4. As shown in Figure 1, let $A B C D-A_{1} B_{1} C_{1} D_{1}$ be a cube, and $M$ be the midpoint of edge $A A_{1}$. Then the size of the dihedral angle $C-M D_{1}-B_{1}$ is equal to $\qquad$ (express using radians or the arccosine function).
4. $\arccos \frac{\sqrt{6}}{6}$.

As shown in Figure 3, let the dihedral angle $C$
$-M D_{1}-B_{1}$ be $\theta$, the dihedral angle $C-M D_{1}-$ $D$ be $\alpha$, and the dihedral angle $B_{1}-M D_{1}-A_{1}$ be $\beta$. Then $\theta+\alpha+\beta$ $=\pi$. Therefore,
$$
\theta=\pi-(\alpha+\beta) \text {. }
$$

Assume the edge length of the cube is 2. Then
$$
\begin{array}{l}
S_{\triangle D M D_{1}}=2, S_{\triangle M D_{1} A_{1}}=1, \\
M B_{1}^{2}=M D_{1}^{2}=5, C M^{2}=9, B_{1} D_{1}^{2}=C D_{1}^{2}=8 .
\end{array}
$$

Using Heron's formula, we get
$$
\begin{array}{l}
S_{\triangle M C D_{1}}=\frac{1}{4} \sqrt{2(5 \times 8+8 \times 9+9 \times 5)-25-64-81} \\
=\frac{12}{4}=3 .
\end{array}
$$

Similarly, we get $S_{\triangle M B_{1} D_{1}}=\sqrt{6}$.
Thus, $\cos \alpha=\frac{S_{\triangle M D D_{1}}}{S_{\triangle M C D_{1}}}=\frac{2}{3}, \sin \alpha=\frac{\sqrt{5}}{3}$;
$$
\cos \beta=\frac{S_{\triangle M M_{1} D_{1}}}{S_{\triangle M B_{1} D_{1}}}=\frac{1}{\sqrt{6}}, \sin \beta=\sqrt{\frac{5}{6}} \text {. }
$$

Therefore, $\cos \theta=-\cos (\alpha+\beta)$
$$
=\sin \alpha \cdot \sin \beta-\cos \alpha \cdot \cos \beta=\frac{\sqrt{6}}{6} \text {, }
$$

which means $\theta=\arccos \frac{\sqrt{6}}{6}$.