## Time dilation in relativity of simultaneity related to twin paradox

Hi All,

I think that I understand (as far as you can!) the ideas behind explaining why two twins have different ages if one travels at near light speed away for a time and then returns to his twin on Earth (the return journey changing the inertial reference frame).

I have a related question where the twin doesn't actually turn around and return home.

If a space ship leaves Earth at 0.8c towards Alpha Centauri 4 light years away then the person on Earth considers that the trip will take t = d/v = 5 years in Earth time. Someone on Alpha Centauri will also agree with this (adjusting for the 4 years time it takes for the light to travel from Earth to AC and so they physically see the ship leave at t=4). Someone on Earth, or on AC, will consider time on the ship to be running reduced by the factor ε= √(1 - v^2/c^2) = 0.6. On arrival the people on AC will therefore expect the space travellers to have aged 0.6 x 5 = 3 years during the time they themselves have aged 5 years.

So the people on AC expect the travellers to arrive at t = 5 having aged 3 years from their departure at t = 0.

For the travellers, time dilation means that the distance they are to travel is εd = 0.6d = 2.4 light years. As they are travelling at 0.8c they consider that they have aged d/v = 3 years during the trip. This agrees with what the people on AC think.

However, from the point of view of the time travellers the clocks of the people on AC are running slow by a factor of 0.6. They therefore expect the AC people to have aged 3 years x 0.6 = 1.8 years.

Can someone please explain the discrepancy between this expected age and the 5 years that the local AC people consider they have aged upon arrival? (or where my algebra has gone wrong!)

Many thanks.

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 At the moment that the travellers accelerate to .8 c, they have changed frames, and they must re-synchronize their clocks to that frame, and upon doing so, they now measure a distance with the length contraction to AC of .6 d = 2.4 light years as you have, right, and a time dilation for the AC observers of .6 also, giving 1.8 years that passes for the AC observers while 3 years pass for themselves, as you also have, but there is one more thing, relativity of simultaneity. Upon accelerating to .8 c and re-synchronizing, they will say that the clock on AC is now set to a time that is d v / c^2 = (4 ly) (.8 c) / c^2 = 3.2 years ahead of their own, so the time that reads upon the clocks of AC when arriving will be 3.2 years + 1.8 years = 5 years.
 Hi Grav-universe - ah I think I follow what you are saying. So if, while on Earth sitting in their spaceship ready to set off, they had a telescope trained on a clock on AC which said t=0, then in the time that they accelerate to 0.8c they see the hands of the clock whip round to t=3.2 and then the clock from that point continues to run slow relative to theirs. Is that right? Assuming so - thanks. That's cleared up a whole lot of confusion for me. For a bit of further information, the situation I described was a simplified one from what was really bugging me. I was thinking of a spaceship (à la tau zero by Poul Anderson) which was constantly accelerating nearer and nearer to c. As they get closer and closer the distance in front shrinks and so they can cross vast distances (from Earth's point of view) in their lifetime. Because their lifetime is so long compared to the people they leave behind they can reach these vast distances but only while vast amounts of time have gone by from the Earth point of view - and the point of view of anyone located a vast distance being approached by the spaceship, assuming they are at rest compared to the Earth. I was therefore trying to reconcile in my mind the fact that the nearer they get to c the slower they see the clocks of the universe around them run. Yet at the same time the local time must be hugely in the future as they move further across the galaxy/universe. So this now makes sense - I think. Because they are constantly accelerating their frame of reference is constantly changing. As they accelerate yes clocks outside the spaceship run slower and slower relative to theirs, but also the constant reassessment of simultaneity means that these external clocks are constantly being moved forward too - and they are moving forward faster quicker because of the changing simultaneity than they are running slower because of the approaching c. So people in the space ship will see the universe around them evolving faster and faster (and not slower and slower as you might assume from considering the approaching c alone). I hope that's it anyway!

## Time dilation in relativity of simultaneity related to twin paradox

Well, if the traveller's look through a telescope immediately after they accelerate to .8 c, they would see the same reading upon a clock from the light coming from AC as they did immediately before accelerating, those photons being the same ones that were there before and are also reaching Earth at that moment, so both Earth observers and the travellers viewing through telescopes that the clock on AC reads 4 years less than their own times. But that is due to the time the light takes to get there, not the current reading on AC.

As you wrote at the end of your post, the travellers would constantly have to re-synchronize their clocks as they accelerate, right, or just reset them when they reach .8 c relative to Earth, and this resetting of the clocks causes the travellers to now measure that the clock on AC is ahead of their own time. To see this more clearly, let's say that the proper length of the traveller's ship is 10 light seconds long. After accelerating to .8 c, Earth observers will now measure it to be 6 light years long, and there is another clock at a space station stationary to Earth 6 light seconds away, so as the back of the ship passes the Earth, the front of the ship will pass the space station according to Earth observers. However, due to relativity of simultaneity, since the clocks on the ship are set differently, the traveller's will not say that the front and back of the ship are at each of these places at the same time.

