Measured Energy, Momentum, and Frequency

Energy and frequency in curved spacetime are local relations between a physical momentum and an observer. For a future-directed unit velocity $u^\mu$, the invariant contraction $-p_\mu u^\mu$ is the particle energy measured where the two vectors meet; for a light wave with wave vector $k^\mu$, the same construction gives $\omega=-k_\mu u^\mu$. The minus signs follow from the timelike norm $u^2=-1$.

Required background. Observers, Rest Spaces, and Measured Components provides observer projectors and measured decompositions; Pseudo-Riemannian Metrics and Signature provides Lorentzian norms and causal classification; Frames, Coframes, and Anholonomy provides dual frames and local Lorentz freedom.

Helpful background. Events, Intervals, and Minkowski Geometry supplies the flat-spacetime limit; Tensor Fields and Natural Operations clarifies tensors versus their components.

Energy from an observer contraction

Let a particle and observer meet at an event. Suppose the particle has future-directed four-momentum $p^\mu$, with

$$p^\mu p_\mu=-m^2,$$

and the observer has future-directed unit four-velocity $u^\mu$. In the observer’s orthonormal tetrad, $e_{\hat0}{}^\mu=u^\mu$, the temporal momentum component is

$$E_{(u)}=p^{\hat0}=-p_\mu u^\mu.$$

To check the sign, choose the observer’s instantaneous rest frame. There $u^{\hat a}=(1,0,0,0)$ and $p^{\hat a}=(p^{\hat0},\mathbf p)$, so $p\cdot u=-p^{\hat0}$. Future-directed causal momenta have $p^{\hat0}>0$, hence the measured energy is positive. This local curved-spacetime formula and its tangent-space limitation are stated in Reall 2013, § 5.2, pp. 51–53, PDF.

The contraction is coordinate invariant but observer dependent. It is meaningful because $p^\mu$ and $u^\mu$ are based at the same event. Contracting a momentum at emission with a velocity at reception is undefined unless a specified propagation or transport law first relates the tangent spaces.

Spatial momentum and the mass shell

The observer’s rest-space projector is

$$h^\mu{}_{\nu}=\delta^\mu{}_{\nu}+u^\mu u_\nu.$$

Define the measured spatial momentum by $\pi_{(u)}^\mu=h^\mu{}_{\nu}p^\nu$. Then

$$p^\mu=E_{(u)}u^\mu+\pi_{(u)}^\mu, \qquad u_\mu\pi_{(u)}^\mu=0.$$

Squaring this decomposition supplies an independent sign and normalization check:

$$-m^2=-E_{(u)}^2+\pi_{(u)}^\mu\pi^{(u)}_\mu, \qquad E_{(u)}^2=m^2+\lvert\boldsymbol\pi_{(u)}\rvert^2.$$

The three components $\pi^{\hat\imath}=e^{\hat\imath}{}_{\mu}p^\mu$ depend on the orientation of the spatial triad, while their Euclidean norm does not. For a massive particle with four-velocity $v^\mu$, $p^\mu=mv^\mu$ and $E_{(u)}=-m\,u\cdot v$. The comoving observer $u=v$ measures $E=m$ and zero spatial momentum.

If another observer has $u'^\mu=\gamma(u^\mu+\beta^\mu)$, where $u\cdot\beta=0$, $\beta^2\lt1$, and $\gamma=(1-\beta^2)^{-1/2}$, then

$$E_{(u')}=\gamma\left(E_{(u)}-\beta_\mu\pi_{(u)}^\mu\right).$$

Here $\beta^\mu$ is the second observer’s spatial velocity relative to $u$, distinct from the particle’s four-velocity $v^\mu$ above. This is the local special-relativistic Doppler/boost law inside one tangent space, not a rule for comparing separated observers. The underlying flat-spacetime energy-momentum construction is reviewed in Carroll 1997, Ch. 1, pp. 26–28.

For $E_{(u)}>0$, the particle’s measured three-velocity can be represented covariantly by $w^\mu=\pi_{(u)}^\mu/E_{(u)}$. Thus

$$p^\mu=E_{(u)}(u^\mu+w^\mu), \qquad w^2=1-\frac{m^2}{E_{(u)}^2}.$$

It is spatial relative to $u$, has norm below one for a massive particle, and reaches unit norm in the massless limit. This relates three distinct readings—energy, momentum, and velocity—without assigning any of them to a coordinate component.

Photon frequency and direction

In geometric optics, let the wave phase be $\Theta$ and $k_\mu=\nabla_\mu\Theta$. For a future-directed ray, $k^\mu$ is null. Along the observer worldline,

$$\frac{d\Theta}{d\tau}=u^\mu\nabla_\mu\Theta=u^\mu k_\mu.$$

With positive frequency and the site’s $e^{-i\omega t}$ convention, the observer measures

$$\omega_{(u)}=-k_\mu u^\mu.$$

Equivalently, if a photon momentum is $p^\mu=\hbar k^\mu$, then $E_{(u)}=\hbar\omega_{(u)}$; here $\hbar=1$ unless restored explicitly. The observer decomposes the null vector as

$$k^\mu=\omega_{(u)}\left(u^\mu+n^\mu\right), \qquad u\cdot n=0, \qquad n^2=1.$$

The null condition checks that the measured spatial wave-vector magnitude equals $\omega_{(u)}$. Rescaling an affine tangent rescales this frequency, so a physical wave phase or photon momentum must fix the normalization; a bare unnormalized null direction does not determine an energy.

Operationally, an ideal localized detector samples phase along its worldline. Successive crests, whose phases differ by $2\pi$, give angular frequency $\omega=2\pi/\Delta\tau$ in the slowly varying monochromatic limit. A single varying phase defines an instantaneous frequency through its derivative, but a general broadband superposition need not have one unique phase or frequency. A real detector returns a spectrum or a time-windowed response. Those instrument effects do not alter the contraction for each resolved phase branch, but they govern how accurately it is inferred.

Local scope and limiting cases

For an inertial observer $u^\mu=(1,0,0,0)$ in Minkowski coordinates, the formulas reduce to $E=p^0$ and $\omega=k^0$. For a particle at rest relative to the observer they reduce to $E=m$. For a photon they give $E=\lvert\boldsymbol\pi\rvert$, confirming both the massless limit and dimensional consistency.

No spacetime symmetry has been assumed. A conserved Killing energy, when one exists, is a different object: it is a constant associated with a symmetry and a trajectory. It becomes a measured local energy only after contraction with a specified observer. Likewise, “the photon loses energy while traveling” is incomplete without naming the sequence of observers used for the comparison.

Exercises

In Minkowski spacetime a photon has $k^\mu=\Omega(1,1,0,0)$ in an inertial frame. An observer moves in the $+x$ direction with speed $v$. Compute the observed frequency.

Solution

The observer velocity is $u^\mu=\gamma(1,v,0,0)$, while $k_\mu=\Omega(-1,1,0,0)$. Therefore

$$\omega=-k\cdot u =\gamma\Omega(1-v) =\Omega\sqrt{\frac{1-v}{1+v}}.$$

The observer moves in the direction of propagation and therefore measures a redshift. At $v=0$ the result is $\Omega$; as $v\to1$ it tends to zero. Reversing the observer velocity gives the corresponding blueshift.

References

• Carroll, Sean M. “Lecture Notes on General Relativity”, arXiv:gr-qc/9712019v1, 1997, Ch. 1, pp. 26–28.
• Reall, Harvey S. General Relativity, PDF, Part III lecture notes, University of Cambridge, 2013, § 5.2, pp. 51–53.