 Solution

ID: past-exam-of-the-kinetic-theory-course-of-the-university-of-oxford/2023/1/a/solution

Write

K_{ij} = K (x_{i}, x_{j})

and

P = \sum_{i} v_{i}

. The unit particle masses make this sum the total momentum. Exchange the two indices and use the symmetry

K_{ij} = K_{ji}

\dot{P} = \sum_{i, j} K_{ij} (v_{j} - v_{i}) = \frac{1}{2} \sum_{i, j} K_{ij} [(v_{j} - v_{i}) + (v_{i} - v_{j})] = 0 .

Thus every interacting pair transfers equal and opposite momentum, even though the velocity alignment dissipates kinetic energy.

For the total kinetic energy

E_{N} = \frac{1}{2} \sum_{i} ∣ v_{i} ∣^{2}

, apply the same exchange of indices:

\dot{E}_{N} = i, j \sum K_{ij} v_{i} \cdot (v_{j} - v_{i}) = \frac{1}{2} i, j \sum K_{ij} [v_{i} \cdot (v_{j} - v_{i}) + v_{j} \cdot (v_{i} - v_{j})] = - \frac{1}{2} i, j \sum K_{ij} ∣ v_{j} - v_{i} ∣^{2} \leq 0 .

The dissipated kinetic energy is that of relative motion. Since the total momentum is constant,

E_{N} - ∣ P ∣^{2} / (2 N) = \frac{1}{2} \sum_{i} ∣ v_{i} - P / N ∣^{2}

also decreases. Equality holds when every pair with

K_{ij} > 0

has the same velocity; the hypotheses permit disconnected interacting groups, so they alone do not guarantee eventual global velocity alignment.

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