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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 ?

Qmechanic
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Frederic Thomas
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  • Note that time dilation factor $\gamma$ is really the minkowski-dot product of the two 4-velocities (akin to cosine of the angle between as the dot product of two unit vectors). – robphy Apr 26 '23 at 14:41
  • Possibly helpful: https://physics.stackexchange.com/q/506106/ – robphy Apr 26 '23 at 15:03
  • You have to distinguish two things: the velocity of the particle under question, and the velocity of the observer who’s doing the measurement. In fact, you don’t really need to talk about observers as actual entities at all. You can simply geometrically speak of “energy of the particle with respect to a direct sum decomposition of the tangent space”. See Why is the first component of the energy-momentum tensor $-p_av^a$. It is essentially “basic” linear algebra. – peek-a-boo Apr 27 '23 at 05:46

1 Answers1

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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.

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?

robphy
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