4. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=$ $A A_{1}=2, A D=2 \sqrt{3}, M$ is a point in the plane $B A_{1} C_{1}$. Then the minimum value of $\overrightarrow{M A} \cdot \overrightarrow{M C}$ is $\qquad$ .
4. $-\frac{16}{7}$.

Take the midpoint $O$ of $A C$. Then
$$
\overrightarrow{M A} \cdot \overrightarrow{M C}=\overrightarrow{M O}^{2}-\overrightarrow{O A}^{2}=|M O|^{2}-4 \text {. }
$$

Thus, we only need to find the distance from point $O$ to the plane $B A_{1} C_{1}$.
Since $A C / /$ plane $B A_{1} C_{1}$, we only need to find the distance $h$ from point $A$ to the plane $B A_{1} C_{1}$.
$$
\begin{array}{l}
\text { By } V_{\text {tri } 1-B A_{1} c_{1}}=V_{\text {tri } C_{1}-A B A_{1}} \\
\Rightarrow \frac{1}{3} S_{\triangle A_{1} B C_{1}} h=\frac{1}{3} S_{\triangle A B A_{1}} B C \\
\Rightarrow h=\frac{2 \sqrt{3}}{\sqrt{7}} .
\end{array}
$$

Therefore, the minimum value of $\overrightarrow{M A} \cdot \overrightarrow{M C}$ is $-\frac{16}{7}$.