A beautiful blue light

In 1-d Newtonian mechanics the energy as a function of momentum for a particle is given by E=12mv2E=\frac{1}{2}mv^2 or, in terms of momentum, E=p2/(2m)E=p^2/(2m). In Einstein's theory of special relativity there is still energy and momentum, and they are still conserved, but the relationship between the two is different. In empty space, the energy of a particle as a function of its momentum p\vec{p} and mass mm is given by E2=c2pp+m2c4E^2=c^2 \vec{p} \cdot \vec{p}+m^2c^4 where cc is the speed of light. The photon, the "particle" of light, has no mass - its energy satisfies E=cpE=c|\vec{p}| in empty space. In a medium like water this can change - photons travel slower in a medium and their energy-momentum relation becomes E=cmpE=c_m |\vec{p}|, where cmc_m is the speed of light in the medium. Our question is the following: A very high energy proton with Eprot2=c2pp+mprot2c4E_{prot}^2=c^2 \vec{p} \cdot \vec{p}+m_{prot}^2c^4 enters a tank of water. What is the minimum magnitude of the proton's momentum, i.e. p|\vec{p}|, in kg m/s, such that the proton can emit a photon with non-zero momentum and still conserve energy and momentum?

Details and assumptions

  • The speed of light in empty space is 3×108 m/s3 \times 10^8~\mbox{m/s}.
  • The mass of the proton is 1.67×1027 kg1.67 \times 10^{-27}~\mbox{kg}.
  • The speed of light in water is 2.3×108 m/s2.3 \times 10^8~\mbox{m/s}.
  • When photon emission begins to occur the initial proton, the final proton, and the emitted photon all travel in the same direction.

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