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Exo-civilization Planetary Feedback

Exo-civilization Planetary Feedback Equation Derivation

We begin with three coupled equations for the interaction between a eco-civilization population , a resource and which they draw energy from ® and the state of the planetary environment (E).

\frac{dN}{dt} =\alpha_N N - \beta_N N - \alpha_{NR} \beta_{RN}(R) N

\frac{dR}{dt} = -\beta_{RN}(N) R + \beta_{ER}(E) R

\frac{dE}{dt} = \gamma_E(E) E + \alpha_{NE}(N) E + \alpha_{RE}(R)

In the equations above \alpha is used for "birth" terms and \beta is used for "death" terms.

In the above,

\alpha_N is the per captia population birth rate.

\beta_N is the per captia population death rate.

\alpha_{NR} is the per captia additional birth rate gained from extraction/consumption of resource R.

\beta_{RN}(R) is the per captia extraction/consumption rate of resource R.

\beta_{NR}(N) is the per unit extraction/consumption rate of resource R.

\beta_{ER}(E) is the "per unit" reduction in resource extraction due to changes in planetary environmental state.

\gamma_E(E) is measure of capacity of planetary environment to return to an pre-civilization equilibrium state (E_o). \alpha_{NE}(N) is per captia forcing of environment from pre-civilization equilibrium state (E_o).

\alpha_{NE}(N) is per unit forcing of environment from pre-civilization equilibrium state (E_o) based on resource extraction.

Now we propose models for terms.

(1) Population

Let the percaptja death rate be dependent on an environmentally dependent carrying capacity K(E)

\beta_N = \alpha_N \frac{N}{K(E)}

Let the per captia additional birth rate gained from extraction/consumption of resource R be,

\beta_{R}(R) = \beta_R R

Thus we our population ODE now becomes a form of the competition predator-prey models

\frac{dN}{dt} =\alpha_N N ( 1 - \frac{N}{K(E)}) - \alpha_{NR} \beta_{R} R N

In what follows we will use

K(E) = \frac{E-E_c}{E_c-E_o}

(2) Resource

Let the per captia and per unit extraction/consumption rate of resource R be the same

\beta_{RN} = \beta_{NR}

and let the "per unit" reduction in resource extraction due to changes in planetary environmental state be

\beta_{RE} = \beta_{RN} N R (\frac{E}{E_c})

where E_c is a "critical" state of the system above which no resource can be extracted.

Thus our resource equation becomes.

\frac{dR}{dt} = \beta_{RN} N R (\frac{E}{E_c}-1)

(3) Environment

Let \gamma_E give a form that acts like logistic growth/decline

\gamma_E(E) = \alpha_E(1- \frac{E}{E_o})

Choose a simple form for per captia environmental driving and resource driving,

\alpha_{NE}(N) = \alpha_{EN} N

\alpha_{ER}(R) = \alpha_{ER} \beta_{RN} R N

We also choose a simple form of the carrying capacity which lowers the capacity as E approaches the critical value

Thus our Environment equation becomes

\frac{dE}{dt} = \alpha_E E (1- \frac{E}{E_o}) + \alpha_{EN} N E + \alpha_{ER} \beta_{RN} R N

Exo-civilization Planetary Feedback Equations

Our final equations are \frac{dN}{dt} =\alpha_N N ( 1 - \frac{N}{\frac{E-E_c}{E_c-E_o}}) - \alpha_{NR} \beta_{R} R N

\frac{dR}{dt} = \beta_{RN} N R (\frac{E}{E_c}-1)

\frac{dE}{dt} = \alpha_E E (1- \frac{E}{E_o}) + \alpha_{EN} N E + \alpha_{ER} \beta_{RN} R N

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

Lets begin with all runs having

Hot Jupiter

M_{p} = M_j

R_{p} = R_j

n_{p} = m_h

T_{p0} = 10^4

Stellar Wind

V_{s} = 100 km/s

#	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

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Results

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