5Phase transitions

II Statistical Physics

5.2 Critical point and critical exponents
Eventually, we want to understand phase transitions in general. To begin, we
will classify phase transitions into different orders. The definition we are using is
a more “modern” one. There is a more old-fashioned way of defining the order,
which we will not use.
Definition
(First-order phase transition)
.
A first-order phase transition is one
with a discontinuity in a first derivative of G (or F ).
For example, if we have a discontinuity in
S
, then this gives rise to latent
heat.
Example.
The liquid-gas phase transition is first-order, provided we stay away
from the critical point.
Definition
(Second-order phase transition)
.
A second-order phase transition is
one with continuous first order derivatives, but some second (or higher) derivative
of G (or F ) exhibits some kind of singularity.
For example, we can look at the isothermal compressibility
κ =
1
v
v
p
T
=
1
v
2
g
p
2
T
.
This is a second-order derivative in
G
. We look at how this behaves at the
critical point, at (T
c
, p(T
c
)).
v
p
CP
Here we have
p
v
T
= 0
at the critical point. Since
κ
is the inverse of this, it diverges at the critical
point.
So the liquid-gas transition is second-order at the critical point.
Example. For water, we have T
C
= 647 K, and p(T
C
) = 218 atm.
Experimentally, we observe that liquids and gases are not the only possible
phases. For most materials, it happens that solid phases are also possible. In
this case, the phase diagram of a normal material looks like
T
p
T
c
solid
liquid
gas
There is clearly another special point on the phase diagram, namely the triple
point. This is the point where all three phases meet, and this is what we
previously used to fix the temperature scale in classical thermodynamics.
Critical point
We are going to spend some more time studying the critical point. Let’s first
carefully figure out where this point is.
If we re-arrange the van der Waals equation of state, we find
pv
3
(pb + kT )v
2
+ av ab = 0. ()
When
T < T
c
, this has three real roots, (
L, G, U
). At
T > T
C
, we have a single
real root, and two complex conjugate roots.
Thus at
T
=
T
C
, there must be three equal real roots of
T
=
T
C
. Hence, the
equation must look like
p
C
(v v
C
)
3
= 0,
Equating coefficients, we find
kT
C
=
8a
27b
, v
C
= 3b, p
C
=
a
27b
2
= p(T
C
).
We introduce some new dimensionless “reduced” quantities by
¯
T =
T
T
C
, ¯p =
p
p
C
, ¯v =
v
v
C
.
At the critical point, all these values are 1. Then the van der Waals equation
becomes
¯p =
8
3
¯
T
¯v
1
3
3
¯v
2
.
We have gotten rid of all free parameters! This is called the law of corresponding
states.
Using this, we can compute the ratio
p
C
v
C
kT
C
=
3
8
= 0.375.
This depends on no external parameters. Thus, this is something we can
experimentally try to determine, which can put our van der Waals model to test.
Experimentally, we find
Substance p
C
v
C
/(kT
C
)
H
2
O 0.23
He, H
2
0.3
Ne, N
2
, Ar 0.29
So we got the same order of magnitude, but the actual numbers are clearly off.
Indeed, there is no reason to expect van der Waals to give quantitative agreement
at the critical point, where the density is not small.
But something curious happens. The van der Waals model predicted all gases
should look more-or-less the same, but we saw van der Waals is wrong. However,
it is still the case that all gases look more-or-less the same. In particular, if we
try to plot
¯
T
against
¯ρ
=
¯v
1
, then we find that all substances seem to give the
same coexistence curve! This is the famous Guggenheim plot. Why is this the
case? There is certainly some pieces we are missing.
Critical exponents
There are more similarities between substances we can talk about. As we tend
towards the critical point, the quantities
v
G
v
L
,
T
C
T
and
p p
C
all tend
to 0. On the other hand, we saw the isothermal compressibility diverges as
T T
+
C
. One natural question to ask is how these tend to 0 or diverge. Often,
these are given by some power law, and the exponents are called the critical
exponents. Mysteriously, it appears that these exponents for all our systems are
all the same.
For example, as the critical point is approached along the coexistence curve,
we always have
v
G
v
L
(T
C
T )
β
,
where β 0.32.
