# Twins paradox and time dilatation

1. May 21, 2014

### Zeor137

I have already read about the twins paradox a lot of times, but what puzzles me is that how the Universe would "now" what twin had to be younger since, if the spaceship travels at high speed in relation to Earth, the planet would move the same with the ship as reference.

2. May 21, 2014

### Staff: Mentor

The two twins can be distinguished because only one of them experiences acceleration. That doesn't mean that the acceleration causes the differential aging (it doesn't), it just means that the two twins have different experiences and the universe can "know" about this difference.

Another way of seeing the difference: There are inertial observers for whom the earth-bound twin is at rest during the entire journey. There are no inertial observers for whom the traveling twin is at rest during the entire journey (although there would be if the traveler didn't turn around, kept on heading out at a constant speed relative to earth).

3. May 22, 2014

### Zeor137

Thank you very much, it's pretty clear now to me. So, according to the material you pointed me, if the twin in the space ship is accelerating, to use him as reference it would be necessary to consider a "pseudo-gravitational field" canceling the acceleration. Am I right?

4. May 22, 2014

### yogi

Einstein published an article in 1918 where he introduced the idea of a pseudo G field to explain the twin time difference - a number of authors including Max Born, have taken the position that the twin time difference can only be resolved by considering the acceleration - others maintain SR is sufficient - To measure an actual clock difference it is not necessary that the twins be reunited - launch a rocket in a straight line to Altair at near c velocity and note the time logged by an on board clock when it arrives and record it. The on-board clock will have logged less time than the clocks in the earth-Altair frame.

5. May 22, 2014

### Staff: Mentor

But it *is* necessary that the clocks at the start and end of the trip, to which the moving clock's reading will be compared (in your example, on Earth and Altair), be synchronized. And that requires a synchronization convention, because the clocks are spatially separated. For the case in which spacetime is flat and the Earth and Altair clocks are at rest relative to each other, there's an obvious synchronization convention that can be used; but it does not generalize to cases where the Earth and Altair clocks are moving relative to each other, much less to cases where spacetime is not flat. Measuring the difference in clock readings when the twins are reunited has none of these problems.

6. May 23, 2014

### PAllen

But that's not the twin paradox at all (your one way trip example). The traveler determines that the Altair clock and the earth clock both elapsed less time than theirs; and that the Altair clock being ahead of theirs on arrival is simply because it was it was synchronized way ahead of theirs. The twin 'paradox' defining feature is that both agree that one has elapsed more time (and no clock synchronization is involved at all).

7. May 23, 2014

### yogi

The Altair clock and the earth clock are in the same frame - little or no relative velocity. The one way trip is a simple example of a very common experiment that is repeated frequently in the lab - the measurement of hi speed pion decay time when ejected during proton-proton bombardment experiments - the pion lives longer because it was initially synchronized to the lab clock in the lab frame - then rapidly accelerated. To measure a real physical time dilation between two events in relatively moving frames - acceleration is involved at some point. For the pion in the lab it was the energy of ejection that imparted the acceleration that lead to the velocity difference and consequently the time difference - in the Altair trip - it was the launch acceleration that distinguished the state of the rocket from the earth frame.

Without some acceleration there is no guarantee that clocks in relatively moving frames will accumulate different amounts of time - ---a clock attached to an object moving past the earth at constant velocity will not necessarily run any slower than an earth clock (ignoring altitude) - while observers on the earth may set up apparatus to measure space and time intervals that indicate the other clock is running slow - the other guy may do the same in his frame - - the problem is in the mensuration process - each regards his own frame at rest so the bogus time difference would have us believe that each clock runs slower than the other - but the situation is symmetrical - so that cant be

launch two satellites in circular polar orbit at the same height traveling in opposite direction - each time they pass their clocks will read the same because the situation is symmetrical - but they have relative velocities - so they measure an apparent time delay of the other clock compared to their own. but its not real - both satellites are in their own inertia system (free fall in orbit) . This is the difference between the reality of GR time dilation and SR time dilation. Einstein introduced a G field as an artifice to explain his own error in Part IV of his 1905 publication.

8. May 23, 2014

### Staff: Mentor

How are the pions accelerated? I thought they were moving at high speed relative to the lab when they were created.

In any case, there are other situations where the created particles *are* moving at high speed relative to the Earth when they are created; see below.

Yes, it will, *with respect to the Earth*. For example, muons are created high in the Earth's atmosphere by cosmic rays. We know how high because we can detect the collisions that create them. Given the altitude at which they are created, the muons should all decay long before they can reach the Earth's surface--yet they are detected at the Earth's surface, with exactly the intensity that they should be if their half-lives are extended by relativistic time dilation.

These muons are not created at rest with respect to the Earth; they are moving relativistically relative to the Earth when they are created, *and* when they are detected. So this is like a version of the Earth-Altair experiment where a rocket flies past the Earth at relativistic speed, records Earth's clock reading as it passes, travels to Altair, and records Altair's clock reading as it passes. If Earth's and Altair's clocks are synchronized, the difference between those two readings will be much larger than the elapsed time on the rocket's clock, even though there is no acceleration at any point.

