Gravitational waves from hydrodynamic instabilities

April 24, 2021, Revision: 1.21

The compressible Navier-Stokes equation with an isothermal equation of state can be written in the form

\frac{∂u}{∂t} = −∇ (h + \frac{1}{2}u^{2}) − ω × u + f + F_{visc}, (1)

\frac{∂h}{∂t} = −u · ∇h − c_{s}^{2}∇ · u,  (2)

where

F_{visc} = ν (∇u^{2} + ∇∇ · u + 2S · ∇ ln ρ) (3)

is the viscous force.

1 Stationary forcing

We adopt a generalized ABC-flow forcing

f = \frac{f_{0}}{N} ( \begin{matrix}C sin kz + σB cos ky \\ A sin kx + σC cos kz \\ B sin ky + σA cos kx \end{matrix} )  (4)

where N^{2} = (A^{2} + B^{2} + C^{2})(1 + σ^{2})/2 is a normalization constant and f_{0} = ⟨f^{2}⟩^{1/2} is the rms value of the forcing. We define the Reynolds number as

Re = u_{rms}/νk_{f}  (5)

where k_{f} = \sqrt{3}k is the effective forcing wavenumber. For σ = 1, we have the standard ABC flow, but we also allow for other values with −1 ≤ σ ≤ 1. The flow has positive (negative) helicity for σ > 0 (< 0) and is fully helical for σ = ±1. For σ = 0, we have the Archontis flow with zero helicity. In particular, for A = 0 and B = C = 0, we have a Beltrami flow when σ = ±1 and the Kolmogorov flow for σ = 0.

In the laminar phase, at small values of Re, the flow is fully helical, i.e., the vorticity ω = ∇ × u is parallel to u with ω = ku. Therefore, ω × u = 0, so the only nonlinearity comes from the dynamical pressure term, u^{2}/2. However, for the ABC flow forcing, u^{2} = const. Therefore, saturation occurs only when u_{rms} = f_{0}/νk_{f}^{2}. The temporal evolution is given by

\frac{d}{dt}u_{rms} = f_{0} − νk_{f}^{2}u_{rms} (laminar).  (6)

The solution is given by

u_{rms}(t) = f_{0} (1 − e^{−νk_{f}^{2}t}) .  (7)

Figure 1: Total kinetic energy (black), kinetic energy in the xy-averaged velocity (red), and kinetic energy in the z-averaged velocity (blue). (paver)

Table 1:

Run  ν  f_{0}  Re
A  5 × 10^{−2}  0.1  25
B  10^{−3}  2 × 10^{−3} 100
B  10^{−3}  5 × 10^{−4} 100
C  5 × 10^{−4} 5 × 10^{−4} 25
C  5 × 10^{−4} 2 × 10^{−4} 25
C  2 × 10^{−4} 2 × 10^{−4} 25
C  5 × 10^{−4} 5 × 10^{−4} 25

Table 2:

Run  ν  f_{0}  Re t_{1}  E_{K}^{max}  E_{GW}^{max}  (ω × u)_{rms}^{max} (∇ · u)_{max}^{2}
E  2 × 10^{−4} 2 × 10^{−4} – 1121 0.017 9.0 × 10^{−13}  0.109  6.0 × 10^{−5}
B  1 × 10^{−3} 5 × 10^{−4} – 894 0.037 2.1 × 10^{−11}  0.177  7.0 × 10^{−4}
H  5 × 10^{−4} 9 × 10^{−4} – 553 0.066 1.3 × 10^{−9}  0.372  7.8 × 10^{−3}
I  5 × 10^{−4} 8 × 10^{−4} – 581 0.059 5.9 × 10^{−10}  0.318  3.7 × 10^{−3}
F  5 × 10^{−4} 7 × 10^{−4} – 638 0.054 6.0 × 10^{−10}  0.259  1.9 × 10^{−3}
C  5 × 10^{−4} 5 × 10^{−4} – 783 0.041 9.9 × 10^{−11}  0.215  7.6 × 10^{−4}
D  5 × 10^{−4} 2 × 10^{−4} – 1348 0.017 5.0 × 10^{−13}  0.070  4.4 × 10^{−5}
K  5 × 10^{−4} 1.4 × 10^{−4} – 1653 0.0112 1.3 × 10^{−13}  0.047  1.1 × 10^{−5}
G  5 × 10^{−4} 1 × 10^{−4} – 2098 0.0076 2.1 × 10^{−14}  0.026  3.2 × 10^{−6}

Figure 2: Evolution of ⟨ω^{2}u^{2}⟩ (black) and it contributions ⟨(ω · u)^{2}⟩ (red) and ⟨(ω × u)^{2}⟩ (blue).

(phel_256e)

References

Podvigina, O., & Pouquet, A. 1994, Phys. D, 75, 471

A Green’s function

Monochromatic

ḧ + k^{2}h = T  (8)

Solution

h(t) = k^{−1} ∫_{0}^{t} sin k(t − t′) T (t′) dt′  (9)

The first derivative is

ḣ(t) = ∫_{0}^{t} cos k(t − t′) T (t′) dt′  (10)

Figure 3: Results

where the derivative on the integral boundary gives zero, because the resulting term is proportional to Second derivative

ḧ(t) = T (t) − k ∫_{0}^{t} sin k(t − t′) T (t′) dt′ (11)

Figure 5: Dependence on noise for the Kolmogorov flow.

