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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-83/3/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 83 / 3 / i / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Take

x^{'} = x - u t

, with the same time coordinate in both frames. The normal fluid transforms as

v_{n}^{'} (x^{'}, t) = v_{n} (x, t) - u

. The field transformation is the unit-modulus Schrödinger boost

ψ^{'} (x^{'}, t) = e^{- i u \cdot x + i ∣ u ∣^{2} t /2} ψ (x, t), A = - i u, B = - \frac{i}{2} ∣ u ∣^{2} .

Writing the phase as

F = - i u \cdot x^{'} - i ∣ u ∣^{2} t /2

makes differentiation at fixed primed position transparent:

\nabla^{'} ψ^{'} = e^{F} (\nabla ψ - i u ψ), \partial_{t}^{'} ψ^{'} = e^{F} (ψ_{t} + u \cdot \nabla ψ - \frac{i}{2} u^{2} ψ) .

It follows that both

- i \partial_{t}^{'} ψ^{'}

and

\nabla^{'2} ψ^{'} /2

gain the same terms

e^{F} [- i u \cdot \nabla ψ - u^{2} ψ /2]

. The density is unchanged,

ρ^{'} = ∣ ψ^{'} ∣^{2} = ρ

, so the nonlinear local term has the same form.

The remaining material derivative transforms as

(\partial_{t}^{'} + v_{n}^{'} \cdot \nabla^{'}) ρ^{'} = [\partial_{t} + u \cdot \nabla + (v_{n} - u) \cdot \nabla] ρ = (\partial_{t} + v_{n} \cdot \nabla) ρ .

Thus the density-damping term acquires only the common phase factor, proving invariance of the complete equation under a Galilean boost. Keeping the normal velocity unchanged in both frames would fail this check; the reference flow must transform along with the condensate.

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