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I think of an NMR experiment, but with a single spin half nucleus initially set to the excited state.

When the nucleus finally returns to its ground state, it will emit a photon. An observer in the lab frame will see a photon in the Larmor frequency. An observer in a rotating frame, very close to the Larmor frequency, will see a photon with a very very low frequency.

My question is: Because of the frequency difference, will it take the detector of the rotating frame observer longer time to tick (i.e., detect the photon) compared to the lab frame observer?

Also, and i know it sounds silly, if there is a different, does this somehow relates to time dilution in general relativity?

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It isn't obvious to me that the rotating observer will see a photon with a low frequency. An observer on the axis of rotation can measure the oscillating electric field and will find it oscillates with the same frequency as the lab observer. – John Rennie Jun 21 '13 at 6:32
Not exactly. In the rotating frame, the Hamiltonian becomes $H=\hbar(\omega-\omega_L)\sigma_z/2$ where $\omega$ and $\omega_L$ are the rotating frame angular velocity and the Larmor frequency respectively. So choosing to rotate at $\omega\approx\omega_L$ should give lower frequency photons – user26047 Jun 22 '13 at 18:44

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