## Thursday, 27 December 2012

### There is no Twin Paradox

For the twin paradox to be considered a true paradox the framing of the scenario must be stringent, that is to say we cannot permit assumptions to be ignored. Therefore I must start with a short description of the twin paradox followed by identification of the inherent assumptions.

I am going to use a variant of the Twin Paradox from EinsteinLight with some very slight editing for the sake of clarity:

Jane and Joe are twins. Jane travels in a straight line at a relativistic speed v to some distant location. She then decelerates and returns. Her twin brother Joe stays at home on Earth. …

Joe observes that Jane's on-board clocks (including her biological one), which run at Jane's proper time, run slowly on both outbound and return leg. He therefore concludes that she will be younger than he will be when she returns. On the outward leg, Jane observes Joe's clock to run slowly, and she observes that it ticks slowly on the return run. So will Jane conclude that Joe will have aged less? And if she does, who is correct? According to the proponents of the paradox, there is … symmetry between the two observers, so, just plugging in the equations of relativity, each will predict that the other is younger. This cannot be simultaneously true for both so, if the argument is correct, relativity is wrong.

The author, Joe Wolfe, goes on to explain that asymmetry resolves the paradox, an explanation that I do not find to be entirely satisfactory.  He uses a flash animated pair of diagrams to support his argument:

Are the space-time diagrams symmetrical? Parts of them are. The first three years of the diagrams for Joe's frame and Jane's departing frame are symmetrical: each twin sends three greetings but only receives one. The last year and a half of Joe's frame and Jane's returning frame are also symmetrical: each sends two greetings and receives four. But the diagrams are not symmetrical in between. Why not?

Look at Jane's diagram. From Jane's point of view, immediately after she has fired her engines (for the return journey), she begins receiving Joe's greetings more frequently. This does not surprise her: she has gone from travelling away from the sender of the greetings and is now travelling towards him.

Jane observes this change as soon as she turns around, which is for her the midpoint of her voyage. (She now receives blue shifted messages instead of red shifted ones. One could apply the same relativistic Doppler factor to the frequency of arrival of the messages.) Joe, on the other hand, doesn't start to receive messages at a higher frequency (blue shifted messages) until considerably after the midpoint between Jane's departure and arrival, simply because the effect of Jane's acceleration and changed reference frame takes a while to get to him: he doesn't see the high frequency arrival of messages until the arrival of the first message that Jane sends after she turns around.

This is a clear example of where the asymmetry of the twins appears. The causes of this asymmetry are the fact that Jane reverses direction and Joe does not, and the finite time that light takes to transmit this information to Joe means that Joe doesn't get the news immediately. Jane leaves one inertial frame and joins another, and she has the effect of that change immediately. Joe, on the other hand, doesn't notice the effects of Jane being in a different inertial frame until much later because she is a long way away from him when it happens. The asymmetry is as simple as that.

There are a few if not hidden, then obscured assumptions, which are perhaps only obvious when one takes time to search for them.

"(S)ome distant location" appears sufficiently vague as to avoid creating problems but an inherent assumption is that this location shares the same frame as Joe.

By placing Joe on Earth we hide (or obscure) the other assumption, which is that we also share the same frame as Joe.

"(Jane) decelerates and returns" is distracting. As the author correctly points out this is a point of asymmetry. However, a similar scenario (to be shown shortly) shows that it doesn't matter which frame undergoes deceleration and a change in direction – that of Jane or the entire universe. It is generally assumed that the period during which Jane changes direction is insignificant enough to ignore.  (Joe Walsh uses the term "immediately" to imply an instantaneous change in Jane's velocity, which we know to be physically impossible - but we can assume ridiculously high accelerations to have a very short turnaround period and just hope that Jane isn't turned into jam during the process.)

Finally, "Jane travels in a straight line at a relativistic speed v" begs the question "relativistic speed v relative to what?" The obscured assumption is "relative to both Joe and the distant location" (and to us, the readers). This is a direct consequence of the assumption that Joe and the distant location share the same frame (and that we also share that frame).

Let me provide a scenario which is analogous to the scenario described in the twin paradox.

Joe floats in space (in a protective space suit) with two clocks (marked as Joe’s).

Jane sits at one end of an extremely long structure which also floats in space, unattached to anything bar Jane and a beacon with another two clocks (marked as Jane’s). At the other end of the structure is the beacon. According to Jane, the structure has a length of L, meaning that the distance between Jane and the beacon (according to Jane) is L. Joe knows this.

Observe that Jane represents "the Earth" and the beacon represents "some distant location" in the twin paradox. The assumption that "the Earth" and "some distant location" have a fixed separation is inherent, but unstated, in the twin paradox.

Jane and Joe are sufficiently distant from any masses as to be considered to be alone in the universe, with no gravitational field in effect. The gravitation exerted by Jane and her structure on Joe is negligible.  (The absence of gravitational effects is another unstated assumption in the twin paradox.)

