# How does $p\cdot u$ relate to observed energy and momentum for a massive particle?

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:

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

• thank you, but what if they are not in the MCRF? Will it be different? – ShoutOutAndCalculate Oct 3 '19 at 20:31
• @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. – 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}

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