- #1
WeiShan Ng
- 36
- 2
Homework Statement
Show that [itex]\Gamma_T[/itex] is maximum at [itex]E_a = \frac{N_aE}{N_a+N_b}[/itex]
Homework Equations
The expression for [itex]\Gamma(E)[/itex] when [itex]N\gg 1[/itex]
[tex]\Gamma_T = C_aC_b exp \left( -\frac{E_a^2}{2N_a\mu_B^2h^2} \right) exp \left[ - \frac{(E-E_a)^2}{2N_b\mu_B^2H^2} \right][/tex]
where [itex]C_a[/itex] and [itex]C_b[/itex] are normalization constants that depends on [itex]N_a[/itex] and [itex]N_b[/itex]
In terms of the entropy:
[tex]\begin{aligned} \frac{1}{k}S_T = \frac{1}{k}(S_a+S_b) &= ln \Gamma_T \\ &= ln(C_aC_b)-\frac{1}{2(\mu_BH)^2} \left[ \frac{E_a^2}{N_a} + \frac{(E-E_a)^2}{N_b} \right]\end{aligned}[/tex]
The Attempt at a Solution
Try to find the maximum of the entropy using the quadratic terms in the square bracket:
[tex]\begin{aligned} \frac{E_a^2}{N_a} + \frac{(E-E_a)^2}{N_b} &= \frac{(N_a+N_b)E_a^2}{N_aN_b} - \frac{2EE_a}{N_b} + \frac{E^2}{N_b} \\ &= \frac{N_a+N_b}{N_aN_b} \left[ E_a^2 - \frac{2N_aN_bEE_a}{N_b(N_a+N_b)} \right] + \frac{E^2}{N_b} \\ &= \frac{(N_a+N_b)}{N_aN_b} \left[ E_a - \frac{N_aE}{N_a+N_b} \right]^2 +\frac{E^2}{N_b} - \frac{N_a^2E^2}{(N_a+N_b)^2} \end{aligned} [/tex]
And the solution manual reduce this equation to
[tex] \frac{(N_a+N_b)}{N_aN_b} \left[ E_a - \frac{N_aE}{N_a+N_b} \right]^2 + \frac{E^2}{N_a+N_b} [/tex]
I can't get the equality of
[tex]\frac{E^2}{N_b} - \frac{N_a^2E^2}{(N_a+N_b)^2}=\frac{E^2}{N_a+N_b}[/tex]
With [tex]\frac{E^2[N_a^2+2N_aN_b+N_b^2-N_bN_a^2]}{N_b(N_a+N_b)^2}=\frac{E^2(N_a+N_b)N_b}{N_b(N_a+N_b)^2}[/tex] I just can't get both side of the equation to be equal...