Kinematics

There is no substitute in this discussion for a little spacetime geometry. I will keep the algebra to the minimum, taking to heart Stephen Hawking's quip that every equation loses you half of your readers, but this will be heavy on spacetime diagrams.

A spacetime diagram. One dimension of space is along the horizontal axis, and time is upwards on the vertical axis. The units are chosen so that the speed of light is 1. For instance, time in years and distance in light-years. Then, light through the origin of the diagram travels along the main diagonals. The dark area is the "interior of the light-cone" and includes all the points in the "causal past" or "causal future" of the origin (the observer). The light area is outside the light cone and it is "spacelike separated" from the observer. Questions in the comments.

Fun fact: in a spacetime diagram, a uniformly accelerating trajectory where acceleration as experienced by the accelerating object is constant looks like a hyperbola:

Hyperbolic motion for various values of the acceleration \(a\). The hyperbolas satisfy \[\Bigl(x+{c^2\over a}\Bigr)^2 - c^2 t^2 = {c^4\over a^2} \] Observe that, near the origin, \[x = {c^2\over a}\Bigl(\sqrt{1 + (a/c)^2t^2} - 1\Bigr) \approx {a\over 2}t^2 + O(t^4)\] which looks like uniform acceleration at \(a\).

The hyperbola is such that the asymptotes in the \((x,t)\) spacetime diagram cross a distance \(-c^2/a\) behind the vertex of the hyperbola where the accelerating object is momentarily at rest. In the case of acceleration at Earth's surface gravity \(g\), this characteristic length is \(c^2/g \approx (0.9684~{\rm lyr})^2\), so assuming that the natural distance \(c^2/g\) is one light-year is good enough for blogging work. Conveniently, therefore, the relationship between distance and (Earth-based) travel time at constant \(1g\) acceleration followed by constant deceleration is as follows (with all times and distances measured in years—or light-years): \[(1 + D/2)^2 - (T/2)^2 \approx 1\] which implies that accelerating at \(1g\) takes you to light-speed so quickly that it takes just under two years longer (Earth-time) than light to get to your destination. But the real boon of relativistic travel is time dilation, thanks to which the ship-measured time is greatly reduced.

The rule for proper time is \[(c{\rm d}\tau)^2 = (c{\rm d}t)^2 - ({\rm d}x)^2 = (c{\rm d}t)^2 \bigl(1 - (v/c)^2\bigr).\] The quantity \(1/\gamma = \sqrt{1-(v/c)^2}\) is the inverse of the so-called Lorentz factor which governs relativistic time dilation and length contraction. Because time dilation is speed-dependent, the proper time for a trip is path-dependent. However, travelling at constant speed maximizes the proper time, so we can give a simple upper bound for on-ship time for hyperbolic travel: \[T_{\rm ship}^2 \le T^2 - D^2 \approx (2 + D)^2 - 4 - D^2 = 4D\] That is, it takes at most \(2\sqrt{D}\) years to travel to \(D\) light years by first accelerating and then decelerating at \(1g\).

In our discussion, the examples of Tau Ceti (12 lyr) and Gliese 581 (22 lyr) featured prominently. Alpha Centauri (4+ lyr) is also of interest. The upper bounds we have found for travel times are:

• Alpha Centauri (4+ lyr): at most 6 years Earth-time and at most 4 years ship-time
• Tau Ceti (12 lyr): at most 14 years Earth-time and at most 7.5 years ship-time
• Gliese 581 (22 lyr): at most 24 years Earth-time and at most 9.5 years ship-time

Propulsion

Among the key laws of physics are the conservation of energy and the conservation of momentum. In relativistic physics they are part of a single conservation of energy-momentum. Energy and momentum are components of a vector, and for an object of rest mass \(m\) they are related by \[E^2 - (p c)^2 = (m c^2)^2\]

Sorry, Albert, that's as close as we're going to get to your obsolete and deprecated formula

