Thermal Physics Reading Assignment 10

docs/physics/thermal/3-interactions-implications.md

@@ -57,3 +57,81 @@ Note that by convention, tabulated entropies include residual entropy due to mol
 Note that to force the integral to converge at absolute zero, we force $\lim_{T \rightarrow 0} C_V = 0$.
 
 Historically, $dS = Q/T$ was the original definition of entropy. This tells us nothing about what entropy is, but is sufficient enough for use in most purposes.
+
+## Section 3.3 - Paramagnetism
+
+Consider a two-state paramagnet, with $N$ spin-$1/2$ particles, in a constant magnetic field $\mathbf{B} = B\hat{\mathbf{z}}$. Then, with each particle acting as a dipole, and neglecting dipole-dipole interactions, we consider the two potential spin values, labeled here as up and down. Then, we can say the energy of any dipole is $\pm \mu B$ for the material's value of $\mu$. If we let a dipole oriented with the field have a negative energy, we can then write the total energy as
+
+$$U = \mu B(N_\darr - N_\uparrow) = U = \mu B((N - N_\uparrow) - N_\uparrow) = \mu B(N - 2N_\uparrow)$$
+
+This holds true as $N = N_\uparrow + N_\darr$. Additionally, this energy can be negative. In this material, we know there is no internal magnetic field, so $B = \mu(H + M)$ (we are not working with the symmetric case) simplifies so that
+
+$$M = \mu(N_\uparrow - N_\darr) = -U/B$$
+
+Now, we can also consider the multiplicity of the system $\Omega$ as $\Omega(N_\uparrow) = \binom{N}{N_\uparrow}$.
+
+A notable consequence of this is if we plot $x = U$ vs $y = S$, the temperature is the reciprocal of the slope (as $1/T = \partial S / \partial U$). We can then normalize this by writing $x = U / \mu B$ (as $B$ is a constant external field) and $y = S / k$. The resulting graph will appear to be a semicircle. Notably, we can find negative temperatures when more than half the dipoles point up.
+
+Analytically, Stirling's approximation tells us that
+
+$$S = k[N \ln N - N_\uparrow \ln N_\uparrow - (N - N_\uparrow) \ln(N - N_\uparrow)]$$
+
+We then see that
+
+$$\frac{1}{T} = (\frac{\partial S}{\partial U})_{N, V} = \frac{\partial N_\uparrow}{\partial U} \frac{\partial S}{\partial N_\uparrow} = -\frac{1}{2\mu B} k \ln(\frac{N - U/\mu B}{N + U / \mu B})$$
+
+We can then solve to see that
+
+$$U = N \mu B \frac{1 - \exp(2 \mu B / kT)}{1 + \exp(2 \mu B / kT)} = -N \mu B \tanh(\frac{\mu B}{kT})$$
+
+Then, we see that $M = N \mu \tanh(\frac{\mu B}{kT})$. Knowing that the heat capacity $C_B$ can be written as $C_B =  (\frac{\partial U}{\partial T})_{N, B}$, we see that
+
+$$C_B = Nk \frac{(\mu B / kT)^2}{\cosh(\mu B / kT)}$$
+
+Note that for an electronic two-state paramagnet, the value of $\mu$ is the Bohr magneton, where $\mu_B = \frac{eh}{4\pi m_e}$. As $x = \mu B / kT \ll 1$ in most cases, we can write $\tanh x \approx x$, so that $M \approx \frac{N \mu^2 B}{kT}$. The fact that $M \propto 1/T$ is known as *Curie's Law*, and holds in the high temperature limit for all paramagnets.
+
+## Section 3.4 - Mechanical Equilibrium and Pressure
+
+Consider two systems separated by a movable partition, in which energy and volume may be exchanged, but the total energy and volume is fixed. Then, at equilibrium, we see that $\frac{\partial S}{\partial U_A} = 0 = \frac{\partial S}{\partial V_A}$. Then, we know that $S = S_A + S_B$, so
+
+$$\0 = \frac{\partial S}{\partial V_A} = \frac{\partial S_A}{\partial V_A} + \frac{\partial S_B}{\partial V_A} = \frac{\partial S_A}{\partial V_A} - \frac{\partial S_B}{\partial V_B} \Rightarrow \frac{\partial S_A}{\partial V_A} = \frac{\partial S_B}{\partial V_B}$$
