1. Dec 20, 2009

### timpeac

Hi, I can see that this topic has been much discussed, but I haven't seen a thread on it with the particular spin I want to give it (just the journey out from Earth).

I understand the traditional view of the twin paradox (I think!): Two twins a and b are on Earth and each has a clock. The clocks are synchronised and one twin travels at, say, 60% the speed of light to a star exactly one light year away, stops and immediately returns. Due to time contraction, during the journey each twin judges that his brother’s clock is running slow – each only showing 0.8 seconds for each second of his own clock - each of which seems, to the holder, to be running normally. On return to Earth when the clocks are compared it is shown that it is the twin who has travelled away and come back whose clock shows less time than his brother’s.

This is the paradox – if you could equally view each twin as the one “at rest” while his brother travels away and back to him, and so both are correct in considering his brother’s clock to be running slow, then how can one of them be shown to really have the slow clock? The answer is that the twin travelling away from Earth has undergone four accelerations, albeit very quick, one when he set off from Earth, one when he arrived at the destination and stopped, one when he accelerated to start back to Earth and one when he decelerated to rest on arrival. This changed his frame of reference and thus changed which times can be considered as simultaneous in each twin’s frame of reference. So far so good – I have no problem with that.

However, my question involves a situation where only half the journey occurs. I can think of two situations (the first of which I can think of explanations for but the second I’d like some help to understand).

a) The first twin sets off as before. At the same time a signal of light is sent from the Earth to the star 1 light year away where someone is waiting to receive it. Since the Earth and the star are at rest with each other they share a frame of reference and can therefore agree on the simultaneity of the journey start. The twin then arrives at the star. Now, for him both the Earth and the star have been moving relative to him at 0.6c and so their clocks have been running slow at 0.8s. Similarly for both the twin on Earth and the person at the star the second twin has been travelling at 0.6c relative to them and they consider that his clock has been running slow (the person waiting at the star just has to remember to add 1 year on to the journey time to get simultaneity because of the time taken for the light signal to arrive).

Now, when the twin has performed this first half of his journey and compares times with the person at the star they can’t both then agree that each other’s clock is slow. On comparing they will see that it is again the twin’s clock which has been running slow. Now, I can understand that because the twin has still undergone two accelerations whereas the person waiting at the star has not. However – in scenario b) below there is no acceleration:

b) Say twin b is not on Earth but travelling from some earlier point at a constant 0.6c past the Earth. As he passes the Earth twin a who is on the Earth sends the light signal towards the star 1 light year away that twin b is also heading directly towards. When twin b approaches the star he doesn’t decelerate to stop but carries on past at the constant 0.6c. However, as he passes the star he and the person their show each other their clocks as they pass. In this situation there has been no acceleration, and again each party can consider the other moving and therefore to have the slow clock. Who will be correct? My suspicion is that again the twin in the spaceship will have the slower clock on comparison – but why? This time there has been no acceleration to change the frame of reference.
Any thoughts? Many thanks.

Last edited: Dec 20, 2009
2. Dec 20, 2009

### Janus

Staff Emeritus
Relativity of Simultaneity. While twin A and the person at the star agree that their clocks read the same time, Twin B does not. According to him, the clock at the Star is actually ahead of Twin A's clock.

So While Twin B measures 1 1/3 years as having passed on his clock during the trip, he will measure 1.06667 years as passing on Twin A's and the clock at at the star. However according to him, the clock at the star reads 0.6 years ahead of Twin A's clock. Thus if Clock A reads 0 at the moment he passes Earth, the star's clock will read .6 years according to him. When he passes the star, his clock reads 1 1/3 years, and the Star's clock will have advanced 1.06667 years to read 1 2/3 years.

3. Dec 20, 2009

### Rasalhague

In the earth/star rest frame frame, the ship takes 1.67 years to travel one light year, so the star clock shows this time when the ship arrives. The ship clock is moving, and so running slow; it shows 1.33 years when it arrives at the star.

In the ship's rest frame, the distance between the earth and the star is only 0.8 light years, so it only takes 1.33 years to arrive. The earth clock and the star clock are synchronised in the earth/star rest frame, but they're out of synch in the ship's rest frame. In the ship's rest frame, when the ship passes earth, the star clock is ahead of the earth clock, due to the relativity of simultaneity, by 0.6 years. In this frame, it's the star's clock that's moving at 0.6c, and so running slow; in the time it takes the ship clock to record 1.33 years, the star clock only records 1.07 years. But the star clock has a head start of 0.6 years when the ship's clock passes earth and is synchronised with the earth clock, and so when the ship reaches the star, the star clock will show 0.6 + 1.07 = 1.67 years. What a coincidence!

