Scattering Angle and Radial Action

The conservative scattering angle is a compact meeting point for otherwise different descriptions of the relativistic two-body problem. It can be read from asymptotic momenta, derived from a Hamiltonian trajectory, obtained by differentiating a radial action, or extracted from an impact-parameter eikonal. Coordinate positions and canonical momenta differ among those descriptions, but the angle at fixed invariant energy and angular momentum does not. That makes it a stringent comparison quantity—provided all calculations use the same masses, asymptotic frame, boundary conditions, and definition of the conservative sector.

Required background. Events, Intervals, and Minkowski Geometry supplies asymptotic Lorentz invariants, and Weak Metric Perturbations and Gauge Freedom supplies the weak-field domain in which PM coefficients are defined.

Helpful background. The Einstein Equation fixes the underlying interaction; Presymplectic Potentials and Currents provides the action and canonical language; and Gravitational Radiation and Waveforms is needed to distinguish a conservative angle from a radiation-reacted one.

Angle from a central Hamilton–Jacobi problem

Assume first a time-independent, rotationally invariant, local conservative Hamiltonian in the center-of-mass frame. At fixed total energy $E=\sqrt{s}$ and angular momentum $J$, solve

$$H\bigl(r,p_r,J\bigr)=E$$

for the radial momentum $p_r(r;E,J)$. Let $r_{\min}$ be the largest positive turning point, chosen by continuing the incoming branch from $r=\infty$. Hamilton’s characteristic function is

$$S=-Et+J\phi+\int p_r(r;E,J)\,dr .$$

Differentiating $S$ with respect to $J$ at fixed $E$ gives the azimuth accumulated along the orbit. The incoming and outgoing halves yield

$$\chi(E,J)+\pi =-2\int_{r_{\min}}^\infty \frac{\partial p_r(r;E,J)}{\partial J}\,dr .$$

The added $\pi$ removes the straight-line change in polar angle. Although the integral of $p_r$ itself diverges at infinity, its $J$ derivative is finite after the free contribution is treated consistently. Damour uses precisely this Hamilton–Jacobi relation to connect the real two-body and effective-one-body scattering functions Damour 2016, §§ III and VI.

The regulated radial action

Free subtraction removes the linear large-distance divergence but leaves a logarithm for a $1/r$ force. Suppose the outgoing radial momentum has the asymptotic form

$$p_r(r;E,J)-p_r^{(0)}(r;E,J) =\frac{\beta(E)}r+O(r^{-2}),$$

where the leading long-range coefficient is independent of $J$. A finite hyperbolic radial action therefore needs an infrared subtraction as well:

$$I_r^{\rm scatt}(E,J) = \lim_{R\to\infty} \left[ 2\int_{r_{\min}}^R p_r\,dr -2\int_{r_{\min}^{(0)}}^R p_r^{(0)}\,dr -2\beta(E)\log\!\left(\frac R{\ell_{\rm IR}}\right) \right],$$

where $p_r^{(0)}$ and $r_{\min}^{(0)}=J/p_\infty$ belong to free motion at the same $E$ and $J\gt0$, and $\ell_{\rm IR}$ fixes the infrared convention. For example, Newtonian scattering has $p_r^2=p_\infty^2+2GM\mu^2/r-J^2/r^2$, so $\beta=GM\mu^2/p_\infty$: the remaining logarithm can be checked without any relativistic machinery. Its subtraction is $J$ independent and therefore cannot affect the angle. Since the derivative of the free radial action is $-\pi$, the free subtraction removes precisely the straight-line angle, giving

$$\boxed{\; \chi(E,J) =-\frac{\partial I_r^{\rm scatt}(E,J)}{\partial J} \;}$$

at fixed $E$ and masses. The action is more informative than the angle: its energy derivative also encodes a regulated time delay once the asymptotic clock and infrared subtraction are fixed. For a long-range gravitational force there is no finite absolute time delay relative to an infinitely distant free trajectory without such a prescription.

At first PM order, write

$$A(E) = \frac{2Gm_1m_2(2\gamma^2-1)} {\sqrt{\gamma^2-1}} .$$

The leading angle is $\chi_{\rm 1PM}=A/J$. Integration with respect to $J$ gives

$$I_{r,{\rm 1PM}}^{\rm scatt} =-A(E)\log\!\left(\frac{J}{J_0(E)}\right).$$

The arbitrary $J_0$ reflects the infrared subtraction; it disappears from $\chi$. This logarithm is the action-space counterpart of the transverse Fourier transform of a massless $1/q^2$ exchange. The same relationship between a radial action and the PM angle persists at higher orders, though iterations and tail regions must be included with their correct boundary conditions Bern et al. 2022, main text around eqs. (5)–(8).

