https://www.tbclrarebooks.com/pages/books/31784/george-turner/image-beloved-son

George Turner was an Australian writer who started writing SF late in life, focusing on genetic engineering and climate change, before it was fashionable. His first SF novel, “Beloved Son” (1978), was published in his 62nd year.

The tale told features a starship, “Columbus VI”, returning after a 44 year trip to Barnard’s Star, in the far off date of 2032. Thus it launched in 1988. A multi-national mission using a monopole magnetic field as an interstellar ramscoop.

Google’s preview lets one read the following discussion between a reporter and the Captain…

“Now, the trip itself…”

“Eight years at one-eighth g will bring us to eighty-seven L – that means eighty-seven per cent of the speed of light – an five-sixths of a light year on our way. Then we’ll coast for about four and third light years – not quite five years travelling – and spend eight years slowing for orbit of the star. Same routine to get home.”

Barnard’s Star is 6 light-years away, so the quoted distances are consistent. However, as “Crowlspace” is about Hard-SF the relativistic trajectory needs to be checked. Acceleration, distance and time are inextricably linked, even in plain old Newtonian mechanics.

To reach a speed v, then slow down to zero again, takes a time, \(t_g\), at an acceleration, g, related by:

$$t_g=\left(\frac {2.v}{g}\right)$$

Conversely the maximum velocity attained, at \(t_g/2\), is:

$$v=\left(\frac{g.t_g}{2}\right)$$

In Relativistic motion the sidereal time required is also a simple relationship between gamma-factor, speed and acceleration:

$$t_g=\left(\frac {2.\gamma.v}{g}\right)$$

Gamma-factor is defined as:

$$\gamma=\sqrt{1-\left(\frac{v}{c}\right)^2} = \sqrt{1+\left(\frac{g.t_g}{2.c}\right)^2}$$

And maximum velocity becomes:

$$v=\frac{\frac{g.t_g}{2}}{\left(1+\left(\frac{g.t_g}{2.c}\right)^2\right)}$$

The “Columbus VI” takes 16 years to reach a gamma factor of 2 at 1/8th gee, then slow down again. But what time is really required?

At 1 gee the time-factor of cee/gee is 0.97 years, thus 1/8 gee is 8 times as long.

The speed is SQRT(3/4) cee and the gamma factor is 2, so the total acceleration time should be 2.(0.97).(8).SQRT(3).

To get to gamma 2 in 16 years, the acceleration has to be (0.97)*SQRT(3) times higher, about 0.21 gee.

Alternatively we can ask ourselves, how far does 1/8 gee really get us in 16 years, followed by 5 years of cruising at the resulting speed?

The Newtonian Displacement (aka Distance) equation, when we substitute in the relevant values is:

$$S= v.t-\left(\frac{v^2}{g}\right)$$

The Relativistic version is similar:

$$S=v.t-\frac{2.c^2}{g}\left(\frac{\gamma-1}{\gamma}\right)$$

Using the definition of the total acceleration time, \(t_g\), above and some algebra, as well as the definition of \(\gamma\) we can eliminate the acceleration:

$$S=v.t\left(1-\frac{t_g}{t}\left(\frac{1}{\gamma+1}\right)\right)$$

If the total time, \(t\), equals \(t_g\), then the equation reduces to:

$$S=v.t\left(\frac{\gamma}{\gamma+1}\right)$$

The values used by George Turner were total time 21 years, acceleration time 16 years, \(\gamma\) of two, and max speed of \(\sqrt{3/4}\) cee. Plugging them in we get:

$$S=\sqrt{3/4}.(21)\left(1-\frac{16}{21}\left(\frac{1}{3}\right)\right)$$

A displacement of 13.57 light-years. But that uses the higher acceleration we computed above.

Alternatively, let’s use the speed that 8 years at 1/8 gee will get us. Accelerating at 1 gee for a Julian Year (365.25 days) is 1.0323 cee, in Newtonian physics. In the relativistic case it’s and a \(\gamma\) factor of 1.437, thus a speed of 0.71825 cee. Plugged into the Displacement equation we get 10.3677 light-years. Interestingly the Newtonian result is almost the same, using Turner’s top speed : 10.4368 light-years

So the displacement George’s fictional starship achieves is entirely doable. What doesn’t work is the acceleration distance of 5/6 light-years. That’s much too short, even in Newtonian mechanics.