Tutorial 3
Problem 1 Consider a classical particle moving in a one-dimensional potential well u(x), as
shown in figure 6.10. The particle is in thermal equilibrium with a reservoir at temperature T,
so the probabilities of it’s various states are determined by Boltzmann statistics.
• Show that the average position of the particle is given by
R∞
xe−βu(x) dx
x̄ = R−∞
∞
e−βu(x) dx
−∞
(1)
where each integral is over the entire x axis.
Answer. Suppose, the particle only take discrete values for positions along the x axis,
namely x1 , x2 , ..xi , ... etc with a spacing ∆x between them. Then at each one of these
positions, the particle will have an energy ui = u( xi ). Identifying, each of these positions
as a state for the particle, one can write down the partition function as,
Z = Σi e−βui = Σi e−βu(xi ) .
(2)
Notice that now the probability of finding the particle at a position is given by,
p(xi ) =
e−βu(xi )
,
Z
(3)
and the expectation value of position is,
hxi = Σi xi p(xi ) =
Σi xi e−βui
.
Σi e−βui
(4)
Now in this case, x is actually not discrete, but can take continuous values. That correspond to the limit ∆x → 0. In this limit the ’sums’ in the above expression become an
integral, and
R∞
xe−βu(x) dx
Σi xi e−βu(xi )
−∞
R
lim
= ∞ −βu(x) .
(5)
∆x→0 Σi e−βu(xi )
e
dx
−∞
• If the temperature is reasonably low (but still high enough for classical mechanics to
apply), the particle will spend most of its time near the bottom of the potential well. In
that case we can expand u(x) in a Taylor series about the equilibrium point x0 :
u(x) = u(x0 ) + (x − x0 )
2 3 du 1
1
2 d u
3 d u
+
(x
−
x
)
+
(x
−
x
)
+ ...
0
0
dx x0 2
dx2 x0 3!
dx3 x0
(6)
Show that the linear term must be zero, and that truncating the series after the quadratic
term result in the trivial prediction x̄ = x0 .
Answer. From the figure it is clear that the function has a minima at x0 . We know
that at all the extrema (in particular at the minima ) of the function, the first derivative
of the function should vanish. therefore,
du = 0.
dx x0
1
(7)
And therefore the linear term in the taylor expansion is zero.
If we keep the expansion to only second order, then
d2 u 1
u(x) = u(x0 ) + (x − x0 )2 2 x0
2
dx
(8)
To obtain x̄, we will use the formula (1).
R∞
−βu(x)
x̄ = R−∞
∞
xe
dx
e−βu(x) dx
−∞
R∞
−∞
=
2u
)
dx2 x0
−β(u(x0 )+ 21 (x−x0 )2 d
xe
R∞
−∞
e
2
−β(u(x0 )+ 12 (x−x0 )2 d u
)
dx2 x0
dx
(9)
dx
One immediately notices that the term e−βu(x0 ) is a constant which comes out of the
2
integrals (and cancels away). Calling b = 21 ddxu2 x0 , and substituting (x − x0 ) = y, the
integral become
R∞
R∞
R∞
2
2
2
(y + x0 ) e−β b y dy
y e−β b y dy + −∞ x0 e−β b y dy
−∞
−∞
R∞
R∞
x̄ =
=
(10)
−β b y 2 dy
e
e−β b y2 dy
−∞
−∞
The first term in the numerator is zero because the integrand is an odd function. Therefore
we have,
R ∞ −β b y2
R∞
2
e
dy
x0 e−β b y dy
−∞
= x̄ = x0 R−∞
= x0 .
(11)
x̄ = R ∞ −β b y2
∞
2
e
e−β b y dy
dy
−∞
−∞
• If we keep the cubic term in the Taylor series as well, the integrals in the formula for
x become difficult. To simplify them, assume that the cubic term is small, so its exponential can be expanded in a Taylor series (leaving the quadratic term in the exponent).
Keeping only the smallest temperature-dependent term, show that in this limit x differs
from x0 by a term proportional to kT. Express the coefficient of this term in terms of the
coefficients of the Taylor series for u(x).
If one keeps the cubic term in the expansion as well,
2 3 1
1
2 d u
3 d u
+ (x − x0 )
.
u(x) = u(x0 ) + (x − x0 )
2
dx2 x0 3!
dx3 x0
(12)
Now if one uses the expansion,
e
3u
dx3 x0
β
(x−x0 )3 d
− 3!
=1−
β
d3 u (x − x0 )3 3 x0 ,
3!
dx
(13)
the average of x become, to this order,
R∞
x̄ =
−∞
2 −β u(x0 )+ 12 (x−x0 )2 d u
2
dx
xe
R∞
−∞
−β
e
2 u(x0 )+ 1 (x−x0 )2 d u
2
2
dx
x0
x0
3u
.
dx3 x0
1
+ 3!