+
+We can state this as $V = V_A + V_B$ being fixed forces $0 = \partial V_A + \partial V_B$ so then $\partial V_B = -\partial V_A$. Then, at equilibrium, $T(\partial S / \partial V)$ is the same for both systems. We then define pressure as $P = T (\frac{\partial S}{\partial V})_{U, N}$. For an ideal gas, with $\Omega = f(N) V^N U^{3N/2}$, we see that $P = \frac{NkT}{V}$ under this definition, which agrees with the ideal gas law.
+
+Now, consider a two-step process. In the first step, energy is changed by $\Delta U$ while volume is fixed. Then, in step two, volume is changed by $\Delta V$ while energy is fixed. Then, $\Delta S = \Delta S_1 + \Delta S_2$. We can write this as $\Delta S = (\frac{\Delta S}{\Delta U})_V \Delta U + (\frac{\Delta S}{\Delta V})_U \Delta V$. As we take the limit, we see that
+
+$$dS = (\frac{\partial S}{\partial U})_V dU + (\frac{\partial S}{\partial V})_U dV = \frac{1}{T} dU + \frac{P}{T} dV$$
+
+**Theorem**. This gives us the *thermodynamic identity*, in which $dU = T dS - P dV$, which is true in any system.
+
+We can then recall that $dU = Q + W$. Then if $W = -P dV$, we can say $Q = T dS$. However, this is only true if this is a quasistatic process, no other work is done, and many variables are held constant. Additionally, in a quasistatic process that is also adiabatic ($Q = 0$), we call the process *isentropic*. Then, $\Delta S = \frac{Q}{T}$.
+
+For constant pressure processes, we can then write $(\Delta S)_P$ = \int_{T_i}^{T_f}\frac{C_P}{T} dT$.
+
+Note that in non-quasistatic processes, $dS > \frac{Q}{T}$, as $W > -P dV$.
+
+## Section 3.5 - Diffusive Equilibrium and Chemical Potential
+
+Now consider two systems, $A$ and $B$, that can also exchange particles. Then, we see that $\partial S_A / \partial N_A = \partial S_B / \partial N_B$ at equilibrium. Then,
+
+$$ -T \frac{\partial S_A}{\partial N_A} = -T \frac{\partial S_B}{\partial N_B}$$
+
+We then define a chemical potential $\mu$ as
+
+$$\mu = -T (\frac{\partial S}{\partial N})_{U, V}$$
+
+Then, we can say that $\mu_A = \mu_B$ at equilibrium. Otherwise, the system with the lesser chemical potential will tend to gain particles, as it will gain more entropy than the other loses. This makes intuitive sense, as the system with more energy tends to have more particles.
+
+Then, we know that
+
+$$dS = (\frac{\partial S}{\partial U})_{N, V} dU + (\frac{\partial S}{\partial V})_{N, U} dV + (\frac{\partial S}{\partial N})_{U, V} dN$$
+
+We can then say that $dS = \frac{1}{T} dU + \frac{P}{T} dV - \frac{\mu}{T} dN$, so that $dU = T dS - P dV + \mu dN$. Then, we tend to associate $\mu dN$ with chemical work.
+
+Then, if we fix $U$ and $V$, the system only has chemical work, so that $0 = T dS + \mu dN$, which implies $\mu = -T (\frac{\partial S}{\partial N})_{U, V}$. Then, if we instead fix $S$ and $V$, we see that $dU = \mu dN$, which implies that $\mu = (\frac{\partial U}{\partial N})_{S, V}$. This tells us that $\mu$ has units of energy.
+
+## Section 3.6 - Summary, A Look Ahead
+
+**Definition**. In Classic Thermodynamics, we see that $dU = T dS - P dV + \mu dN$, where $\frac{1}{T} = (\frac{\partial S}{\partial U})_{V, N}$, $\frac{P}{T} = (\frac{\partial S}{\partial V})_{U, N}$, and $-\frac{\mu}{T} = \frac{\partial S}{\partial N}_{U, V}$.
+
+Additionally, we can find info for systems such as paramagnets using statistical mechanics.