ASTR/ATOC-5410: Fluid Instabilities, Waves, and Turbulence September 30, 2016, Axel Brandenburg

Handout 11: p-modes for the isentropic case

1 Derivation for isentropic case

The continuity equations is D ln ρ/Dt = −∇ · u, and from Ds/Dt = 0 we obtain D ln p/Dt = −γ∇ · u. Our linearized equations are then

ρ̇_{1} + u_{1} · ∇ρ_{0} + ρ_{0}∇ · u_{1} = 0,  (1)
ṗ_{1} + u_{1} · ∇p_{0} + γp_{0}∇ · u_{1} = 0,  (2)
ρ_{0}u̇_{1} + ∇p_{1} + ρ_{1}gẑ = 0.  (3)

Assume ξ̇ = u and integrate the continuity equation in time, i.e.,

ρ_{1} + ξ_{1} · ∇ρ_{0} + ρ_{0}∇ · ξ_{1} = 0,  (4)
p_{1} + ξ_{1} · ∇p_{0} + γp_{0}∇ · ξ_{1} = 0,  (5)
ρ_{0}ξ̈_{1} + ∇p_{1} + ρ_{1}gẑ = 0.  (6)

Divide by ρ_{0}, use ∇p_{0} = −ρ_{0}gẑ, γp_{0} = c_{s}^{2}ρ_{0}, replace gẑ = c_{s}^{2}ẑ/H_{ρ} = −c_{s}^{2}∇ ln ρ_{0}, and insert the above expressions for p_{1} and ρ_{1}, so

ρ_{1} + ξ_{1} · ∇ρ_{0} + ρ_{0}∇ · ξ_{1} = 0,  (7)
p_{1} − ξ_{1z}ρ_{0}g + c_{s}^{2}ρ_{0}∇ · ξ_{1} = 0,  (8)
ξ̈_{1} + \frac{1}{ρ_{0}} ∇p_{1} + \frac{ρ_{1}}{ρ_{0}} gẑ = 0.  (9)

Insert p_{1} and ρ_{1} into the momentum equation, so

ξ̈_{1} + \frac{1}{ρ_{0}} ∇ [ρ_{0}(ξ_{1z}g − c_{s}^{2}∇ · ξ_{1})] − (ξ_{1} · ∇ ln ρ_{0} + ∇ · ξ_{1})gẑ = 0,  (10)

or

ξ̈_{1} + \frac{1}{ρ_{0}} ∇ [ρ_{0}(ξ_{1z}g − c_{s}^{2}∇ · ξ_{1})] + (ξ_{1z}/H_{ρ} − ∇ · ξ_{1})gẑ = 0,  (11)

or, using the fact that gH_{ρ} = c_{s}^{2}, we get

ξ̈_{1} + \frac{1}{ρ_{0}} ∇ [ρ_{0}(ξ_{1z}g − c_{s}^{2}∇ · ξ_{1})] + (ξ_{1z}g − c_{s}^{2}∇ · ξ_{1})ẑ/H_{ρ} = 0,  (12)

so

ξ̈_{1} + ∇ (ξ_{1z}g − c_{s}^{2}∇ · ξ_{1}) = 0.  (13)

Assuming that the displacement is a potential field, i.e., ξ_{1} = ∇Φ, we have (Bogdan & Cally, 1995)

Φ̈ + g∂Φ/∂z − c_{s}^{2}∇^{2}Φ = 0.  (14)

This leads to an eigenvalue problem for eigenvalue ω^{2},

ω^{2}Φ = c_{s}^{2}k^{2}Φ + g \frac{∂Φ}{∂z} − c_{s}^{2} \frac{∂^{2}Φ}{∂^{2}z}.  (15)

At the bottom, we assume Φ = 0, which means that the last data point has to be omitted from the matrix. At the top, we require (k^{2} − d^{2}/dz^{2})Φ = 0, which means that we replace the eigenvalue problem on the boundary by

ω^{2}Φ = g \frac{∂Φ}{∂z}  (16)

Note that Equation (15) is valid even in the non-isothermal case. In Figure 1, we compare solutions for the isothermal and non-isothermal (n = 3/2) cases. Note that in the latter case the p-modes bend down and the difference between qualitative difference between the f- and p-modes diminishes.

Figure 1: kω diagram for isothermal case (left) and various isentropic cases.

2 Reduction to the isothermal case

It is instructive to see how to obtain the dispersion relation for the isothermal case from Equation (15). Replacing ∂_{z} → ik_{z}, we have

ω^{2} = c_{s}^{2}k^{2} + ik_{z}g + c_{s}^{2}k_{z}^{2} = 0,  (17)

Split k_{z} = k′_{z} + ik′′_{z} into real and imaginary parts, so we get

ω^{2} = c_{s}^{2}k^{2} + (ik′_{z} − k′′_{z})g + c_{s}^{2}(k_{z}^{′2} − k_{z}^{′′2} + 2ik′_{z}2k′′_{z}) = 0,  (18)

and solve for real and imaginary parts:

ω^{2} = c_{s}^{2}k^{2} − k′′_{z}g + c_{s}^{2}(k_{z}^{′2} − k_{z}^{′′2}) = 0,  (19)
gk′_{z} + 2k′_{z}2k′′_{z}c_{s}^{2} = 0,  (20)

So now replace k′′_{z} = −g/2c_{s}^{2} into Equation (19) and get

ω^{2} = c_{s}^{2}k^{2} + g^{2}/2c_{s}^{2} + c_{s}^{2}(k_{z}^{′2} − g^{2}/4c_{s}^{4}) = 0,  (21)

so

ω^{2} = c_{s}^{2}k^{2} + g^{2}/4c_{s}^{2} + c_{s}^{2}k_{z}^{′2} = 0.  (22)

References

Bogdan, T. J., & Cally, P. S., “Jacket Modes: Solar Acoustic Oscillations Confined to Regions Surrounding Sunspots and Plage,”

Astrophys. J. 453, 919-928 (1995).

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