Photon rocket

This article is about using photons for spacecraft propulsion. For the null dust solution to the Einstein field equations, see William Morris Kinnersley.

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

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.

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:

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

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

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

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

${P}_i = \begin{pmatrix} \frac{{m}_i c^{2}}{c} \\ 0 \\ 0 \\ 0 \end{pmatrix}$

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

${P}_f = \begin{pmatrix} \ {\gamma}{m}_i c \\ {\gamma}P \\ 0 \\ 0 \end{pmatrix}$

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

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

See also

• Beam-powered propulsion
• Laser propulsion
• Nuclear photonic rocket
• Photonic laser thruster

External links

• Whatever happened to Photon Rockets?