For a relativistic electron, the radiated power grows as γ⁴, where γ is the Lorentz factor.
Let us calculate with the non-relativistic Larmor formula. The acceleration is
a = v² / r
= ω² r.
According to the Larmor formula, the radiated power is
P = e² ω⁴ r² / (6 π ε₀ c³).
The cycle time is t = 2 π / ω.
The energy per a single cycle is
E = e² ω³ r² / (3 ε₀ c³).
The energy of a single photon is
E' = h f = h ω / (2 π).
The number of photons per cycle is
n = 2/3 π e² ω² r² / (h ε₀ c³)
= 2/3 π e² v² / (h ε₀ c³)
= 0.031 * v² / c².
That is, if the electron is mildly relativistic (v ~ c), it takes ~ 33 cycles to emit one photon. The number 33 is not very far from 1. Is it a coincidence that a relatively small number of cycles is able to produce a single photon?
If the velocity v = 0.01 c, like for the electron in the hydrogen atom, it takes 330,000 cycles to produce a photon.
The energy of a single photon for a mildly relativistic electron is
E'' = h ω / (2 π)
~ h c / (2 π r)
= h / t,
where t is the cycle time. This is the energy - time uncertainty relation:
ΔE Δt ≥ h.
We can derive the uncertainty relation by building a wave packet where we assume that the photon energy of a wave of a frequency f is h f.
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