4. Dec 20, 2009

### yuiop

According to travelling twin b, the star clock is ahead of the Earth clock by a factor L*v/c^2 at the start of the journey, while according to Earth twin a, the Earth clock and the star clock are sycronised. L is the distance between the Earth and the star in the twin a's frame. The Earth/star system syncronises its clocks by sending a signal from Earth at time t1 = 0 to the star and then setting the star clock to t2 = L/c to allow for the signal travel time L/c. According to the travelling twin's observations the distance between the Earth and star is L/y due to length contraction, where y is the gamma factor 1/sqrt(1-v^2/c^2). The time for the signal to travel from the Earth to the star in twin b's frame is L/(y(c-v)) and taking time dilation into account this equates to L/(y^2(c-v)) for how much time the Earth clock advances during the signal travel time. So when the star clock t2 is set to L/c the Earth clock t1 reads L/(y^2(c-v)) and the difference between the two readings (t2 - t1) is L*v/c^2. It is this lack of simultaneity of a's clocks from b's point of view, that allows b to reason that a's clocks are running slower but read a greater total ellapsed time due the "head start" programmed into the star clock t2 during the synchronisation process. This difference of opinion between observers as to what is simultaneous, is known as the "relativity of simultaneity".

5. Dec 21, 2009

### Jorrie

I think it is not strictly correct to say this. In a purely inertial scenario, there is no "ship clock that is moving, and and so running slow". In this purely inertial "half-twin-paradox", no measurement can tell which clock "runs slow". In order to make a definite prediction (and measurement), at least one of the clocks must be accelerated, so that they can at least pass each other for a second time.

6. Dec 21, 2009

### Al68

The traveling twin will have less elapsed time between the events for the exact same reason as the standard twins paradox.

The second event is defined a priori as 1 ly from earth in earth's frame. If you were to instead define the second event as 1 ly from the ship in the ship's frame (buoy trailing the ship?), you would get the opposite result, ie the elapsed time between the events would be less on earth's clock than the ship clock.

7. Dec 21, 2009

### Al68

None of the clocks were accelerated in the scenario presented. But the elapsed time between events is easy to determine and measure in both frames.

8. Dec 21, 2009

### yogi

AI68 and I have posted several explanations of the TP in prior years - based upon reducing the problem to a one way trip and doubling the result, The one way trip analysis is the easy way to break the problem down and eliminate accelerations that confound most explanationss - starting with and initial speed (an on the fly reading as the clocks pass to start) and the same on the fly reading on arrival, eliminates any weasel wording that depends upon a false application of a pseudo G field as Einstein proposed in1918 The one way trip is simply an application of the principle of interval invariance in all frames

9. Dec 21, 2009

### Jorrie

Agreed, but IMO it is false to compare the propertime interval of the inertially 'moving twin' with the coordinate time interval of the inertial 'home twin' for this situation and then declare that the 'moving twin' has aged less than the 'home-twin' during the away trip. To do so would mean giving preference to one inertial frame above another. Only the 'away twin' measures the invariant interval between the departure and arrival events directly.

10. Dec 21, 2009

### Rasalhague

It's the sort of language often used in these contexts, as in the popular motto "a moving clock runs slow", and movement is defined by the frame, so there's no ambiguity (the ship clock is moving in the earth/star rest frame; the earth and star clocks are moving in the ship's rest frame). But I agree that when these ideas as introduced in the form of a scenario that contrasts a rocket or train or other vehicle with a planet or station or some other thing that our non-relativistic intuition tells us is stationary, it can beguile those instincts into thinking of motion as absolute, and obscure the equivalence. Perhaps the best way, pedagogically, is to introduce the basic idea of relativity using two identical space ships. Then there isn't that distraction for beginners of wondering whether it's significant that one of the frames is centred on a vehicle and the other on something that our non-relativistic instincts tell us is inherently "not moving".

Last edited: Dec 21, 2009
11. Dec 21, 2009

### timpeac

Thank you all for your replies - they make perfect sense. I was forgetting that just because both twins agree that passing the Earth is simultaneous, and twin a and the person at the star agree that the event is simultaneous this does not necessarily imply that twin b and the person at the star will also agree the same thing. In twin b's opinion the person at the star has a "head start".

