This explanation is inspired by Paul Davies most excellent book "About Time"
Usually the first place where common sense and the theory of relativity come into glaring conflict is the twins paradox. This is not really a paradox, but it has been given that name because it is usually oversimplified, it is mistakenly thought to be a justification of the theory and it really is contrary to common experience.
The scenario:
Firstly suppose the special theory of relativity is correct:
There are two twins, one goes on a journey in a fast space ship away from Earth then returns to Earth whilst the other twin stays on Earth. When the travelling twin meets up with the Earthbound one, the traveller is younger. This is because the theory says if you look at someone going fast then their time will appear to you to run slower than yours
Now some people think that this is the paradox, but it is not, we are assuming the theory of relativity to be correct and it says if you go fast then time slows, so this is simply an example of the prediction of the theory.
The apparent paradox comes from the second principal of relativity, that there is no absolute frame of reference. This means that the space ship twin can validly say that it is the Earth that has travelled away from her and then back again and that therefore the twin on the speeding planet should be the younger when they meet again. The apparent paradox is that each twin is at the same time older and younger than her sibling. The resolution of the so called paradox has to do with the word "validly".
Albert Einstein is reported to have said "Strive to make explanations as simple as possible, but never simpler than that" and this is a classic case of not explicitly stating some fairly important information in order to make the story simpler.
The first important bit is that the situation is not symmetric, the travelling twin will have undergone acceleration and deceleration while the Earth bound one will not have. To put this more relativisticly, the Earthbound twin has been in one frame of reference all the time, whilst the space traveller has been in two (well more than two, but we will leave out the accelerating and turn around ones)
Secondly we have invoked time dilation but not distance dilation in the explanation. To put this more clearly, let us have a thought experiment on a scale which is reasonably able to be recreated and reasonably able to be observed. We must also remove any acceleration from the measurements so that they comply with the assumption that no acceleration is going on, the whole thing still works but is much harder to calculate if acceleration has to be taken into account.
Suppose we have several trillion dollars and use it to build two space buoys and a fast space probe and two really powerful magnetic rail guns to accelerate and decelerate the probe. We set the buoys up 8 light seconds (2,400,000 km) apart and stationary with respect to each other. The buoys will be called A and B, our probe is a small robot one that is accelerated and decelerated by the two magnetic rail guns, it has no power of its own (so it can't accelerate or decelerate and therefore can't cheat), the rail gun can accelerate the probe to .8 c (240,000 km per second). The probe has a clock that can be activated or deactivated by a signal from either buoy.
The probe is loaded into one gun and accelerated to .8c it zips past A, as it does so A starts its clock and signals the probe to activate its clock too. The probe continues until it passes B, when it is abreast of B, B starts its clock and sends a signal to the probe to de activate its clock. The probe with its clock turned off enters the second rail gun, which brings it to a stop then accelerates it to .8c back out the way it came. This time when abreast B the signal is sent to activate the probe clock, and B's clock is deactivated the probe flies past A and is signalled to deactivate its clock, A's clock is turned off too. The probe decelerates to a stop relative to A, B and the rail guns, and the whole apparatus is collected and brought together. This temporal ballet means that the probe's clock is only on when it is not accelerating and is between A and B, whilst the clocks on A and B operate in such a way that , later on when all the bits are retrieved, the time the probe took to go between them can be calculated.
The time on A's clock is the time it took for the probe to pass A twice first into the experiment zone and then out , the time on B's clock is the time it took for the probe to pass it twice too, first out of the experiment zone and then into it. The probe was travelling at a constant velocity between A and B in each direction so the difference between the times recorded by A and B is the experimental flight time. This should be 20 seconds, because the probe was travelling at .8c and the distance between A and B (which are in the same frame of reference, stationary with respect to each other) is 8 light seconds, thus 10 seconds flight time each way. This is a simple calculation which is the same for Newtonian mechanics because the measurements are taken in the same reference frame.
