6. Let $M$ be the midpoint of the height $D D_{1}$ of the regular tetrahedron $A B C D$. Then the dihedral angle $A-M B-C$ in radians is $\qquad$
6. $\frac{\pi}{2}$.

Let the edge length $AB=a$. It is easy to get the height $D D_{1}=\frac{\sqrt{6}}{3} a$.
Thus, $M A=\sqrt{M D_{1}^{2}+A D_{1}^{2}}$
$$
=\sqrt{\left(\frac{D D_{1}}{2}\right)^{2}+\left(\frac{A B}{\sqrt{3}}\right)^{2}}=\frac{\sqrt{2}}{2} a \text {. }
$$

Notice that,
$$
\begin{aligned}
M B & =M A \Rightarrow M A^{2}+M B^{2}=A B^{2} \\
\Rightarrow & \angle A M B=\frac{\pi}{2} .
\end{aligned}
$$

Therefore, $\angle B M C=\angle C M A=\angle A M B=\frac{\pi}{2}$, which means $M-A B C$ is a right trihedral angle.
Hence, the radian measure of the dihedral angle $A-M B-C$ is $\frac{\pi}{2}$.