# Photon rocket

A photon rocket is a hypothetical rocket that uses thrust from emitted photons for its propulsion.[1] The standard textbook case of such a rocket is the ideal case where all of the fuel is converted to photons which are radiated in the same direction. In more realistic treatments, one takes into account that the beam of photons is not perfectly collimated, that not all of the fuel is converted to photons etc., see e.g. Nuclear photonic rocket.

## Speed

The speed an ideal photon rocket will reach in the absence of external forces, depends on the ratio of its initial and final mass:

$v = c \frac{\left(\frac{m_{i}}{m_{f}}\right)^{2}-1}{\left(\frac{m_{i}}{m_{f}}\right)^{2}+1}$

where $m_{i}$ is the initial mass and $m_{f}$ is the final mass.

The gamma factor corresponding to this speed has the simple expression:

$\gamma = \frac{1}{2}\left(\frac{m_{i}}{m_{f}} + \frac{m_{f}}{m_{i}}\right)$

## Derivation

We denote the four-momentum of the rocket at rest as $P_{i}$, the rocket after it has burned its fuel as $P_{f}$, and the four-momentum of the emitted photons as $P_{\text{ph}}$. Conservation of four-momentum implies:

$P_{\text{ph}} = P_{i} - P_{f}$

squaring both sides (i.e. taking the Lorentz inner product of both sides with themselves) gives:

$P_{\text{ph}}^{2} = P_{i}^{2} + P_{f}^{2} - 2P_{i}\cdot P_{f}$

According to the energy-momentum relation, the square of the four-momentum equals the square of the mass, and $P_{\text{ph}}^{2}=0$ because all the photons are moving in the same direction. Therefore the above equation can be written as:

$0 = m_{i}^{2} + m_{f}^{2} - 2 m_{i}m_{f}\gamma$

Solving for the gamma factor gives:

$\gamma = \frac{1}{2}\left(\frac{m_{i}}{m_{f}} + \frac{m_{f}}{m_{i}}\right)$

## References

1. ^ McCormack, John W. "5. PROPULSION SYSTEMS". SPACE HANDBOOK: ASTRONAUTICS AND ITS APPLICATIONS. Select Committee on Astronautics and Space Exploration. Retrieved 29 October 2012.