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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