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.


  1. What does $E=-p\cdot u$ stand for in this context? (for massive particle) Is it really energy?

  2. Further, how was $E$ relate to $\vec{p}$ and $\vec{v}$? Especially, how was $\vec{p}$ and $\vec{v}$ calculated?


What does $E=−p⋅u$ stand for in this context? (for massive particle) Is it really energy?

Let $c=1$ and signature metric be $(-,+,+,+)$.

In a momentarily comoving reference frame, $\mathbf u=(u^0,u^1,u^1,u^3)=(1,0,0,0)$ and hence $$-\mathbf p\cdot\mathbf u=-p^0u^0\eta_{00}=p^0=E$$ It is a four-vector expression and is preserved under Lorentz transformations. Hence, it indeed represents the energy.

Further, how was E related to p and v? Especially, how were p and v calculated?

Notice that $(\mathbf p\cdot \mathbf u)^2 = E^2 = m^2+p^2$ and $(\mathbf p \cdot \mathbf p)^2=-m^2$. Upon plugging them you derive, $$[(\mathbf p\cdot \mathbf u)^2+ (\mathbf p \cdot \mathbf p)^2]^\frac{1}{2} =[m^2+p^2-m^2]^\frac{1}{2} =|\vec p|$$

Also, $|\vec p| = m\gamma(v_x,v_y,v_z)$ and hence $\dfrac{|\vec p|}{E}=\dfrac{|\vec p|}{p_0} = \dfrac{m\gamma(v_x,v_y,v_z)}{m\gamma}=\vec v$

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  • $\begingroup$ thank you, but what if they are not in the MCRF? Will it be different? $\endgroup$ – ShoutOutAndCalculate Oct 3 '19 at 20:31
  • $\begingroup$ @user9976437 These are four-vectors and are invariant under Lorentz transformation. Yes, it will hold even in other inertial frames too. To derive directly in a general frame of reference, you can make use of the rapidity and $E = m\gamma$. It is easier to derive expressions in MCRF and if they are four-vectors, it stands true in other inertial frames too. $\endgroup$ – Abhay Hegde Oct 4 '19 at 12:57

This answer is meant to complement exp ikx's answer. I'm just trying to get you to see these expressions in various ways.

Let $\bf \hat u$ be the 4-velocity of the observer. Let $\bf p$ be the 4-momentum of the particle.
With the $(-,+,+,+)$ convention, $\bf \hat u\cdot \hat u=-1$ and ${\bf p\cdot p}=-m^2$

So, think of $\bf -p \cdot \hat u$ as the observer measuring the time-component of the particle's 4-momentum [in the observer frame, of course]. Introducing the rapidity $\theta$ between their 4-velocities $\bf \hat u$ and ${\bf \hat p}={\bf p}/m$, we have $$-{\bf p \cdot u}=m\cosh\theta=\gamma m=(\mbox{relativistic energy} E) $$ Note \begin{align} \bf p &=\bf p_{\textrm{parallel to u}}+p_{\textrm{perp to u}}\\ &=\bf (-p\cdot u)\hat u + (p-(-p\cdot u)\hat u)\\ &=\bf (\textrm{$E$})\hat u+ (\textrm{$p_x$})\hat u_{\bot}\\ &=\bf \textrm{$m$}\cosh\theta\ {\hat u} + \textrm{$m$}\sinh\theta\ { \hat u_{\bot}}\\ &=\bf \textrm{$m$}\cosh\theta( \hat u + \tanh\theta\ \hat u_{\bot})\\ &=\bf \textrm{$m$}\gamma( \hat u + \beta\ \hat u_{\bot})\\ \end{align}

$$\beta=\tanh\theta=\frac{p_x}{E}=\frac{m\sinh\theta}{m\cosh\theta}$$ $$-m^2=-E^2+p^2=-(m\cosh\theta)^2+(m\sinh\theta)^2=-m^2(\cosh^2\theta-\sinh^2\theta)$$

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