Energy Component 0 of 4 Momentum Vector P

[kent davidge]

The energy is the 0-th component of the four momentum vector $p^\alpha$. How is called the component $p_0 = g_{0\alpha}p^\alpha$?

 

[robphy]

The energy is the 0-th component of the four momentum vector $p^\alpha$. How is called the component $p_0 = g_{0\alpha}p^\alpha$?

I don't think it has a special name (or any special interpretation)
since its value depends on the signature convention of the metric...
...and since "energy [in that frame]" as $p^0$ is already defined.
So, I'd call it "the 0th-component of $p_\mu$".

 

[martinbn]

Just to point out that the energy is not a component of a vector, that would not be an invariant quantity. It is the inner product of the timelike Killing vector and the momentum, and it is equal to the zero-th component of momentum in coordinates in which the Killing vector is $(1,0,0,0)$.

 

[kent davidge]

energy is not a component of a vector, that would not be an invariant quantity

I thought neither energy nor spatial momentum needed to be separetely invariant.

 

[andresB]

No.
Indeed energy doesn't need to be a Lorentz invariant, it's a frame dependent quantity.

 

[PeterDonis]

energy is not a component of a vector

It depends on your choice of terminology. Many sources do indeed use the term "energy" to mean the 0th component of 4-momentum. Which means, as noted, that "energy" under this interpretation is not an invariant.

It is the inner product of the timelike Killing vector and the momentum

This only works in spacetimes that have a timelike Killing vector field; not all spacetimes do. The most commonly encountered example of a spacetime that doesn't is FRW spacetime.

 

[robphy]

Given an observer Alice's 4-velocity $\hat u^a$ and the 4-momentum $p^a$ of a particle,
the energy of that particle according to Alice is $E=g_{ab} \hat u^a p^b$ (in the +--- convention).
This is a statement at an event of spacetime when and where Alice and the particle meet.
It is an algebraic relation. The energy of the particle according to Alice is
the (Alice's time)-component of the particle's 4-momentum.

It is an invariant in the sense that "E is the energy that that Alice measures from the particle".
(Certainly different observers will measure different energies
(since measuring observers have different 4-velocities)...
but
all observers agree that "Alice measured that $E=g_{ab} \hat u^a p^b$", a dot-product.)

With additional structures and constructions, one can generalize the situation... but those aren't necessary for the statement above. The above holds with or without killing vectors, with or without additional structures or conditions.

 

[vanhees71]

Well, for a point particle the four-momentum is
$$p=(E/c,\vec{p}),$$
i.e., here the energy of the particle indeed occurs as a 0-component of a vector. The momentum obeys the constraint ("on-shell condition")
$$p_{\mu} p^{\mu}=(E/c)^2-\vec{p}^2=m^2 c^2,$$
where $m$ is the mass (the one and only adequation definition of math by the way, i.e., the invariant mass!).

Sometimes it's of advantage to define energy with respect to a special frame of reference which is somehow defined by the physical situation. An important example is a gas in thermal equilibrium, where you measure all intrinsic quantities like temperature, pressure, etc. in the rest frame of the gas. Then it makes sense to define the energy with respect to this rest frame $E^*$ as a scalar quantity. To see that it is a scalar, you simply have to introduce the four-velocity of the fluid $u^{\mu}=\gamma(1,\vec{v}/c)$. In the rest frame $u^{*\mu}=(1,0,0,0)$, i.e., you have
$$E^*/c=u_{\mu}^* p^{*\mu}=u_{\mu} p^{\mu},$$
which shows that it is indeed a scalar, independent of the reference frame. Correspondingly the phase-space distribution function (which is a scalar quantity by definition in the modern relatistic formulation of thermodynamics) reads (Maxwell-Boltzmann-Jüttner distribution)
$$f(\vec{p})=\frac{1}{(2 \pi \hbar)^3} \exp[-(c u_{\mu} \cdot p^{\mu}-\mu)/(k_{\text{B}} T)],$$
where $\mu$ is the chemical potential and $T$ the (absolute) temperature, both of which are scalar (!) quantities in the modern formulation of relativistic thermodynamics (statistical physics).

 

[PAllen]

Since most sources treat 4-velocity exclusively as a vector, it is often convenient to treat 4-momentum as a one form (covariant, not contravariant) to take its inner product directly with an observer 4-velocity, yielding observed energy. Further, relativistic treatments of Lagrangians and Hamiltonians that I've seen always use momentum as a one form, leading to force as a one-form. Some authors even argue that 4-momentum as a vector is 'incorrect' (I don't go this far).

 

[robphy]

Since most sources treat 4-velocity exclusively as a vector, it is often convenient to treat 4-momentum as a one form (covariant, not contravariant) to take its inner product directly with an observer 4-velocity, yielding observed energy. Further, relativistic treatments of Lagrangians and Hamiltonians that I've seen always use momentum as a one form, leading to force as a one-form. Some authors even argue that 4-momentum as a vector is 'incorrect' (I don't go this far).

In my opinion, the mathematical formulation is a model of physical phenomenon...
in the sense that the structures in our theories are motivated
--not by convenience--but by the physical properties we wish to capture
(and without inadvertently introducing structures that would suggest unphysical properties).

4-velocity is a vector (visualized as an arrow) seems appropriate since it is the tangent to a curve [worldline].
Momentum as a one-form in phase space seems appropriate...
but I would need better motivation for 4-momentum for a particle as a one-form in spacetime.
For me, a one-form suggests a visualization of linear approximations of level surfaces
of a scalar field (parallel planes, approximating equipotentials).
To me, it's not clear what that would be for a particle in spacetime.

Using the spacetime metric, one could form the metric-dual of a 4-momentum vector... and use that for convenience... but the physics is fundamentally in the 4-momentum vector.

My $0.02.

 

[vanhees71]

Since most sources treat 4-velocity exclusively as a vector, it is often convenient to treat 4-momentum as a one form (covariant, not contravariant) to take its inner product directly with an observer 4-velocity, yielding observed energy. Further, relativistic treatments of Lagrangians and Hamiltonians that I've seen always use momentum as a one form, leading to force as a one-form. Some authors even argue that 4-momentum as a vector is 'incorrect' (I don't go this far).

Well in spaces with fundamental form (as Minkowski space is) there's a natural, i.e., coordinate independent mapping between vectors and covectors, and usually you identify them. I'd thus not say it's incorrect to say to take (canonical) momenta as one-forms only, but indeed the natural structure is to take it as a one-form, because
$$p_{\mu}=\frac{\partial L}{\partial \dot{q}^{\mu}},$$
where the dot stands for a derivative wrt. an arbitrary world-line parameter (which most conveniently is chosen as the proper time if you consider a massive particle).