Why the angle survives canonical changes

A canonical transformation changes the one-form by an exact differential,

$$p_i\,dq^i=P_i\,dQ^i+dF .$$

Consequently, an action integral changes by endpoint terms. If the transformation approaches the identity at the incoming and outgoing boundaries, or if its endpoint terms are independent of $J$, then $-\partial I_r^{\rm scatt}/\partial J$ is unchanged. This is the practical canonical-invariance statement behind comparisons among harmonic-coordinate, ADM, isotropic, amplitude-derived, and EOB Hamiltonians.

The qualifications matter. A transformation that rotates the asymptotic frame, changes the identification of $J$, or carries a $J$-dependent boundary term is not innocuous. Likewise, quoting a coordinate turning radius is not equivalent to quoting $\chi(E,J)$. A clean comparison specifies the invariant energy, the total canonical angular momentum, the masses and spins, and whether radiative or time-symmetric boundary conditions were used.

The angle also connects directly to the impulse. For elastic planar scattering,

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

Thus Impulse and Momentum Transfer supplies a vector check, while the radial action supplies a scalar generating function. Agreement of both exposes sign errors or missing longitudinal iterations that an angle-only comparison can hide.

Test-body and limiting checks

Take $m_1=\mu\ll m_2=M$ and hold the test body’s specific energy $\gamma$ fixed. Since $J=\mu j+O(\mu^2)$, the general 1PM result becomes

$$\chi_{\rm 1PM}^{\rm test} = \frac{2GM}{j} \frac{2\gamma^2-1}{\sqrt{\gamma^2-1}} .$$

This is the weak-deflection expansion of timelike Schwarzschild geodesic scattering. For $\gamma=1+v_\infty^2/2+O(v_\infty^4)$ and $j=bv_\infty+O(v_\infty^3)$, it gives $2GM/(b v_\infty^2)$. For $\gamma\to\infty$ and $j=b\sqrt{\gamma^2-1}$, it gives $4GM/b$, the leading null deflection. One formula therefore checks both the Newtonian and ultrarelativistic limits without a change of PM order.

The radial-action representation is valid while a real scattering turning point exists and the weak-field series can be continued from infinity to it. It breaks down at capture, when the turning point merges with a separatrix, and it becomes nonuniform near such a merger. For spinning motion that is not planar or integrable, more actions and scattering angles are required. For hereditary conservative tails, the full on-shell functional is nonlocal in time; an “effectively local” $p_r(r)$ reconstructed on hyperbolic orbits need not continue unchanged to bound motion. This limitation is central to Bound-Orbit Information from Scattering Data.

Exercises

Reconstruct the 1PM action

Suppose an independently computed scattering angle is $\chi(E,J)=A(E)/J+O(G^2)$. Determine the $J$-dependent part of the radial action and explain the integration ambiguity.

Solution

Using $\chi=-\partial_J I_r^{\rm scatt}$,

$$\frac{\partial I_r^{\rm scatt}}{\partial J} =-\frac{A(E)}{J}+O(G^2).$$

Integration at fixed $E$ gives

$$I_r^{\rm scatt} =-A(E)\log J+C(E)+O(G^2) =-A(E)\log\!\left(\frac{J}{J_0(E)}\right)+O(G^2).$$

The function $C(E)=A(E)\log J_0(E)$ depends on the infrared subtraction and affects a convention-dependent time delay through its energy derivative. It cannot affect the scattering angle because it is independent of $J$.

Check the two test-body limits

Use the test-body formula above to recover both Newtonian massive deflection and null deflection.

Solution

At small velocity, $\sqrt{\gamma^2-1}=v_\infty+O(v_\infty^3)$, $2\gamma^2-1=1+O(v_\infty^2)$, and $j=bv_\infty+O(v_\infty^3)$. Therefore

$$\chi_{\rm 1PM}^{\rm test} =\frac{2GM}{b v_\infty^2} +O\!\left(\frac{GM}{b}\right).$$

At large $\gamma$, $(2\gamma^2-1)/\sqrt{\gamma^2-1}=2\gamma+O(\gamma^{-1})$ and $j=b\gamma+O(\gamma^{-1})$, so

$$\chi_{\rm 1PM}^{\rm test} =\frac{4GM}{b}+O(\gamma^{-2}).$$

The first limit tests the long interaction time of a slow particle; the second tests the energy-independent leading bending of a null ray by a fixed Schwarzschild mass.

References

• Bern, Zvi, Julio Parra-Martinez, Radu Roiban, Michael S. Ruf, Chia-Hsien Shen, Mikhail P. Solon, and Mao Zeng. “Scattering Amplitudes, the Tail Effect, and Conservative Binary Dynamics at $\mathcal O(G^4)$.” Physical Review Letters 128, 161103 (2022). DOI. Open PDF.
• Damour, Thibault. “Gravitational Scattering, Post-Minkowskian Approximation and Effective One-Body Theory.” Physical Review D 94, 104015 (2016). DOI. Open PDF.

Further reading

• Kälin, Gregor, and Rafael A. Porto. “From Boundary Data to Bound States II: Scattering Angle to Dynamical Invariants (with Twist).” Journal of High Energy Physics 2020, 120 (2020). DOI. Open PDF.