YE OLDE TWIN PARADOX

Version 1; 1 July 1994

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PRECIS

Since Albert Einstein invented the special theory of relativity in 1905, there has been a great deal of hot air blown about the so-called Paradox of the Twins. Here is some more.

My thesis is that there is no paradox at all, and that the whole wuzzle is caused by failure to pay attention to the way the problem is set up.

This article is set out in ten sections:

PRECIS, which you are reading;

BASICS, which briefly describes relatively moving coordinate reference frames, events, and the Lorentz transforms which enable one to calculate the coordinates of events in one frame given the coordinates in another frame. Length contraction and time dilation are derived;

SIMULTANEITY, SEQUENTIALITY, AND CAUSALITY, which discusses these topics and the strange meanings they assume in Lorentzian spacetime;

AGE, which gives the definition of how observers in various frames of reference age between two events;

WUZZLE, which states the twins problem and describes the observations one observer in relative motion with respect to another can make which can lead to the conclusion that "the other guy is always younger;"

TWINS, which and shows why the Twins Paradox isn't a paradox;

NUMERICAL EXAMPLE, which discusses a journey between Earth and Proxima Centaurus at v=0.99c;

QUESTIONS, which discusses the various questions which arise in most discussions of the twins problem; and

MORE NUMERICAL EXAMPLES, which describes return journeys of various kinds, and is included for those who are not convinced by the previous discussion;

SUMMARY, which briefly restates the conclusions of this article.

None of the mathematics is beyond high school level. All of the special relativity is standard stuff which can be found in any decent textbook.

If you find any errors, typographical or otherwise, please let me know.

BASICS

Lorentz Reference Frames and Spacetime Coordinates

In Lorentz reference frames, the spacetime coordinates of events are given by (x,y,z,t), where x,y, and z are space coordinates, and t is the time coordinate.

S and T are two Lorentz frames in relative motion. The space axes of both frames are co-planar. The x axes are co-linear. The velocity of T with respect to S is v, and of S with respect to T is -v.

Event coordinates in S will be given as (x,y,z,t), and individually as x, y, z, and t. Where necessary for absolute clarity, (x,y,z,t)S will be used to indicate that the coordinates refer to the S frame. Likewise event coordinates in T will be given as (x',y',z',t'), or (x',y',z',t')T, and individually as x', y', z', and t'.

Whenever a variable is dashed, e.g., w', its values are to be measured in the T frame. The values of un-dashed variables are to be measured in the S frame.

Frames S and T are chosen so that the origins (0,0,0,0)S and (0,0,0,0)T are the same event.

The dilation factor is called k, and k=sqrt(1-v^2/c^2). Note that v0 then k<1.

sqrt() means the positive square root of the quantity in brackets.

a^b means a raised to the power b.

a*b means a times b. The * is omitted when context allows, so ab will usually mean a times b.

a<< b means a is less than b, a>b that a is greater than b.

a<=b means a is less than or equal to b, a>=b that a is greater than or equal to b.

Occasionally it is necessary to refer to two different times or positions measured in the same frame. This is done by using, for example, t1 to mean the first time, and t2 to mean the second, or t1' and t2' if the times are measured in the T frame.

Lorentz Transforms

The Lorentz transforms describe the basic properties of Lorentzian spacetime. They enable one to calculate the coordinates (x',y',z',t')T from the coordinates (x,y,z,t)S of the same event in S. The transforms are:

x'= (x-vt)/k and, inversely, x = (x'+vt')/k

y' = y and, inversely, y = y'

z' = z and, inversely, z = z'

t' = (t-vx/c^2)/k and, inversely, t = (t'+vx'/c^2)/k

Since the y and z transforms are identities, and all the work in this article will concern events on the x axes, I will usually shorten event coordinates to (x,t).

(x,t)S <-> (x',t')T means that event (x,t) in S maps to (x',t') in T, and vice versa. (x,t)S <-> (x',t')T if and only if they are the coordinates of the same event.

Events and Things

An event is something which happens at a particular place (x,y,z) and at a particular time t. This definition applies no matter whether we are considering events in Euclidian space and time or Lorentzian spacetime. In both cases an event will have unique coordinates in each relatively moving coordinate system, or frame of reference. In Euclidian space and time, the coordinates of an event will always have the same time coordinate, no matter what the speed between the reference frames its coordinates are measured in. In Lorentzian spacetime, both the space and time coordinates of the same event will usually be different in relatively moving frames.

It is important not to confuse events with things. Things are sets of events which persist through time, though not all sets of events which persist through time are things. What a thing looks like depends on the relative speed between the thing and the person making the observation.

Length Contraction

For example, consider a rigid rod with its left hand end at x=0 in S, and its right hand end at x=1 in S. From the perspective of an observer in S, that rod looks like a line segment 1 unit long lying at rest on the x axis. From the perspective of an observer in T, that thing looks like a moving line segment. The length contraction effect is that the observer in T will see the rod as shorter than one unit long.

