# Beyond the Twin Paradox

1. Feb 9, 2008

### keithh

This thread is another extending from the Schwarzschild Metric and Climbing out of a Black Hole. I hope the members of the physics forum will not object to me testing out some ideas against their wider knowledge in this subject area.

Anybody reading about relativity bumps into the Twin paradox sooner or later. It relates to two twins, one takes a return trip to a star at near light speed, while his sibling remains on Earth. If everything is relative, then either twin could claim that the other was moving relative to them and therefore should experience time dilation due to velocity, i.e. SR. The paradox arises as both twin are arguing that the other will be younger at the end of the trip.

The standard response to the paradox is that only one twin feels the force of acceleration and deceleration with respect to a common starting point on Earth. As such, the twin who travels to the star is younger because he can translate the acceleration into a higher relative velocity with respect to the stay at home twin. QED.

However, if you extend the twin paradox to a triplet paradox, things get a bit more interesting. One of the triplets stays at home, while the other two head off towards two stars (A & B). These stars are in the opposite direction to each other but on returning, each will pass Earth at near light speed and go on to the other star before eventually returning to Earth. We can clarify their paths as follows:

Triplet-1: Earth (E)
Triplet-2: E-A-E-B-E
Triplet-3: E-B-E-A-E

Having visited the first star, the two space-faring triplets speed past each other, as well as their Earth bound sibling on the way to the second star. At this point, all three could argue that the others are moving at near light speed with respect to them .

What is the relative age of each triplet on finally returning to Earth?

I would argue that the two space-faring triplets have to be the same age, but younger than their Earth bound brother, otherwise you just create another paradox. However, this answer raises an interesting issue, at least, I thought so. Time dilation due to special relativity is not just about relative velocity; it has to account for the relative starting point. If so, two galaxies that are speeding away from each other at near light speed, along radial paths, from a common point in spacetime, they must experience time and space against a common tick of the clock.

Would this mean that time is relative to the expansion of the universe?

Again, would be interested in any other thoughts.

Last edited: Feb 9, 2008
2. Feb 9, 2008

### pervect

Staff Emeritus
Our thoughts are that we don't like too much speculation, but prefer to teach standard aproaches. See the PF guidelines on "overly speculative posts". There are several standard approaches to the twin paradox, see for instance the sci.physcis.faq

So - does your approach reduce to any of these standard approaches (in the above FAQ) or some other standard approach that you can find in a textbook or a peer reviewed paper (note that arxiv papers may not be peer reviewed)?

If so, which one does it reduce to? Give details.

If your approach does not reduce to one of these standard approaches, we would encourage you to start learning the standard approaches first, and NOT spend a lot of time developoing your own, personal, idosyncratic approach. We find that criticizing personal idosyncratic approaches like yours is counterproductive, because it encourages people to put energy into defending them, rather than learning the standard approaches.

Teaching the standard approaches is our goal here, so we want to encourage you to start reading what other people have written, rather than focusing too much on your own ideas.

3. Feb 9, 2008

### JesseM

Sure, that's the standard answer you'll get if you analyze this problem in any inertial frame in SR.
Why do you say that? In any inertial frame, if a clock's speed is v in that frame, it's rate of ticking is slowed by a factor of $$\sqrt{1 - v^2/c^2}$$; so, if you want to figure out the total time elapsed on a clock between two events which happen at time $$t_0$$ and $$t_1$$ in the time coordinate of your chosen frame, and the speed of the clock as a function of coordinate time is v(t) in this frame, then the time elapsed on the clock will be $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$. This should give the correct answer for any problem in SR, including your triplet scenario.

4. Feb 10, 2008

### keithh

First, let me address the concerns of pervect. The intention of my original post was simply to raise a question; it was not intended to be read as a speculative theory. It was based on the standard response to the twin paradox, but extrapolated to the triplet scenario in order to highlight an issue concerning time dilation with respect to the two frames of reference moving relative to each other. I have read (again) the rules of the forum and apologise if I transgressed the limit of an overly speculative post. I realise that in using the phrase “I would argue that the two space-faring triplets have to be the same age” I was forwarding an opinion rather than a question, as was my intention. However, I believe the question, if not the opinion, to be legitimate.

