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Binary Star Fall Back Disk

Binary Star Fall Back Disk Variable Set

Here are variables which go into the HAFBAD problem (Highly Abstracted Fall Back Disk)

Stars

Primary Mass (solar masses): M_p

Binary Mass ratio: q=M_s/M_p

Orbital Separation: a

Outflow in the form of a brief Shell

Velocity of Shell (< escape velocity but >> sound speed: V_s

Density of Shell: \rho_s

Temperature of Shell (with sound speed constraint: T_s

Rotation of Shell: \Omega_s

Duration of shell ejection \Delta t_s

Simulation

Resolution defined in terms of injection radius R_0: N = R_0/\delta x

Planetary Wind Problem

Problem Set Up

We have 3 conditions to initialize: initial ambient medium (not important); the conditions in the base of the Hot Jupter (HJ) atmosphere; the conditions in the stellar wind at the orbital radius (a) of the HJ.

Initial Ambient Medium

The ambient medium is just a place holder to allow the planetary wind to expand into.

\rho_{a0} = m_h \mu n_{0a}

T_{a0}

c = [\frac{(k_mT_{a0}}{(\mu m_h)} ]^{1/2}

HJ Ex0base

\rho_{p0} = m_h \mu n_{pa}

T_{p0}

Stellar Wind at orbital radius a

\rho_{p0} = m_h \mu n_{pa} =\frac{\dot{M_s}}{(4\pi V_s(a) a^2)}

T_{s0}

V_{s0}

Test Runs

04/04/15

Test Run A: standard. 200x400. 1 level of refinement. Tmax = 1.0

Test Run b: standard. 100x200. 0 level of refinement. Tmax = 1.0

• seeing weird behavior?

Test Run C: standard. 100x200. 0 level of refinement. Tmax = 4.9; 50 frames

• seeing weird behavior.
• Wind does not expand uniformly

Test Run D: standard. 100x200. 1 level of refinement. Tmax = 4.9; 50 frames

• no change from C

Test Run E: standard. 200x400. 1 level of refinement. Tmax = 4.9; 50 frames

• no change from C, just better resolution
• is it the \gamma=5/3 or something about timing? Almost looks like the wind is turning off.

Grid of Runs

#	lambda	Solar Wind Mach #	n_s/n_p
1	1.01	0.8	10-4
2	1.01	0.8	10-4
3	1.3	0.8	10-4
4	1.3	0.8	10-4
5	1.01	5.0	10-4
6	1.01	5.0	10-4
7	1.01	5.0	10-6
8	1.01	5.0	10-3
9	1.3	5.0	10-4
10	1.01	0.8	10-3

Fourier Transforms

Fourier Transforms

The basic equation for the fourier transform.

\Xi[f] = F(k) = \int^{+\infty}_{-\infty} f(x) e^{(-i 2 \pi k x)} dx

with the inverse

\Xi[F]^{-1} = f(x) = \int^{+\infty}_{-\infty} F(k)) e^{(i 2 \pi k x)} dk

And of course all the transform really means is this sum of intergrals.

F(k) = \int^{+\infty}_{-\infty} f(x) cos(2\pi kx) dx - i\int^{+\infty}_{-\infty} f(x) sin(2\pi kx) dx

Here we want to solve the signup function which is runs from -1 to 1 with a step at x=a

f(x) = sgn(x-a)

Use the derivative property of \Xi[f]

\Xi[f ] = \frac{1}{2\pi i k} \Xi[ \frac{df}{dx} ]

Since derivative of f is the direct delta function

\frac{df}{dx} = \delta(x-a)

and

\Xi[ \delta(x-a) ] = e^{-i2\pi k a}

Thus we have

\Xi[f ] = F(k) = \frac{1}{2\pi i k} e^{-i2\pi k a}

The next step is to define the Energy Spectral Density (ESD)

We use Parcivals Thm which tell us "energy" under the curve f(x)

\int^{+\infty}_{-\infty} | f(x)|^2 dx = \int^{+\infty}_{-\infty} | F(k)|^2 dk

Thus the ESD which we write as E(k) is

E(k) = | F(k)|^2

which for the sgn function is

E(k) = (\frac{1}{2\pi i k} e^{-i2\pi k a})^* (\frac{1}{2\pi i k} e^{-i2\pi k a})

E(k) = \frac{1}{4\pi^2 k^2}

So for a step function

E(k) \propto k^{-2}

which is exactly what you get in compressible turbulence because every shock is a sgn function.

Description of Problem

Results

movie

Results

movie