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

Take ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{u}t$, with the same time coordinate in both frames. The normal fluid transforms as ${\mathbf{v}}_n^{\prime }({\mathbf{x}}^{\prime },t)={\mathbf{v}}_n(\mathbf{x},t)-\mathbf{u}$. The field transformation is the unit-modulus Schrödinger boostWriting the phase as $F=-i\mathbf{u}\cdot {\mathbf{x}}^{\prime }-i|\mathbf{u}{|}^{2}t/2$ makes differentiation at fixed primed position transparent:It follows that both $-i{\partial }_t^{\prime }{\psi }^{\prime }$ and ${\nabla }^{\prime 2}{\psi }^{\prime }/2$ gain the same terms $e^F[-i\mathbf{u}\cdot \nabla \psi -u^2\psi /2]$. The density is unchanged, ${\rho }^{\prime }=|{\psi }^{\prime }{|}^2=\rho$, so the nonlinear local term has the same form.

The remaining material derivative transforms asThus 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.

Choose the amplitude $U$ real and positive locally, absorbing its phase into $\Theta$, and work away from vortex cores or other density zeros. Put $\rho =U^2$ and $\theta =\Theta /ϵ$. Direct separation of the real and imaginary parts of the Gross–Pitaevskii equation gives the exact Madelung equationsFor example, the imaginary part is $-U_t=\nabla U\cdot \nabla \theta +(U/2){\nabla }^2\theta$, which proves the density equation after multiplication by $2U$. The real damping coefficient affects the phase balance rather than adding a particle source.

Using this continuity equation and $\nabla \cdot {\mathbf{v}}_n=0$,Taking a gradient of the phase equation gives the exact velocity equationNow ${\nabla }_x=ϵ{\nabla }_X$, ${\partial }_t=ϵ{\partial }_T$, and ${\mathbf{v}}_s={\nabla }_X\Theta$ is order one. The inertial and chemical-potential gradients are $O(ϵ)$; the double-gradient bulk term is $O({ϵ}^2)$; and the quantum potential force contains three slow derivatives and is $O({ϵ}^3)$, provided $U$ stays bounded away from zero. Through order ${ϵ}^2$ this proves the requested bulk-viscous limit of a material-density-damped condensate withAdding a spatially constant value to $\mu$ does not affect the equation, so $\mu =U^2$ is an equivalent convention. The density and viscosity coefficients are in the equation's dimensionless units. The bulk viscosity term is the gradient of the divergence of the relative mass flux, not a shear-viscosity term. At density zeros the slow-amplitude expansion and the division by $U$ fail; there one must retain the original field equation.

A constant-amplitude plane wave has no density derivative and therefore no damping. The complete family iswith arbitrary constant phase $\chi$. The background in the question is the member $A=1$.

Use the exact amplitude–phase equations derived above with ${\mathbf{v}}_n=0$. To first order in the perturbation, $\rho =1+2ϵR$ and ${\mathbf{v}}_s=\mathbf{k}+ϵ\nabla \Theta$, soThese equations retain the quantum potential contribution because no long-wavelength truncation is assumed in this subpart. In particular damping contains $R_t$, not the advective derivative along the moving condensate: its reference is the stationary normal fluid.

For the proposed periodic perturbation, write $q=\mathbf{k}\cdot \mathbf{l}$ and $L=l^2$. The amplitude coefficients satisfyA nonzero perturbation therefore requires the plane-wave dispersion of a material-density-damped condensateor equivalently ${\sigma }^2+(2iq+\zeta l^2)\sigma +l^2+l^4/4-q^2=0$. At zero background flow the roots are $-\zeta l^2/2\pm \sqrt{{\zeta }^2{l}^4/4-l^2-l^4/4}$, whose real parts are negative for $l\ne 0$. As $\zeta \to 0$, $\sigma =-iq\pm i\sqrt{l^2+l^4/4}$ recovers the advected Bogoliubov spectrum with long-wavelength sound speed one in these units. This differs from question 1's time normalization, whose sound speed is $1/\sqrt{2}$.