Yes it can, because "time" is relative, and so is simultaneity. You can't analyze this scenario just looking at time dilation; you have to look at relativity of simultaneity as well.

That's another reason, btw, why the standard "twin paradox", where the two twins come back together to compare clocks, is easier to analyze; you don't have to worry about relativity of simultaneity or any other conventions. The difference in the twins' clock readings when they meet again is a direct, local observable.

Yes. But both clocks will have *different* elapsed times than a clock that "hovers" above one of the Earth's poles (I'm assuming that's where the satellites meet on every orbit) at the same altitude. Why do you think that is?

In each one's local inertial frame, yes. But which local inertial frame each satellite is in changes as they go around in orbit. You can't analyze this situation using SR.

No; both satellites are moving inertially (in free fall), but that does *not* mean you can set up a single inertial frame for each one in which it is always at rest. You can only do that locally, in a small spacetime neighborhood of a particular event. When gravity is present, spacetime is curved, and in a curved spacetime there are no inertial frames covering large spacetime regions.

No, it's the difference between flat and curved spacetime. You could set up the same situation in flat spacetime; the two satellites moving in circles would be accelerated, not moving inertially, but their motion would still be symmetric and their clocks would still show the same elapsed time each time they met again. And they would both show *different* elapsed times than a clock that was just sitting at rest at the point where they meet on every orbit.

No, he introduced it because you can't model gravity with flat spacetime. His 1905 paper didn't include gravity, and didn't need to. But he knew his 1905 theory couldn't be complete if it didn't cover gravity, so he spent the next ten years figuring out how to make a relativistic theory of gravity.

9. May 23, 2014

### xox

The Haefele Keating experiment proves that the above is false. So does GPS. If you can you do the math describing the above, you could find your errors by yourself.

Last edited: May 23, 2014
10. May 23, 2014

### yogi

Muons created by collision in the earth's atmosphere are moving very fast - so wrt the average velocity of planets like the earth.... relative to the CBR ... you can say with confidence, it is the muon that will experience time slowing - but when the earth is passed by another massive object there are no experiments to determine which one has been accelerated. When acceleration is involved - the situation is not reciprocal.

Point 2 - you can always apply the principle of the "invariance of the interval in all reference frames" and arrive at a value for the realistic time dilation using SR - this gives you the answer that follows from considering your own frame to be at rest - but it will be incorrect if there is a real permanent time dilation since only one clock can be slower. If the two clocks are initially synced while at rest in the same frame ... then every SR problem will involve an acceleration in order to develop a relative velocity - and since we know which one accelerated we know which clock runs slow. Real time dilation involves physical changes in the energy of one frame wrt another - time dilation in SR is a problem of mensuration.

Point 3: The hovering clock is in the acceleration field of the earth

More later

11. May 23, 2014

### Staff: Mentor

The same is true for the muons. If you think there's a difference, then please tell me where the big particle accelerator in the sky is that accelerates the muons.

This is true; proper acceleration is different, physically, from free fall. But that certainly doesn't cover all possible situations where proper time along different worldlines is different. For an example, see below.

Sure, once you've specified the interval--i.e., once you've specified the specific pair of events you are going to use to define the invariant spacetime interval whose elapsed time you are calculating. But that assumes that there is one unique pair of events that can be used to do that. In your Earth-Altair scenario, there isn't; that is, there is no one pair of events in spacetime that all three clocks (the Earth clock, the Altair clock, and the rocket clock) pass through. That's why you need a simultaneity convention.

How are you defining "real permanent time dilation"? For example, are you claiming that the rocket in the Earth-Altair example experiences "real permanent time dilation" compared to Earth and Altair? Remember that there is no unique pair of events that all three clocks in this scenario pass through, as above.

In flat spacetime, yes, *if* there is a unique pair of events that both clocks pass through. If not, you need a simultaneity convention to make this determination.

In curved spacetime, no, you cannot assume that the accelerated clock always runs slower, even if there *is* a unique pair of events that both clocks pass through. I already gave a counterexample: the "hovering" clock over the North Pole runs *faster* than the orbiting satellite clocks, even though the hovering clock is accelerated and the orbiting clocks are in free fall.

More precisely, it has nonzero proper acceleration so it's not in free fall. Well, if that makes a difference to you, consider another clock, which happens to be flying radially outward over the North Pole so that it passes by the two satellites just as they cross there, and whose upward velocity relative to the Earth at that instant is just right for it to free-fall upward until it comes to a stop, and then free-fall back downward, just in time to again pass the two satellites as they meet again over the North Pole on the next orbit. Now there is no proper acceleration anywhere in the scenario; yet the clock that free-falls radially will show *more* elapsed time than the two satellite clocks.

