Wald's definition of energy of a massive particle

Question

In R.Wald's book on "General Relativity" appears on p.61 (section 4.2) a rather peculiar definition of the energy of a massive particle:

$$ E = -p_a v^a \tag{4.2.8}$$

I guess the minus sign comes from the use of the east-coast metric in Wald's book $\eta=diag(-1,1,1,1)$. Right before he first defines the momentum 4-vector $p^a$ in (4.2.7) as:

$$ p^a = m u^a \tag{4.2.7}$$

where $m$ is the (rest) mass of the particle and $u^a$ are the components of a tangent vector to the (time-like) curve along which the particle moves.

He goes on saying that

The ${\it{energy}}$ of the particle as measured by an observer -- present at the site of the particle -- whose 4-velocity $v^a$ is defined by

$$ E = -p_a v^a \tag{4.2.8} $$

Thus, in special relativity the energy is recognized to be the "time component" of the vector $p^a$. For a particle at rest with respect to the observer (i.e. $v^a = u^a$), equation (4.2.8) reduces to the familiar formula $E=mc^2$ (in our units with $c=1$).

Actually I am quite confused by this statement. Why 2 velocities intervene ? It seems that in a more general case it could be even $v^a\neq u^a$. Is there is somebody which could explain to me formula (4.2.8), how it is meant ?

Answer 1

The 4-momentum is the timelike tangent-vector to the particle worldline, and its magnitude is the invariant [rest] mass of the particle. The 4-velocity is the timelike-unit tangent-vector.

The tangent vector determines the instantaneous "time-axis" of the particle.

While the slope of the 4-velocity is related to the "relative spatial velocity", I think it's best to first think of the 4-velocity vector as defining the unit-vector along the time-axis of the particle.

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So, if our observer has 4-velocity $\hat v^a$, the time-component of a 4-vector $ Q^a$ according to $\hat v^a$ is the gotten with the dot-product $-g_{ab}\hat v^aQ^b $ using the $(-+++)$ convention. (The analogue in Euclidean geometry to find the x-component of a vector $\vec W$ is to calculate $\hat x \cdot \vec W$, which implicitly uses the Euclidean metric.)

For the 4-momentum of a particle with rest-mass $m$, we have $p^a=m\hat u^a$. Time-component of that 4-momentum is relativistic energy, according to that observer.

Thus, our observer $\hat v^a$ determines the relativistic energy of $p^a$ to be $$E=-g_{ab}\hat v^a (m\hat u^b)=-m g_{ab}\hat v^a \hat u^b=m\cosh\theta=m\gamma,$$ which can be expressed in terms of the relative-rapidity $\theta$ between their 4-velocities (between their time-axes) and in terms of the time-dilation factor.

The particle itself determines the relativistic energy of its 4-momentum to be $$E'=-g_{ab}\hat u^a (m\hat u^b)=-m g_{ab}\hat u^a \hat u^b=m,$$ the rest energy.

For further points, go to my answer in How does $p\cdot u$ relate to observed energy and momentum for a massive particle?