4. As shown in Figure 1, in the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the size of the dihedral angle $B-A_{1} C-D$ is $\qquad$ .
4. $120^{\circ}$.

Let the edge length of the cube be 1.
As shown in Figure 4, draw $B E \perp A_{1} C$ at point $E$, and connect $D E$.
By the properties of the cube, we know that
$$
\triangle A_{1} D C \cong \triangle A_{1} B C \text {. }
$$

Then $D E \perp A_{1} C, \angle B E D$
is the plane angle of the dihedral angle $B-A_{1} C-D$,
and
$$
\begin{array}{l}
B E=D E=\frac{\sqrt{2}}{\sqrt{3}}, \\
B D=\sqrt{2} .
\end{array}
$$

Thus, $\cos \angle B E D=\frac{\frac{2}{3}+\frac{2}{3}-2}{2 \times \frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{2}}{\sqrt{3}}}=-\frac{1}{2}$
$\Rightarrow$ the size of the dihedral angle $B-A_{1} C-D$ is $120^{\circ}$.