How the Born term and the fluctuation term
cancel each other out in liquids
Corrado Rainone
May 26, 2016
The expression for the infinite frequency shear modulus for a generic system with an interaction potential U (X), where X ≡ (r 1 , r 2 , . . . , r N )1 is a 3N
dimensional vector that identifies the microstate of the system, is
2
µ = µBorn − βV [ σ 2 − hσi ]
(1)
which in term of canonical averages can be rewritten as
2 2
Z
Z
d U
dU
β
1
−βU (X)
dX
dX
e
−
e−βU (X) ,
µV =
Z
dγ 2
Z
dγ
(2)
where we have used the fact that the average stress in the liquid is zero, hσi = 0.
We consider here a simple shear transformation with a strain tensor whose only
non-zero component is γxy = γ. In this case, the derivative over γ can be written
as
N
X
d
∂
yi
=
.
(3)
dγ
∂x
i
i=1
Plugging this into the first term of the expression above we find
1
Z
Z
dX
d2 U
dγ 2
e−βU (X) =
Z
N
X
1
∂
dU
dX yi
e−βU (X) ;
Z
∂x
dγ
i
i=1
(4)
we now integrate by parts over xi each term in the sum; since we are integrating over the whole space of configurations without any constraint (equilibrium
liquids are ergodic), the boundary terms vanish, and we get
1
Z
Z
dX
d2 U
dγ 2
−βU (X)
e
Z
N
X
β
dU
∂U (X) −βU (X)
=
dX
yi
e
,
Z
dγ
∂xi
i=1
which, using equation (3) backwards, can be rewritten as:
2
Z
β
dU
dX
e−βU (X) ,
Z
dγ
(5)
(6)
which is equal to the fluctuation term: we have proven that the Born and
fluctuation terms cancel each other out and consequently µ is zero in equilibrium
liquids, as expected. Some comments:
1r
i
≡ (xi , yi , zi ).
1
1. Notice how the proof works only for derivatives with respect to off-diagonal
components of the strain tensor: in the case of, say dγdxx , we would have
had
N
X
d
∂
=
,
xi
dγxx
∂x
i
i=1
and the integration by parts would have yielded an extra term. This means
that the argument does not apply to the bulk compression moduli, which
are of course non-zero even in liquids.
2. The vanishing of the boundary term would not take place if we were not
integrating over the whole space of configurations in the thermodynamic
limit, i.e. the proof works only if ergodicity is preserved. So it does not
apply to glasses or crystals (as it should be), wherein ergodicity is broken,
and which are equipped with non-zero shear moduli.
2