N-body Microscopic Heat Current Expression

Li Ju

May 30, 1998

Suppose the Hamiltonian of a collection of particles is written as

H = ∑_{i} K_{i} + ∑_{i<j} V_{2}(r_{i}, r_{j}) + ∑_{i<j<k} V_{3}(r_{i}, r_{j}, r_{k}) + .. + ∑_{i_{1}<i_{2}<..<i_{n}} V_{n}(r_{i_{1}}, r_{i_{2}}, .., r_{i_{n}}) + .., (1)

where K_{i} is the kinetic energy of each particle:

K_{i} = \frac{1}{2}m_{i}|v_{i}|^{2},  (2)

and the n-body potential V_{n}(r_{i_{1}}, r_{i_{2}}, .., r_{i_{n}}) is invariant with respect to any permutation of particles,

r_{i_{α}} ⇀ ↽ r_{i_{β}}, 1 ≤ α, β ≤ n,  (3)

i.e., the functional form puts no extra emphasis on any specific particle, then the division of H

H = ∑_{i} E_{i}  (4)

into “single particle energies” E_{i} is intuitively clear:

E_{i} = K_{i} + ∑_{pairs} V_{2}(r_{i}, r_{j})/2 + ∑_{triplets} V_{3}(r_{i}, r_{j}, r_{k})/3 + .. + ∑_{n−lets} V_{n}(r_{i}, r_{i_{2}}, .., r_{i_{n}})/n + .. (5)

where “pairs”, “triplets” etc. refer to all interactions that this ith particle participates in, and it is reasonable to say that it “owns” 1/n of all the V_{n} interactions that it took parts in.

The net heat current of the system, J, is defined as

J = \frac{d}{dt} (∑_{i} E_{i}r_{i})
= ∑_{i} E_{i}v_{i} + Ė_{i}r_{i}
= ∑_{i} E_{i}v_{i} + J̃.  (6)

It is easy to show that J̃ is linearly superpositionable, in the sense that the influence of any n-let interaction (via V_{n}) directly adds onto J̃,

J̃ = ∑_{i<j} J̃_{ij}^{2} + ∑_{i<j<k} J̃_{ijk}^{3} + .. + ∑_{i_{1}<i_{2}<..<i_{n}} J̃_{i_{1}i_{2}..i_{n}}^{n} + ..  (7)

which is true even if V_{n} does not have permutation symmetry. One underlying reason is that

K̇_{i} = F_{i} · v_{i},  (8)

but F_{i} is the linear sum of all interactions,

F_{i} = ∑_{pairs} F_{ij}^{i} + ∑_{triplets} F_{ijk}^{i} + .. + ∑_{n−lets} F_{ii_{2}i_{2}i_{n}}^{i} + ..  (9)

in which, for instance, F_{ijk}^{i} is the force contribution of any ijk-triplet interaction to particle i.

So now we can focus on a specific n-let interaction, and simply add everything together in the end onto J̃. Let me denote the particles involved r_{1}, r_{2}, .., r_{n}, and the part that this n-let interaction contributes to by Δ. Because of Eqn. (5), ΔĖ_{i} contains two parts: a kinetic energy part and a potential energy part. With ΔK̇_{i} = ΔF_{i} · v_{i} and

ΔV = ΔV (r_{1}, r_{2}, .., r_{n}),  (10)

there is

Δ (\frac{V̇}{n} ) = \frac{1}{n} ∑_{j=1}^{n} \frac{∂ΔV}{∂r_{j}} · v_{j}

= −\frac{1}{n} ∑_{j=1}^{n} ΔF_{j} · v_{j},  (11)

as by definition,

ΔF_{j} = −\frac{∂ΔV}{∂r_{j}} .  (12)

So, there is

ΔJ̃ = ∑_{i=1}^{n}(ΔĖ_{i})r_{i}
= ∑_{i=1}^{n} (ΔF_{i} · v_{i} − \frac{1}{n} ∑_{j=1}^{n} F_{j} · v_{j}) r_{i}.  (13)

One can see from Eqn. (13) that ΔJ̃ depends only on the relative separations of r_{1}, r_{2}, .., r_{n} and not on the origin of the coordinate frame, i.e., it is frame-independent, because for any uniform shift: r_{i} → r_{i} + a, where a is a constant,

Δ(ΔJ̃) = ∑_{i=1}^{n} (ΔF_{i} · v_{i} − \frac{1}{n} ∑_{j=1}^{n} F_{j} · v_{j}) a
= (∑_{i=1}^{n} ΔF_{i} · v_{i} − \frac{1}{n} ∑_{i=1}^{n} ∑_{j=1}^{n} ΔF_{j} · v_{j}) a
= 0.  (14)

Thus, in evaluating Eqn. (13), we are free to pick any coordinate origin that we like. One further simplification is

∑_{i=1}^{n}(ΔF_{i}) = 0,  (15)

which can be derived from the translational invariance of V_{n}.

• Two-body interaction: For any particle pair (i, j) that falls into the interaction range,

ΔV = V_{2}(r_{i}, r_{j}),  (16)

and there is

ΔF_{i} = −ΔF_{j}.  (17)

Because I am free to choose any coordinate frame origin, I can let r_{i} = 0, and so only one term in Eqn. (13) (the n = j term) contributes,

ΔJ̃_{2} =  (ΔF_{j} · v_{j} − \frac{1}{2}(ΔF_{i} · v_{i} + ΔF_{j} · v_{j})) r_{j}
= \frac{1}{2} (ΔF_{j} · (v_{j} + v_{i})) r_{j},  (18)

where we have made use of Eqn. (17). If one remembers that the present frame origin is on particle i, the expression must be

ΔJ̃ = \frac{1}{2} (ΔF_{j} · (v_{j} + v_{i})) (r_{j} − r_{i})  (19)

in other frames.

• Three-body interaction:

Now,

ΔV = V_{3}(r_{i}, r_{j}, r_{k}),  (20)

and

ΔF_{i} = −ΔF_{j} − ΔF_{k}.  (21)

I still can choose r_{i} = 0, then

ΔJ̃_{3}
=  (ΔF_{j} · v_{j} − \frac{1}{3}(ΔF_{i} · v_{i} + ΔF_{j} · v_{j} + ΔF_{k} · v_{k})) r_{ji} +
(ΔF_{k} · v_{k} − \frac{1}{3}(ΔF_{i} · v_{i} + ΔF_{j} · v_{j} + ΔF_{k} · v_{k})) r_{ki}  (22)
= \frac{1}{3} \{(ΔF_{j} · (2v_{j} + v_{i}) + ΔF_{k} · (v_{i} − v_{k})) r_{ji} + (ΔF_{k} · (2v_{k} + v_{i}) + ΔF_{j} · (v_{i} − v_{j})) r_{ki}\} ,

where

r_{ji} = r_{j} − r_{i}, r_{ki} = r_{k} − r_{i}.  (23)