# Thermal Equilibrium and Longitudinal Relaxation

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1. Sep 12, 2015

### Shinnobii

1. The problem statement, all variables and given/known data

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

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

2. Relevant equations

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

where $m_s$ is the magnetic quantum number.

3. The attempt at a solution

Use the fact that $kT >> \hbar w_o$,

let $\alpha = \frac{\hbar w_o}{kT}$.

Then $e^{\alpha m_s} \approx 1 + \alpha m_s$.

$M_o = \rho_o \gamma \hbar \frac{\sum m_s(1 + \alpha m_s)}{\sum (1 + \alpha m_s)}$,

but $\sum m_s = 0$, therefore

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

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

If so, I get,

$M_o = \frac{\rho_o \gamma^2 \hbar^2 B_o}{kT} s(2s+1)(s+1)$,

since $w_o = \gamma B_o$.

Clearly this is not the correct form, where have I gone wrong?

2. Sep 12, 2015

### TSny

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
3. Sep 13, 2015

### Shinnobii

Thanks, rookie mistakes. . .

$\sum 1 = (2s + 1)$, since summing from -s to s.

and I forgot a factor a 3

$\sum m_s^2 = \frac{s(2s+1)(s+1)}{3}$ .

Thanks for pointing those out to me!