Statistical and Thermal Physics - Summary of the contents - Part - 2

Let us start with the more abstract mathematical formalism of the statistical mechanics by introducing some new concepts. Let us start with defining $P_i$ is the probability of finding the system in some specific state "i". It is not the combination of outcomes for a particular event but the probability of a single outcome i.e. "i" refers to one particular microstate with energy $E_i$ 
So, the probability for the system to be found in this state is, $$ P_i = C\, N(E_0 - E_i) = C\, N(E)$$ where $ E_i + E = const. = E_0$ You may ask why we don't treat this event as the combination of two independent events and write the probability, as we did previous in two systems S,S' with energy E, E' as P(E) = C N(E) N(E') 
because now we consider the first one as a single microstate. 
Now, we assume E act as a reservoir for $E_i$, and it gives, on Taylor expansion and with suitable approximation, $$ \ln{N(E_0 - E_i)} = \ln{N(E_0)} - \beta{E_i} \\ N(E_0-E_i)= N(E_0) e^{-\beta{E_i}}$$ or simply, $$ P_i = C e^{-\beta{E_i}} = \sum_E N(E) e^{-\beta{E}}$$ We know in addition, the sum of all probabilities should give unity, $$ \sum_i P_i = C \sum_i e^{-\beta{E_i}} = 1 $$ or $$ C = \frac{1}{\sum_i e^{-\beta{E_i}}} \\ \rightarrow\,\,\,\,\,\, P_i = \frac{e^{-\beta{E_i}}}{\sum_i e^{-\beta{E_i}}}$$ Note that, the probabiliy depends exponentially on the value of energy E. 
Defining a new term called Partition function and denoting with a new symbol, $$ X = \sum_i e^{-\beta{E_i}} $$ we try to express all other variable using this X,
The average of Energy is given by, $$ \langle{E}\rangle = \frac{\sum_i e^{-\beta{E_i}E_i}}{\sum_i e^{-\beta{E_i}}} $$ but using X, $$ \sum_i e^{-\beta{E_i}}E_i = -\frac{\partial{X}}{\partial{\beta}}$$ and $$ \langle{E}\rangle = -\frac{1}{X} \frac{\partial{\ln{X}}}{\partial\beta}$$ The generalized force is given by (similar procedure), $$ f = \frac{1}{\beta} \frac{\partial\ln{X}}{\partial{x}}$$ with the same notations used in the previous post. In particular, f = p when x = V. When we talk about pressure, we mean the average pressure. 
Since, X is function of both $\beta$ and x, we get $$d(\ln{X}) = \beta{dW} - E d\beta \\ d\left(k\ln{X} + k\beta{E}\right) = \frac{dQ}{T} \\\rightarrow\,\,\,\,\,\, S = k\left(\ln{X} + \beta{E}\right)$$ where S - entropy. 
The partition for an ideal gas can be calculated as, $$ X = \frac{V^N}{h^{3N}} \left[\int_{-\infty}^{\infty}e^{-\beta{\frac{p^2}{2m}}} \,\,d^3p_i\right]^{3N} $$ The gaussian integral gives, $$\int_{-\infty}^{\infty} e^{-\alpha{x^2}} \,dx = \sqrt{\frac{\pi}{\alpha}} $$
So, $$\int_{-\infty}^{\infty} e^{-\frac{\beta}{2m}p^2}\,dp = \sqrt{\frac{2m\pi}{\beta}}$$
And so, $$ X = V^N \sqrt{\frac{2\pi{m}}{\beta{h^2}}}^{3N} \\ \ln{X} = N \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{m}}{h^2}) - \frac{3}{2} \ln{\beta}\right] $$
And the mean energy is calculated to be $$ E = \frac{3}{2}kT$$
with this, the Entropy is calculated to be, $$ S = k (\ln{X}+\beta{E}) = k N \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{m}}{h^2}) - \frac{3}{2} \ln{\beta}+ \frac{3}{2}\right] \\~\\= kN \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{mk}}{h^2}) - \frac{3}{2} \ln{T} + \frac{3}{2}\right]$$ But, it has some small correction due to Gibbs paradox, and it is corrected as, $$ X' = \frac{X}{N!} $$ So that, our equation will get modified to (using sterling's approximation), $$ \ln{X'} = \ln{X} - \ln{N!} = \ln{X} - N \ln{N} + N $$ $$ S = kN \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{mk}}{h^2}) - \frac{3}{2} \ln{T}+\frac{3}{2} - \ln{N} + 1 \right] \\~\\= kN \left[\ln{\frac{V}{N}} + \frac{3}{2} \ln(\frac{2\pi{mk}}{h^2}) + \frac{3}{2} \ln{T} + \frac{5}{2}\right]$$
which is the final result for Entropy. 
