Boltzmann Picture

• What could happen? If no thermal equilibrium issues all spins will line up with main magnetic field
  

  $$M_{0,max} = \frac{\gamma \hbar \rho_{0}}{2}$$ (6.1)

  
  
  would be like Iron, extremely magnetized. Could be dangerous.

  
  
  
  
  
  
  
  

• This situation only occurs near $T = 0$. Magnetization is going to be mediated by thermal energy that can cause spins to transition from state to state.
  
  

• What is relation between difference between spins states relative to thermal energy? Thermal energy characterized by $kT$.

  
  
  
  
  
  
  
  

• 
  

  $$\frac{1/2 \hbar \omega_{0}}{k T} \equiv \hspace{2.0in} (at 1.0T)$$ (6.2)

  
  

• Boltzmann said that the probability of system being in state $\epsilon$ if your in a big bath at temperature $T$ is related to the two energies according to the following
  

  $$P(\epsilon) = \frac{e^{-\epsilon/kT}}{Z}$$ (6.3)

  
  

• where like the last chapter need to normalize which is what $Z$ does
  

  $$Z = \sum_{\epsilon} e^{-\epsilon/kT}$$ (6.4)

  
  

  
  
  
  
  
  
  
  

• Look at probability of more spins in one state, rather than the other. Notice ahead of time that $\hbar \omega_{0}<< kT$ for a person at 37C
  

  $$N_{excess\;aligned} = \frac{N \cdot\left[ P(down) - P(up) \right]}{ Z}$$

  $$= \frac{N \cdot \left[ e^{-\hbar \omega_{0}/2/k/T} - e^{\hbar \omeg... ...ght]} {\left[ e^{-\hbar \omega_{0}/2/k/T} - e^{\hbar \omega_{0}/2/k/T} \right]}$$

  $$\approx= N\cdot \frac{1-\hbar \omega_{0}/2/k/T - 1 - \hbar \omega_{0}/2/k/T} {1-\hbar \omega_{0}/2/k/T + 1 - \hbar \omega_{0}/2/k/T}$$

  $$\approx N \cdot \left( -\hbar \omega_{0}/2/T \right)$$

  $$\approx N \cdot \frac{6.63E-34 (42.57e6)}{1.38E-23 (310)} \cdot B_{0}$$

  $$\approx N \cdot 3.3\times 10 ^{-6} B_{0} \;\;\; (in\; T)$$ (6.5)

  
  

• Only a few excess spins out of a million will be aligned with $B_0$ (IN CLASS PROBLEM)

  
  
  
  
  
  
  
  

• Can also calculate the magnetization of the sample by including magnetization in the numerator of the previous equation
  

  $$M_{0} = \rho_{0}\sum_{m=-s}^{s} P(\epsilon (m)) \mu_{z}(m)$$

  $$= \frac{N \gamma \hbar}{V} \frac{\sum_{m= -s}^{s} m e^{mu}} {\sum_{m= -s}^{s}e^{mu}}$$ (6.6)

  
  
  with
  

  $$u \equiv \frac{\hbar \omega_{0}}{k T}$$ (6.7)

  
  

• Again, make the assumption that the thermal energy is much greater than the energy associated with a spins state, i.e., $u << 1$

  
  
  
  
  
  
  
  

• complete the sum for $m= \pm 1/2$ to get
  

  $$M_{0} = \gamma \hbar \rho_{0} \sum_{m \pm 1/2} {m P(\epsilon_{m}) }$$

  $$= \gamma \hbar \rho_{0} \frac{1}{2}\frac{(1+u/2) - (1-u/2)}{(1+u/2) + (1-u/2)}$$

  $$= \gamma \hbar \rho_{0} B_{0} \frac{ \gamma^{2} \hbar^{2}}{4 k T}$$

  $$= \propto B_{0}$$ (6.8)

  
  

• Magnetization is proportional to static field.