 2\pi \nu_m t$ …….(7)
Making use of the trigonometric relation $\sin x \sin y = \frac{1}{2} [\cos (x-y) -\cos (x+y) ]$ this equation reduces to:
$\mu= \alpha_0 E_0 \sin 2\pi \nu t + \frac {1}{2} \frac {d\alpha}{dq} q_0 E_0 [\cos 2\pi (\nu - \nu_m) t - \cos 2\pi (\nu+\nu_m) t]$ ……(8)
Thus, we find that the oscillating dipole has three distinct frequency components:

1• The exciting frequency $\nu$ with amplitude $\alpha_0 E_0$
2• $\nu - \nu_m$
3• $\nu + \nu_m$ (2 & 3 with very small amplitudes of $\frac {1}{2} \frac {d\alpha}{dq} q_0 E_0$. Hence, the Raman spectrum of a vibrating molecule consists of a relatively intense band at the incident frequency and two very weak bands at frequencies slightly above and below that of the intense band.

If, however, the molecular vibration does not change the polarizability of the molecule then $(d\alpha / dq )=0$ so that the dipole oscillates only at the frequency of the incident (exciting) radiation. The same is true for the molecular rotation. We conclude that for a molecular vibration or rotation to be active in the Raman Spectrum, it must cause a change in the molecular polarizability, i.e., $d\alpha/dq \ne 0$ …….(9)

Homonuclear diatomic molecules such as $\mathbf {H_2 \, N_2 \, O_2}$ which do not show IR Spectra since they don’t possess a permanent dipole moment, do show Raman spectra since their vibration is accompanied by a change in polarizability of the molecule. As a consequence of the change in polarizability, there occurs a change in the induced dipole moment at the vibrational frequency.