# Four-momentum conservation: a trip to Alpha Centauri

Required math: calculus

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 3; Problem 3.6.

As a rather fanciful example of using the conservation of four-momentum, suppose we have a spaceship of total mass (including fuel) ${M}$, initially at rest on Earth. The fuel consists of matter/anti-matter which when mixed, produces photons that are ejected out of the back of the ship. If the ship burns enough fuel to accelerate to ${v=0.95}$, then travels to some star system such as Alpha Centauri, then decelerates to zero for a landing, then, after some time at its destination it reverses the trip by again accelerating to ${v=0.95}$, returning to Earth, and decelerating to rest, what is its final mass as a fraction of its initial mass?

We can assume that all motion takes place along the ${x}$ axis, and treat the problem in the Earth’s frame. Then the initial momentum is

 $\displaystyle \mathbf{p}_{0}$ $\displaystyle =$ $\displaystyle \left[M,0\right] \ \ \ \ \ (1)$

After accelerating, the combined momentum of the ship + ejected photons is

$\displaystyle \mathbf{p}_{1}=\gamma m_{1}\left[1,0.95\right]+E_{1}\left[1,-1\right] \ \ \ \ \ (2)$

where ${\gamma=1/\sqrt{1-v^{2}}=3.2}$, ${E_{1}}$ is the energy of the ejected photons and ${m_{1}}$ is the mass of the ship after burning the fuel needed to accelerate.

By the conservation of momentum, we have ${\mathbf{p}_{1}=\mathbf{p}_{0}}$ so

 $\displaystyle \gamma m_{1}+E_{1}$ $\displaystyle =$ $\displaystyle M\ \ \ \ \ (3)$ $\displaystyle 0.95\gamma m_{1}-E_{1}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (4)$

Adding these 2 equations, we get

$\displaystyle m_{1}=\frac{M}{1.95\gamma}=0.16M \ \ \ \ \ (5)$

Now to decelerate the ship, we eject the photons ahead of the ship, and we start with a momentum of ${\gamma m_{1}\left[1,0.95\right]}$. After deceleration, the mass is now ${m_{2}}$ and the ship is at rest, so we must have

$\displaystyle \mathbf{p}_{2}=\left[m_{2},0\right]+E_{2}\left[1,1\right] \ \ \ \ \ (6)$

Again, conservation of momentum requires ${\mathbf{p}_{2}=\gamma m_{1}\left[1,0.95\right]}$, so

 $\displaystyle \gamma m_{1}$ $\displaystyle =$ $\displaystyle m_{2}+E_{2}\ \ \ \ \ (7)$ $\displaystyle 0.95\gamma m_{1}$ $\displaystyle =$ $\displaystyle E_{2} \ \ \ \ \ (8)$

Subtracting these equations we get

 $\displaystyle m_{2}$ $\displaystyle =$ $\displaystyle 0.05\gamma m_{1}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{0.05}{1.95}M \ \ \ \ \ (10)$

On the return trip, we go through exactly the same procedure, except we now start with a mass ${m_{2}}$ rather than ${M}$. Thus on the return to Earth, the ship’s mass will be

 $\displaystyle m_{E}$ $\displaystyle =$ $\displaystyle \frac{0.05}{1.95}m_{2}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{0.05}{1.95}\right)^{2}M \ \ \ \ \ (12)$

Thus the ship’s initial mass is

$\displaystyle M=1521m_{E} \ \ \ \ \ (13)$

Virtually all the initial mass is fuel.

Incidentally, if we do this calculation for an arbitrary velocity ${v}$, we get

 $\displaystyle m_{1}$ $\displaystyle =$ $\displaystyle \frac{M}{\left(1+v\right)\gamma}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{1-v}{1+v}}M\ \ \ \ \ (15)$ $\displaystyle m_{2}$ $\displaystyle =$ $\displaystyle \frac{1-v}{1+v}M\ \ \ \ \ (16)$ $\displaystyle m_{E}$ $\displaystyle =$ $\displaystyle \left(\frac{1-v}{1+v}\right)^{2}M \ \ \ \ \ (17)$

Thus each acceleration or deceleration multiplies the previous mass by a factor of ${\sqrt{\frac{1-v}{1+v}}}$.