Now, we can calculate Maxwell's velocity distribution law using these ideas and by assuming the molecules to be non-interacting classical ideal particles.
The probability of finding the particle in the position range r and r+dr and momentum range p and p+dp is given by the canonical distribution as we used in the partition function, we can convert it into position and velocity and obtain equation, $$ M(r,v) d^3r \,d^3v = C e^{-\frac{\beta{mv^2}}{2}} d^3r \, d^3v $$ and integrating through all over the space and for all velocity range, we get the total number of molecules , $$ N = C V \left[\int_{-\infty}^{\infty} e^{-\frac{\beta{mv_i^2}}{2}}dv_i \right]^3 \\~\\= CV \left(\frac{2\pi}{\beta{m}}\right)^{\frac{3}{2}} = N $$ 
And so,we get the value of C , and the Maxwell's velocity function (as it does depend only on "v"), $$ M(v) \,d^3r\,d^3v = \frac{N}{V} \left(\frac{m}{2\pi{kT}}\right)^{\frac{3}{2}} e^{-\frac{mv^2}{2kT}} \,d^3r\,d^3v $$ Once we get this, we can ask for the various ways of velocity distribution. 
We can calculate the old school stuff - the three kinds of velocities - but now with the correct mathematical formalism.
Average velocity: To calculate the average of anything, first we need the probability density function. With our equation, the density function is calculated by taking small volume in the velocity space, $$ m(v) dv = 4\pi{M(v)} v^2 dv $$ and the function should be normalized, and so, $$ \int_0^\infty m(v) dv = \frac{N}{V}$$
gives the mean velocity as, $$ \langle{v}\rangle = \frac{V}{N} \int_0^\infty m(v) v dv = \frac{4V\pi}{N} \int_0^\infty M(v) v^3 dv$$
Applying the value of M(v), we get, $$ \langle {v}\rangle = 4\pi \left(\frac{m}{2\pi{kT}}\right)^{\frac{3}{2}} \int_0^\infty e^{-\frac{mv^2}{2kT}}v^3 \,dv  = \sqrt{\frac{8kT}{\pi{m}}}$$
Similarly, the mean square speed is $$ \langle{v^2}\rangle = \sqrt{\frac{3kT}{m}}$$
Finally the most probable speed is calculated by finding the maximum of the function m(v). $$ \frac{dm}{dv} = 0 $$
Apart from the constants, $$ m(v) = c v^2 e^{-\frac{mv^2}{2kT}} = kv^2 e^{(\,\,)} $$ and so the maximum condition gives, $$ 2v e^{(\,\, )} - \frac{m}{kT} v^3 e^{(\,\,)} = 0$$ $$ v_{(mp)} = most \,\,probable\,\,speed\,\,=\sqrt{\frac{2kT}{m}} $$
Specifically, if the particles are moving in one dimension (let us say "x"), then we get, $$\frac{dm_x}{dx} = 0 $$
In one dimension, the factor $4\pi{v^2}$ would not come in the equation of m(v), so it is just $$ m(v) = c e^{-mv^2}{2kT} $$ and maximum condition gives, $$ c \left(-\frac{m}{kT}\right) v = 0 \\\rightarrow \,\,\,\,\,\, v_{(mp)} = 0 $$ 
So, the most probable speed and also the mean speed in one dimension is calculated to be zero. Since, the distribution curves attain maximum at v = 0 and it is also symmetrical about v = 0.

Statistical and Thermal Physics - Summary of the contents - Part - 1

I just want to make a brief summary of the mathematics and concepts involved in Statistical Mechanics. 
Microstate is the state of the system where the phase space is split into unit cells and labelled by some indices.
Instead of analyzing the states of all single particles, we introduce the concept of Ensemble, where we take a large number of identical particles but at different states characterized by various corresponding parameters like spin, pressure, angular momentum, etc. And we ask for the probability of the particular value of that parameter - and so the probability arguments comes into the play.  

To make a probability argument, first we need to define the system and its behaviour as a postulate. For example, in an event of throwing the dice, we are intrinsically assuming the dice is a perfect one so that no outcome is preferred than any other outcome.
