A beautiful blue light

As the noted TV personality/string theorist Brian Greene says, "energy is the most convertible currency in the universe". At the core, this question is about the economics of photon production.

In order to create a photon of momentum $p$ in the tank of water we must pay the photon factory with the energy $pc_m$ . I.e. the cost of photon momentum (in the denomination of energy) is given by $\displaystyle \frac{\partial E_{\gamma}}{\partial p_{\gamma}} = c_m$ .

Let us assume that the photon is produced at no loss. Under this assumption, the only source of momentum or energy for the photon are the commensurate losses in the slowing proton, therefore any momentum lost by the proton shows up in the proton and, likewise, any momentum lost by the proton shows up in the photon. This will lead us to the minimum momentum the proton must have to radiate a photon.

For every unit of momentum lost by the particle, the loss in energy is given by

$\displaystyle \frac{\partial E_{\mbox{p}}}{\partial p_{\mbox{p}}} = -\frac{\partial}{\partial p} \sqrt{p^2c^2+m_0^2c^4} = -\frac{pc^2}{E}$

In other words, the maximum price that slowing proton can pay to the photon factory is $\displaystyle \frac{pc^2}{E}$ units of energy per unit of momentum.

Let's play with this expression a bit. The energy of the proton $E$ can also be written as $E = m_0\gamma c^2$ where $\gamma$ is the Lorentz factor $\sqrt{1-v^2/c^2}$ (not to be confused with the subscript I used to denote the photon above). Another way to write the proton momentum is given by $p = m_0 v \gamma$ .

Now we can see that the price that the slowing proton can afford to pay for photons is given by

$\displaystyle \frac{\partial E_{\mbox{p}}}{\partial p_{\mbox{p}}} = \frac{m_0v\gamma c^2}{m_0\gamma c^2} = v$

Putting it altogether, the fixed price of photon momentum in the denomination of energy is given by $c_m$ . The maximum price that the proton can afford to pay is given by its velocity, $v$ . A fast proton can afford to pay more than a slow proton. Obviously, the proton can't make any photons if it can't afford $c_m$ so at the very least it must be able to pay $c_m$ , i.e. it must be traveling with the velocity $c_m$ , at the very least, with respect to the water tank. This shows that $v \geq c_m$ .

The minimum momentum required to radiate the blue light is then $\displaystyle p(c_m) = m_0c_m\gamma(c_m) = \frac{m_0 c_m}{\sqrt{1-c_m^2/c^2}} \approx \boxed{ 6\times 10^{-19}\mbox{ kg m/s}}$ .

*Note that this cutesy (maybe too cute) mapping of photon production onto the buying of photon momentum at the market for energy dollars is really just the combined principles of the conservation of momentum and energy. Matching the loss of energy relative to momentum loss in the proton to the gain in energy relative to momentum gain in the photon ensures that both energy and momentum are conserved in the reaction.

All the motion is in the same straight line, so this becomes a scalar, not vector, problem. Suppose that the momentum of the proton is $p$ initially and $q$ finally, and that the momentum of the photon is $r$ , where $p,q,r > 0$ . Then conservation of energy and momentum tell us that $\begin{array}{rcl} \sqrt{c^2p^2 + m^2c^4} & = & \sqrt{c^2q^2 + m^2c^4} + c_mr \\ p & = & q + r \end{array}$ If we write $\gamma = \frac{c_m}{c}$ , then $\begin{array}{rcl} \sqrt{p^2 + m^2c^2} & = & \sqrt{q^2 + m^2c^2} + \gamma(p-q) \\ \gamma^2(p-q)^2 & = & \big[\sqrt{p^2+m^2c^2} - \sqrt{q^2+m^2c^2}\big]^2 \\ & = & p^2 + q^2 + 2m^2c^2 - 2\sqrt{(p^2+m^2c^2)(q^2+m^2c^2)} \end{array}$ and so $4(p^2+m^2c^2)(q^2+m^2c^2) \; = \; \big[p^2 + q^2 - \gamma^2(p-q)^2 + 2m^2c^2\big]^2$ which simplifies to $(p-q)^2\left\{ (1-\gamma^2)\big[(p+q)^2 - \gamma^2(p-q)^2\big] - 4\gamma^2m^2c^2\right\} \; = \; 0$ We are interested in solutions where $p-q=r >0$ , and so we need to solve $(1-\gamma^2)\big[(p+q)^2 - \gamma^2(p-q)^2\big] \; = \; 4\gamma^2m^2c^2$ This is the equation of a rotated hyperbola in the $(p,q)$ plane, and we are interested in solutions for which $p > q > 0$ . Thus we require $p > p_\mathrm{min}$ , where $p_\mathrm{min} > 0$ and $\begin{array}{rcl} (1-\gamma^2)\big[(p_\mathrm{min} + p_\mathrm{min})^2 - \gamma^2(p_\mathrm{min} - p_\mathrm{min})^2\big] & = & 4\gamma^2m^2c^2 \\ (1-\gamma^2)p_\mathrm{min}^2 & = & \gamma^2m^2c^2 \end{array}$ (the line $p=q$ meets the hyperbola in the first quadrant at the point $(p_\mathrm{min},p_\mathrm{min})$ ), and hence $p_\mathrm{min} \; = \; \frac{\gamma m c}{\sqrt{1-\gamma^2}} \; = \; \frac{mcc_m}{\sqrt{c^2 - c_m^2}}$ This gives us the value $p_\mathrm{min} = 5.98 \times 10^{-19}$ kg m/s.