Jane and Joe pass each other twice, at relativistic velocities of v and -v. Joe and the beacon pass each other twice, also at relativistic velocities of v and -v (Jane and the beacon are fixed to the same structure and hence share the same frame).

Four noteworthy events take place:

1. Joe and Jane are collocated as they pass for the first time. Their clocks begin measuring time elapsed.

2. Joe and the beacon are collocated as they pass for the first time. Joe's clocks are paused and the beacon sends a message to Jane's clocks to pause.

3. Joe and the beacon are collocated as they pass for the second time. Joe's clocks restart measuring time elapsed and the beacon sends a message to Jane's clocks to resume measuring time elapsed.

4. Joe and Jane are collocated as they pass for the second time. Their clocks stop measuring time elapsed and Joe and Jane exchange clocks such that they both have one marked Joe’s and one marked Jane’s. Neither consults the other as they each attempt to work out what the other's clock will read.

Observe that I quite specifically do not say who reverses direction. For the purposes of the thought experiment, we can say that both Joe and Jane were anaesthetised while one of them reversed direction, so that neither knows which has changed direction relative to any third observer (such as the reader). By virtue of the scenario, both clocks are paused while any acceleration takes place and therefore no acceleration affects the measured time elapsed.

There is an asymmetry in this scenario, but Jane and Joe cannot determine on whose part that asymmetry lies.

Jane's calculations:

Joe is in motion relative to Jane. Jane calculates that the total time elapsed between events 1 and 2 and events 3 and 4 will have been 2L/v and her clock confirms that this is the case.  Assuming that the clock is sophisticated enough, Jane will be able to see that the time elapsed between events 1 and 2 was (L/v + L/c).  This is because the signal to stop timing would have taken a period of L/c to reach Jane’s clocks.  The time measured for the return trip, events 3 and 4, was (L/v - L/c), again due to the time taken for the signal to reach Jane’s clocks (a start signal this time).

Jane further calculates that because Joe is in motion, his clocks will run slow and will show a time elapsed of (2L/v) / γ where γ = 1 / √(1-v2/c2) (see The Lightness of Fine Tuning (Part 2) for an explanation of how this value of gamma (γ) is calculated, look up “time dilation” or here for confirmation that t’ = t / γ).

Jane can check Joe’s clock and see:

time elapsed (event 1 – event 2) = (L/v + L/c) / γ;

time elapsed (event 3 – event 4) = (L/v - L/c) / γ; so

total time elapsed = (L/v + L/c) / γ + (L/v - L/c) / γ = (2L/v) / γ

Therefore:

Jane’s clock, according to Jane = 2L/v

Joe’s clock, according to Jane = (2L/v) / γ

Joe's calculations:

Jane is in motion relative to Joe. Joe therefore calculates that Jane's structure is foreshortened by a factor of 1/γ. Therefore the time elapsed while the entirety of the structure passes twice will be (2L/v) / γ. Sure enough, Joe checks his clock and sees that this is the case.

Working out what Jane's clock will read is a little more complex. Joe knows that not only is Jane's structure foreshortened, but that Jane's clocks will also run slow by a factor of γ.

The first period elapsed can therefore be calculated as follows (noting that Jane's relative motion is in the same direction as the message from the beacon to Jane's clocks):

t1         = γ.(γ.L/v + γ.L/(c-v))

= γ2.(L/v + L/(c-v))

= γ2.(L/v.(c2-v2)/(c2-v2) + L(c+v)/(c2-v2))

= γ2.(c2.L/v - Lv + Lc + Lv)/(c2-v2)

= γ2.(c2.L/v + Lc)/(c2-v2)

but since γ2 = 1 - v2/c2 = (c2 - v2)/c2,

t1         = (c2 - v2)/c2 . (c2.L/v + Lc)/(c2-v2)

= (c2.L/v + Lc)/c2

= L/v + L/c

The same process can be used to calculate that the second period elapsed is (L/v - L/c). The total time elapsed on Jane's clock, as calculated by Joe, will be 2L/v - precisely the same as calculated by Jane and as shown on the clock labelled as Jane’s.

Therefore:

Joe’s clock, according to Joe = (2L/v) / γ

Jane’s clock, according to Joe = 2L/v

If both Jane and Joe agree about what the other’s clock should read, and this agreement is confirmed by measurement, then there is no paradox.

I believe also that my scenario demonstrates pretty conclusively that neither the acceleration nor the subsequent change to Jane’s inertial frame in Joe Wolfe’s scenario has a direct impact, since (in my scenario) there is no indication as to which inertial frame changed.

------------------------

Please note that Joe Wolfe clearly states that there is no paradox here and nothing in this article should be taken as implying that Joe is a proponent of the Twin Paradox as a paradox.

#### 1 comment:

1. cool post. I have heard of the twin paradox but never knew the details in much depth. Now I feel like I have some sense of what it is and why it isn't really a paradox.

Feel free to comment, but play nicely!

Sadly, the unremitting attention of a spambot means you may have to verify your humanity.