Impulse and Momentum Transfer

The impulse is the finite change between a body’s incoming and outgoing four-momentum. Unlike an acceleration at a chosen coordinate time, it is an asymptotic observable: a permissible gauge transformation may distort the trajectory in the interaction region but cannot change the measured momenta of freely separated bodies. At leading PM order the impulse is purely transverse to both incoming worldlines. Beyond leading order, transverse and longitudinal components must be assembled together so that each outgoing momentum remains on shell and total momentum, including radiation, is conserved.

Required background. Events, Intervals, and Minkowski Geometry supplies mass-shell kinematics, and Weak Metric Perturbations and Gauge Freedom supplies the linear field and the distinction between local gauge variables and asymptotic observables.

Helpful background. The Einstein Equation fixes the interaction, Presymplectic Potentials and Currents motivates the charge interpretation, and Gravitational Radiation and Waveforms supplies the missing momentum channel in an inelastic encounter.

The asymptotic impulse

For body $a$, define

$$\Delta p_a^\mu \equiv \lim_{\tau_a\to+\infty}p_a^\mu(\tau_a) - \lim_{\tau_a\to-\infty}p_a^\mu(\tau_a).$$

The limits require asymptotic flatness, freely separated bodies, and a common asymptotic Poincaré frame. They also require a labeling convention when the bodies are identical. Under those conditions $\Delta p_a^\mu$ is invariant under gauge transformations that approach the identity sufficiently rapidly at infinity. It is not invariant under an arbitrary change of asymptotic frame, but it transforms covariantly as a four-vector.

For elastic conservative scattering,

$$\Delta p_1^\mu+\Delta p_2^\mu=0.$$

For retarded scattering, the correct balance instead reads

$$\Delta p_1^\mu+\Delta p_2^\mu+P_{\rm rad}^\mu=0,$$

where $P_{\rm rad}^\mu$ is the momentum carried by gravitational radiation. This relation follows directly from translation invariance in the amplitude-based observable construction Kosower, Maybee, and O’Connell 2019, §§ 3.2–3.5, pp. 8–16.

Covariant derivation at first PM order

Use the incoming straight lines

$$x_a^\mu(\tau_a)=b_a^\mu+u_a^\mu\tau_a, \qquad \gamma=-u_1\!\cdot u_2, \qquad b^\mu=b_1^\mu-b_2^\mu ,$$

with $b\!\cdot u_1=b\!\cdot u_2=0$ and $b=\sqrt{b^2}$. At order $G$, the linearized field sourced by body 2 may be inserted into body 1’s geodesic equation. Equivalently, integrating out that field gives a reduced interaction

$$W_1 = \frac{2Gm_1m_2(2\gamma^2-1)} {\sqrt{\gamma^2-1}} \log\!\left(\frac{b}{b_0}\right).$$

The two proper-time integrals that precede this expression enforce

$$q\!\cdot u_1=q\!\cdot u_2=0$$

for the exchanged momentum $q^\mu$. The four-dimensional propagator integral therefore collapses onto the two-dimensional spacelike plane orthogonal to both incoming velocities. On that plane, the Fourier transform of $1/\boldsymbol q^2$ is logarithmic, which explains the $\log b$ in $W_1$. This straight-worldline reduction and its invariant Jacobian are worked out explicitly in Damour 2016, § V and eqs. (38)–(53).

Hamilton’s endpoint variation, or a direct integral of the geodesic force, now gives

$$\Delta p_1^{\mu(1)} =-\frac{\partial W_1}{\partial b_\mu} =-\frac{2Gm_1m_2(2\gamma^2-1)} {\sqrt{\gamma^2-1}} \frac{b^\mu}{b^2},$$ $$\Delta p_2^{\mu(1)}=-\Delta p_1^{\mu(1)}.$$

Because $b^\mu$ points from body 2’s line to body 1’s line, the minus sign is attraction. The infrared scale $b_0$ drops out. The answer depends only on asymptotic invariants, even though the intermediate field and trajectory depend on gauge. In an amplitude derivation, the same formula arises by weighting the momentum transfer in the outgoing quantum state and then taking a wave-packet classical limit; the required order of the classical and narrow-packet limits is described in Kosower, Maybee, and O’Connell 2019, §§ 4.1–4.3, pp. 18–28.

From momentum transfer to deflection

In the center-of-mass frame, let $\boldsymbol p^-$ and $\boldsymbol p^+$ be body 1’s spatial momenta. Elasticity implies $\lvert\boldsymbol p^+\rvert=\lvert\boldsymbol p^-\rvert=p_\infty$. Hence

$$\lvert\Delta\boldsymbol p_1\rvert =\lvert\boldsymbol p^+-\boldsymbol p^-\rvert =2p_\infty\sin\frac{\chi}{2}.$$

At first PM order $\chi=O(G)$, so the difference between $2p_\infty\sin(\chi/2)$ and $p_\infty\chi$ starts at $O(G^3)$. Substitution of the leading impulse and $J=bp_\infty$ gives

$$\chi_{\rm 1PM} =\frac{2Gm_1m_2}{J} \frac{2\gamma^2-1}{\sqrt{\gamma^2-1}}.$$

The transverse component controls the deflection at linear order, but it is not the whole higher-order impulse. If the incoming momentum is along $\hat{\boldsymbol z}$ and the outgoing one is rotated by $\chi$ in the scattering plane, then

$$\Delta\boldsymbol p_1 =p_\infty\sin\chi\,\hat{\boldsymbol b}_{\rm in} -2p_\infty\sin^2\frac{\chi}{2}\,\hat{\boldsymbol z},$$

up to the sign chosen for the inward transverse direction. The longitudinal term begins at $O(\chi^2)$. Discarding it at 2PM or beyond violates the mass-shell constraint.

