15\. A rocket propels itself rectilinearly by emitting radiation in the direction opposite to its motion. If \(V\) is its final velocity relative to its initial rest frame, prove \(a b\) initio that the ratio of the initial to the final rest mass of the rocket is given by $$ \frac{M_{\mathrm{i}}}{M_{\mathrm{f}}}=\left(\frac{c+V}{c-V}\right)^{1 / 2} $$ and compare this with the result of Exercise 5 above. [Hint: equate energies and momenta at the beginning and at the end of the acceleration, writing \(\Sigma h v\) and \(\Sigma h v / c\) for the total energy and momentum, respectively, of the emitted photons.]

The given variables and constants are: - V: the final velocity of the rocket relative to its initial rest frame - \(M_i\): the initial rest mass of the rocket - \(M_f\): the final rest mass of the rocket - c: the speed of light

The total energy of the emitted photons is given by \(\Sigma h\nu\). The total momentum of the emitted photons is given by \(\Sigma h\nu/c\). We denote these as \(E_\text{ph}\) and \(p_\text{ph}\)

Initially, the total energy of the rocket is contained in its mass: \(\frac{M_i c^2}{\sqrt{1 - \frac{0^2}{c^2}}} = M_i c^2\). After the acceleration, the total energy of the rocket is the sum of the mass-energy of the remaining rocket and the energy of the emitted photons: \(\frac{M_f c^2}{\sqrt{1 - \frac{V^2}{c^2}}} + E_\text{ph}\). Applying the conservation of energy: $$M_i c^2 = \frac{M_f c^2}{\sqrt{1 - \frac{V^2}{c^2}}} + E_\text{ph}$$

Initially, the momentum of the rocket is 0, since it is at rest. After the acceleration, the momentum of the rocket is given by the remaining rocket mass times its final velocity: \(\frac{M_f V}{\sqrt{1 - \frac{V^2}{c^2}}}\). Equating this with the total momentum of the emitted photons, \(p_\text{ph}\): $$0 = \frac{M_f V}{\sqrt{1 - \frac{V^2}{c^2}}} - p_\text{ph}$$ But, \(E_\text{ph} = c p_\text{ph}\), so \(p_\text{ph} = \frac{E_\text{ph}}{c}\). Substituting this into the above equation: $$0 = \frac{M_f V}{\sqrt{1 - \frac{V^2}{c^2}}} - \frac{E_\text{ph}}{c}$$

We will solve the equations from steps 3 and 4 simultaneously to find the ratio \(\frac{M_i}{M_f}\). Multiply the expression from step 4 by c, $$0 = M_f V - E_\text{ph}$$ From the expression from step 3, $$E_\text{ph} = M_i c^2 - \frac{M_f c^2}{\sqrt{1 - \frac{V^2}{c^2}}}$$ Substitute this into the expression from step 4, $$0 = M_f V - M_i c^2 + \frac{M_f c^2}{\sqrt{1 - \frac{V^2}{c^2}}}$$ Now, isolate the ratio \(\frac{M_i}{M_f}\): $$\frac{M_i}{M_f} = \frac{M_f V + \frac{M_f c^2}{\sqrt{1 - \frac{V^2}{c^2}}}}{M_f c^2}$$ Simplifying, $$\frac{M_i}{M_f} = \frac{V + \frac{c^2}{\sqrt{1 - \frac{V^2}{c^2}}}}{c^2}$$ Now, multiply the numerator and denominator by \(\sqrt{1 - \frac{V^2}{c^2}}\): $$\frac{M_i}{M_f} = \frac{(V + c^2) \sqrt{1 - \frac{V^2}{c^2}}}{c^2 (1 - \frac{V^2}{c^2})}$$ Simplifying further, $$\frac{M_i}{M_f} = \frac{(c + V)(1 + \frac{V}{c})}{c^2 - V^2}$$ Now, noting that \((c^2 - V^2) = c^2 (1 - \frac{V^2}{c^2})\), $$\frac{M_i}{M_f} = \frac{(c + V)(1 + \frac{V}{c})}{c^2 (1 - \frac{V^2}{c^2})}$$ Simplifying one last time, $$\frac{M_i}{M_f} = \left(\frac{c + V}{c - V}\right)^{1/2}$$ The ratio of the initial to the final rest mass of the rocket is thus given by: $$\frac{M_i}{M_f} = \left(\frac{c+V}{c-V}\right)^{1 / 2}$$ This is the desired result of the exercise.