When people make a simple observation of a thing, they see all of it at the same time. I say "simple observation" because people also talk about observing more complicated things which consist of a sequence of events through time, like a car moving along a road, or a discussion between other people. When the observer in S looks at the rod, say at time t=0, he sees the ends as events (0,0)S and (1,0)S. The distance between the ends is the difference between the x coordinates, which is 1.

When an observer in T looks at the rod, he is looking at a moving object. Suppose he is looking at time t'=0. His observation of the left hand end will be the event (0,0)T, because the Lorentz transforms give the same numerical coordinates of this event in both frames. His observation of the right hand end is another event with t' coordinate equal to zero. He will see the right hand end of the rod where it was when t'=0. This is not where it was when t (in the S frame) was 0, because the rod is moving from the T observer's perspective, and because time coordinates for events as seen from the two frames relate to each other through the Lorentz transform.

We can calculate the x' coordinate of the right hand of the rod event at t'=0 as follows:

We know that the right hand end of the rod is always at x=1 in S.

We know that the time the observer in T is looking at the rod is t'=0 in T.

We wish to find x' in T.

The Lorentz transform connecting these three is

x = (x'+vt')/k

so we substitute x=1, t'=0 and get

x'=k

Now we know that the observer in T sees the left end of the rod as event (0,0)T and the right end as (k,0)T. Subtracting the space coordinates gives the length of the rod as seen from T, namely k.

This is the famous length contraction effect. The observer in T sees the rod at rest in S as shorter than it appears to be in S. How much shorter depends on the relative speed with which T is moving with respect to S, speed rather than velocity, because the velocity term plus or minus v appears in k as a squared term, which is always positive.

This argument is reversible. If we had started with a rod of length 1 in the T frame, we would have concluded that observers in S will see it as of length k, shorter than 1.

The most interesting thing about the length contraction effect is that it arises because the two observers are not comparing the same events when they "observe" the length of the rod. The S observer is comparing (0,0)S and (1,0)S. According to the Lorentz transforms, (0,0)S <-> (0,0)T, and (1,0)S <-> (1/k,- v/(kc^2))T, but the T observer is not looking at (1/k,- v/(kc^2))T when he observes the right hand end of the rod. He is looking at (k,0)T, so it is not too surprising that he thinks the rod has a different length.

Length Expansion (an aside)

One may ask what happens if the observer in T tries to make exactly the same observation as the observer in S does. He is then comparing (0,0)T with (1/k,-v/(kc^2))T. These are not simultaneous events for him, but that does not always matter when measuring distance.

If instead of thinking of (0,0)S and (1,0)S as being events at the ends of a rod, one thought of them as being flashes emitted by two fireflies one unit apart in the S frame, then the observer in T could ask himself what the distance between the two fireflies was in the T frame, and get the perfectly reasonable answer 1/k by subtracting the distance coordinates of the events. 1/k > 1, so the observer in T would see the distance between the fireflies as larger than the observer in S does. Using observations made in this way, one could develop an argument for length expansion due to relative motion instead of the conventional arguments for length contraction.

Time Dilation

The time dilation effect can be derived from the properties of Lorentzian spacetime in much the same way as we derived the length contraction effect. Where an observer in S considers a time interval which starts at (0,0)S and ends at (0,1)S, an observer at x'=0 in T will see an interval which starts at (0,0)T and ends at (0,t')T. We may calculate t' from the Lorentz transform connecting x'=0, t=1, and t' which we wish to find. The appropriate transform is t = (t'+vx'/c^2)/k, and substitution gives t'=k. The observer in T sees time intervals which require 1 time unit in S to require only k time units in T.

This argument too is reversible. For observers in S, time appears dilated in the T frame.

Again the interesting thing is that the two observers are not comparing the time interval between the same events. (0,1)S <- > (-v/k,1/k)T, and (0,k)T <-> (v,1)S. The observer in S is comparing (0,0)S and (0,1)S, while the observer in T is comparing (0,0)S with (v,1)S.

Time Expansion (another aside)

As with the length expansion argument above, one may ask what happens if the observer in T tries to make exactly the same observation as the observer in S does. He is then comparing (0,0)T with (-v/k,1/k)T, corresponding to the Lorentz transformations of (0,0)S and (0,1)S respectively. In T the time difference between these two events is 1/k. 1/k > 1, so time appears to expand.

Again a firefly analogy may illustrate what is happening here. This time there is one firefly which flashes twice, first at (0,0)S and then at (0,1)S. As the Lorentz transform shows, the observer in T sees an elapsed time of 1/k between the flashes instead of the time of 1 which the observer in S sees.

Observations

The above discussion describes how the length contraction and time dilation effects of special relativity in Lorentz spacetime come about. They also point up the difficulties inherent in discussing observations of the same events from two relatively moving Lorentz frames.

