THE TWIN PHENOMENON

by David N. Jamieson

for 640-245 Special Relativity

This is a greatly expanded monologue on the Twin Phenomenon, based on problem 11 on Sheet 1 where it is called `The Clock Paradox'.

Refer to the accompanying Minkowski diagram on page 5 of these notes for the spacetime location of the events discussed here.

1) A spaceship, equipped with a signal laser, departs from Earth at O on a round trip to a star which is L = 24 light-years away. The spaceship travels at speed v = 0.96c and can accelerate and decelerate in negligible time (!) A twin of the spaceship pilot stays on Earth. When the spaceship reaches the star, it turns around and returns to Earth at the same speed. When the spaceship twin compares the elapsed time with the Earth twin, BOTH find that the spaceship twin is younger. How can this be possible given that the time dilation formula does not depend on sign?

The answer is that the situation is not symmetrical! The spaceship turns around, but the Earth does not. In this discussion, the situation is analysed and it is shown how both twins see time dilation of the other's clocks, yet both agree on the elapsed time on both clocks at the conclusion of the journey.

The signal laser is used to let the Earth twin know how the clock on the spaceship is ticking, but is otherwise irrelevant to the phenomenon. Laser flashes are sent to Earth by the spaceship at the moment of its departure, halfway out towards the star, at the turnaround and halfway back to Earth.

2) Let the spaceship reference frame be on the outward journey and on the return journey. Let the Earth reference frame be S.

3) At time = 7 years, by the clock on the spaceship, the spaceship reaches the star and turns around to return to Earth.

On the Minkowski diagram, the Worldline of the spaceship is the -axis, since this is the path of the origin of the spaceship reference frame seen by the observer on Earth. This passes through the common origins of the Earth frame and spaceship frame, defined to be the point O when the spaceship started its journey.

4) The Worldline of the first laser flash from the spaceship at the start of its journey has equation x = ct. The -axis is symmetrically placed about this line from the -axis.

5) Halfway along the outward journey the spaceship emits the second laser flash back to Earth. This is received on Earth just before the halfway time of the journey, T as measured by the Earth observer.

6) The Worldline of the spaceship intersects with the Worldline of the star at time on the spaceship clock. This is time T on the Earth clock. This is the turnaround point.

7) Immediately before the turnaround point, the line of simultaneity of the spaceship, which is a line parallel to the -axis, passing through the point on the -axis, intersects with the Worldline of the Earth at time . In the spaceship frame, events on the Earth that occur at time are observed to be simultaneous with the arrival of the spaceship at the star at time . and are related by the Lorentz time dilation formula:

This is because the spaceship sees the earth clock receeding at speed v and hence running slow.

Likewise, the Earth observer sees the spaceship clocks running slow, hence the Earth clock measured turnaround time, T, is related to the spaceship measured turnaround time, , by the time dilation formula:

8) Immediately after the turnaround, the spaceship heads back to Earth at the same speed v. Now the spaceship is in a completely new reference frame, since the velocity vector points in the opposite direction. The time axis of the new reference frame, , is just the Worldline of the spaceship on the return journey.

The third flash of laser light emitted from the spaceship at the turnaround point arrives back at Earth just before the spaceship, since the spaceship is chasing it, seen by the Earth observer, at speed v which is a significant fraction of the speed of light.

9) The line of simultaneity of the spaceship immediately after the turnaround point, which is the -axis, is symmetrically placed about the turnaround light flash to the -axis.

The -axis intercepts the Earth Worldline at time S , so events on Earth at this time are seen by the spaceship to be simultaneous with the time immediately after the turnaround.

Notice that the spaceship does not see the elapse of time on Earth from S to S because of its change in reference frame.

10) On the return journey, the twin on the spaceship observes the clock on Earth advance from to 2T. Therefore, the time that elapses on Earth, seen by the spaceship twin, between immediately after the turnaround and the spaceship arrival back on Earth, is the time between S and 2T. This time interval in the spaceship frame, , is the same as that measured on the outward journey, by symmetry, so = . Once again, application of the time dilation formula relates the time measured on the spaceship to the time seen to elapse on Earth:

As before, this is because the spaceship sees the earth approaching at speed v and hence sees the Earth clocks running slow.

Likewise, the Earth observer sees the approaching spaceship clocks running slow, hence the time for the return journey on the spaceship is related to the time for the return journey on Earth by:

11) The fourth laser flash emitted by the spaceship when it is halfway back arrives on Earth just ahead of the spaceship itself.

12) The Earth Twin's Story: On the arrival of the spaceship back on Earth, the Earth twin and the spaceship twin compare their time readings for the journey. The Earth twin records a total time of T for the outward journey and a total time of T for the return journey, giving a total of 2T for the complete journey.

Because of the effects of time dilation, the Earth twin expects the spaceship twin to record a time of = T/ for both the outward and return journey for a total journey time of 2T/ . Hence the Earth twin expects the spaceship twin will be younger by the factor .

13) The Spaceship Twin's Story: The spaceship twin times both the outward and return journey to be hence the total time will be 2 . The spaceship twin observes that time S elapsed on Earth during the outward journey, which is less than because of time dilation. Application of the time dilation formula gives:

Now, immediately after turnaround, the time on the Earth clock, as seen by the spaceship twin located at = , at time = 0, will be given by:

where = is the position of the Earth clock, seen by the spaceship twin. This is just an application of the Lorentz transformation. It has also been assumed that the spaceship twin, when momentarily at rest relative to the star, and hence Earth, observes that the Earth clocks read T. is the time jump, seen on the Earth clock by the spaceship twin, when the spaceship enters the reference frame for the return journey.

Since = = L/ and T = L/v then the previous equation gives:

This is half the missing time between and that is not seen by the spaceship twin because of the turnaround. Hence the total time that the spaceship twin sees elapse on Earth during the journey is:

Hence:

Hence the time the spaceship twin observes to elapse on Earth is the same time recorded by the Earth twin.

It is simple to show, from the length contraction of the distance from the Earth to the star, that the spaceship twin also observes that . Hence the observations made by the twin on the spaceship also show that the spaceship twin will be younger than the earth twin.

14) The two twins agree on the times recorded on each others clocks, the spaceship twin is really younger than the Earth twin. The spaceship twin has traveled in time because of the speed of the journey!

The situations of the two twins are not symmetrical because the Earth twin does not change reference frames half way along the journey. The Earth twin stays reference frame S for the duration, but the spaceship twin changes from to at the turnaround. It was the change of reference frame by the spaceship twin which caused the spaceship twin to return younger than the Earth twin.

15) There is no paradox, this is just a phenomenon of Special Relativity.

16) It might be argued that this problem should be treated with General Relativity, since the twin on the spaceship is in an accelerated reference frame during the turnaround. It can be shown that Special Relativity is relevant to the problem despite this by considering a slightly different, but equivalent, scenario.

Imagine that when the spaceship reaches the star, it does not change its velocity. Instead it continues on a straight path but communicates the time reading of its clock to a second spaceship that is already heading towards Earth with the same speed and is therefore already in the reference frame . Now the total time for the journey is the sum of the time recorded by the original spaceship in the reference frame with the time recorded by the second spaceship in the reference frame. This will be the same as the result already discussed.

17) The actual numbers in the problem were = 7 years and v = 0.96c.

This gives = 7 0.96c = 6.71 light-years, which is the distance to the star seen by the spaceship twin.

Also

Therefore L = = 6.71 3.57 = 24 light-years which is the distance to the star seen by the Earth twin. Then

Which gives

T = 3.57x14 = 50 years.

That is, the spaceship twin ages 14 years during the journey while the twin on Earth ages 50 years.