Thermal Equilibrium and Longitudinal Relaxation

[Shinnobii]

Homework Statement

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

Homework 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.

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?

 

[TSny]

Welcome to PF!

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}.

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

Am I correct that \sum m_s^2 = s(2s+1)(s+1)?

Almost. You're off by a simple numerical factor. See http://mathschallenge.net/library/number/sum_of_squares

 

[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!