Most accounts of relativity describe these difficulties in terms of the disagreements which relatively moving observers have concerning simultaneity. That approach seems to me to miss the fundamental point, which is that two good observers ought to compare like with like. If they both make the same observation, then their records of the observation should concern the same events as related by the Lorentz transforms. Traditional accounts of relativity do not discuss observations of the same events in this sense: they discuss similar types of observations, or observations of different event patterns which are close to each other in spacetime. This similarity and closeness are not the same as identity. The difference is what gives rise to the confusing and often spurious discussions of the effects of length contraction and time dilation in discussions of the twins problem.

SIMULTANEITY, SEQUENTIALITY, AND CAUSALITY

Simultaneity is not a well defined concept in relativistic spacetime. The intuitive meaning of simultaneity is as follows: two events are simultaneous if the values of their time coordinates are numerically equal. Given two events E1 and E2, with coordinates (x1,t1)S and (x2,t2)S, observers in S will claim that the events are simultaneous if t1=t2.

This definition will work for all Euclidian frames of reference, because these are related by Galilean transforms which transform time coordinates from one Euclidian frame to another as an identity. Hence it is fair to say that simultaneity is well defined for Euclidian space and time: simultaneity is invariant from one frame to another.

Simultaneity is not well defined in Lorentzian spacetime. Observers at rest in one Lorentz frame may consistently use the Euclidian definition of simultaneity, provided they restrict themselves to considering events identified only by spacetime coordinates applicable to their rest frame. It is not possible to translate two different simultaneous events in one Lorentz frame to simultaneous events in any another Lorentz frame. The Lorentz transforms to not preserve this kind of simultaneity as we have seen in the explanation of length contraction and time dilation above. It follows that one cannot use the Euclidian definition of simultaneity when discussing events from the perspectives of different Lorentz frames.

The situation is in fact worse than this. In spacetime the concepts of "before" and "after" do not always transform either, because the time transform from one frame to another depends on position as well as time, so that an event which succeeds another in one frame may well precede it in a different frame.

The upshot of all this is that "normal" thinking about the succession of events and about simultaneity must be completely abandoned when discussions about relativity are undertaken. This comes as a deep psychological shock to most people.

New concepts are needed for spacetime which roughly correspond to the ideas of "simultaneity," "before," and "after" in Euclidian space and time.

If we consider two events E1 (x1,t1)S <-> (x1',t1')T, and E2 (x2,t2)S <-> (x2',t2')T and enquire into the conditions necessary for t1' to be <= t2', using the Lorentz transforms we find that t1' <= t2' whenever v(x2-x1)/(t2-t1) <= c^2. If we impose the further condition that t1 <= t2, we have the requirements for preserving the "before," and "after" concepts, though we cannot use these terms for them because, first, the new concepts will not apply to all events as they do in Euclidian space and time, and, second, the new concepts will not apply to all Lorentz frames, only to a subset of them.

I use the following concepts:

Sequentiality:

Two events E1 (x1,t1)S <-> (x1',t1')T and E2 (x2,t2)S <-> (x2',t2')T are sequential in frames S and T if and only if t1 <= t2 and t1' <= t2'. This occurs if and only if v(x2-x1)/(t2- t1) <= c^2, or, equivalently, if and only if v(x2'-x1')/(t2'- t1') <= c^2.

Simultaneity in frame S:

E1 and E2 are simultaneous in S if and only if t1=t2.

Simultaneity in frames S and T:

E1 and E2 are simultaneous in frames S and T if and only if they are the same event.

The above definitions of simultaneity require that the frames in which events are judged to be simultaneous be clearly specified.

Causality:

In addition to the problems of simultaneity and sequentiality, causal relations are extremely important in physics. Causal relations between two events are only possible when light signals, or signals which travel slower than light, can travel between the two events. Light can only travel between E1 and E2 when c^2(t2-t1)^2 - (x2-x1)^2 >= 0, or, equivalently, when c^2(t2'-t1')^2 - (x2'-x1')^2 >= 0. (It should be noted that y and z coordinates, in whatever frame, are irrelevant, because they transform as identities.)

Those familiar with relativity will recognise the last relationship as the square of the interval or separation between two events. Whenever the interval is real, one event may (or may not) have a causal effect on the other. When the interval is imaginary, the events cannot have causal effects on each other.

Some care must be taken with the causality concept.

First, two events which cannot have causal effects on each other may both be the result of a cause at some other event.

Second, events which may have a causal effect on each other may not necessarily do so. A causal relationship between E1 and E2 requires that there be physical laws connecting the two events, such as the law that distance equals speed times time. In order to determine that two events which may be causally related by the above rules are in fact so related, the physical laws involved must be specified. To claim a causal relationship between events E1 and E2 therefore requires that the events meet the c^2(t2-t1)^2 - (x2-x1)^2 >= 0 condition, and that the physics which connects E1 and E2 be specified and shown to imply that E2 follows from E1.

One of the postulates of relativity is that physical laws have the same form in all frames. It is only necessary to show that a physical law applies in one frame to imply that it applies in all frames.

