Pair Creation Thresholds

Pair creation requires enough invariant energy and a channel that conserves momentum, charge, and the other relevant quantum numbers. The familiar $2mc^2$ rest-energy cost is not a universal laboratory photon threshold: recoil and collision angle matter. Threshold kinematics tells us when a process is possible; an interaction calculation is still needed to find whether its rate is appreciable.

Required background. The Energy–Momentum Relation defines the total invariant mass; Four-Vectors supplies four-momentum conservation.

The invariant energy available to a final state

Let $P_{\mathrm{tot}}^\mu=(E_{\mathrm{tot}}/c,\mathbf P_{\mathrm{tot}})$ be the total incoming four-momentum, and define

$$s=c^2P_{\mathrm{tot}}^\mu P_{{\mathrm{tot}},\mu} =E_{\mathrm{tot}}^2-c^2|\mathbf P_{\mathrm{tot}}|^2.$$

Here $s$ has units of energy squared. When a center-of-momentum frame exists, $\sqrt{s}$ is the energy available there. For free outgoing particles of masses $m_a$,

$$\sqrt{s}\geq\sum_a m_ac^2.$$

For massive final particles, their relative momenta vanish at the kinematic threshold in that frame. The accessible continuum phase space shrinks to zero at threshold; the inequality does not imply a finite rate at equality. Bound final states, target excitation, or additional outgoing particles can change the minimum final invariant mass and must be included explicitly. An arbitrarily soft extra photon need not raise that minimum.

One real photon cannot create an isolated free pair

An isolated on-shell photon has $k^2=0$, however large its energy in a particular frame. If it converted in vacuum into two free particles of mass $m$, four-momentum conservation would give

$$0=c^2k^2=c^2(p_++p_-)^2\geq4m^2c^4,$$

which is impossible for $m>0$. The obstruction is simultaneous energy and momentum conservation, not merely insufficient photon energy.

A second photon, a recoiling target, or a prescribed background can provide the missing momentum balance. An off-shell photon in a larger process also need not have $k^2=0$; it is not the isolated real photon considered here.

Two-photon production and collision angle

For two real photons with energies $E_1,E_2$ and angle $\theta$ between their momenta,

$$s=2E_1E_2(1-\cos\theta).$$

The channel $\gamma\gamma\to e^-e^+$ therefore requires

$$E_1E_2(1-\cos\theta)\geq2m_e^2c^4.$$

For a head-on collision, $\theta=\pi$, this reduces to $E_1E_2\geq m_e^2c^4$. For photons moving in the same direction, $\theta=0$, the invariant mass remains zero and no free massive pair is possible. For a small collision angle the required energy product grows as $4m_e^2c^4/\theta^2$.

For example, a head-on photon of energy $1\,\mathrm{eV}$ requires the other photon to have at least

$$E_{1,\mathrm{thr}}= \frac{(0.511\,\mathrm{MeV})^2}{1\,\mathrm{eV}} \simeq2.61\times10^{11}\,\mathrm{eV}.$$

The large laboratory asymmetry is compatible with a pair produced near rest in the center-of-momentum frame, which itself moves rapidly relative to the laboratory. Breit and Wheeler’s original calculation supplies the dynamics of this allowed channel; the threshold follows directly from the invariant.

Photon production with a recoiling target

Consider

$$\gamma+T\longrightarrow T+e^-+e^+,$$

where the initial target is free and at rest, and its rest mass $M$ is unchanged in the final state. Neglect binding and internal excitation. Then

$$s=M^2c^4+2Mc^2E_\gamma, \qquad s_{\mathrm{thr}}=(M+2m_e)^2c^4.$$

Equating them yields the exact kinematic threshold under these assumptions:

$$E_{\gamma,\mathrm{thr}} =2m_ec^2\left(1+\frac{m_e}{M}\right).$$

For a very heavy nucleus, recoil adds a small correction to $2m_ec^2$. For a free electron target, $M=m_e$, the threshold is $4m_ec^2$, the triplet production threshold. The two final electrons are identical; their antisymmetrized amplitude matters for the rate, but does not change this minimal invariant-mass calculation.

The threshold configuration is not one in which all final particles are at rest in the laboratory. They share a common velocity there because the incoming photon carries momentum. Their relative motion vanishes only in the center-of-momentum frame. Omitting this recoil is precisely what loses the factor $m_e/M$.

Background fields and critical scales

An ideal constant electric field is not an incoming on-shell photon. Its source supplies work and momentum, and quantized matter can undergo vacuum pair production in that prescribed classical background. The scale

$$E_{\mathrm{cr}}=\frac{m^2c^3}{|q|\hbar}$$

characterizes the exponential suppression in the ideal constant-field calculation. It is not a hard onset below which the rate is exactly zero. Finite spatial extent, finite duration, pulse frequency, and field invariants affect whether and how particles are produced. A potential drop or a work-over-distance estimate is useful only within a stated field model.

Schwinger’s field-theory calculation is the appropriate source for that vacuum-instability statement. Kinematics alone cannot produce its rate. In particular, a large static magnetic field by itself does not provide the electric work mechanism of the constant-electric-field example.

Exercises

1. Two equal-energy photons meet at $90^\circ$. Find the threshold energy of each photon and compare it with a head-on collision.

Solution

$s=2E^2$ at $90^\circ$, so $E\geq\sqrt2m_ec^2$. Head-on photons have $s=4E^2$ and need only $E\geq m_ec^2$ each. Less opposition between the momenta leaves more laboratory energy in the motion of the final center of mass.

2. A stationary target of mass $M$ becomes an excited target of mass $M'$. Derive the modified free-pair threshold.

Solution

The initial invariant remains $M^2c^4+2Mc^2E_\gamma$, while the final minimum is $(M'+2m_e)^2c^4$. Hence

$$E_{\gamma,\mathrm{thr}} =\frac{(M'+2m_e)^2-M^2}{2M}c^2,$$

for the endothermic channel considered here. The unchanged-target formula is recovered at $M'=M$.

3. Why can increasing the laboratory energy of one isolated real photon never overcome the vacuum two-body obstruction?

Solution

Its invariant remains $k^2=0$ at every energy. A massive pair has positive invariant mass in every frame. Increasing a frame-dependent energy without adding another momentum-bearing participant does not change this mismatch.

References

• G. Breit and J. A. Wheeler, “Collision of Two Light Quanta,” Physical Review 46, 1087–1091, 1934, doi:10.1103/PhysRev.46.1087 — two-photon pair production.
• S. Navas et al. (Particle Data Group), “Kinematics,” in Review of Particle Physics, Physical Review D 110, 030001, 2024, kinematics review — invariant energy, thresholds, and phase space.
• J. Schwinger, “On Gauge Invariance and Vacuum Polarization,” Physical Review 82, 664–679, 1951, doi:10.1103/PhysRev.82.664 — pair production in a constant electric background.