become, respectively,

$$\frac{\partial{e'}}{\partial{y}} = 0 \label{ymom1},$$

$$\nabla_{\mu}\bigg{[}e'U^{\mu}\bigg{]}+\frac{1}{3}e'\nabla_{\mu}U^{\mu} = \frac{8}{27}\frac{e'}{\rho'\kappa}\bigg{(}\frac{\partial{S}}{\partial{y}}\bigg{)}^2 \label{gasen1},$$

$$\bigg{(}\rho'+\frac{4}{3}e'\bigg{)}U^{\mu}\nabla_{\mu}S+\frac{1}{3}\Gamma\frac{d{e'}}{d{z}} = \frac{8}{27}\frac{\partial}{\partial{y}}\bigg{[}\frac{e'}{\rho'\kappa}\frac{\partial{S}}{\partial{y}}\bigg{]} \label{zmom1},$$

$$\nabla_{\mu}\bigg{[}\rho'U^{\mu}\bigg{]} = 0 \label{masscont1}.$$

where

$$S = \ln\bigg{(}\Gamma{v_z}+\sqrt{1+\Gamma^2{v_z}^2}\bigg{)} = \text{arcsinh}(\Gamma{v_z}) \label{Seq}.$$

The first of these shows that the radiation energy density, and consequently the pressure, is constant across the boundary layer. In the inviscid limit, equation can be transformed to give

$$U^{\mu}\nabla_{\mu}\bigg{[}\Gamma\bigg{(}1+\frac{4}{3}\frac{e'}{\rho'}\bigg{)}\bigg{]} = 0,$$

which, as we mentioned, is the relativistic generalization of the Bernoulli equation. We will also assume that the ambient energy density is independent of $z$, i.e., that $e'(z) = e'_0$, with $e'_0$ a constant.

Because it will be convenient, we will change variables from $y$ to $\tau$, where $\tau$ is given by

$$\tau \equiv \int_0^{y}\rho'(\tilde{y},z)\,\kappa\,d\tilde{y} \label{taueq},$$

which is related to the optical depth across the boundary layer as measured from the axis. In terms of this variable, equations and become, respectively,

$$\frac{4}{3}e'U^{\mu}\nabla_{\mu}\bigg{(}\frac{1}{\rho'}\bigg{)} = \frac{8}{27}e'\kappa\bigg{(}\frac{\partial{S}}{\partial\tau}\bigg{)}^2 \label{gasen2},$$

$$\bigg{(}1+\frac{4}{3}\frac{e'}{\rho'}\bigg{)}U^{\mu}\nabla_{\mu}S = \frac{8}{27}e'\kappa\frac{\partial^2S}{\partial\tau^2} \label{zmom2},$$

where we have used the assumption that $e'(z) = e'_0$. It should also be noted that the $y$-component of the convective derivative is now with respect to $\tau$, not $y$, i.e.,

$$U^{\mu}\nabla_{\mu} = U^{z}\frac{\partial}{\partial{z}}+\rho'\kappa{\,}U^{y}\frac{\partial}{\partial\tau}.$$

In the ensuing section we will seek self-similar solutions to these equations.

Self-similar approximation
--------------------------

We can immediately solve 