[Link to ADS entry](http://adsabs.harvard.edu/abs/1978ApJ...226..650I)

Loading library and setting up environment


```python
import sympy
sympy.init_printing()
```

Equation 5 - Conservation of mass

$\frac{\partial \rho}{\partial t} + \nabla \left(\rho v \right) = 0$

Equation 6 - Momentum conservation

$\frac{d v}{d t} = \frac{J \times B}{4 \pi \rho} - \frac{\nabla P}{\rho}$

Equation 7 - Ideal Ohm's law

$E + \frac{v}{c} \times B=0$

Equation 8 - Ampere's law

$\nabla \times B = \frac{4 \pi}{c} J$

Equation 9 - Faradays's law

$\frac{1}{c}\frac{\partial B}{\partial t} = -\nabla \times E$

Equation 10 - Ideal gas law

$P = n_e k T_e$

Equation 11 - Eliminating the current density from the momentum equation using Ampere's law

$4 \pi \rho \frac{d v}{d t} = \left(\nabla\times B\right) \times B - 4 \pi \nabla P$

Equation 12 - Eliminiating the electric field from Faraday's equation using the Ideal Ohm's law

$\frac{\partial B}{\partial t} = \nabla \times \left(v \times B\right) = \left(B \cdot \nabla \right) v - \left(v\cdot \nabla \right) B - B \left(\nabla \cdot v \right)$

In steady state $\partial v / \partial t = \partial B / \partial t = 0$. The author introduces a new variable: the displacement, which relates to the velocity through $v = d \xi / d t$. Linearising the momentum equation yields equation 13

$4 \pi \rho \frac{\partial^2 \xi}{\partial t^2} = \left(\nabla \times B'\right) \times B + \left(\nabla \times B\right) \times B' - 4 \pi \nabla P'$

Equation 14 can be obtained from equation 12 by replacing the velocity with the displacement

$B' = \left(B \cdot \nabla \right) \xi - \left(\xi \cdot \nabla \right) B - B \left(\nabla \cdot \xi\right)$

To derive equation 15, we start from the perturbed conservation of mass

$\rho' = -\xi \nabla \rho - \rho \nabla \xi$

Substituting in the ideal gas equation yields

$\frac{P'}{P} = \frac{\rho'}{\rho} = -\xi \frac{\nabla \rho}{\rho}- \nabla \xi = - \chi_p \xi - \nabla \xi$

Equations 16 - 18

$\left(\nabla \times B'\right) \times B + \left(\nabla \times B\right) \times B' = \left( B \cdot \nabla \right) B' + \left(B' \cdot \nabla \right) B - \nabla \left(B \cdot B' \right)$

We only need to expand the first term

$ \left(B \cdot \nabla \right) B'= \left(B \cdot \nabla \right) \left[ \left(B \cdot \nabla \right) \xi - \left( \xi \cdot \nabla \right) B - B \left( \nabla \cdot \xi\right) \right]$

The remainder of the derivation is straightforward

$4 \pi \rho \frac{\partial^2 \xi}{\partial t^2} - \left(B \cdot \nabla\right)^2 \xi +\nabla \left[4 \pi P' + B' \cdot B \right] - \left(B' \cdot \nabla \right) B + \left(B \cdot \nabla\right) \left[\left(\xi \cdot \nabla\right) B + B \left(\nabla \cdot \xi\right) \right] = 0$

For equation 19, Here's a sketch to understand the geometry



When the ambient magnetic field is constant equation 16 is greatly simplified

$4 \pi \rho \frac{\partial^2 \xi}{\partial t^2} - B^2 \frac{\partial^2 \xi}{\partial x_{\parallel}^2} +\nabla \left[4 \pi P' + B' \cdot B \right] + \hat{x}_{\parallel} B^2 \frac{\partial}{\partial x_{\parallel}} \left(\nabla \cdot \xi \right)= 0$

Substituting the perturbed mode yields

$\epsilon \xi_r = \frac{d}{d x_r} \left(4 \pi P'+B' \cdot B\right)$

$\epsilon \xi_{\perp} = i k_{\perp} \left(4 \pi P' + B' \cdot B\right)$

$\epsilon \xi_{\parallel} = i k_{\perp} \left(4 \pi P' + B'\cdot B\right) + i B^2 k_{\parallel} \left(i k_{\parallel}\xi_{\parallel} + i k_{\perp} \xi_{\perp} + \frac{d \xi_{r}}{d \xi_r} \right)$

where $\epsilon = 4 \pi \rho \omega^2 - k_{\parallel} B^2$

Equation 24

$\frac{4 \pi P'+ B \cdot B'}{4 \pi P+B^2} = \frac{-4 \pi P \left(\chi_p \cdot \xi +\nabla \xi\right)+ B \cdot \left[ \left(B \cdot \nabla \right) \xi - \left(\xi \cdot \nabla \right) B - B \left(\nabla \cdot \xi\right) \right]}{4 \pi P + B^2} = - \nabla \cdot \xi +\frac{ -4 \pi P \chi_p \cdot \xi + B^2 \frac{d \xi_{\parallel}}{d x_{\parallel}} }{4 \pi P + B^2} = - \left(i k_{\parallel} \xi_{\parallel} + i k_{\perp} \xi_{\perp} + \frac{d \xi_r}{d x_r} \right ) +\frac{ -4 \pi P \chi_p \cdot \xi + i k_{\parallel} B^2 \xi }{4 \pi P + B^2} = $

$= -i k_{\perp} \xi_{\perp}- \frac{d \xi_r}{d x_r} - \frac{P}{P+B^2/4\pi} i k_{\parallel} \xi_{\parallel} - \frac{P \chi_p \cdot \xi}{P+B^2/4 \pi}$

This is almost the same expression as the right hand side of equation 24, but with a superfluous extra term at the end.