Similarly, here we assume that, when an isolated system is in equilibrium (the probability of the state of the particles are independent of time), the system has equal probability for any of its accessible states. 
The accessible states are determined by the initial conditions which can be imposed arbitrarily by the observer - the part where Experimental physicist like to play very much.   
Let us consider an ensemble of particles with energy ranges from E and E+$\delta{E}$ and N(E) denote the total number of possible states of the system in this range (similar to the number of possible combination of the outcomes for an event). Out of these states, some "N(E,s_j)" number of states correspond to some other event corresponding to a physical quantity with the value "$s_j$".   Then the probability of getting the value of the physical quantity "$s_j$", $$ P(s_j) = \frac{N(E,s_j)}{N(E)}$$
[Note: The reason I am using different notation for each time is to practice with the flexibility of our mind - So, don't stick with the notation. Everytime, it will be stated the meaning of the representation] 
Macroscopic system which consists of many microscopic states and characterized by external parameters like Pressure, Volume, etc.
Let us considering two systems S and S' in two different case where in the first place it is allowed only the exchange energy due thermal interactions, and in the second part the energy is transferred purely on mechanical interactions. When we talk about large number of particles, the mean Energy is implied.
First case gives, $$\langle{E}\rangle + \langle{E'}\rangle = constant \\~\\ Heat\,\,\,\,\rightarrow\,\,\,\,\,\,Q + Q' = 0 $$ 
Second case gives, Work done by S = - Work done by S' ,$$W + W' = 0$$
In the general case where both Energy and external parameters are changed, the change in mean energy is given by, 
Differential Change in Mean Energy (dU = d$\langle{E}\rangle$)= differential change in mean energy due to external parameters (mechanical work done on the system "W") + small amound of Heat given to the system "Q"
Which gives us, 
dU = dQ + W(on the system) 
or in reverse, using W (mechanical work on the system) = -W(mechanical by the system), we get $$ dQ = dU + W $$ which is known to be the first law of Thermodynamics. 
If we considered two systems S, S' with energy E, E', then the number of states for the energy E is N(E) and for E' is N(E'). Then the probability of  happening two at the same time (where both are independent), so $$P(A\cap{B}) = P(A).P(B)$$
So, the probability for the event with total energy E + E' = const. is, $$P(E) = C N(E) N(E')$$ where C - proportionality constant. 
To get the maximum probability w.r.t. Energy, we get, $$ \frac{\partial{P}}{\partial{E}} = 0 \\~\\ \frac{\partial\ln(N(E))}{\partial{E}} = \frac{\partial\ln(N(E'))}{\partial{E'}} $$
For the mathematical purpose, we take log, so that function will become more smooth. And we denote, $$\beta(E) =\beta(E') \\~\\ \beta(E) = \frac{\partial\ln(N(E))}{\partial{E}} $$ where we define this quantity as the inverse of temperature, so that $$ \beta = \frac{1}{kT} $$ and we define another new and useful quantity named Entropy as, $$ S = k \ln(N(E)) $$
And thus we can see that, when thermal equilibrium is attained between any two states, they should have the same $\beta$ value or the same temperature. Here comes the zeroth law of thermodynamics, states that if two system A, B are in thermal equilibrium with a third system C then all the three system will be in thermal equilibrium with each other i.e. If A with C and B with C implies A with B. All of them will have same temperature characterized by $\beta$. This temperature has a special name which is Absolute Temperature.  
On a special occasion, if we consider one system as a heat reservoir then we have for the reservoir (if we give some heat Q') then using taylor expansion, $$ f(x) = f(a) + f'(a) (x-a) + \frac{f''(a) (x-a)^2}{2!}+... $$ 
Then, if I replace f(x) = ln(N(E+Q)) where x = E+Q and a = E $$ \ln {N(E+Q)} = \ln(E) + \frac{\partial\ln{N(E)}}{\partial{E}} Q + \frac{1}{2}\frac{\partial^2\ln{N(E)}}{\partial{E^2}}Q^2 +...\\~\\ = \beta{Q} + \frac{1}{2}\frac{\partial\beta}{\partial{E}}Q^2 + ...$$
From the definition of reservoir, we can neglect higher order terms and we get, $$ \ln(N(E+Q)) - \ln(N(E)) = \beta{Q}= \frac{Q}{kT} \\~\\ \delta{S} = \frac{Q}{T} $$ If we take infinitesimal amount of heat energy $Q \,\,\rightarrow\,\, dQ$ then we get, $$ ds = \frac{dQ}{T}$$ which is the formal definition of Entropy. The third law is mostly based on the physical properties of this Entropy when the absolute temperature goes to zero. 