12. Dec 21, 2009

### matheinste

Yes. I think that this is a problem with the train/embankment scenario when first seen by beginners. It seems obvious that the train is moving and the embankment is not. That's the way the world appears to be.

Matheinste.

13. Dec 21, 2009

### yuiop

Good to see you grasped the concept so quickly. These sort of threads normally go on for hundreds of posts.

It might be worth mentioning in conclusion, that when the calculations are done correctly as shown for example by Janus and Rasalhague, then it is true that there is no way to actually determine which clock if any is "really" running slower than the other in a one way non-accelerating experiment.

14. Dec 21, 2009

### timpeac

But doesn't the situation I described in b) in my original post show that both clocks are indeed running slow in relation to each other without acceleration taking place? Twin b's clock is indeed slower than the person at the star's on comparison when he arrives and the person at the star's clock - while being ahead of twin b's in absolute terms - is still less than it would have been if time hadn't contracted from twin b's point of view (he had a 0.6 year head start when twin a passed the Earth and this has reduced to 1/3 years by the time twin b arrives).

15. Dec 21, 2009

### yuiop

Yes, both observer's measure the other observer's clocks to be running slower than their own clocks and in the "hard core relativistic" or traditional interpretation that means both clocks are "really" running slower than each other. Personally I believe it is physically impossible and logically inconsistent for two clocks to physically run slower than each other and I prefer to think that in this situation the physical reality is undetermined and we can not say in an absolute sense which clock is running slower than the other, or if they are running at the same rate. It is only when one observer turns around and returns, that proper elapsed times can be compared in a way that all observers can agree on. That is my personal view, but it probably does not represent the formal accepted view.

16. Dec 23, 2009

### Jorrie

For the all-inertial case, I agree. There is however an interesting twist in the logic when we consider an all-inertial, three-clock (ABC) experiment, where a third inertial clock (C) does the return trip at -v relative to A. By this setup and Lorentz invariance:

$$\Delta\tau_B = \Delta\tau_C = \gamma\frac{\Delta\tau_A}{2}$$

where the $\Delta\tau$ are the propertime intervals between the three events making up the Minkowski triangle. Although this does not prove anything, it is extremely suggestive that clocks B and C must record time 'slower' than clock A. However, the fact that all three are inertial observers, makes the conclusion invalid, not so?

17. Dec 26, 2009

### yogi

The one way experiment is easily done in a lab using pions -a hi speed pion is generated and passes a detector with a clock #1 attached - a few feet away is a second clock with a detector #2 positioned at the point where the particle disintegrates. These two clocks are in the same room and they are always in sync - but the difference between the time measured by the two clocks is the proper time difference between the two events - the pion carries its own clock in the form of a decay period - this pion clock is not going to read the same as the difference between #1 clock and #2 clock

There is no acceleration - the pion is up to its travel speed at the time it reaches #1 clock.
Its important to keep your eye on the experiment - there is a space interval in the pions frame and space interval in the lab frame - there is a time difference logged by the pion's clock and there is a time difference logged by the difference between #1 and #2 clocks
The space time interval must be the same in both frames

Last edited: Dec 26, 2009
18. Dec 26, 2009

### Jorrie

Yes, but the pion's space interval is zero, that's why its "own clock" observes the propertime interval, which then equals the spacetime interval. The lab clock does not measure the propertime interval, but rather the larger lab coordinate time interval. Therefore we should not say that the pion clock is "running slower" than the lab clock (which is the reason behind the current discussion).

I do not quite understand what you mean by "... there is a time difference logged by the difference between #1 and #2 clocks".

19. Dec 26, 2009

### Al68

The current discussion is about which frame has less elapsed time between the same two events, not which clock ran "slower".

yogi's scenario is analogous to the OP's in that the pion clock shows less elapsed coordinate time than the lab frame clocks between the two defined events.

20. Dec 26, 2009

### shyamalshukla

According to the theory of relativity only force free objects can be considered true reference frames. In the case of the spaceship moving at a constant velocity, it would have certainly accelerated at some point in the past so as to achieve its current motion, which renders it unsuitable as a reference frame.

Hence, it is the stationary twin and the stationary star who are true reference frames and therefore it is the travelling twin whose watch runs slower.

Fairly new to this subject and hence ask for suggestions in case you find the above out of place.

Thanks