# A beautiful blue light

In 1-d Newtonian mechanics the energy as a function of momentum for a particle is given by $$E=\frac{1}{2}mv^2$$ or, in terms of momentum, $$E=p^2/(2m)$$. 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 $$\vec{p}$$ and mass $$m$$ is given by $$E^2=c^2 \vec{p} \cdot \vec{p}+m^2c^4$$ where $$c$$ is the speed of light. The photon, the "particle" of light, has no mass - its energy satisfies $$E=c|\vec{p}|$$ in empty space. In a medium like water this can change - photons travel slower in a medium and their energy-momentum relation becomes $$E=c_m |\vec{p}|$$, where $$c_m$$ is the speed of light in the medium. Our question is the following: A very high energy proton with $$E_{prot}^2=c^2 \vec{p} \cdot \vec{p}+m_{prot}^2c^4$$ enters a tank of water. What is the minimum magnitude of the proton's momentum, i.e. $$|\vec{p}|$$, 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 $$3 \times 10^8~\mbox{m/s}$$.
• The mass of the proton is $$1.67 \times 10^{-27}~\mbox{kg}$$.
• The speed of light in water is $$2.3 \times 10^8~\mbox{m/s}$$.
• 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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