8—Multivariable Calculus 199This, I can differentiate. Maximizing (8.36) is the same as maximizing its logarithm, and that’s easier
to work with.
maximizef= ln(N!)−ln(n 1 !)−ln(n 2 !)−ln(n 3 !)
subject ton 1 +n 2 +n 3 =N and n 1 E 1 +n 2 E 2 +n 3 E 3 =E
There are two constraints here, so there are two Lagrange multipliers.
∇
(
f−λ 1 (n 1 +n 2 +n 3 −N)−λ 2 (n 1 E 1 +n 2 E 2 +n 3 E 3 −E)
)
= 0
Forf, use Stirling’s approximation, but not quite. The termln
√
2 πnis negligible. Fornas small as
106 , it is about 6 × 10 −^7 of the whole. Logarithms are much smaller than powers. That means that I
can use
∇( 3
∑
`=1(
−nln(n) +n`
)
−λ 1 n−λ 2 nE`
)
= 0
This is easier than it looks because each derivative involves only one coordinate.
∂
∂n 1
→−lnn 1 −1 + 1−λ 1 −λ 2 E 1 = 0, etc.
This is
n=e−λ^1 −λ^2 E, `= 1, 2 , 3
There are two unknowns here,λ 1 andλ 2. There are two equations, forN andE, and the parameter
λ 1 simply determines an overall constant,e−λ^1 =C.
C
∑^3
`=1e−λ^2 E`=N, and C
∑^3
`=1Ee−λ^2 E=E
The quantityλ 2 is usually denotedβin this type of problem, and it is related to temperature by
β= 1/kT where as usual the Lagrange multiplier is important on its own. It is usual to manipulate
these results by defining the “partition function”
Z(β) =
∑^3
`=1e−βE` (8.38)
In terms of this functionZyou have
C=N/Z, and E=−
N
Z
dZ
dβ
(8.39)
For a lot more on this subject, you can refer to any one of many books on thermodynamics or statistical
physics. There for example you can find the reason thatβis related to the temperature and how the
partition function can form the basis for computing everything there is to compute in thermodynamics.
Especially there you will find that more powerful versions of the same ideas will arise when you allow
the total energy and the total number of particles to be variables too.