(x−x0 )3 d
3 + 1 (x−x0 )3 d u
3
3!
dx
dx
(14)
x0
dx
Now using equation (12), we can write,
2 −β u(x0 )+ 21 (x−x0 )2 d u
2
R∞
x̄ =
dx
xe
−∞
R∞
−∞
e
2 −β u(x0 )+ 12 (x−x0 )2 d u
2
dx
2
x0
3 1 − 3!β (x − x0 )3 ddxu3 x0 dx
x0
1−
β
(x
3!
−
3
x0 )3 ddxu3 x0
dx
(15)
1 d2 u 1 d3 u Making the substitutions: a = u(x0 ), b =
and
c
=
and y = x − x0 , we
2 dx2 x0
3! dx3 x0
can re write the integrals as
R∞
2
(y + x0 ) e−β b y (1 − c y 3 ) dy
−∞
R∞
(16)
x̄ =
e−β b y2 (1 − β c y 3 ) dy
−∞
Again using the fact that odd integrals vanish, this integral simplifies to
R∞
R∞
R∞
2
4 −β b y 2
−β b y 2
dy
dy
−
c
β
y
e
c β y 4 e−β b y dy
x
e
0
−∞
−∞
−∞
R∞
R∞
= x0 −
(17)
x̄ =
e−β b y2 dy
e−β b y2 dy
−∞
−∞
We explicitly calculate the integrals below,
Z ∞
Z ∞
2
4 −β b y 2
cβy e
dy = 2c β
y 4 e−β b y dy
−∞
Z0 ∞
du
u2 e−β b u √
= 2c β
2 u
Z 0∞
u3/2 e−β b u du
= cβ
0
Z ∞
cβ
=
x3/2 e−x dx
5/2
(βb)
0
cβ
Γ(5/2)
=
(βb)5/2
√
cβ 3 π
=
(βb)5/2 4
(18)
(19)
(20)
(21)
(22)
(23)
notice the substitutions u = y 2 and x = βbu in between the steps. We have also used the
definition of a gamma integral in equation (21-22). Similarly we also have the Gaussian
integral,
√
Z ∞
π
−β b y 2
e
(24)
dy = √
βb
−∞
Using the above two results, we get,
R∞
2
c β y 4 e−β b y dy
3c
−∞
R∞
(25)
= 2
2
−β
b
y
4b β
e
dy
−∞
Substituting this result in (17) we get,
R∞
2
c β y 4 e−β b y dy
3c
−∞
R∞
x̄ = x0 −
=
x
−
0
2
4b2 β
e−β b y dy
−∞
(26)
We see that when the third order term in the Taylor expansion is added, the answer
deviates from the previous case by a term proportional to β1 (or the temperature T ).
• The interaction of noble gas atoms can be modelled using the Lennard-Jones potential,
x 6 x0 12
0
u(x) = u0
−2
(27)
x
x
Sketch this function, and show that the minimum of the potential well is at x = x0 , with
depth uo . For argon, x0 = 3.9A and u0 = 0.010 eV. Expand the Lennard-Jones potential
in a Taylor series about the equilibrium point, and use the result of part (c) to predict
the linear thermal expansion coefficient (see Problem 1.8) of a noble gas crystal in terms
of u0 . Evaluate the result numerically for argon, and compare to the measured value
α = 0.0007k −1 (at 80 K).
3
Answer: We start with defining,
∆L = x̄ − x0 = −
3c kB T
.
4b2
(28)
The coefficient of linear thermal expansion is defined as,
α=
1 ∆L
L ∆T
(29)
In this case We have,
∆L
3c kB
=− 2
∆T
4b
(30)
Using the definitions of b and c,
b=
1 d2 u 36 u0
1 d3 u 252u0
=
,
and
c
=
=− 3 ,
2
x
x
2
3
0
0
2 dx
x0
3! dx
x0
(31)
we get,
−
3c kB
7 x0 kB
=
2
4b
48 u0
(32)
Therefore,
α=
1 ∆L
1 7 x0 kB
7 kB
=
=
= 0.0013K −1 .
L ∆T
x0 48 u0
48 u0
(33)
The answer is a factor of two of the measured value.
Problem 2 Use the Maxwell distribution to calculate the average value of v 2 for the molecules
in an ideal gas. Check that your answer agrees with equation 6.41.
Answer: The Maxwell’s distribution is given by,
m 32
mv 2
p(v) = 4π
v 2 e− 2kT
2πkT
(34)
Method 1.
v¯2 =
∞
Z
v 2 p(v) dv
(35)
0
∞
m 32
mv 2
2
and a =
.
v 4 e−av dv, where, c = 4π
2πkT
2kT
0
Z ∞
d2
2
c. 2
e−av dv
da 0
r
d2 1 π
c. 2
da 2 a
√ π −1
−3
c.
a−5/2
2
2
2
5/2
√ 3 m 23
2kT
π
.4π.
