# Mathmatical Formalism¶

This chapter details the mathematical formalism on which GYRE is built.

## Physical Formulation¶

### Fluid Equations¶

The basis for most subsequent equations are the fluid equations, comprising the conservation laws for mass

$\pderiv{\rho}{t} + \cdot \nabla \left( \rho \vv \right) = 0$

and momentum

$\rho \left( \pderiv{}{t} + \vv \cdot \nabla \right) \vv = -\nabla P - \rho \nabla \Phi;$

the heat equation

$\rho T \left( \pderiv{}{t} + \vv \cdot \nabla \right) S = \rho \epsnuc - \nabla \cdot \vF;$

and Poisson’s equation

$\nabla^{2} \Phi = 4 \pi G \rho.$

Here, $$\rho$$, $$p$$, $$T$$, $$S$$ and $$\vv$$ are the fluid density, pressure, temperature, specific entropy and velocity; while $$\Phi$$ is the gravitational potential, $$\epsnuc$$ is the specific nuclear energy generation rate and $$\vF$$ the energy flux.

The energy flux is the sum of the radiative ($$\vFrad$$) and convective ($$\vFcon$$) fluxes,

$\vF = \vFrad + \vFcon;$

$\vFrad = \frac{c}{3\kappa\rho} \nabla (a T^{4}),$

where $$\kappa$$ is the opacity and $$a$$ the radiation constant.

### Thermodynamic Relations¶

The fluid equations are augmented by the thermodynamic relationships between the four state variables ($$p$$, $$T$$, $$\rho$$ and $$S$$). Only two of these are required to uniquely specify the state (we assume that the composition remains fixed over an oscillation cycle). In GYRE, $$p$$ and $$S$$ are adopted as these primary variables, and the other two are presumed to be derivable from them:

$\rho = \rho(p, S), \qquad T = T(p, s).$

### Equilibrium State¶

In a static equilibrium state the fluid velocity vanishes. The momentum equation then becomes the hydrostatic equilibrium equation

$\nabla P = - \rho \nabla \Phi.$

Assume the equilibrium state is spherically symmetric, this simplifies to

$\deriv{p}{r} = - \rho \deriv{\Phi}{r}.$

Poisson’s equation can be integrated once to yield

$\deriv{\Phi}{r} = \frac{G}{r^{2}} \int 4 \pi \rho r^{2} \, \diff{r} = \frac{G M_{r}}{r^{2}},$

and so the hydrostatic equilibrium equation becomes

$\deriv{p}{r} = - \rho \frac{G M_{r}}{r^{2}}.$

The heat equation in the equilibrium state is

$\rho T \pderiv{S}{t} = \rho \epsnuc - \nabla \cdot \vF.$

If the star is in thermal equilibrium then the left-hand side vanishes, and the nuclear heating rate balances the flux divergence term.

### Linearized Equations¶

Applying an Eulerian (fixed position, denoted by a prime) perturbation to the mass and momentum conservation equations, they linearize about the static equilibrium state as

$\rho' + \nabla \cdot ( \rho \vv' ) = 0,$
$\rho \pderiv{\vv'}{t} = - \nabla P' + \frac{\rho'}{\rho} \nabla P - \rho \nabla \Phi',$
$\nabla^{2} \Phi' = 4 \pi G \rho'$

Likewise applying a Lagrangian (fixed mass element, denoted by a $$\delta$$) perturbation to the heat equation and the thermodynamic relations, they linearizes about the equilibrium state as

$T \pderiv{\delta S}{t} = \delta \epsnuc - \delta \left( \frac{1}{\rho} \nabla \cdot \vF \right).$
$\frac{\delta \rho}{\rho} = \frac[1}{\Gamma_{1}} \frac{\delta p}{p} - \upsilon_{T} \frac{\delta S}{c_{p}}$
$\frac{\delta T}{T} = \nabla_{\rm ad} \frac{\delta p}{p} + \frac{\delta S}{c_{p}}$

The absence of either a prime or a $$\delta$$ denotes an equilibrium quantity. No $$\vv'$$ terms appear because the equilibrium state is static. The thermodynamic partial derivatives are defined as

$\Gamma_{1} = \left( \pderiv{\ln p}{\ln \rho} \right)_{S} \qquad \upsilon_{T} = \left( \pderiv{\ln \rho}{\ln T} \right)_{p} \qquad c_{p} = \left( \pderiv{S}{\ln T} \right_{p} \qquad \nabla_{\rm ad} = \left( \pderiv{T}{p} \right)_{S}$

### Separation¶

With a separation of variables in spherical-polar coordinates $$(r,\theta,\phi)$$, and assuming an oscillatory time ($$t$$) dependence with angular frequency $$\sigma$$, solutions to the linearized fluid equation can be expressed as

$\xir(r,\theta,\phi;t) = \operatorname{Re} \left[ \sqrt{4\pi} \, \txir(r) \, Y^{m}_{\ell}(\theta,\phi) \, \exp(-\ii \sigma t) \right],$
$\vxih(r,\theta,\phi;t) = \operatorname{Re} \left[ \sqrt{4\pi} \, \txih(r) \, r \nablah Y^{m}_{\ell}(\theta,\phi) \, \exp(-\ii \sigma t) \right],$
$f'(r,\theta,\phi;t) = \operatorname{Re} \left[ \sqrt{4\pi} \, \tf'(r) \, Y^{m}_{\ell}(\theta,\phi) \, \exp(-\ii \sigma t) \right]$

Here, $$\xir$$ is the radial component of the displacement perturbation vector $$\vxi$$, and $$\vxih$$ is the corresponding horizontal (polar and azimuthal) part of this vector; $$\nablah$$ is the horizontal part of the spherical-polar gradient operator; $$Y^{m}_{\ell}$$ is the spherical harmonic with harmonic degree $$\ell$$ and azimuthal order $$m$$; and $$f$$ stands for any perturbable scalar. The displacement perturbation vector is related to the velocity perturbation via

$\vv' = \pderiv{\vxi}{t}$

## Oscillation Equations¶

The oscillation equations follow from substituting the above solution forms into the linearized equations:

$\trho' + \frac{1}{r^{2}} \deriv{}{r} \left( r^{2} \txir \right) - \frac{\ell(\ell+1)}{r} \rho \txih = 0,$
$-\sigma^{2] \rho \txir = - \deriv{\tP'}{r} + \frac{\trho'}{\rho} \deriv{P}{r} - \rho \deriv{\tPhi'}{r},$
$-\sigma^{2} \rho r \thxi = - \tP' - \rho \tPhi',$