Example 4 Given two points $M(-1,0), N(1,0)$, and point $P$ such that $\overrightarrow{M P} \cdot \overrightarrow{M N}, \overrightarrow{P M} \cdot \overrightarrow{P N}, \overrightarrow{N M} \cdot \overrightarrow{N P}$ form an arithmetic sequence with a common difference less than zero, find the curve that is the locus of point $P$. (2002 National College Entrance Examination Question)
Let $P(x, y)$, from $M(-1,0), N(1,0)$ we get
$$
\begin{array}{l}
\overrightarrow{P M}=-\overrightarrow{M P}=(-1-x,-y), \quad \overrightarrow{P N}=-\overrightarrow{N P}=(1-x,-y), \\
\overrightarrow{M N}=-\overrightarrow{N M}=(2,0) . \quad \overrightarrow{M P} \cdot \overrightarrow{M N}=2(1+x), \\
\overrightarrow{P M} \cdot \overrightarrow{P N}=x^{2}+y^{2}-1, \quad \overrightarrow{N M} \cdot \overrightarrow{N P}=2(1-x) .
\end{array}
$$

Thus, $\overrightarrow{M P} \cdot \overrightarrow{M N}, \overrightarrow{P M} \cdot \overrightarrow{P N}, \overrightarrow{N M} \cdot \overrightarrow{N P}$ forming an arithmetic sequence with a common difference less than zero is equivalent to
$$
\left\{\begin{array}{l}
x^{2}+y^{2}-1=\frac{1}{2}[2(1+x)+2(1-x)], \\
2(1-x)-2(1+x)<0,
\end{array}\right.$$ Therefore, the trajectory of point $P$ is the right half of a circle centered at the origin with a radius of $\sqrt{3}$.