ASTR/ATOC-5410: Fluid Instabilities, Waves, and Turbulence September 2, 2016, Axel Brandenburg

Handout 4b: Rayleigh–Bénard problem (Part IIb)

Let us recall that the three boundary conditions yield^{1}

∑_{i=1}^{3} cosh q_{i}/2 = 0.  (1)

∑_{i=1}^{3} q_{i} sinh q_{i}/2 = 0.  (2)

∑_{i=1}^{3} (q_{i}^{2} − k_{⊥}^{2})^{2} cosh q_{i}/2 = 0.  (3)

We write this in matrix form, which then yields

det ( \begin{matrix}cosh q_{1}/2 & cosh q_{2}/2 & cosh q_{2}/2 \\ q_{1} sinh q_{1}/2 & q_{2} sinh q_{2}/2 & q_{3} sinh q_{3}/2 \\ (q_{1}^{2} − k_{⊥}^{2})^{2} cosh q_{1}/2 & (q_{2}^{2} − k_{⊥}^{2})^{2} cosh q_{2}/2 & (q_{3}^{2} − k_{⊥}^{2})^{2} cosh q_{3}/2 \end{matrix} ) = 0  (4)

Making use of the fact that q_{1} is purely imaginary, and thus defining q_{0} = iq_{1}, and using the fact that i q_{0} sinh i q_{0}/2 = −q_{0} sin q_{0}/2, we have

det ( \begin{matrix}cos q_{0}/2 & cosh q_{2}/2 & cosh q_{2}/2 \\ −q_{0} sin q_{0}/2 & q_{2} sinh q_{2}/2 & q_{3} sinh q_{3}/2 \\ (q_{0}^{2} + k_{⊥}^{2})^{2} cos q_{0}/2 & (q_{2}^{2} − k_{⊥}^{2})^{2} cosh q_{2}/2 & (q_{3}^{2} − k_{⊥}^{2})^{2} cosh q_{3}/2 \end{matrix} ) = 0  (5)

Next, we divide all rows by row 1,

det ( \begin{matrix}1 & 1 & 1 \\ −q_{0} tan q_{0}/2 & q_{2} tanh q_{2}/2 & q_{3} tanh q_{3}/2 \\ (q_{0}^{2} + k_{⊥}^{2})^{2} & (q_{2}^{2} − k_{⊥}^{2})^{2} & (q_{3}^{2} − k_{⊥}^{2})^{2} \end{matrix} ) = 0  (6)

We recall that, using Eq.(10) of Handout 4,

q_{i}^{2} − k_{⊥}^{2} = Ra^{1/3} k_{⊥}^{2/3} × \{ \begin{matrix}−1 & for i = 1 \\ \frac{1}{2} + \frac{i}{2} \sqrt{3} & for i = 2 \\ \frac{1}{2} − \frac{i}{2} \sqrt{3} & for i = 3 \end{matrix} ,  (7)

so we have

det ( \begin{matrix}1 & 1 & 1 \\ −q_{0} tan q_{0}/2 & q_{2} tanh q_{2}/2 & q_{3} tanh q_{3}/2 \\ 1 & \frac{1}{2}(i\sqrt{3} − 1) & −\frac{1}{2}(i\sqrt{3} + 1) \end{matrix} )  (8)

Subtracting now row 1 from from 3, recalling that q_{3} = q_{2}^{∗}, and dividing by −\sqrt{3}/2 we have

det ( \begin{matrix}1 & 1 & 1 \\ −q_{0} tan q_{0}/2 & q_{2} tanh q_{2}/2 & q_{2}^{∗} tanh q_{2}^{∗}/2 \\ 0 & \sqrt{3} − i & \sqrt{3} + i \end{matrix} )  (9)

This yields

Im [(\sqrt{3} + i)q_{2} tanh q_{2}] + q_{0} tan q_{0}/2.  (10)

Instead of solving this equation, we just draw a contour plot; see Figure 1. The smallest value of Ra is reached at k_{⊥} = 3.12 and gives Ra(k_{⊥}) = 1708.

1Note the q_{i} term in front of sinh, which I forgot last time!

Figure 1: Determinant in the plane Ra versus k_{⊥} = 3.12.

Figure 2: Similar to Figure 1, but for a larger range.

References

Chandrasekhar, S. Hydrodynamic and Hydromagnetic Stability. Dover Publications, New York (1961).

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