## Mathematical Recreations

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#### Interstellar Travel

If we suppose that we eventually have the ability to harness enormous resources, but do not uncover new laws of physics, then it will always take individual humans years to travel between the stars. The problem is that we can't accelerate faster than our bodies can survive. So, if we assume that the passengers want to experience the journey at an acceleration of 1 g, then how much travel time do they experience on an interstellar journey?

The difficulty that we have to work through is that the traveler isn't in an inertial frame of reference. That is, v keeps changing. The traveler starts at rest and undergoes a constant rate of acceleration g (in the traveler's frame of reference). What is the traveler's velocity (relative to the original frame of reference) at any time?

Let's define some coordinates. The position of the traveler in the original frame of reference is (x, t). (Here I'm using "position" to refer to both space and time.) The velocity of the traveler as measured in the original frame of reference is v. (The traveler sees the same velocity, but has a different distance scale...) The cumulative elapsed time that the traveler has experienced is τ. We want to define the relationship between these coordinates. To do so, we define two additional sets of coordinates: The coordinates in the traveler's inertial frame of reference are (x1, t1). The traveler doesn't really have an inertial frame of reference, since he/she is accelerating constantly, but this is the inertial frame that the traveler would be in if the acceleration were turned off briefly. The traveler is at position x1 = 0. When we envision turning off the acceleration briefly, we will take that moment to correspond to t1 = 0. At that moment, we will also want to consider another set of coordinates (x0, t0) in the Earth's inertial frame of reference. These coordinates are defined by the Lorentz transformation:

(1)
 x1  =  γ (x0 – v t0) t1  =  γ (t0 – v x0 / c2) x0  =  γ (x1 + v t1) t0  =  γ (t1 + v x1 / c2)
where   γ  =  √[1 – v2 / c2] –1

There will be an offset between x and x0 and between t and t0, but while we imagine the acceleration to be turned off, the offest is constant. That is, dx = dx0 and dt = dt0. Similarly, there is an offset between τ and t1, but dτ = dt1.

To make the physics perfectly clear, let's consider the acceleration to be a series of discrete boosts in velocity. That is, the traveler instantaneously shifts from the frame (x1t1) into another frame of reference (x2t2). The change in velocity dv1 happens at intervals dτ, such that dv1 = g dτ = g dt1. So, the relative velocity of the coordinate system (x1t1) with respect to the coordinate system (x0t0) is v, and the relative velocity of the coordinate system (x2t2) with respect to the coordinate system (x1t1) is dv1, but what is the relative velocity of the coordinate system (x2t2) with respect to the coordinate system (x0t0)?

We find

(2)
 x2 =  γ1 (x1 – v t1) =  γ1 γ (1 + v dv1 / c2) [ x0  – (v + dv1) c2 c2 + v dv1 t0 ]

 t2 =  γ1 (t1 – v x1 / c2) =  γ1 γ (1 + v dv1 / c2) [ t0  – v + dv1 c2 + v dv1 x0 ]

where   γ1  =  √[1 – (dv1 / c)2] –1

Now we can recognize that this must also be a Lorentz transformation. Therefore, we must have

(3)
 v2  = (v + dv1) c2 c2 + v dv1

γ2  =  γ1 γ (1 + v dv1 / c2)

where   γ2  =  √[1 – (v2 / c)2] –1

A little painful algebra verifies that the third equation follows from the first two. If dv1 is actually an infinitesimal change in velocity, then we only care about the first order term.

(4)
 v2 ≈  v  +  (1 – v2 / c2) dv1 dv =  (1 – v2 / c2) dv1

Since dv1 = g dτ, we can integrate to find the velocity at any time τ.

(5)
 ∫ dv 1 – v2 / c2
 = ∫ g dτ

 c tanh–1 (v / c)
 =  g τ

 v
 =  c tanh (g τ / c)

This gives the velocity in terms of the traveler's elapsed time. From here it is a simple process to integrate this equation and find the position as a function of time. It is not, however, a trivial process, because the velocity is not equal to dx / dτ. The velocity is equal to dx / dt, so we need to determine the relationship between dt and dτ before we can integrate. We want to consider this relationship from the point of view of the traveler at x1 = 0, since that is where Equation 5 applies. Looking at Equation 1, we see that dt0 = γ dt1. Since dt = dt0 and dτ = dt1, we find

(6)
 dt  =  γ dτ

Now we can use Equation 5 to write γ = cosh(g τ / c), and it is trivial to integrate Equation 5.

(7)
 ∫ dx =  ∫ v γ dτ x =  ∫ c sinh (g τ / c) dτ =  (c2 / g) [cosh (g τ / c) – 1]

This is very close to the formula that we want. We want to know the value of τ when the traveler has made it halfway to the destination, because then the deceleration starts. If the total distance is X, then the total travel time T is given by

(8)
 X / 2 =  (c2 / g) [cosh (0.5 g T / c) – 1] T =  (2 c / g) cosh–1 (1 + 0.5 g X / c2)

If X = 4.3 light-years, then T = 3.6 years. Dozens of stars could be reached in five to six years. In fact, a traveler could even go the Andromeda galaxy in under 29 years if a constant acceleration could be maintained. But how much energy would that require? That's another topic...

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Thanks,
Steve