be independent of $z$. This is equivalent to requiring that the total momentum contained in the outflow be conserved. Integrating $e'\Gamma^2v_z^2$ from $y = -\infty$ to $\infty$, using equations and , taking the ultrarelativistic limit and changing variables from $y$ to $\xi$, we find

$$\dot{P} \simeq \bigg{(}\frac{z}{z_0}\bigg{)}^{\frac{1}{2}-2m}\int_{-\infty}^{\infty}\left(\frac{df}{d\xi}\right)^2d\xi.$$

Since the integral in this equation is a constant that is greater than zero, we find that $m = 1/4$ if the momentum flux is conserved.

With this value of $m$, the $z$-component of the four-velocity scales as

$$\Gamma{v_z} = \Gamma_jv_j\bigg{(}\frac{z}{z_0}\bigg{)}^{-1/4}\frac{df}{d\xi} \label{vzsim2},$$

and the self-similar variable $\xi$ is given by

$$\xi = \tau\,\sqrt{\frac{9\Gamma_j^2v_j^2}{2\rho'_0\kappa{z_0}}}\bigg{(}\frac{z}{z_0}\bigg{)}^{-3/4} \label{xieq}.$$

We [then]{} find from equation that the stream function must satisfy

$$\psi = \sqrt{\frac{2}{9}\rho'_0\kappa{z_0}}\bigg{(}\frac{z}{z_0}\bigg{)}^{1/2}f.$$

Equation then gives the $y$-component of the four-velocity:

$$\rho'\kappa\Gamma{v_y} = -\sqrt{\frac{2}{9}\frac{\rho'_0\kappa}{z_0}}\bigg{[}\frac{1}{2}\bigg{(}\frac{z}{z_0}\bigg{)}^{-1/2}f+z_0\bigg{(}\frac{z}{z_0}\bigg{)}^{1/2}\frac{\partial\xi}{\partial{z}}f_{\xi}\bigg{]} \label{vysim},$$

[where, both here and in future equations, a subscripted $\xi$ on the functions $f$ and $g$ denotes differentiation with respect to $\xi$ and the number of subscripts indicates the number of derivatives, i.e., $f_{\xi} = df/d\xi$, $f_{\xi\xi} = d^2f/d\xi^2$, etc.]{}

Inserting equations , and into equations and , changing variables from $\tau$ to $\xi$ and performing a bit of algebra, we find that they become, respectively,

$$-\frac{1}{2}\Gamma{f}g_{\xi} = \zeta\bigg{(}\bigg{(}\frac{\partial{S}}{\partial\xi}\bigg{)}^2-v_z\frac{\partial^2{S}}{\partial\xi^2}\bigg{)} \label{ss1},$$

$$-\frac{1}{4}g\bigg{(}(f_{\xi})^2+2ff_{\xi\xi}\bigg{)} = \frac{\partial^2S}{\partial\xi^2} 