# Photon rocket

A photon rocket is a hypothetical rocket that uses thrust from emitted photons (radiation pressure by emission) for its propulsion.[1]

Photons could be generated by onboard generators, as in the nuclear photonic rocket. 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, and so on. A large amount of fuel would be required and the rocket would be a huge vessel.[2][3]

In the beamed laser propulsion, the photon generators and the spacecraft are physically separated and the photons are beamed from the photon source to the spacecraft using lasers.

In the photonic laser thruster, collimated photons are reused by mirrors, multiplying the force by the number of bounces.

## 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:

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

where ${\displaystyle m_{i}}$ is the initial mass and ${\displaystyle m_{f}}$ is the final mass.[4]

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

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

## Derivation

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

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

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

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

According to the energy-momentum relation (${\displaystyle E^{2}-(pc)^{2}=(mc^{2})^{2}}$), the square of the four-momentum equals the square of the mass, and ${\displaystyle P_{\text{ph}}^{2}=0}$ because photons have zero mass. Therefore, the above equation can be written as:

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

We can now solve for the gamma factor, where beta represents the change in speed relative to the initial rest frame, by noting that the two four–vectors are:

${\displaystyle {P}_{i}={\begin{pmatrix}{\frac {{m}_{i}c^{2}}{c}}\\0\\0\\0\end{pmatrix}}.}$

As we start in the rest frame (i.e. the zero momentum frame) of the rocket the final–four vector is

${\displaystyle {P}_{f}={\begin{pmatrix}\ {\gamma }{m}_{i}c\\{\gamma }P\\0\\0\end{pmatrix}}}$

with P being the final momentum vector. Therefore, taking the Minkowski inner product (see four-vector), we get the following after rearranging:

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

## Maximum speed limit

Standard theory says that the speed limit of a photon rocket is below the speed of light. Haug has recently, in Acta Astronautica,[4] suggested a maximum speed limit for an ideal photon rockets that is just below the speed of light. This speed he has suggested is a function of the heaviest subatomic particle in the rocket. The maximum velocity can based on this be calculated to be

${\displaystyle v_{max}=c{\sqrt {1-{\frac {l_{p}^{2}}{{\bar {\lambda }}^{2}}}}}.}$

where ${\displaystyle l_{p}}$ is the Planck length and ${\displaystyle {\bar {\lambda }}}$ is the reduced Compton wavelength of the subatomic fundamental particle. This velocity is for known subatomic particles above what currently can be achieved at the Large Hadron Collider, but below the speed of light. Based on the relativistic rocket equation this also means two Planck masses of fuel are needed for every subatomic particle in payload in the ideal photon rocket to reach maximum velocity.