Can an experiment reach $\alpha \chi^{2/3}$ above one before radiation losses
In plain words
To test the breakdown, an electron must feel a field about a thousand times the critical field in its own frame, but it loses energy to radiation while entering the field. The question is whether any machine can get electrons there intact.
Precise statement
Reaching $\chi \sim \alpha^{-3/2} \sim 1600$ needs $\gamma E/E_S \sim 1e3$ ($E_S = m^2 c^3/(e \hbar) \sim 4.4e13\,\mathrm{statvolt}/\mathrm{cm}$), but radiation reaction lowers gamma before the peak field is reached unless the interaction time is shorter than the radiation time. Determine whether any configuration, such as collisions of compressed $100\,\mathrm{GeV}$-class electron beams (Yakimenko et al. 2019), electron collisions with attosecond or tightly focused laser pulses, or aligned-crystal fields, yields a measurable number of events at $\alpha \chi^{2/3} \ge 1$ with an identified observable that distinguishes nonperturbative from perturbative predictions. An answer is a start-to-end design with event rates, or a proof that radiation losses forbid it.
What would settle it
Start-to-end simulations including radiation reaction showing a signal above background in a buildable configuration, followed by the measurement.
Status in the literature
Unverified note
Heinzl, Ilderton and King (Phys. Rev. Lett., 2021) showed that perturbation theory already fails in classical electrodynamics through radiation reaction at lower fields and must be resummed, which bears on any approach to $\alpha \chi^{(2/3)} \sim 1$; beam-beam and crystal schemes remain proposals as of 2026.