Derivation of equation 25


```python
epsilon = sympy.Symbol('epsilon')
xi_parallel = sympy.Symbol(r'\xi_{\parallel}')
xi_perp = sympy.Symbol(r'\xi_{\perp}')
k_perp = sympy.Symbol(r'k_{\perp}')
k_parallel = sympy.Symbol(r'k_{\parallel}')
tilde_P_prime = sympy.Symbol(r"\tilde{P}'")
dxirdxr = sympy.Symbol(r'd \xi_r/d x_r')
B = sympy.Symbol('B')
eqn_22 = sympy.Eq(epsilon*xi_parallel, 4*sympy.pi*sympy.I*k_parallel*tilde_P_prime+
                 sympy.I*B**2*k_parallel*(sympy.I*k_parallel*xi_parallel+sympy.I*k_perp*xi_perp+dxirdxr))
eqn_22
```


```python
P = sympy.Symbol('P')
eqn_24 = sympy.Eq(tilde_P_prime/(P+B**2/4/sympy.pi),
                 - dxirdxr-sympy.I*k_perp*xi_perp-sympy.I*k_parallel*xi_parallel*(P/(P+B**2/4/sympy.pi)))
eqn_24
```


```python
eqn_21 = sympy.Eq(epsilon*xi_perp,
                 4*sympy.pi*sympy.I*k_perp*tilde_P_prime)
eqn_21
```


```python
alpha = sympy.Symbol('alpha')
rho = sympy.Symbol('rho')
omega = sympy.Symbol('omega')
eqn_26 = sympy.Eq(alpha, 1+2*sympy.pi*rho*beta*omega**2/(4*sympy.pi*rho*omega**2-beta*k_parallel**2*B**2/2))
eqn_26
```


```python
eqn_23 = sympy.Eq(epsilon, 4*sympy.pi*rho*omega**2-k_parallel**2*B**2)
eqn_23
```


```python
beta = sympy.Symbol('beta')
temp = eqn_24
temp = temp.subs(sympy.solve(eqn_22,xi_parallel,dict=True)[0])
temp = temp.subs(sympy.solve(eqn_21,tilde_P_prime,dict=True)[0])
temp = temp.subs(P, beta*B**2/8/sympy.pi)
temp = temp.subs(sympy.solve([eqn_23, eqn_26], [rho, k_parallel], dict=True)[1])
temp = sympy.solve(temp, xi_perp)[0].simplify()
eqn_25 = sympy.Eq(xi_perp, temp)
eqn_25
```

Equation 28

$\epsilon \xi_r = 4 \pi \frac{d}{d x_r} \left(\frac{\epsilon \xi_{\perp}}{4 \pi i k_{\perp}} \right ) = \frac{1}{k_{\perp}} \frac{d}{d x_r} \left(\epsilon \frac{\alpha k_{\perp} B^2}{\alpha k_{\perp} B^2 - \epsilon} \frac{d \xi_r}{d x_r} \right)$

Equation 41 makes no sense to me. I don't see how the change of variables in equation 41 gets you from equation 30 to equation 41. Also, I can give an example where the two equations give a different answer. For simplicity, let's define a dimensionless variable $x = x_r k_{\perp}$, and assume $\epsilon \propto x^2$. Equation 30 becomes

$x^2 \frac{d^2 \xi}{d x^2} + 2 x \frac{d \xi}{d x} - x^2 \xi = 0$

and the solution is

$\xi \propto \frac{\exp \left(\pm x\right)}{x}$

The new variable is $\psi = x^2 / \frac{d}{d x} \left(x^2\right) = \frac{1}{2} x$, and the solutions to equation 41 are modified Bessel functions. The substitution from $x$ to $\psi$ will not turn them to exponents.

Regardless, equation 30 can be solved for both the case where $\epsilon$ is constant and when it varies linearly with $x$. When $\varepsilon$ is constant the solutions are exponents, and when it is linear the solutions are modified Bessel functions.


```python

```

142e8f385fd69ac583f9aa62474be9a70999e913
ipynb
Jupyter Notebook
companion_to_ionson_1978.ipynb
bolverk/cepheid_x_ray
3400d7ea0a3b5578e3ad43b0d32437e6a4518b09
MIT
companion_to_ionson_1978.ipynb
bolverk/cepheid_x_ray
3400d7ea0a3b5578e3ad43b0d32437e6a4518b09
MIT
2017-11-08T20:46:51.000Z
2017-11-09T16:43:20.000Z
companion_to_ionson_1978.ipynb
bolverk/cepheid_x_ray
3400d7ea0a3b5578e3ad43b0d32437e6a4518b09
MIT
Qwen/Qwen-72B
1. YES
2. YES