Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 65 / 1

Differentiating the velocity potential gives $u=a\omega e^{kz}\cos (kx-\omega t)$ and $w=a\omega e^{kz}\sin (kx-\omega t)$. Let $A=a{e}^{kz}$ and $\vartheta =kx-\omega t$. To first order in the wave steepness $ka$, evaluate these velocities at the initial parcel position and integrate from time zero:These expressions satisfy both initial conditions and distinguish initial and mean parcel labels in a surface wave. Their actual short-time limits areThe printed expressions omit the integration constants and therefore cannot be the small-time displacement from the prescribed initial point. For example, at $x=0$, they give $Z(0)-z=a{e}^{kz}\ne 0$. This is a genuine inconsistency, rather than a missing step in the calculation.

The intended oscillatory orbit is recovered by using mean parcel coordinates instead. Set $x_c=x+A\sin (kx)$ and $z_c=z-A\cos (kx)$ to this order. ThenThus a deep-water gravity wave produces approximately circular parcel orbits with radius $a{e}^{k{z}_c}$, decaying exponentially with depth. The approximation is one of small amplitude over a wave cycle, not an assertion that these oscillatory coordinates vanish at time zero.

The leading Stokes drift comes from evaluating the oscillatory velocity at the displaced particle position. Taylor expansion givesHere $u_x=-ak\omega e^{kz}\sin \vartheta$ and $u_z=ak\omega e^{kz}\cos \vartheta$. Inserting the initial-position displacements from the preceding solution yieldsIn particular the instantaneous difference is zero at $t=0$, as it must be. The unaveraged constant equality in the PDF is incompatible with its initial labels. Averaging over a period $\mathcal{T}=2\pi /\omega$ removes the oscillatory term:Equivalently, using mean parcel labels and the purely oscillatory displacements gives $a^{2}k\omega e^{2kz}({\sin }^2\vartheta +{\cos }^2\vartheta )$ at this order. This recovers the intended constant result. The period-mean drift is in the wave-propagation direction and decreases as $e^{2kz}$. The corresponding second-order vertical difference is $a^{2}k\omega e^{2kz}\sin (\omega t)$ for initial labels and has zero mean. The resulting horizontal displacement per cycle is $u_s\mathcal{T}$; closed first-order circles do not imply zero second-order transport.

Use the nonrotating linear Boussinesq approximation with reference density ${\rho }_0$ and $N^2=-(g/{\rho }_0)d\bar{\rho }/dz>0$:Let $K_h^2=k^2+l^2>0$, $K^2=K_h^2+m^2$ and $E=e^{i(kx+ly+mz-\omega t)}$. The horizontal momentum equations give $u^{\prime }=k{p}^{\prime }/({\rho }_0\omega )$ and $v^{\prime }=l{p}^{\prime }/({\rho }_0\omega )$. Continuity requires $k{u}^{\prime }+l{v}^{\prime }+m{w}^{\prime }=0$. Thus the internal-wave polarization isPhysical perturbations are the real parts. The density is in temporal quadrature with the vertical velocity, because buoyancy responds to the vertical fluid displacement. Substitution into the vertical momentum equation determines the internal gravity wave dispersion relation,A nontrivial propagating wave requires a nonzero horizontal wave vector and a nonzero frequency. The formulas are not to be divided by $K_h^2=0$: incompressibility excludes a nonzero oscillatory vertical velocity for a purely vertical wave vector.

For the positive-frequency internal gravity wave branch, $\omega =N{K}_h/K$. Differentiating with respect to the components of the wave vector gives the group velocity,The wavefront-normal phase velocity is ${\mathbf{c}}_p=\omega \mathbf{k}/K^2$. Directly,The negative-frequency branch changes both propagation signs but not their orthogonality. Another proof is that $\omega$ is homogeneous of degree zero in the wave vector, so Euler's homogeneous-function identity gives $\mathbf{k}\cdot {\nabla }_{\mathbf{k}}\omega =0$. Energy propagates along constant-phase surfaces, rather than normal to them. When $m=0$, $\omega =N$ and ${\mathbf{c}}_g=0$; the scalar-product result remains true, but there is no nonzero group-velocity direction to describe geometrically.

Use the attenuation coefficient read from the PDF,where here $k$ is the beam wave-number magnitude and the angle convention has $\sin \theta >0$ for the chosen forward attenuation coordinate. The omitted denominator in the TeX is essential both physically and dimensionally. Work in the prescribed along-beam kinematic model, with small parcel excursion compared with ${\lambda }^{-1}$.

The first-order along-beam fluid displacement, initially zero, is $\delta \xi =A\sin (\omega t)/\omega$. Therefore the leading displacement correction to the along-beam velocity isThis oscillatory drift of an attenuated internal-wave beam is oscillatory, soIndeed the scalar parcel equation $\dot{\Xi }=U_0{e}^{-\lambda \Xi }\cos \omega t$ can be integrated exactly:For excursions small enough that the right side stays positive, the parcel returns to its initial along-beam coordinate after each period. This demonstrates the zero net drift in this supplied model. It is not a claim that all components of a viscous beam, or a separately generated Eulerian mean flow, vanish; the question specifies the along-beam component only.