How to calculate the twin paradox

The Web is full of pages about Special Relativity and how it is responsible for slowing clocks and creating the twin paradox. But what if a star ship accelerates during the first half of its journey and slows down during the second half?

WAIT A MINUTE! Shouldn’t we switch to general relativity when dealing with accelerated systems? Not necessarily. If you accelerate and decelerate along a straight trajectory that joins two star systems and ignore the curvature of space caused by other objects, you are fine with special relativity.

This article tells you how to calculate the time spent by a subluminal (i.e., no warp drives!) star ship constantly accelerating half of the way towards its destination and then constantly decelerating during the second half of its voyage. Its purpose is to support Science Fiction writers who need to write about interstellar travel. I adapted the formulae from an article from the University of California Riverside and, for fun, I re-obtained them from the standard Lorentz transformations for length and time. Initially, I thought I would also explain how it is done, but it would have been a bit too complicated for most people. If you are curious, I found a paper from the University of Leipzig and another article from UCR to be useful.

First of all, let’s define some terminology:

• ‘a’ is the acceleration of the ship measured on the ship itself. Technically called the proper acceleration of the ship, which is the acceleration felt by the passengers. That is, what an accelerometer placed on the ship will measure.
  
  
• ‘D’ is the distance between the point of departure and point of arrival, measured when the ship is moving at a speed much lower than the speed of light and with its engines off. Basically, you can take it as the distance we would measure from Earth.
  
  
• ‘T’ is the time needed by the ship to make its journey, as measured on Earth. Earth orbits the Sun at 30km/s and the Sun moves at 370 km/s with respect to the cosmic microwave background. But we can ignore these speeds, because they only represent some 0.1% of the speed of light. On Earth, we are also subjected to its gravity and both Earth and the Sun move on curved trajectories, but these accelerations can also be ignored for our purposes. I just read that special-relativity effects slow down the clocks on GPS satellites (orbiting at 20,000km above sea level and travelling at one orbit per 12 hours, or 3.83km/s) by 7μs/day, while general-relativity effects (Earth’s gravitational force is much weaker up there) speed them up by 45μs/day.
  
  
• ‘t’ is the time needed by the ship to make its journey, as measured on the ship.

Here we go. Let’s start with the time measured on Earth.  This is given by:

T = 2 sqrt[(D/2/c) ² + D/a]

To make our life easier, we will measure time in y (year), distances in ly (light year, the distance light covers in one year), and speeds as fractions of c (the speed of light ~300,000km/s). In this units, g (gravitational acceleration on Earth’s surface, 9.81 m/s²) turns out to be 1.03 ly/y². With this choice of units, c disappears from the above formula, because c = 1.

For example, let’s suppose that we want to reach Proxima Centauri, the nearest star to our solar system (4.24 ly) and that our ship can sustain the acceleration of 0.1g. For the people left back on Earth, the journey will take:

T = 2 * sqrt[(4.24/2) ² + 4.24/(0.1 * 1.03)] = 13.51y

With an acceleration of 1g (ten times higher), still from the point of view of Earth-bound people, the journey would take 5.87y (you only need to remove the 0.1 from the above expression).

The time measured on the ship is given by:

t = c / a * 2 * arcsinh[a*T/c/2]

With c = 1, the formula becomes:

t = 2 * arcsinh[a*T/2] / a

arcsinh is the inverse function of the hyperbolic sine. You’ll probably find it in Excel (haven’t checked). I have it in the calculator application on my Mac when I set it to scientific mode.

So, how older do the passengers of our ship become when they travel to Proxima at 0.1g and 1g? You only need to plug a and T into the formula and obtain 12.61y and 3.55y.

Not a big deal, is it? In case you are wondering, the top speed, when the ship is half a way to Proxima and switches from 1g of acceleration to 1g of deceleration, is given by:

v = a* T / 2 / sqrt[1 + (a*T/2/c) ²] = 1.03 * 5.87 / 2 / sqrt[1 + (1.03*5.87/2) ²] = 0.95c

If you go to Tau Ceti, a star similar to ours that is 11.9 ly away, you get, for 1g acceleration:

T = 23.98y

t = 6.23y

v = 0.9967c

Now the differences become more significant. Still, you would have imagined a more dramatic difference, wouldn’t you? I did.

OK. Let’s look at the planet HD 40307g. It is the latest Earth-like planet discovered. It might have a gravity twice as strong as Earth’s, but it orbits a star slightly cooler than ours with a 200-day period. It also seems that it rotates on its axis, which would imply a day-and-night cycle. It could have liquid water and be able to sustain life. Its distance from us is 42 ly.

T = 43.90y

t = 7.40y

v = 0.9990c

Well, here the twin paradox is definitely dramatic. After a round trip, the twin on Earth would be 2 * (43.9 – 7.4) = 73y older.

Now, what type of propulsion could possibly accelerate a ship at 1g for more than seven years? You tell me!