The Energy-momentum vector \((E,p c)\) is proportional to the tangent vector to the trajectory, \((c{\rm d}t,{\rm d}x)\): \[(E, p c){\rm d}\tau = mc(c{\rm d}t,{\rm d}x)\] and this means that the speed of motion is \(v/c = pc/E\). The point of all this is that, if the spaceship accelerates changing \(v\), since energy-momentum is conserved there must have been some transfer of energy-momentum to the environment. This may or may not result in a change in the rest mass of the ship, too. For instance, if the ship slows down by friction with the interstellar medium and some particles from the medium stick to the ship, the mass of the ship will have increased. Or if the ship is firing its rockets it's ejecting mass (and the lost mass is carrying away the necessary energy-momentum). Or, as Dodo pointed out in the other thread, if the spaceship is carrying a mirror light may have bounced off of it, imparting it some momentum but leaving its mass unchanged. Therefore, the same path in space, at the same speed and acceleration, will result in widely different payloads reaching the destination.

Light sailing

As Dodo pointed out, "using starlight" allows the entire initial mass of the ship to be delivered as payload at the destination. Starlight is "used" by bouncing it off a rear-facing mirror. In practice, the "mirror" is a huge reflective sail that's deployed from the ship, like so:

The mechanics of light sailing are illustrated in the following diagram:

Dynamics of light sailing. This diagram is in momentum coordinates \((pc,E)\) rather than in spacetime coordinates \((x,ct)\). The dashed hyperbola \(E^2 - (pc)^2 = (mc^2)^2\) is called the mass hyperboloid. The solid arrows are the energy-momentum vectors before and after accelerating, and the dashed segments represent the energy-momentum of incoming and reflected light. Note that the energy of the light changes on reflection: it is redshifted (becomes less energetic) if it shines from behind and accelerates the ship, and it is blueshifted (becomes more energetic) if it shines from the front and slows it down.

We assume the the energy-momentum change comes entirely from the incoming and reflected light

Light sailing won't provide uniform acceleration, though. It will accelerate you away from the nearest star but the acceleration rate will be proportional to the star's apparent brightness, so you will accelerate more at the start of the trip, stop accelerating around the middle of the trip, and then start braking at an increasing pace as you get closer to your destination. You can adjust this somewhat by changing the size and orientation of your sail, but there's a limit to how large the sail can be.

In fact, it turns out that the amount of light pressure needs to be larger in the middle section of the flight if you intend to go at constant acceleration, and that is the opposite of what would happen with starlight.

Light rocketing

Alternatively, we could design the ship as a rocket. The rocket principle is that energy is shed at high speed, so that the ship accelerates even if the combined energy-momentum of the ship and the ejecta stays constant. It turns out that the optimal rocket in terms of fuel and payload (highest thrust per mass, lowest mass loss for given thrust, or highest delivered mass for a given trajectory) is a light-rocket: the rocket consists basically of a powerful light source. This can be seen in the following diagram:

Dynamics of rocketing. The diagram is analogous to the light-sailing diagram above: the solid arrows are the energy-momentum vectors before and after accelerating. The dashed segment joining the tips of the two arrows represents the energy-momentum of the rocket ejecta. It can be seen that the smallest loss of energy (or of rest mass) is achieved when the ejecta is at 45 degrees in the diagram that is, it is massless (i.e., radiation). Such a diagram can be used to show that the payload to fuel ratio of a light-rocket depends only on the maximum speed attained in the trip.

Something that is relatively easy to deduce from this is just how expensive in terms of deliverable payload it is to accelerate to relativistic speeds with a rocket. The maximum speed attained accelerating and then decelerating at \(1g\) on a trip to a distance of \(D\) light-years is \({v\over c} = {T\over 2+D}\), attained at the midpoint of the trip. It can then be seen that the delivered payload is \(\bigl({2\over 2+D+T}\bigr)^2\) times the starting mass. As this is an upper bound on the payload for any rocket accelerating a \(1g\), we see that the delivered payloads are, to use proper spacecraft engineering jargon, piss-poor:

• Alpha Centauri (4+ lyr): at most 1:35 payload to fuel
• Tau Ceti (12 lyr): at most 1:195 payload to fuel
• Gliese 581 (22 lyr): at most 1:575 payload to fuel