On the other hand, if we approach the critical point along the critical isotherm
T = T
C
, then we get
p p
C
(v v
C
)
δ
,
where δ 4.8.
Finally, we can vary the isothermal compressibility
κ =
1
v
v
p
T
along v = v
C
and let T T
+
C
, then we find
κ (T T
C
)
γ
,
where γ 1.2.
These numbers
β, γ, δ
are the critical exponents, and appear to be the same
for all substances. The challenge is to understand why we have this universal
behaviour, and, more ambitiously, calculate them.
We start by making a rather innocent assumption. We suppose we can
describe a system near the critical point using some equation of state, which
we assume is analytic. We also assume that it is “non-degenerate”, in the sense
that if there is no reason for the derivative at a point to vanish, then it doesn’t.
If this equation of state gives the same qualitative behaviour as van der
Waals, i.e. we have a critical point
T
C
which is a reflection point on
T
=
T
C
,
then we have
ρ
v
T
=
2
p
v
2
T
= 0
at the critical point. Therefore, using the non-degeneracy condition, we find
p p
C
(v v
C
)
3
.
near T = T
C
. So we predict δ = 3, is wrong.
But let’s continue and try to make other predictions. We look at how
γ
behaves. We have
p
v
T
(T, v
C
) a(T T
C
)
for some a near the critical point. Therefore we find
κ (T T
C
)
1
.
if v = v
C
and T T
+
C
, and so γ = 1. This is again wrong.
This is pretty bad. While we do predict universal behaviour, we are always
getting the wrong values of the exponents! We might as well work out what
β
is,
which is more complicated, because we have to figure out the coexistence curve.
So we just compute it using van der Waals along the coexistence curve
¯p =
8
¯
T
3¯v
L
1
3
¯v
2
L
=
8
¯
T
3¯v
G
1
3
¯v
2
G
.
Rearranging, we find
¯
T =
(3¯v
L
1)(3¯v
G
1)(¯v
L
+ ¯v
G
)
8¯v
2
G
¯v
2
L
. ()
To understand how this behaves as
T T
C
, we need to understand the coexis-
tence curve. We use the Maxwell construction, and near the critical point, we
have
¯v
L
= 1 + δ¯v
L
, ¯v
G
= 1 + δ¯v
G
.
In the final example sheet, we find that we get
δ¯v
L
= δ¯v
G
=
ε
2
for some ε. Then using (), we find that
¯
T 1
1
16
ε
2
= 1
1
16
(¯v
G
¯v
L
)
2
.
So we find that
v
G
v
L
(T
C
T )
1/2
.
This again doesn’t agree with experiment.
Why are all these exponents wrong? The answer is fluctuations. Near the
critical point, fluctuations are large. So it is not good enough to work with mean
values hvi, hpi. Recall that in the grand canonical ensemble, we had
N
2
=
1
β
N
µ
T,V
=
1
β
N
p
T,V
p
µ
T,V
.
We can try to work out these terms. Recall that we had the grand canonical
potential
Φ = E T S µN,
which satisfies the first law-like expression
dΦ = S dT p dV N dµ.
We also showed before, using extensivity, that
Φ = pV.
So if we plug this into the first law, we find
V dp = S dT N dµ.
Therefore we know that
dp
dµ
T,V
=
N
V
.
So we find
N
2
=
N
βV
N
p
T,V
.
To figure out the remaining term, we use the magic identity
x
y
z
y
z
x
z
x
y
= 1.
This gives us
N
2
=
N
βV
1
p
V
T,N
V
N
T,p
=
N
βV
N
V
T,p
V
p
T,N
=
κN
β
N
V
T,p
.
Recall that the density is
ρ(T, p, N) =
N
V
,
which is an intensive quantity. So we can write it as ρ(T, p). Since we have
N = ρV,
we simply have
N
V
T,p
= ρ.
Therefore we have
N
2
=
kN
β
ρ.
So we find that
N
2
N
2
=
κkT
V
.
This is how big the fluctuations of N are compared to N.
Crucially, this is proportional to
κ
, which we have already seen diverges
at the critical point. This means near the critical point, we cannot ignore the
fluctuations in N .