Addressing the post of JessM. I agree with you, but the triplet scenario raised a question in my mind. At the point that two of the triplets pass each other and their sibling on Earth, they are all moving at near light speed relative to each other. As far as I can see your equation would give the same net result for both moving frames of reference, i.e. the perception of distance travelled and time elapsed is the same at the end of their journey. As such, they are the same age, but younger than their stationary sibling, even though they were always moving at a relative velocity to each other. However, I feel I am being asked to refrain from any further speculation on this topic.

5. Feb 13, 2008

### Bork

Hi, Keith. I'm sure the provided FAQ has all the answers you need, but I figure I should throw in my own approach to the problem, as I feel it can help to clarify this paradox for you immensely. An insight from one of my intro physics textbooks opened the door for me to understand this paradox in a clear and simple manner, without any need to refer to Doppler shifts and complicated calculations.

The first thing you need to understand is that what you claim is a "standard response" to the paradox, is not a response at all. Special relativity is fully equipped to handle the problem, even if we assume instantaneous accelerations (i.e. if two spaceships were travelling in opposite directions and synchronized clocks as soon as they swapped positions, their effective measurements of time would be the same as for a single ship making an instantaneous 180 degree turn). So basically, there is no need whatsoever to refer to acceleration or General Relativity to solve this paradox, as such references only tend to complicate the matter (as far as understanding the essence of the problem is concerned). It is true Einstein called this problem a paradox as on the surface it seems to be one, but he understood its resolution well even in 1905, and used it as a way of illustrating the consequences of his theory.

Now for the solution: the key thing to notice is that, in fact, this is not a symmetric problem. That is to say, the length and time transformations the astronaut sees while watching Earth are not the same as the transformations seen by the Earth while watching the astronaut. Let's say the two twins agree on a distant marker/beacon in space, like the star Alpha Centauri, and that the twin travelling into space will reverse course once they reach this star. Now from the perspective of the Earth frame, the trip occurs over a distance of 4 light years. However, from the astronaut's perspective, because they are travelling very fast towards this star, the distance they believe they have travelled will be much less, i.e. 2 light years. So from the Earth's point of view, the astronaut travels 4 light years in each direction, whereas from the astronaut's point of view, the Earth travels 2 light years in each direction. If you do the calculations from either reference frame, using the full Lorentz transformations which include both a contraction and a time translation (relativity of simultaneity), you will get consistent results from either point of view- the astronaut ages less than their stationary twin.

So the key ingredient which breaks the symmetry of the two reference frames is the fact that the beacon they use to determine their trip is stationary relative to the Earth and moving relative to the astronaut. One more thing to note carefully is that the beacon itself must be in an inertial frame (it is a marker for an inertial coordinate system), which is what allows us to differentiate whether the Earth is making the trip or the astronaut, since the beacon must be moving with constant velocity and cannot reverse direction. With this understanding of the Twins Paradox, I think your more complicated scenario need not be any cause for concern, as the same line of thinking and calculating should resolve it, and give the simple answer you indeed expect. Hope this helps!

Last edited: Feb 13, 2008
6. Feb 13, 2008

### JesseM

What do you mean by "both moving frames of reference"? The standard time dilation equation of SR, along with the integral for calculating total elapsed time given v(t) that I posted above, only work in inertial (non-accelerating) frames. The two twins who travel away from Earth and then turn around do not each remain at rest in a single inertial frame--each one's rest frame during the outbound leg is different from their rest frame during the inbound leg. Now, if their speeds are perfectly symmetrical in the middle (inertial) twin's frame, then the rest frame of the left twin during the left twin's outbound leg will be the same as the rest frame of the right twin during the inbound leg, and vice versa. So in this frame, left twin starts out at rest before the turnaround, the middle twin is moving at constant speed to the right, the right twin starts out moving at even greater velocity to the right before his own turnaround. It's important to note that in this frame, the event of the left twin turning around is not simultaneous with the event of the right twin turning around, even though the two turnarounds were simultaneous in the middle twin's frame--this is the relativity of simultaneity. So, in this frame when the left twin turns around, the right twin is still on the outbound leg, so their speeds will be identical (and greater than the Earth's) for a while; then later the right twin will turn around, coming to rest in this frame. Finally, the middle will catch up with the right twin, and the left twin (moving at greater speed than the middle twin) will catch up with the two of them at the same moment. If you add up the amount each twin aged both before and after the turnaround, you will still find that the two non-inertial twins are the same age when they meet, and younger than the inertial middle twin.

