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Consider an ideal 2-state paramagnet. The entropy is given by:

S = k [N ln N - Nu ln Nu - Nd ln Nd],

where Nu and Nd are related to total number N, internal energy U, magnetic field B and magnetic moment μ by

Nu = (N - U/μB)/2, Nd = (N + U/μB)/2

(a) Use the relation T-1 = (∂S/∂U)N,B to derive U/μB = -N tanh x (where x = μB/kT)

(b) Show that S can be written as

S = Nk [ln (2 cosh x) - x tanh x]

(c) Find the low and high temperature limits (T → 0 and T → ∞) of S.

I have no idea how to derive this at all. I am unsure as to how to do parts a, b, or c. HELP!!!

(a)

S = k [N ln N - Nu ln Nu - Nd ln Nd],

T-1 = (∂S/∂U)N,B
1/T = (∂S/∂U)N,B = k ∂/∂U [ - Nu ln Nu - Nd ln Nd],

Note: Since N is a constant, ∂N/∂U = 0
Note: ∂( ln X )/∂U = 1/X ∂( X )/∂U

Using these,

1/kT = ∂/∂U [ - Nu ln Nu - Nd ln Nd]

= - Nu ∂( ln Nu )/∂U - ln Nu * ∂( Nu )/∂U - Nd ∂( ln Nd )/∂U - ln Nd * ∂( Nd )/∂U

∂( ln Nu )/∂U = 1/ Nu ∂( Nu )/∂U

1/kT = - Nu (1/ Nu ∂( Nu )/∂U) - ln Nu * ∂( Nu )/∂U - Nd (1/ Nd ∂( Nd )/∂U)- ln Nd * ∂( Nd )/∂U

1/kT = - ∂( Nu )/∂U) - ln Nu * ∂( Nu )/∂U - ∂( Nd )/∂U)- ln Nd * ∂( Nd )/∂U

But, Nu = (N - U/μB)/2, Nd = (N + U/μB)/2

∂( Nu )/∂U) = -1/ 2μB and ∂( Nd )/∂U) = 1/ 2μB
Substituting back,

1/kT = ∂( Nu )/∂U) * [-1 - ln Nu] - ∂( Nd )/∂U) [1 + ln Nd ]
1/kT = -1/ 2μB [-1 - ln Nu] - 1/ 2μB [1 + ln Nd ]

1/kT = 1/ 2μB [ln Nu - ln Nd]

= 1/ 2μB [ln Nu/Nd]

1/kT = 1/ 2μB [ln (N - U/μB)/ (N + U/μB)]

Re arranging

[ln (N - U/μB)/ (N + U/μB)] =2μB/kT = 2x

(N - U/μB)/ (N + U/μB) = e 2x

Solve for U.

Trick to solve this kind of problem: Use the following method

a/b = c/d is same as a+b/a-b = c+d/c-d

(N - U/μB)/ (N + U/μB) = e 2x

2N / 2U/μB = (e 2x + 1)/ (e 2x - 1)

U = N μB (e 2x -1)/ (e 2x + 1) = N μB (e -x - e x)/ (e x + e-x) = - N μB tanh x

(b)

S = k [N ln N - Nu ln Nu - Nd ln Nd],

S /k = [N ln N - Nu ln Nu - Nd ln Nd],

From part (a),
U = - N μB tanh x

Also we know,

Nu = ½ (N - U/μB) = ½ [N + N tanh x]= N (ex/2coshx)
Nd = 1/2(N + U/μB) = ½ [N -N tanh x] = N (e -x/2coshx)

Substituting these on S/k,

S /k = [N ln N - N (ex/2coshx) ln Nu - N (e -x/2coshx) ln Nd],

S /k = N ln N - ½ N ex/coshx * ln Nu - ½ N (e -x/coshx) ln Nd],

S /k = N ln N - ½ N [(ex/coshx) ln Nu + (e -x/coshx) ln Nd]

Lets find ln Nu and ln Nd and find the term (ex/coshx) ln Nu + (e -x/coshx) ln Nd and then substitute above.

Nu = N (ex/2coshx)

ln Nu = ln N + x - ln (2coshx)

Nd = N (e -x/2coshx)

ln Nd = ln N - x - ln (2coshx)

(ex/coshx) ln Nu + (e -x/coshx) ln Nd

= (ex/coshx) (ln N + x - ln (2coshx) + (e -x/coshx) (ln N - x - ln (2coshx)

= [(ex/coshx) + (e -x/coshx) ] [ln N - ln (2coshx)] + x ( ex/coshx - e -x/coshx)

(ex/coshx) ln Nu + (e -x/coshx) ln Nd = 2 [ln N - ln (2coshx)] + 2x tanh x ---------(1)

Substituting back on S/k from (1)

S /k = ln N - ½ [2 [ln N - ln (2coshx)] + 2x tanh x]

S /kN = - ½ [ -2 ln (2coshx)] + 2x tanh x]

S /kN = ln (2coshx)] - x tanh x]

S = Nk [ln (2 cosh x) - x tanh x]

(c)

At T reaches zero, x reaches infinity.

lets see what happens to cosh x and tanh x at this limit.

cosh x = (ex + e-x) / 2

as x reaches infinity ex reaches infinity and + e-x reaches zero.

as x reaches infinity , cosh x reaches infinity

tanh x = sinhh x /cosh x = (ex - e-x) / (ex + e-x) = (1 - e-2x) / (1 + e-2x)
as x reaches infinity tanh x reaches one.

as x reaches infinity S = Nk [ln (2 cosh x) - x tanh x] reaches infinity.

As T reaches infinity, x reaches zero. At this limit, cosh x = 1 and tanh x = 0

S = Nk [ln (2 cosh x) - x tanh x] = Nk ln 2