Let us analyze the ideal gas.. We know that for an ideal gas, $$ N \propto V^N \Phi(E)$$ or $$ \ln{N} = N\ln{V} + ln{\Phi(E)} + const. $$ and it is defined the generalized force, $$ f = \frac{1}{\beta}\frac{\partial\ln{N}}{\partial{s_i}}$$ where s- external parameter, and f-generalized force. When x=V we get the mean pressure as,  $$ \langle{p}\rangle = \frac{1}{\beta}\frac{\partial\ln{N}}{\partial{V}} $$ Using this, $$ \langle{p}\rangle = \frac{N}{\beta} \frac{\partial\ln{V}}{\partial{V}} $$ other terms don't depend on "V" and give zero.
We finally get, $$ \langle{p}\rangle = \frac{N}{\beta{V}} = \frac{NkT}{V} \\ \langle{p}\rangle V = nRT $$ where k-boltzmann constant.
N - number of particles
R = Avogadro number * boltzmann const. 
and n- number of moles = N/Avogadro number
and specifically $\beta$ depends only on energy $\Phi(E)$ and Temperature depends only on $\beta$ which gives us, for an ideal gas Energy is a function of Temperature. 
Now, we define a new quantity named specific heat capacity - that is the measure of the heat required to raise the temperature of the system to a unit value keeping a parameter at a constant value, $$ C_s = \left(\frac{dQ}{dT}\right)_s$$ 
Let us look at the relation between $ C_p = \left(\frac{dQ}{dT}\right)_p $ and $ C_v = \left(\frac{dQ}{dT}\right)_v$ where it is measured at constant pressure and volume. Using this in the first law we get, $$ dQ = dU + pdV $$ at constant volume $$dQ_v = dU = c_v dT $$ similarly from the equation of state of an ideal gas at constant pressure, $$ p\, dV = nR\, dT$$ and $$ dQ_p = c_p dT $$ Combining the equations, $$ c_p = c_v + nR$$ and if we change to molar specific heat capacity we get, $$ c_p - c_v = R $$  
We can also measure the microscopic calculation of specific heats from our assumption of monatomoic ideal gas, where the interaction between the particles is negligible. It implies the number of states can be written as, $$ N(E,V) = C V^N E^{\frac{3N}{2}}$$ 
where N is the number of particles and 
C - proportionality constant.
V - volume,
E - energy,
N - number of states. 
Then, $$ \ln{N(E,V)} = \ln{C} + N \ln{V} + \frac{3N}{2} \ln{E}$$ but we know, $$\beta = \frac{\partial\ln{N}}{\partial{E}} = \frac{3N}{2E}$$ and it gives us $$ E = \frac{3N}{2\beta} = \frac{3}{2}NkT = \frac{3}{2} n RT$$ where "n" is the number of moles.
In addition, the molar specific heat at constant volume is given by, $$ C_v = \frac{1}{n} \left(\frac{\partial{U}}{\partial{T}}\right)_v$$ where at constant volume $ dQ = dU$ so, it gives $$ C_v = \frac{3}{2}R$$  for an ideal gas. And so we can find $ C_p,\,\,and\,\,\, \gamma = \frac{C_p}{C_v}$
We can explore more with these. Now various situation may arise depending on which two variables are changed. Let us start with the first law, $$ dQ = dU + pdV$$ Using the relation of entropy, $$ dU = T dS - pdV $$ where the independent variable is S, V.
Now, if we have the independent variables as S, P then we can make use of the legendre transform as, $$ dU = T dS - d(pV)+ VdP \\ d\left(U + pV\right) = T dS + VdP $$ where we call H = U + pV = Enthalpy.
$$ dH = T dS + VdP$$ Next, if we take chose T and V as the independent variables then, $$ dU = TdS - pdV = d(ST) - pdV - SdT \\ d \left(U-TS\right) = - pdV - SdT \\ dF = -SdT - pdV$$ where F = U - TS is called the Helmholtz free energy. 
Finally if we take the independent variables T, P then $$ dU = d(TS) - SdT - d(pV) + VdP \\ d \left(U-TS+pV\right) = -SdT + VdP \\ dG = -SdT + VdP$$ where G = U -TS +pV = F + pV is called the Gibbs Free energy.