on both variables $x$ and $y$. If we consider the eddy depicted in Fig. \[fig:wx\], then we see that the difference in momentum in the $x$ direction, across the eddy, is given by $$\rho(w(x+s)-w(x-s)) \sim 2\rho s w_x,$$ for $y$ fixed, where $w_x=\frac{d}{dx}w$.

![The eddy of radius $s$ and the variation in the fluctuations across it in the $x$ (streamwise) direction.[]{data-label="fig:wx"}](Figure_1){width=".3\textwidth"}

This means that the total turbulent stress, across a vertical surface at $x$, denoted by a dotted line on Fig. \[fig:wx\] for an eddy of radius $s\sim y$, is $$\tau_0 = \tau_t+\tau_x,$$ where $\tau_x= 2\kappa_\tau \rho y w_x v_y$, analogous to formula Eq. (\[eq:shear-stress\]) above. Then we get, using Eq. (\[eq:e-viscosity\]) and $$\epsilon =  (\tau_t+\tau_x)(u'+w_x) \rho,$$ that $$\tau_t+\tau_x= \kappa^2 \rho I^{3/4} y^2(u'+w_x)^2,$$ where prime denotes the derivative with respect to $y$, and $$\begin{aligned}
(\tau_t+\tau_x)^{1/2} &=& \kappa \rho^{1/2}I^{3/8}y(u'+w_x) {\nonumber}\\
&=& \langle \tau_0 \rangle^{1/2}+ \kappa \rho^{1/2}I^{3/8}y|w_x|,\end{aligned}$$ since both parts must be positive. The derivation is completely analogous to the derivation in Sec. \[sec:fluct\], but here with $w$ varying in the $x$ direction and $w_y=0$. This gives that for $y$ fixed, $$\begin{aligned}
\tau_0^{1/2}-\langle \tau_0 \rangle^{1/2} &=& (\tau_t+\tau_x)^{1/2}-\langle \tau_0 \rangle^{1/2} {\nonumber}\\
&=& \kappa \rho^{1/2}I^{3/8}y|w_x|.\end{aligned}$$ Considering the leading order $\log(y/2Re\sqrt{f})$ term in Eq. (\[eq:w2der\]) 