Let $G$ be a connected, reductive group over a field $k$. Assume $G$ is quasi-split. Let $A_0$ be a maximal split torus of $G$ with centralizer $T$, and let $B$ be a minimal parabolic (Borel) subgroup of $G$ containing $T$. Let $W = N_G(T)/T$ be the absolute Weyl group, and $W_0 = N_G(A_0)/T$. Each element of $W_0$ has a $k$-rational representative, and in fact we have $$W_0 = W(k) = N_G(A_0)(k)/A_0(k) = N_{G(k)}(A_0(k))/A_0(k)$$
There are two Bruhat decompositions of $G$. First, we have
$$G(\overline{k}) = \coprod\limits_{w \in W} B(\overline{k})wB(\overline{k})$$
and second,
$$G(k) = \coprod\limits_{w \in W_0} B(k)wB(k)$$ where each representative $w \in W_0$ is chosen to be $k$-rational. I'm pretty sure that $B(k)$ is Zariski dense in $B(\overline{k})$, and so $\overline{B(k)wB(k)} = \overline{B(\overline{k})wB(\overline{k})}$. Also, it is generally true that $G(k)$ is Zariski dense in $G(\overline{k})$, so we have
$$\coprod\limits_{w \in W} B(\overline{k})wB(\overline{k}) = G(\overline{k}) = \overline{G(k)} = \bigcup\limits_{w \in W_0} \overline{B(\overline{k})wB(\overline{k})} \tag{1}$$
Now, each $\overline{B(\overline{k})wB(\overline{k})}$ is a union of the cells $B(\overline{k})w' B(\overline{k})$, where $w' \in W$ can be obtained from $w$ by taking a reduced decomposition of $w$ (in the Coxeter system corresponding to $B$ and $T$), and removing generators.
So, where does the long element of $W$ appear on the right hand side? In general, I think the long element of $W$ need not lie in $W_0$, and it seems in this case that the length of every double coset in (1) is strictly less than the long element of $W$.