(As an aside, the condition that there be a physical law connecting two events does play hob with experimental relativistic physics. If one is experimenting to find a new law, then to say that a law must obtain before one can compare two events as to whether they are causally related is rather a restriction. The experimental physicist may escape this dilemma by adopting the approach of hypothesising his physical laws. Truth will out by the usual means.)

To summarise Simultaneity, Sequentiality, and Causality:

Two events are simultaneous in any frame S if and only if they have the same time coordinate.

Two events are simultaneous in frames S and T (which have an non zero relative velocity) if and only if they are the same event.

Two events are sequential in frames S and T if and only if the event which occurs later in time in S also occurs later in time in T.

Two events may be causally related if it is possible to send a light signal, or a signal travelling slower than light, between one event and the other in all frames.

Two events are causally related if they may be causally related and the physical law connecting the events is clearly specified.

AGE

Much controversy arises over the definition of age.

If two events E1 and E2 occur, then the elapsed time between them depends on the coordinates of the events in whatever reference frame they are measured. If the events have coordinates (x1,t1)S and (x2,t2)S, then the elapsed time in S is t2-t1. Any observers at rest in S will age by t2-t1 between the events. This argument applies to any frame. In T, the event coordinates will be (x1',t1')T and (x2',t2')T, and all observers at rest in T will age by t2'-t1' between the two events.

The ageing of an observer of a physical experiment is also determined by the difference between the time coordinates of the end event and the start event in the observer's rest frame of reference. The reason for this is simple. The elapsed time between the start event and the end event of an experiment are related by the simultaneity rules which apply to that frame and by the physical laws which require the end event to occur when and where it does. Observers at rest in different frames will note different elapsed times. These elapsed times are a consequence of the structure of Lorentzian spacetime as determined by the Lorentz transforms which define that structure.

How much one ages between the occurrences of two events equals the elapsed time between the events in the frame of reference in which one is at rest. This is the only consistent definition of ageing allowed by either the general or special theories of relativity. Time passes "at the same rate" everywhere in a single inertial frame. It does not matter how far apart in space (the difference between the x coordinates) is: Einstein clocks will measure the same time difference between the two events from any position in the same frame.

WUZZLE

The Twins Problem is usually stated something as follows:

Two twins are growing up together, side by side, at rest in the same frame of reference. One twin embarks on a journey and travels for a while. During this journey, both twins observe each other to be travelling under contracted time. The question is, which twin is aging more slowly?

To model this problem mathematically, suppose the twins spend their time together at x=0 in frame S. To keep the arithmetic simple, they separate at event (0,0)S <-> (0,0)T, one moving to a rest position at x'=0 in frame T, the other remaining at rest at x=0 in frame S. It also simplifies things to assume that they slipped out of the birth canal at exactly the same time and spent an infinitesimal amount of time together before the journey began, so that they were both aged zero (i.e., just born) when the journey began.

The physical law governing the journey, or experiment, is the distance = speed * time relationship.

It is at this point in the problem when the furious arguments begin. These arguments revolve around the fact that each twin observes time to be dilated in his sibling's frame. All observations are symmetrical, so it appears impossible to tell which twin is aging more slowly.

The arguments are made worse by the difficulty in getting either of the twins to agree what is meant by the term "observation," given that there are a number of ways human beings can make observations, or to agree as to exactly what events should be observed as was discussed above. Further complications arise because of the interconnection of space and time which leads to numerical values of event coordinates as measured in different frames in ways which go counter to everyday expectations.

If we stop thinking-- something all to easy to do at this stage in the discussion, there is nothing ahead of us except endless, unfruitful debate. The problem has to have an endpoint: at what spacetime event does one want to compare the twins ages?

TWINS

It is clear that twins separating from a common starting event will each "observe" the other to be aging more slowly, and to be shorter along their x dimension than he is himself. Relativity states that moving observers age more slowly and have their lengths contracted than observers at rest. So, since either twin can argue that it his sibling rather than himself who is in motion, how is it possible to decide which twin will be the younger when the journey is over?

The answer is simple: it depends on the event which defines the journey's end. Once that event is agreed, it will have coordinates in both reference frames, and the age of each twin can be uniquely determined.

If the "journey's end" event is (x,t)S <-> ((x-vt)/k,(t- vx/c^2)/k)T according to the Lorentz transforms. The twin at rest in S will have aged by t, and the twin at rest in T will have aged by (t-vx/c^2)/k. Since the "journey's start" event was (0,0) in both frames, the elapsed time is obtained by subtracting zero from the "journey's end" event time coordinate.

This argument is reversible. If the "journey's end" event is given in coordinates from the T frame, (x',t')T, we have (x',t')T <-> ((x'+vt')/k,(t'+vx'/c^2)/k). The twin at rest in T will have aged by t', while the twin at rest in S will have aged by (t'+vx'/c^2)/k.

NUMERICAL EXAMPLE

In the 1993-94 discussions of the twins problem on the Science / Maths CompuServe forum, the twins have usually been taken to have been born on Earth at (0,0). One twin departs for Proxima Centaurus, the nearest star, at rest with respect to Earth, and four light years away from Earth. Both Earth and Proxima Centaurus are at rest in the S frame, so their x coordinates are respectively 0 and 4 light years in the S frame. The travelling twin has been taken to be at rest at x'=0 in the T frame.