Mass shell, iteration, and independent checks

Since both endpoint momenta describe the same body,

$$(p_a^-+\Delta p_a)^2=(p_a^-)^2=-m_a^2,$$

and therefore

$$2p_a^-\!\cdot\Delta p_a+\Delta p_a^2=0.$$

At order $G$, the impulse is orthogonal to $p_a^-$ and the equation is satisfied. At order $G^2$, the square of the 1PM transverse impulse sources a longitudinal projection. This is why Born iterations or iterated lower-order trajectories are part of the classical answer rather than optional corrections. In amplitude language, apparently singular “superclassical” pieces cancel or combine into precisely these iterations; an explicit on-shell cross-check appears in Kosower, Maybee, and O’Connell 2019, § 5.2.6, pp. 44–46.

Three checks should accompany any PM impulse. First, verify the mass shell separately for both bodies through the retained order. Second, verify total four-momentum balance, adding $P_{\rm rad}^\mu$ for retarded boundary conditions. Third, project the elastic result into the center-of-mass frame: its total spatial magnitude must equal $2p_\infty\sin(\chi/2)$, while the component perpendicular to the incoming momentum has magnitude $p_\infty\lvert\sin\chi\rvert$. These agree only at leading small angle. Agreement of only the transverse component is insufficient beyond leading order.

The validity conditions are asymptotic flatness, freely separated endpoints, weak deflection, and impact parameters large compared with gravitational radii and relevant material scales. The construction fails for a bound binary without free asymptotic states, after an encounter crosses a capture threshold, or when finite-size response becomes nonperturbative. It also does not make a coordinate-space “force” observable. The invariant content is the endpoint map. Scattering Angle and Radial Action packages the same endpoint map into a scalar function, while Eikonal Methods and High-Energy Deflection explains its impact-parameter representation.

Exercises

Determine the longitudinal completion

Suppose an elastic impulse is known at leading order as a transverse vector $\boldsymbol Q=O(G)$ in the center-of-mass frame. Find the longitudinal component through $O(G^2)$.

Solution

Write the incoming spatial momentum as $p_\infty\hat{\boldsymbol z}$ and

$$\Delta\boldsymbol p=\boldsymbol Q+O(G^2)_{\perp} +\alpha\hat{\boldsymbol z}, \qquad \boldsymbol Q\!\cdot\hat{\boldsymbol z}=0, \qquad \alpha=O(G^2).$$

Elasticity requires

$$\lvert p_\infty\hat{\boldsymbol z}+\Delta\boldsymbol p\rvert^2=p_\infty^2.$$

The leading transverse vector determines the mass-shell condition only through $G^2$: the unknown 2PM transverse piece contributes at $G^3$ when squared with $\boldsymbol Q$. Hence

$$\alpha=-\frac{\boldsymbol Q^2}{2p_\infty}+O(G^3).$$

If the full transverse impulse $\boldsymbol Q_{\rm full}$ were known, elasticity would instead give $\alpha=-\boldsymbol Q_{\rm full}^2/(2p_\infty) +O(\boldsymbol Q_{\rm full}^4/p_\infty^3)$. The distinction between an expansion in the exact angle and an expansion in $G$ prevents an unjustified claim of 3PM accuracy from 1PM input.

Verify the Newtonian impulse

Take the low-velocity limit of the 1PM result and show that its magnitude is $2Gm_1m_2/(b v_\infty)$.

Solution

For $v_\infty\ll1$, $\sqrt{\gamma^2-1}=v_\infty+O(v_\infty^3)$ and $2\gamma^2-1=1+O(v_\infty^2)$. Therefore

$$\lvert\Delta\boldsymbol p_1\rvert =\frac{2Gm_1m_2}{b v_\infty} +O\!\left(\frac{Gm_1m_2v_\infty}{b}\right).$$

The same result follows from integrating the transverse Newtonian force along the unperturbed path:

$$\int_{-\infty}^{+\infty} \frac{Gm_1m_2 b\,dt}{(b^2+v_\infty^2t^2)^{3/2}} =\frac{2Gm_1m_2}{b v_\infty}.$$

This independently fixes both the normalization and the attractive sign.

References

• Damour, Thibault. “Gravitational Scattering, Post-Minkowskian Approximation and Effective One-Body Theory.” Physical Review D 94, 104015 (2016). DOI. Open PDF.
• Kosower, David A., Ben Maybee, and Donal O’Connell. “Amplitudes, Observables, and Classical Scattering.” Journal of High Energy Physics 2019, 137 (2019). DOI. Open PDF.

Further reading

• Herrmann, Enrico, Julio Parra-Martinez, Michael S. Ruf, and Mao Zeng. “Radiative Classical Gravitational Observables at $\mathcal O(G^3)$ from Scattering Amplitudes.” Journal of High Energy Physics 2021, 148 (2021). DOI. Open PDF.