Gravitation by Charles W. Misner, Kip Throne and John Wheeler page 65 Exercise 2.5.
The book defined "energy" for a photon $E=-\mathbf p\cdot \mathbf u$ for subsection 2.8, which later explained as a "coordinate-free contacts". This was still understandable.(Notice that $c=1$ as usual, and both $p$ and $u$ here represent 4 momentum. )
However, in Exercise 2.5, the book used $E=-\mathbf p\cdot \mathbf u$ again for a particle of non-zero rest mass, where $\mathbf p$ was the particle's momentum with rest mass $m$, and $\mathbf u$ was the observer's 4 velocity.
Further, the book claimed that $|\vec{p}|=[(-\mathbf p\cdot \mathbf u)^2+(-\mathbf p\cdot \mathbf p)]^{1/2}$ was the momentum measured by the observer, with $|\vec{v}|=\frac{|\vec{p}|}{E}$ the ordinary velocity measured by the obsever.
Question:
What does $E=-p\cdot u$ stand for in this context? (for massive particle) Is it really energy?
Further, how was $E$ relate to $\vec{p}$ and $\vec{v}$? Especially, how was $\vec{p}$ and $\vec{v}$ calculated?