In the forum discussions, the x axes have been aligned to connect Earth and Proxima, with the positive direction being from Earth towards Proxima. v has been taken to be 0.99c in the S frame, -0.99c in the T frame, which makes k = 1/7.09, about one seventh.

Numerical work in this section will usually be accurate to two decimal places. Distances will be given in light years, and times in years.

According to the laws of physics in the S frame, the travelling twin's journey of 4 light years at 0.99c will require 4.04 years so that his arrival at Proxima will be the event (4,4.04)S, units being light years for x distances, and years for time. The Lorentz transform of this "journey's end" event to the T frame, in which the travelling twin is at rest is (0,0.57)T. Between the start and the end of the journey, 4.04 years have elapsed in the S frame, and 0.57 years (around 7 months) have elapsed in the T frame.

The conclusion is clear. The twin who has travelled from Earth to Proxima is 0.57 years old at the journey's end, because only 0.57 years have elapsed between the start and the end of the journey in his frame. The stay at home twin is 4.04 years old.

QUESTIONS

Four questions often arise at this juncture:

The journey was specified according to the laws of physics in the S frame. What did the journey look like from the perspective of the twin in the T frame?

What happened to the famous symmetry of observation argument, whereby either twin could legitimately claim that his sibling was in motion and therefore aging more slowly?

The way I have stated the problem, neither twin underwent any acceleration: one "jumped" instantly from the S frame into the T frame. Doesn't acceleration have something to do with the problem?

The twins problem is usually stated in terms of one twin travelling from Earth to Proxima, and then returning, comparing their ages when they are reunited. Why hasn't that been considered here, and what happens if the problem is considered this way?

Consider these questions in turn.

The journey was specified according to the laws of physics in the S frame. What did the journey look like from the perspective of the twin in the T frame?

The usual answer is that at t'=0, Proxima appears to the traveller to be a star 0.56 light years away rushing towards x'=0 in the T frame at v=-0.99c. The distance, speed, and time relationship indicate that the journey will require 0.56/0.99 = 0.57 years. This is correct, but it begs some interesting questions.

What is Proxima Centaurus?

It is a star, so far as we know the nearest star to the sun, about four light years away from the sun, and for practical purposes at rest in the S frame, so that in the S coordinate system, Proxima is the set of all events at (4,t)S, for any time t at all. This is definitely not the case for an observer in T, for whom Proxima is moving. As time in T changes, Proxima will approach and then recede from x'=0.

What are we seeing when we look through a telescope and "observe" Proxima?

Light from Proxima takes four years to reach the Earth, so that if an observer on Earth at x=0 took a look at Proxima at t=0, he would see the photons from an event which happened at (4,-4)S. Since (4,-4)S <-> (56.43,-56.43)T it must be the case that the travelling observer, at x'=0 in the T frame, also looking through his telescope at t'=0, when he was coincident with the Earth observer, would see the photons from the same event. In the T frame this event has coordinates (56.43,- 56.43)T, and so was 56.43 light years away from the observer in T, and occurred 56.43 years ago, according to T frame reckoning. (There is no way to escape this conclusion: both observers must be observing the same group of photons.)

When on Proxima, did the journey begin?

The journey began at t=0 in S and t'=0 in T. Observers in S reckon simultaneity in such a way that the event (4,0)S on Proxima is simultaneous with (0,0)S on Earth. (4,0)S <-> (28.36,-28.07)T, so to the S frame way of reckoning simultaneity, the journey begins simultaneously with the event (28.36,-28.07)T.

Unfortunately, simultaneity is reckoned differently in the T frame. Since the traveller left Earth at t'=0, the simultaneous event on Proxima must also have time coordinate t'=0. Since lengths are contracted by factor k, the four light years in the S frame contracts to 0.56 light years in the T frame, and the event on Proxima simultaneous with the traveller's departure is (0.56,0)T. Note that (0.56,0)T <-> (4,3.96)S.

As was discussed earlier, it is a theorem of special relativity that simultaneity is reckoned differently in different Lorentz frames, and that no two different events which are simultaneous in one frame can be simultaneous in any other frame. But very different things are happening on Proxima at (4,0)S and at (4,3.96)S, so the difference in reckoning might really mean something. Does it? How could we tell?

In physics, the only way to discover whether a difference is meaningful or not is to ask whether it has causal implications. If there are causal implications, then the difference is meaningful. If there are no causal implications, then the difference is of no consequence. So how can we decide whether there are possible causal implications arising from the difference in reckoning of the twins separation time from the point of view of an observer on Proxima?

Fortunately, there is an easy answer to this question. Two events can only be causally linked if light speed or sub light speed signals can travel between them. A difficulty with causality could arise if one of the twins could discover an event on Proxima, and pass information about this event to his sibling so that the information arrived before the event happened in the sibling's frame.