Let me give a numerical example. Suppose that in the middle twin's frame, the other two twins go away in opposite directions at 0.6c, then 10 years later they turn around, head back at 0.6c for another 10 years, and all meet when the middle twin is 20 years older. The time dilation factor for a speed of 0.6c is squareroot(1 - 0.6^2) = 0.8, so the left and right twin should be 0.8*20 = 16 years older when they meet, in this frame.

Now take the perspective of the inertial frame where the left twin starts out at rest. In this frame the middle twin is moving to the right at 0.6c, and using the velocity addition formula, we can figure out the the right twin must start out moving at (0.6c + 0.6c)/(1 + 0.6*0.6) = 0.882353c. So, the middle twin is aging at 0.8 the normal rate in this frame, while the right twin is aging at squareroot(1 - 0.882353^2) = 0.470588 the normal rate. Call the time

Call the time when they departed from a common location t=0 in this frame. Then 8 years later in this frame, at t=8, the left twin turns around and is no moving the same speed as the right twin, at 0.882353c. Then at t=17 years in this frame, the right twin turns around, and comes to rest in this frame. Finally, at t=25 in this frame, they all reunite.

In this frame, the left twin was at rest the first 8 years, so he aged 8 years before turning around; then he was moving at 0.882353c for another 25 - 8 = 17 years, so using the time dilation factor, he aged 0.470588*17 = 8 years between turning around and reuniting with the other two. The middle twin was moving at a constant speed of 0.6c the whole time, so after 25 years, he's aged 0.8*25 = 20 years. And the right twin was moving at 0.882353c for 17 years before turning around, so he aged 0.470588*17 = 8 years before the turnaround; then he came to rest and stayed that way for another 25 - 17 = 8 years, so he aged another 8 years between turning around and reuniting with the other two. So, lo and behold, we get exactly the same prediction in this frame as the other frame! The two non-inertial twins will have aged 16 years between the time all three departed and the time all three reunited, and the middle twin will have aged 20 years.

You can see these numbers make sense in terms of distances too. If in this frame the left twin starts out at x=0 light years, then after 8 years of rest he'll still be at x=0, while the middle twin has been moving at 0.6c so he's at x=8*0.6=4.8 light years. Then the left twin turns around and travels at 0.882353c for another 17 years, so at t=25 he'll be at x=17*0.882353=15 light years. Meanwhile the right twin travels at 0.882353 for the first 17 years, so at the turnaround he'll also be at x=17*0.882353=15 light years, and the middle twin will be at x=17*0.6=10.2 light years, so the right twin is also 4.8 light years away from the middle twin at the moment he turns around (note that in the middle twin's frame the other two twins are 6 light years away when they turn around, and the lorentz contraction factor when going to the second frame is 0.8, and 6*0.8=4.8). After t=17 years the right twin comes to rest, remaining at x=15 light years, so he'll still be there at t=25. Finally, the middle twin is always moving at 0.6c, so at t=25 he'll be at x=25*0.6=15 light years.

7. Feb 13, 2008

### yogi

Kieth: "two galaxies that are speeding away from each other at near light speed, along radial paths, from a common point in spacetime, they must experience time and space against a common tick of the clock.

Would this mean that time is relative to the expansion of the universe?"

Ed Harrison discusses this in his book "Cosmology, the Science of the Universe" to wit: "In SR we learn than nothing moves through space faster than light. How is it possible for recession to be faster than light? The answer is that galaxies are not moving through space at all but are wafted apart by the expansion of intergalactic space ..."