According to the Earth observers, the back of the ship coincides at Earth and the front of the ship at the space station at the same time, say T=0. Before accelerating, the clocks on the ship also read T = 0, so all clocks read the same. After accelerating, the travellers must re-synchronize their clocks. We want to know the time that passes in relation to a traveller coming from Earth, so we will set the clock at the back of the ship which coincides with Earth to T=0 also. In the new frame of the ship travelling at .8 c, then, in order for light to be measured at c in the new frame, the clock at the front of the ship will have to be set to T = - d v / c^2, where d is the proper length of the ship, giving T = - (10 ls) (.8 c) / c^2 = -8 seconds. So since the front of the ship that coincides with the space station now reads T = -8 seconds while directly viewing the clock on the space station to read T = 0, using the time of their own frame according to the new settings of their clocks, the travellers say that the time of the space station is, while the same before, now 8 seconds ahead of their own time.

I used a somewhat different scenario for this because it is simpler to look at things from the perspective of Earth's frame for what I showed this time rather than the traveller's frame as we examined before, but some of the measurements would be different due to the difference in simultaneity between the two frames and I didn't want to confuse you. I just wanted to give you an idea about what is happening with the clocks and why the traveller's say AC's clock reading jumps forward. Upon re-synchronizing in the new frame, the new readings upon the clocks of the travellers becomes the official time for their own frame, so from what they now observe, clocks ahead of them in Earth's frame are set ahead of their own, further ahead with greater distance.

 Thank you both for your replies. I will think on what you have said. Just one quick follow up question at the moment - While still on Earth at t=0 the travellers will see light coming from AC showing that their clocks say t=-4, and they will therefore infer that "now" is t=0 also for AC since that light has taken 4ly to arrive. Immediately on accelerating they will still see the photons that show t=-4 on the clocks coming from AC, but they now consider these to have originated from a distance of 2.4ly. The inference is now that "now" for AC is t=-1.6ly - is that right?

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 Quote by timpeac Thank you both for your replies. I will think on what you have said. Just one quick follow up question at the moment - While still on Earth at t=0 the travellers will see light coming from AC showing that their clocks say t=-4, and they will therefore infer that "now" is t=0 also for AC since that light has taken 4ly to arrive. Immediately on accelerating they will still see the photons that show t=-4 on the clocks coming from AC, but they now consider these to have originated from a distance of 2.4ly. The inference is now that "now" for AC is t=-1.6ly - is that right?
No. If you look on the second graph, you can see that AC was 12 light-years away at the time the -4 year signal was sent.

The question of "now" only makes sense for a remote clock that is at rest with respect to the observer for two reasons: first if there is relative motion then the two clocks tick at different rates so if there ever was a moment that could be identified as "now" for both of them, it would only apply for that one instant in time. Secondly, since the way to synchronize two clocks is by sending a round trip signal between them, the remote clock will have moved during the process and how could you identify which position "now" applied to?

You should realize that the question of "now" between two separated clocks is not one that we can derive from nature. We inject into nature what "now" means and the way Einstein says to do that is to send a round trip synchronizing signal between the two clocks. That is going to take at least 8 years in this example. Now you might say that we really don't have to do that since we know how far away the other clock is but there is no way to even know that short of measuring it with a round trip signal so we're in the same situation.

So how do the travelers know that AC has moved from 4 light-years away to 2.4 light-years away after they start their trip without cheating and calculating it just like we did here? Well, they would have to send that round trip signal. Look at the graph. A signal sent by the travelers at time 0 will arrive at AC when their clocks read 4. (Follow the thin black line going up and to the right from the zero location at a 45 degree angle.) Now AC responds with a signal back to the travelers which the travelers will detect when their clocks are at 2.667 years and they are very close to AC. So they will conclude that AC was 1.333 light-years away and it was at the half way point between the start and end of their test. But what they wanted to know is how far away it was when they started their trip. Can you figure out how they would make this assessment?

 Recognitions: Gold Member It might be interesting to see how earth determines how far away AC is in the second IRF. If they send the same signal that I discussed in the previous post that arrives at AC when their clock reads 4 years, then track how the response gets back to earth, I think you can see that it will arrive on earth when earth's clock is at 8 years. So earth will conclude that AC is one-half that distance the light has traveled or 4 years. And you can see that earth will see that their clocks are synchronized because the return signal claims that it is 4 years earlier. We are assuming that the signal include the time the signal was sent. In the same way, AC can determine that earth is 4 light-years away and that their clocks are synchronized by the signal it sends when its clock was at -4 years and receives the response back from earth at +4 years saying that the earth time was zero.