Lorentzian spacetime is designed so that this can't happen: signals originating from one event and sent to another place always arrive at the same event regardless of the frame in which the signal travel is analysed.

The proof is straightforward.

Suppose the traveller signals Earth from (0,t')T. Applying the Lorentz transforms, we have (0,t')T <-> (vt'/k,t'/k)S, so, in S we must analyse a signal which is travelling from x=vt'/k to x=0, leaving at time t=t'/k. The distance the signal must cover is vt'/k. At c, this takes S frame time vt'/(kc), so the signal arrives at S frame time t'*sqrt( (c+v)/(c-v) ), and the signal arrival event is ( 0, t'*sqrt( (c+v)/(c-v) ) )S.

In T, Earth is moving with velocity -v. When the signal is emitted, Earth is distance vt' from the traveller. The signal takes a time w' to reach the Earth: v(t'+w')=cw', so w'=vt'/(c-v). In the T frame, the arrival time t'+w' is t'(c/(c-v)). At this time, Earth has arrived at position - vt'(c/(c-v)). Applying the Lorentz transforms to the signal arrival event as analysed in T gives

(-vt'(c/(c-v)),t'(c/(c-v)))T <-> (0,t'*sqrt((c+v)/(c-v)))S

which is exactly the same result as when the signal was analysed in S.

The interesting case in our example is where the travelling twin reckons that he is departing simultaneously with event (4,3.96)S, which is 3.96 years later than his Earth twin reckons. If the departing traveller could discover, say, the result of a horse race on Proxima, which took place any time between t=-4 years and t=3.96 (S frame reckoning), he could pass this information to his twin in the S frame who could place a bet about events on Proxima at time 0 (S frame reckoning) and win, because he had successfully predicted an event which happened before a light signal from Proxima could notify other Earthmen of its occurrence.

At t'=0, the travelling twin is receiving signals from Proxima which were sent at t'=-56.43 years when Proxima was at x'=56.43 light years, as was noted in the discussion of telescopic observations of Proxima: (56.43,-56.43)T <-> (4,- 4)S. The traveller is receiving exactly the same information from Proxima as the stay at home is.

Information from Proxima about events at, or prior to (4,3.96)S <-> (0.56,0)T will not reach the traveller until t'=0.56, when his coordinates are (0,0.56)T and he has almost reached Proxima. If he then tries to forward this information to his sibling on Earth, it will not arrive until (-55.86, 56.43)T <-> (0,7.96)S, which is exactly when the Earth twin would expect to receive it, 4 years S frame time after it was sent.

The analysis for signals which travel at sub luminal speeds is similar, and leads to the same conclusions.

The second question was:

What happened to the famous symmetry of observation argument, whereby either twin could legitimately claim that his sibling was in motion and therefore aging more slowly?

The answer here again depends on stating the "end of journey" event for the problem. The entire argument about what the traveller does and what the stay at home does can be turned on its head by considering a journey by the Earth twin to "anti Proxima," an imaginary destination at rest at (-4,t')T in the T frame. The twin who used to be the traveller can now argue that his sibling is travelling from (0,0)T to (-4,4.04)T. (The negative sign is necessary because the Lorentz frames are receding from each other in opposite directions.)

The symmetry is preserved, because the former stay at home will arrive at anti-Proxima after 0.57 years, and all the other arguments can be reversed in the same manner, care having only to be taken to use -v for +v in the calculations, so that the former traveller is 4.04 years old when the arrival at anti-Proxima occurs.

This is where the fun begins.

In terms of the twins' original adventure, where the twin in the T frame (call him the T twin, or twin T for short) was going to Proxima, the physics of the journey required that he age continuously from t'=0 to t'=0.57 during the journey. While the journey went on, the twin in the S frame (S twin, or twin S) aged continuously from t=0 to t=4.04. When we try to consider the S twin's journey to anti-Proxima coincidentally with the T twin's journey to Proxima, we seem to hit on a nice contradiction: according to the simultaneity rules for the S frame, the S twin completes his journey to anti-Proxima long before (4.04-0.57 = 3.47 years before) he (the S twin) considers that the T twin's journey to Proxima is complete. Yet, at this arrival at anti-Proxima event, according to the simultaneity rules which apply to the T frame, the T twin must consider himself to be 4.04 years old. According to the physics of the original journey (T twin going to Proxima) this means that the journey to Proxima must be complete, beca use it ended when t'=0.57.

For logicians, the essential point here is that simultaneity is not defined across relatively moving frames of reference in Lorentzian spacetime, except for identical events. This makes the argument invalid: there is no contradiction; spacetime does not obey the rules which make what seems like a contradiction into a real contradiction.

For people not familiar with mathematical logic, the coincidental consideration of the two journeys, T twin to anti-Proxima and S twin to Proxima, still "feels" like a contradiction. How can one twin's journey end when he is 4.04 years old when he has not completed another journey, which started at the same time, and which only takes 0.57 years?

