Momentum balance

The second equation in Eq.(2.22) has the following star interaction terms:

$$\frac{\partial u_{\star}}{\partial t}+u_{\star} \frac{\parti... ...} = \bigg( \frac{\delta u_{\star}}{\delta t}\bigg)_{\rm drag}$$ (27)

The interaction term is due to the decelerating force at which stars that move inside the gas are subject to. As we shall see, an estimate for the force is given by Eq.(). Explicitely, it is

$$\bigg( \frac{\delta u_{\star}}{\delta t}\bigg)_{\rm drag} = - X_{\rm drag}\frac{1}{\rho_{\star}} (u_{\star}-u_{\rm g})$$ (28)

where we have introduced the following definition:

$$X_{\rm drag} \equiv -C_{D} \frac{\pi r_{\star}^2}{m_{\star}}\rho_{\star} \rho_{\rm g} \sigma_{\rm tot},$$ (29)

with $\sigma_{\rm tot}^2=\sigma_{\rm r}^2+\sigma_{\rm t}^2+(u_{\star}-u_{\rm g})^2$ To the end of the calculation of the logarithmic variable version of the equation, we multiply Eq.(2.33) by $\rho_{\star} r/p_{\rm r}$:

$$\frac{\rho_{\star} r}{p_{\rm r}} \bigg( \frac{\partial u_{\s... ...rm r}})=-X_{\rm drag} \frac{r}{p_{\rm r}}(u_{\star}-u_{\rm g})$$ (30)