Water masses across a front: density, compensation and double diffusion
Two CTD stations are occupied 40 km apart across a frontal zone in the Southern Ocean (45°S). At 500 m, Station North reports potential temperature θ = 12.0 °C and salinity S = 35.60; Station South at the same depth reports θ = 8.0 °C and S = 35.10. The North water is observed to lie above the South water. Use α = −(1/ρ)(∂ρ/∂θ) = 1.7 × 10^−4 K^−1 and β = (1/ρ)(∂ρ/∂S) = 7.6 × 10^−4 (psu)^−1 at these temperatures, with a reference density ρ0 = 1027 kg m^−3. A colleague argues that because the northern water is saltier it must be the denser of the two and that the section is statically unstable.
1. Calculate the density difference between the two stations at 500 m and decide who is right.
Read the worked answer
Linear equation of state, going from North to South: Δρ/ρ0 = −αΔθ + βΔS with Δθ = 8.0 − 12.0 = −4.0 K and ΔS = 35.10 − 35.60 = −0.50.
−αΔθ = −(1.7e−4)(−4.0) = +6.8e−4 βΔS = (7.6e−4)(−0.50) = −3.8e−4 Net: (+6.8 − 3.8)e−4 = +3.0e−4
Δρ = 3.0e−4 × 1027 ≈ 0.31 kg m^−3, with South the denser.
The colleague is wrong. The temperature contrast dominates: 4 K of cooling outweighs 0.5 psu of freshening. Because the lighter (warm, salty) water lies above the denser (cold, fresh) water, the water column is statically stable at this interface. Note that α itself is strongly temperature dependent (~0.6e−4 at 0 °C, ~2.5e−4 at 25 °C), so a linearised calculation with a fixed α taken at 12 °C overstates the thermal contribution for the colder parcel - the true density difference is slightly larger than 0.31 kg m^−3, not smaller. Correct sign, right conclusion.
2. Compute the stability ratio Rρ for this interface and say what mixing process you would expect, and what you would look for in a profile.
Read the worked answer
Rρ = αΔθ / (βΔS) = 6.8e−4 / 3.8e−4 = 1.79.
Interpretation:
- Rρ > 1 means the density field is stable (temperature stratifies, salinity destabilises but not enough to overturn).
- 1 < Rρ < 2 with warm and salty above cold and fresh is the classic salt-fingering regime: a parcel displaced downward loses heat faster than it loses salt (molecular diffusivity of heat ≫ that of salt), so it becomes denser still and keeps sinking, while a displaced upward parcel freshens faster than it cools and keeps rising.
What to look for in the profile: a well-mixed "staircase" of alternating homogeneous layers a few to tens of metres thick separated by sharp interfaces, with θ–S points lying on a line of slope dθ/dS = β/α ≈ constant, and intrusive fine structure. If instead Rρ were between 0 and 1 the water would be doubly unstable and would overturn convectively. Because Rρ = 1.79 sits in the middle of the fingering window (not just above 1), the expectation is robust.
3. At what salinity contrast would the front become density-compensated, and what would the section look like then?
Read the worked answer
Compensation requires the two contributions to cancel:
βΔS = αΔθ → ΔS_crit = (α/β)Δθ = (1.7e−4 / 7.6e−4)(4.0) = 0.895 psu, i.e. S_South ≈ 34.71 at the same 500 m.
With the observed ΔS = 0.50 the South is 0.31 kg m^−3 denser; at ΔS = 0.895 the difference is only: −αΔθ + βΔS = 6.8e−4 − 6.80e−4 ≈ 0 → thermally compensated.
Consequences:
- The horizontal density gradient vanishes, so there is no geostrophic shear to support a front: the two water masses can interleave as intrusions rather than stack.
- Because the density difference is zero but both T and S gradients persist, the interface becomes the most favourable possible site for double-diffusive interleaving (Rρ → 1).
- Caution: α is not constant; it is larger in the warmer northern water, so the true compensation salinity is slightly higher than the linear estimate. Any answer that treats α, β as global constants should state this as a known approximation.