The only answer is to control one's feelings. Spacetime is not structured in a way which allows the contradiction to be real. True conclusions can only be reached from true premises, and one of the premises in the argument is false: the one about the commonsense notion of simultaneity.

It may help some readers to re-consider the section of this article about simultaneity, sequentiality, and causality. They should note that the events (0,0.57)T, the T twin's arrival at Proxima, and (0,0.57)S, the S twin's arrival at anti-Proxima, are not sequential, and are not causally related, though they have a common cause in the event (0,0)S <-> (0,0)T.

The third question was:

The way I have stated the problem, neither twin underwent any acceleration: one "jumped" instantly from the S frame into the T frame. Doesn't acceleration have something to do with the problem?

The answer is no. Acceleration has nothing to do with the problem. The easiest way to see this is to consider greater and greater accelerations, until in the mathematical limit the transition from the S frame to the T frame happens instantaneously. The only effect this has on the calculations is to gradually decrease the time required for the journey in both frames to the minimum where the traveller spends the entire journey at maximum velocity.

A detailed analysis may be obtained by deriving an expression for the acceleration dv/dt in S relative to the co-moving frames through which the accelerating traveller passes on his way to maximum velocity. This can easily be integrated up to obtain distances from Earth as a function of elapsed time in the Earth rest system, velocities of the traveller as a function of that elapsed time, and so on.

There are two reasons why people arguing about the twins bring acceleration into the problem. One reason is to distinguish the traveller from the stay at home, on the grounds that the traveller is the one who has accelerated. This is unnecessary. The traveller is the one who is journeying between places at rest in the stay at home frame. Acceleration is not required to make this distinction. The other reason is to bring general relativity into the problem, possibly because few people feel capable of arguing about general relativity because they can't cope with the rather torturous mathematics involved. This reason is entirely spurious. Special relativity is part of general relativity; simple algebra and calculus apply to both. General relativity must agree with any conclusions reached from special relativity. The special theory is a subset of the general, and nothing accomplished by using the special theory is invalidated by the existence of the general theory.

The final question was:

This question has been left till last because it is not strictly relevant to the solution of the problem. The twin who is travelling between places at rest in another reference frame arrives younger than the twin who stayed at home. Less time has passed in the travelling frame than passed in the other frame. If the travelling twin arrives at Proxima and then returns to Earth, the same thing will happen: he will age less during the journey than inhabitants of the frame in which the point of departure and point of arrival are defined.

Readers who require more mathematical precision will possibly not be happy with the second sentence in the previous paragraph. A more precise expression would be to say that the twin for whom the position coordinate x of the start and end event are the same will age less than the twin for whom the position coordinate of the two events are different.

MORE NUMERICAL EXAMPLES

This section is included for those who like numerical discussions because they are not convinced by algebra alone. Three problems are considered. In all cases A is the observer who remains at rest on Earth in S for the initial stages of the problem.

First, what happens when a traveller B goes from Earth to Proxima, and then returns in a frame U which is initialised so that (0,0)U <-> (0,0)S and which is moving at relative velocity v = -0.99c relative to S?

Second, what happens when B goes to Proxima and then jumps to the S frame, A goes to anti Proxima and then returns in frame T to B, now at rest in frame S?

Third, what happens when B goes from Earth to Proxima, transfers to U, and A goes to anti Proxima, transfers to T, and the pair meet?

The idea here is to give an analysis of each problem from the point of view of an observer in each of the different reference frames, and to show that the various analyses give the same results.

Before we start to discuss these problems, let us clarify the coordinates of the relevant events in the different frames. In each case the coordinates of an event in the various frames are obtained from the Lorentz transforms. T is moving with respect to S with v=0.99c. U is moving with respect to S with v=-0.99c. T is moving with respect to U with velocity u=0.999949c, which velocity is obtained by adding |v| and |-v| using the relativistic formula for the addition of velocities, u= (v1+v2)/(1+v1v2/c^2). This formula has not been discussed in this article, and the reader is referred to any good textbook on relativity for the derivation.

To extend the notation adopted at the beginning of this article, coordinates which are defined in the U frame are indicated by x", t" etcetera.

The event which corresponds to B leaving Earth in the T frame for Proxima, A (who stays on Earth) leaves B in the S frame for anti Proxima, and the U frame is initialised, is (0,0) in all three frames.

The event which corresponds to B's arrival at Proxima is (4,4.04)S <-> (0,0.57)T <-> (56.71,56.71)U.

The event which corresponds to A's arrival at anti Proxima is (0,0.57)S <-> (-4,4.04)T <-> (4,4.04)U.

Another Aside

Introducing the third frame U provides an opportunity to look at the various journeys from a new point of view. B's journey to Proxima may be considered as a journey from (0,0)U to (56.71,56.71)U at v=0.999949c. During the journey, B will still age 0.57 years, but an observer at rest in U will age 56.71 years.

