4. In space, four lines satisfy that the angle between any two lines is $\theta$. Then $\theta=$ $\qquad$ .
4. $\arccos \frac{1}{3}$.

As shown in Figure 2, the four lines are translated to the same point $O$. By symmetry, we can take one point $A$, $B$, $C$, and $D$ on each of the four lines, such that
$$
\begin{array}{l}
O A=O B=O C \\
=O D=1 .
\end{array}
$$

Thus, $O$ is the center of the regular tetrahedron $A B C D$.
Clearly, $A O \perp$ plane $B C D$, and let the foot of the perpendicular be $E$. Then $E$ is the center of $\triangle B C D$.
Let $O E=x$. Then $B C=A B=\sqrt{3\left(1-x^{2}\right)}$. By the Pythagorean theorem, we have
$$
\begin{array}{l}
A E=\sqrt{2\left(1-x^{2}\right)} \\
\Rightarrow 1+x=\sqrt{2\left(1-x^{2}\right)} \Rightarrow x=\frac{1}{3} \\
\Rightarrow \cos \theta=\frac{1}{3} \Rightarrow \theta=\arccos \frac{1}{3} .
\end{array}
$$