## EOS_Polytrope

Ian Hawke

22/4/2002

EOS_Polytrope

### 1 The equations

This equation provides a polytropic equation of state to thorns using
the CactusEOS interface found in EOS_Base. As such it’s a fake, as
EOS_Base assumes that, e.g., the pressure is a function of both density and
speciﬁc internal energy. Here the pressure is just a function of the
density, and is set appropriately (the speciﬁc internal energy is always
ignored).

The two ﬂuid constants are
$K$ (eos_k)
and $\Gamma $
(eos_gamma), which default to 100 and 2 respectively. The formulas that are
applied under the appropriate EOS_Base function calls are

$$\begin{array}{rcll}P& =& K{\rho}^{\Gamma}& \text{(1)}\text{}\text{}\\ \mathit{\epsilon}& =& \frac{K{\rho}^{\Gamma -1}}{\Gamma -1}& \text{(2)}\text{}\text{}\\ \rho & =& \frac{P}{\left(\Gamma -1\right)\mathit{\epsilon}}& \text{(3)}\text{}\text{}\\ \frac{\partial P}{\partial \rho}& =& K\Gamma {\rho}^{\Gamma -1}& \text{(4)}\text{}\text{}\\ \frac{\partial P}{\partial \mathit{\epsilon}}& =& 0.& \text{(5)}\text{}\text{}\end{array}$$

To calculate the units of the Cactus quantities and back, remember that
$G=c={M}_{\odot}=1$
in Cactus.

Here is one example how to calculate densities:

$${\rho}_{\text{Cactus}}=\frac{{G}^{3}{M}_{\odot}^{2}}{{c}^{6}}\cdot \rho \approx 1.6167\cdot 1{0}^{-21}\frac{{\text{m}}^{3}}{\text{kg}}\cdot \rho =1.6167\cdot 1{0}^{-18}\frac{{\text{cm}}^{3}}{\text{g}}\cdot \rho $$ | (6) |

and one example for calculating
$K$ (for
$\Gamma =2$):