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}}}.

Leave a Reply

Your email address will not be published. Required fields are marked *