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Thermal Equilibrium and Longitudinal Relaxation

  1. Sep 12, 2015 #1
    1. The problem statement, all variables and given/known data

    Problem 6.2 from Magnetic Resonance Imaging: Physical Principles and Sequence Design.

    Show that [itex]M_o \approx \rho_o \frac{s(s+1)\gamma^2\hbar^2}{3kT}B_o[/itex]

    2. Relevant equations

    [itex] M_o = \rho_o \gamma \hbar \frac{\sum m_s e^{m_s(\hbar w_o / kT)}}{\sum e^{m_s(\hbar w_o / kT)}} [/itex],

    where [itex] m_s [/itex] is the magnetic quantum number.

    3. The attempt at a solution

    Use the fact that [itex] kT >> \hbar w_o[/itex],

    let [itex] \alpha = \frac{\hbar w_o}{kT} [/itex].

    Then [itex] e^{\alpha m_s} \approx 1 + \alpha m_s [/itex].

    [itex] M_o = \rho_o \gamma \hbar \frac{\sum m_s(1 + \alpha m_s)}{\sum (1 + \alpha m_s)} [/itex],

    but [itex] \sum m_s = 0 [/itex], therefore

    [itex] M_o = \rho_o \gamma \hbar \frac{\alpha \sum (m_s^2)}{1 + 0} [/itex].

    Assuming the above is correct. Am I correct that [itex] \sum m_s^2 = s(2s+1)(s+1) [/itex]?

    If so, I get,

    [itex] M_o = \frac{\rho_o \gamma^2 \hbar^2 B_o}{kT} s(2s+1)(s+1) [/itex],

    since [itex] w_o = \gamma B_o [/itex].

    Clearly this is not the correct form, where have I gone wrong?
     
  2. jcsd
  3. Sep 12, 2015 #2

    TSny

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    Homework Helper
    Gold Member

    Welcome to PF!

    You did not evaluate the sum in the denominator correctly. ##\sum 1 \neq 1##

    Almost. You're off by a simple numerical factor. See http://mathschallenge.net/library/number/sum_of_squares
     
    Last edited: Sep 12, 2015
  4. Sep 13, 2015 #3
    Thanks, rookie mistakes. . .

    [itex] \sum 1 = (2s + 1) [/itex], since summing from -s to s.

    and I forgot a factor a 3

    [itex] \sum m_s^2 = \frac{s(2s+1)(s+1)}{3} [/itex] .

    Thanks for pointing those out to me!
     
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