The probe's clock will have been turned on as it passed A, it will not be accelerating at this point so as far as the probe is concerned, it is A that flashes by at .8c in the opposite direction, sometime later B flashes by too, the probe duly turning on and off its clock as instructed. According to relativity the distance between two objects in the direction of travel reduces with relative speed, for a speed of .8c this factor is .6, so the probe does not measure the distance from A to B to be 8 light seconds, instead it measures 8 X .6 or 4.8 light seconds, which at .8c will take 6 seconds, similarly the reverse B to A trip will take 6 seconds, total time 12 seconds. So, after all the bits are collected into the same frame of reference the duration of the experiment for the probe probe is 12 seconds, 8 seconds less than the buoys experienced.
The probe's observations are qualitatively different to those of the buoys. The probe sees two objects flash by it twice, the buoys each see one object flash by twice. The probe experiences two frames of reference whilst its clock is active travelling in one direction at .8c then the reverse direction at .8c (and it knows that the frame of reference has changed because of the unavoidable and absolute acceleration required to move from one frame to the other or the appearance of the background stars). The buoys experience no acceleration and remain in one frame of reference for the whole experiment.
Note that I have not mentioned time dilation, neither the probe nor the buoys experience dilation themselves, they see it or more accurately calculate it as occurring in a frame of reference other than their own, this factor is again .6 for a speed of .8c, and this effect is mutually calculated for the probe viewing the buoys and vice versa, however they see something different, what is seen depends on the time dilation effect and on the time it takes for light to travel from one place to another. For example suppose that when any clock in the experiment turns on or off, it broadcasts the fact in all directions. When the probe passes A heading towards B a signal is sent to B from A effectively stating that the probe is now at A. Now we are in the A and B frame here so that signal won't reach B for 8 seconds the probe takes 10 seconds to make the distance. So B sees the probe take 2 seconds to make the journey from A to itself, the probe's clock reads 6 seconds as it reaches B, so B sees the probe time speeded up by a factor of 3. On the other hand when the probe passes B, A will receive the information 8 seconds later, the information will say "the probe is level with B its clock reads 6 seconds", this message is sent 10 seconds after A's clock is started so A gets the information "the probe's clock reads 6 seconds" 18 seconds after the probe passes A, thus time is seen to run 3 times slower on the probe than at A. The return journey has the reverse effect, so what is actually seen in the A,B frame is that the probe's time is 3 times faster travelling towards the observing buoy and three times slower when receding from it.
A traveller on the probe (okay so that would take many more trillions of dollars) as it passes B on the out journey would see her clock registering 6 seconds, At this moment A's clock is 8 light seconds away (in A's reference frame), and the probe has taken 10 seconds (again according to A) to get to B. So B got the message "The probe is passing A" 10 - 8 or 2 seconds before the probe got to B, a second clock on B, started when this information was received would be showing "2 seconds since the experiment started" as the probe passes. Thus according to the probe 2 seconds of B's time is seen to have elapsed whilst 6 seconds of probe time has passed, thus the probe sees B's time as being three time faster than its own. Recall that A does not see the probe pass B until 18 seconds of A time into the experiment.
The emphasis in the last two paragraphs is on the word SEE, what is happening NOW depends on whose now you mean, that is the nub of it, time is different and nows are different in different frames of references, the clocks read different but the situation is still logically consistent under the assumptions of relativity. If you accept the premise of the theory then the probe travels 4.8 X 2 or 9.2 light seconds at .8c taking 12 seconds measured by itself and 8 X 2 or 16 light seconds also at .8c taking 20 seconds as measured by A or B
These hypothetical gymnastics should help explain, not the paradox, but how there is no paradox when sufficient information is taken into account, the experience of a twin on the probe and another on a buoy are qualitatively different and so it should not be so surprising that they are quantitatively different too.
This document was created on 23 August 1995
last modified on 23 August 1995
and is written by and copyright to btaylor@taylormade.com.au