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Time and horizontal translational invariance at the interface conserve $\omega$ and $k$. For every component, horizontal momentum givesThe background density has the same value on both sides, so pressure continuity implies continuity of horizontal velocity. Normal velocity is also continuous: no fluid accumulates at the interface.

Use the particle-displacement amplitudes and polarization directions defined in part (a). For downward incident and transmitted waves and an upward reflected wave, continuity of $u$ and $w$ givesSolving the two equations proves the internal-wave transmission across a buoyancy-frequency jump formulaThe reflected wave has the same frequency and wavelength as the incident wave and opposite vertical wavenumber. The transmitted wave hasIts group velocity is downward at angle ${\theta }_2$; the formula in part (a), with $N_2,m_t$, gives its speed.

If instead one uses signed vertical-displacement amplitudes, the reflected coefficient has the opposite sign: $(m_1-m_2)/(m_1+m_2)$. This follows because the reflected polarization's vertical component changes sign, and is not a difference in the physical reflected field. As an energy check, the reflected fraction is $[(m_1-m_2)/(m_1+m_2){]}^2$ and the transmitted fraction is $4m_1{m}_2/(m_1+m_2{)}^2$; they sum to one.

When $N_2<\omega <N_1$, take $m_t=-i\kappa$, $\kappa =k\sqrt{1-N_2^2/{\omega }^2}$, so $e^{i{m}_{t}z}$ decays as $z\to -\infty$. The transmitted field is evanescent, carries no mean vertical energy flux, and the reflected wave has unit amplitude magnitude. A real lower-layer ray angle and wavelength are then inapplicable.