Similarly A's journey to anti Proxima may be considered as a journey from (0,0)U to (4,4.04)U at v=0.99c. Numerically, A as seen from U going to anti Proxima is making the journey as B seen from S going to Proxima. A ages 0.57 years, and an observer in U ages 4.04 years during the journey.

The Three Problems

Our first problem was, what happens when a traveller B goes from Earth to Proxima, and then returns to Earth in a frame U which is initialised so that (0,0)U <-> (0,0)S and which is moving at relative velocity v = -0.99c relative to S?

The analysis requires B, aged 0.57 by virtue of the time he has spent in T going to Proxima, to jump instantly to frame U, joining it at (56.71,56.71)U, and then resting there until x"=56.71 in U meets x=0 in S. This occurs when x"=56.71 (because B has stayed at rest in the U frame for the return journey) and at t"=57.28, i.e., at event (56.71,57.28)U <-> (0,8.08)S which may be confirmed by the Lorentz transformation. B has thus aged 57.28 - 56.71 = 0.57 years in the U frame. Add that to the time he spent in the T frame going to Proxima, also 0.57 years, and the conclusion is that B is 0.57 + 0.57 = 1.14 years old when the twins reunite. A on the other hand is 8.08 years old. B has travelled, A has not, so B is the younger.

Our second problem was, what happens when B goes to Proxima and then jumps to the S frame, A goes to anti Proxima (which is the same thing as staying on Earth for 0.57 years) and then returns in frame T to B, now at rest in frame S at Proxima?

B arrives at Proxima at (0,0.57)T <-> (4,4.04)S, aged 0.57 years and jumps to the S frame at (4,4.04)S. From then on B is at rest at x=4 in S.

A jumps to the T frame from Earth at anti Proxima at (- 4,4.04)T <-> (0,0.57)S when he is 0.57 years old. He travels towards (4,t)S, at 0.99c, which requires 4.04 years of S time, or 0.57 years of T time. A arrives at (4,4.04+4.04)S = (4,8.08)S <-> (-4,4.04+0.57)T = (-4,4.61)T. On reuniting with B, A is 1.14 years old, having spent 0.57 years in the S frame, and 0.57 years in the T frame. B is also 1.14 years old, having spent first 0.57 years in the T frame and then 0.57 years in the S frame while A travelled to him.

Both twins have travelled the same distance at the same speed, so they are the same age.

The third problem was, what happens when B goes from Earth to Proxima, transfers to U, and A goes to anti Proxima, transfers to T, and the pair meet?

This is the most complicated problem, but the solution is straightforward.

In S, A is starting from event (0,0.57)S travelling at +0.99c, and B from (4,4.04)S, travelling at -0.99c. They will meet halfway, i.e., when t = 0.5*(4/0.99+0.57+4.04) = 4.33 to two decimal places. Using the Lorentz transforms, we find that this occurs at x = 3.72 for both A and B, so the reuniting event is (3.72,4.33)S <-> (-4,4.57)T <-> (56.71,56.75)U.

Analysed either from A's point of view in the T frame, or B's point of view in the U frame, the velocity of approach is +/- 0.999949c, but the conclusion is the same. They meet at (- 4,4.57)T <-> (56.71,56.75)U <-> (3.72,4.33)S.

In T, B departs (0,0.57)T and makes a 4 ly journey to A at x'=4, travelling at v = -0.999949c. This requires 4.0002 years of T time, so the arrival event is indeed (4,4.57)T. k=99.02 for |v| = 0.999949c, so B sees this time contracted to 0.04 years.

In U, A departs (4,4.04)U and makes a 52.71 ly journey to B at x"=56.71 travelling at v = 0.999949c. This requires 52.71 years of U time, so the arrival event is (56.71,56.75)U. Again A sees this time as contracted, in this case to 0.53 years.

When they reunite, A has spent 0.57 years in S and 4.57-4.04 = 0.53 years in T, and so is 1.10 years old. B has spent 0.57 years in T, and 56.75-56.71 = 0.04 years in U, and so is 0.61 years old.

Note that both A and B are younger than an observer who stayed at rest in S between (0,0)S when the twins parted, and (3.72,4.33)S when they reunited.

There is a lot of fun to be had playing around with this third example.

SUMMARY

The twins problem is not a paradox. A traveller between two events whose coordinates are (x1,t1)S and (x2,t2)S who is at rest in another inertial frame T will always observe less time to pass during the journey than inhabitants of the S frame. If the coordinates of these events in the T frame are (x1',t1')T and (x1',t2')T, then t2'-t1' will always be a shorter time interval than t2-t1.

The reasons why the twin problem seems paradoxical reside in the imprecise use of language when the problem is described. Two imprecisions are involved. One concerns the comparison of observations made from relatively moving frames of reference. Statements of the problem which make it seem paradoxical usually make use of comparable observations which are not observations of the same sequence of events. The other imprecision concerns the confusion of notions of simultaneity, which is not a well defined concept in Lorentzian spacetime.

When the problem is analysed in terms of the events which mark the changes of frames, no paradoxical elements are encountered.

\TWINZZ

JW 1/7/94