= Internal-wave ray in uniform vertical shear
{title2=$m(t)=m_0-kst,\quad z(t)=(\omega-\widehat\omega(t))/(ks)$}
For a uniform vertical <shear flow> $U=sz$ with $s>0$, constant <buoyancy frequency> $N>0$ and horizontal <wavenumber> $k>0$, choose the initially rising branch $m_0<0$. The ground-frame <angular frequency> is the conserved ray Hamiltonian
$$
\omega=ksz+\widehat\omega,\qquad\widehat\omega=\frac{Nk}{\sqrt{k^2+m^2}}.
$$
The <internal-wave ray tracing> equations give $\dot k=0$ and $\dot m=-ks$, hence $m=m_0-kst$ and $z=(\omega-\widehat\omega)/(ks)$. As $t\to\infty$, the vertical <wavenumber> tends to negative infinity, the <intrinsic frequency> tends to zero and the height approaches the <critical level> $z_c=\omega/(ks)$. For $\theta=\arctan(|m|/k)$, the laboratory ray slope is
$$
\frac{dx}{dz}=\tan\theta+\frac{ksz}{N\sin\theta\cos^2\theta}.
$$
The second term is mean <advection> divided by vertical <group velocity>. The zero-shear case has no finite $z_c$ and must instead be treated as propagation in a uniform current.
Back to article page