Figure 6: Results for the Kolmogov flow.

Let us now assume that T (t) is given by u^{2} using Eq. (7), so that

h(t) = \frac{f_{0}^{2}}{k} ∫_{0}^{t} sin k(t − t′) (1 − e^{−2νk_{f}^{2}t′}) dt′ (12)

Figure 7: Time dependence of instability on Gaussian noise (kinetic drive)

Figure 8: Error dependence on time step (kinetic drive)

ḣ(t) = f_{0}^{2} ∫_{0}^{t} cos k(t − t′) (1 − e^{−2νk_{f}^{2}t′}) dt′ (13)

Using Ptolemy’s identities,

sin k(t − t′) = sin kt cos kt′ − cos kt sin kt′, (14)
cos k(t − t′) = cos kt cos kt′ − sin kt sin kt′, (15)

and the integrals

∫_{0}^{t} cos kt′e^{−2νk_{f}^{2}t′} dt′ = \frac{e^{−2νk_{f}^{2}t}}{\sqrt{ω^{2} + 4ν^{2}k_{f}^{4}}} cos(ωt − ϕ)

(16)

∫_{0}^{t} sin kt′e^{−2νk_{f}^{2}t′} dt′ = \frac{e^{−2νk_{f}^{2}t}}{\sqrt{ω^{2} + 4ν^{2}k_{f}^{4}}} sin(ωt − ϕ)

(17)

where ϕ = −1/(1 + ω^{2}/4ν^{2}k_{f}^{4})^{1/2}. Using again Ptolemy’s identities, we find

h(t) = \frac{f_{0}^{2}}{k} (1 − \frac{e^{−2νk_{f}^{2}t}}{\sqrt{ω^{2} + 4ν^{2}k_{f}^{4}}} sin ϕ)) , (18)

ḣ(t) = −\frac{f_{0}^{2}}{k} \frac{e^{−2νk_{f}^{2}t}}{\sqrt{ω^{2} + 4ν^{2}k_{f}^{4}}} cos ϕ.  (19)

Figure 9: Evolution of h (solid black and dashed blue) and ḣ (solid yellow and dashed red) for γ = 0.1, T_{0} = 0.1, and k = 1. The solid lines indicate the numerical solution and the dashed lines indicate the analytic solution.

B Exponential growth

Let us now assume that T (t) is given by T_{0}(k) e^{γt}, so that

h(t) = \frac{T_{0}(k)}{k} ∫_{0}^{t} sin k(t − t′) e^{γt′} dt′  (20)

Using sin ϕ = (e^{iϕ} − e^{−iϕ})/2i, we have

h(t) = \frac{T_{0}(k)}{2ik} ∫_{0}^{t} [e^{(γ−ik)t′+ikt} − e^{(γ+ik)t′−ikt}] dt′

(21)

Integrating between the two boundaries yields

h(t) = \frac{T_{0}(k)}{2ik} [\frac{e^{γt} − e^{ikt}}{γ − ik} − \frac{e^{γt} − e^{−ikt}}{γ + ik} ] (22)

Expanding this yields

h(t) = \frac{T_{0}(k)}{2ik(γ^{2} + k^{2})}[(γ + ik)(e^{γt} − e^{ikt})

−(γ − ik)(e^{γt} − e^{−ikt})] (23)

or

h(t) = \frac{T_{0}(k)}{2ik(γ^{2} + k^{2})}[γ(e^{γt} − e^{ikt}) + ik(e^{γt} − e^{ikt})

−γ(e^{γt} − e^{−ikt}) + ik(e^{γt} − e^{−ikt})] (24)

or

h(t) = \frac{T_{0}(k)}{2ik(γ^{2} + k^{2})}[ − γ(e^{ikt} − e^{−ikt}) + 2ike^{γt}

−ik(e^{ikt} + e^{−ikt})] (25)

or

h(t) = \frac{T_{0}(k)}{k(γ^{2} + k^{2})} (−γ sin kt + ke^{γt} − k cos kt)

(26)

and therefore

h(t) = \frac{T_{0}(k)}{γ^{2} + k^{2}} (e^{γt} − cos kt − \frac{γ}{k} sin kt) (27)

with the derivative being

ḣ(t) = \frac{T_{0}(k)}{γ^{2} + k^{2}} [γ (e^{γt} − cos kt) + k sin kt] (28)

Delta correlated T (k) ∼ δ(k) leads to Sp(T) ∼ k^{2}, and δ(k)/k leads to Sp(T) ∼ k^{−2}. In simulations with chiral magnetic effect, we have γ = ημ_{0}^{2}, so the critical k is k ∼ ημ_{0}^{2}/c.

Figure 10: Solid line: last time, dotted lines: last 20 times, dashed lines: early times during growth phase, dashed-dotted lines: early decay phase.

C Numerical solution

Figure 10 shows a numerical solution for an initial spectrum of the source ∝ k^{4}/[1 + (k/k_{f})^{6}], where k_{f} = 20 was chosen. The time evolution of the source was T = t for 0 < t < 1 and 1/[1+(t−1)/τ]^{2} for t > 1.

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Figure 4: Results 5