8 2πkT
m
√
3
kT
3kT
√
.4 2 =
.
m
4 2 m
Z
= c.
=
=
=
=
=
4
(36)
Method 2.
v¯2 =
=
=
=
=
=
=
Z
∞
v 2 p(v) dv
0
m 23 Z ∞
mv 2
4π
v 4 e− 2kT dv
2πkT
0
m 23 Z ∞
mx
dx
x2 e− 2kT √
4π
2πkT
2 x
0
m 23 2kT 52 Z ∞
dy
y 3/2 e−y
4π
2πkT
m
2
0
5
3
m 2 2kT 2 1
Γ(5/2)
4π
2πkT
m
2
m 23 2kT 52 1 3√π
4π
2πkT
m
2 4
3kT
m
(37)
(38)
(39)
(40)
(41)
(42)
(43)
Notice the change of variables and the definition of a gamma integral used again.
Problem 3 Imagine a world in which space is two dimensional but the laws of physics are
otherwise the same. Derive the speed distribution formula for an ideal gas of non relativistic
particles in this fictitious world, and sketch this distribution. Carefully explain the similarities
and differences between the two dimensional and three dimensional cases. What is the most
likely velocity vector? What is the most likely field ?
Probability(v, ..v + δ v) = D(v) dv
(44)
D(v) ∝ (prob. molecule. vel. v) × (num. of vec ~v corresp. to v)
(45)
We can also write,
The first term is proportional to boltzmann factor,
(prob. molecule. vel. v) ∝ e−mv
2 /2kT
.
(46)
To compute the second factor in the above equation, think about a 2 dimensional velocity space,
in which each point represent a velocity vector. The set of velocity vectors corresponding to
any given speed v lies on a circle with radius v. The larger v is, bigger the circle and more
possible velocity vectors there are. Therefore the second factor in above equation is nothing
but the perimeter of this circle.
(num. of vec ~v corresp. to v) = 2πv.
(47)
Putting this degeneracy factor together with the Boltzmann factor, we obtain,
D(v) = C.2π.v.e−mv
2 /2kT
,
where C is a constant of proportionality, to be fixed by normalization.
Z ∞
m
2
1=C
2π.v.e−mv /2kT ⇒ C =
2πkT
0
5
(48)
(49)
Therefore,
D(v) =
mv −mv2 /2kT
e
kT
(50)
The most likely velocity vector is zero. But most likely speed is v for which D(v) is maximum.,
Demanding the first derivative vanish yields this speed as,
mv 2
p
dD(v)
me− 2kT (kT − mv 2 )
=0⇒
= 0 ⇒ v = kT /m
2
2
dv
k T
(51)
Problem 4. Some advanced textbooks define entropy by the formula
S = −kΣs P(s) ln P(s),
(52)
where the sum runs over all microstates accessible to the system and P(s) is the probability of
the system being in microstate s.
• For an isolated system P(s) = 1/Ω for all accessible states s. Show that in this case the
preceding formula reduces to our familiar definition of entropy.
1
1 1
S = −kΣs P (s) ln P (s) = −kΣs ln = k ln Ω Σs = k ln Ω.
Ω Ω
Ω
(53)
• For a system in thermal equilibrium with a reservoir at temperature T, P(s) = e−E(s)/kT /Z.
Show that in this case as well, the preceding formula agrees with what we already know
about entropy.
S = −kΣs P (s) ln P (s) =
=
=
=
−E(s)β e−E(s)β
e
−kΣs
ln
Z
Z
e−E(s)β
[ −E(s)β + ln Z ]
−kΣs
Z
E(s)e−E(s)β
kln Z kβ Σs
−
Σs e−E(s)β
Z
Z
F
U
kβU + k ln Z = − .
T
T
(54)
(55)
(56)
(57)
Problem 5 Show explicitly from the results of this section that G = N µ for an ideal gas.
Answer: We start with the relation,
G = U − TS + PV = F + PV
(58)
Using the results obtained in section 6.7, we have,
N kT
,
V
vZint
µ = −kT ln
where,
N vQ
3
h
vQ = √
.
2πmkT
P =
6
(59)
(60)
(61)
Using eqn(6.90) Schroeder,
F = −N kT ( ln V − ln N − ln vQ + 1) + Fint + N kT,
(62)
G = F + PV
= F + N kT
= −N kT ( ln V − ln N − ln vQ + 1) + Fint + N kT
V Zint
= −N kT ln
= N µ.
N vQ
(63)
(64)
(65)
we have
Therefore for an ideal gas, G = N µ.
7
(66)