12. May 23, 2014

### jkl71

These aren't in conflict, as long as gravity is negligible SR is sufficient. The proper time is just the length of the world line and the twin that accelerates will have a shorter (as measured by Lorentzian metric) world line, so less time elapsed on their clock. My take is acceleration is important in the sense that it causes the world line of one twin to not be a geodesic, but there's no need to talk about pseudo gravity.

13. May 24, 2014

### stevendaryl

Staff Emeritus
I like the analogy with planar Euclidean geometry:
1. Pick coordinate axes: an x-axis and a y-axis.
2. Relative to these axes, we can compute the slope for a curve at some point along the curve.
3. The slope is defined as the ratio: $s = \dfrac{\delta y}{\delta x}$.
4. The length of any curve is found in terms of the slopes along the curve as follows:$L = \int \sqrt{1 + s^2} dx$
5. Given two points $A$ and $B$, the shortest curve connecting those points is the one with constant slope.
6. Even though the formula for length doesn't involve the rate of change of the slope, it's still the case that the shortest curve has constant slope.

Similar points hold for SR:
1. Pick coordinate axes: an t-axis and a x-axis.
2. Relative to these axes, we can compute the velocity for a trajectory at some point along the trajectory.
3. The velocity is defined as the ratio: $v = \dfrac{\delta x}{\delta t}$.
4. The length of any curve is found in terms of the slopes along the curve as follows:$\tau = \int \sqrt{1 - v^2} dx$ (in units where c=1).
5. Given two events $A$ and $B$, the longest (in proper time) trajectory connecting those points is the one with constant velocity.
6. Even though the formula for proper time doesn't involve the rate of change of the velocity, it's still the case that the longest trajectory has constant velocity.

The analogies are:
• t-axis → x-axis
• y-axis → x-axis
• trajectory → curve
• velocity → slope
• proper time → length
• change in velocity → change in slope
• $\sqrt{1-v^2}$ → $\sqrt{1+s^2}$

The weird part is the occurrence of the minus sign in the proper time formula, while the corresponding equation for length has a plus sign. But most of the other peculiarities of the twin paradox have their analogy in Euclidean geometry: Even though slope is relative to a choice of coordinate axes, the length that is computed using the slope is a number that is independent of that choice. Even though velocity is a relative to a choice of rest frame, the proper time that is computed using the velocity is independent of that choice.

14. May 25, 2014

### jkl71

Me too, IMO analogies like this can be very helpful

15. May 25, 2014

### yogi

While both SR and GR can be used to arrive at the correct answer for the classical twin problem, there is more to the issue. Lets take the one way trip of a clock to Altair - there are two ways to begin - first we consider the spaceship with the clock already in motion as it passes the earth at cruising speed and sync on the fly - this avoids all acceleration and we wish to know if the traveling clock aboard the spaceship will accumulate less time - applying SR we consider the earth frame at rest and tentatively conclude yes - But since no acceleration is involved after sync, the spaceship is an equally valid inertial frame and so the on board clock will have accumulated more time that any of the clocks in the earth-Altair frame when the spaceship arrives at Altair. In this fact situation a comparison of the clocks would not show any difference since one inertial frame is no better than another - the situation is apparently symmetrical - but is it?

Next we consider all 3 clocks (earth, Altair and the on board clock) synced at rest then the space ship is blasted from earth so there is an acceleration phase - we know the answer to this question - this is the laboratory muon - initially it is boosted to near c velocity and definitely exhibits time dilation. This is a non symmetrical situation.

It is often asserted that since the acceleration phase can be ignored if the cruising time of the trip is long compared to the acceleration phase - but this reveals the flaw in the analysis - once the acceleration phase ceases - we still need to account for the greater part of time difference between the two clocks - but if we simply use SR from the point where the acceleration ended-you then have two equally valid inertial frames falsely computing time dilation in the other frame - at arrival, the only net difference in the time will be the result of the initial acceleration.

Einstein attempted to introduce a pseudo G field at the turn around - yes - it gives the correct result - but it does not answer the problem posed -

16. May 25, 2014

### Staff: Mentor

Sure, it would; but which difference it would show depends on how you compare the clocks, i.e., which simultaneity convention you adopt. If you adopt the Earth-Altair simultaneity convention, then the spaceship clock will show less elapsed time than the Earth and Altair clocks. If you adopt the spaceship simultaneity convention, then the Earth and Altair clocks will show less elapsed time than the spaceship clock. There is no contradiction because you are using two different simultaneity conventions, hence you are making two different comparisons. Drawing a spacetime diagram makes all this obvious.

Yes, if you define "symmetrical" as "one inertial frame is no better than another". Any attempt to show an asymmetry in this scenario amounts to privileging one inertial frame over others.

In other words, you are privileging the Earth-Altair frame because the only objects in this version of the scenario that experience no acceleration are the Earth and Altair, which are at rest in this frame. This is fine for this scenario--but as I've already pointed out, this way of picking which frame to privilege does not generalize well.

No, you don't; you just have, to a good enough approximation, the first version of the scenario, where the spaceship never changes speed--it flies by Earth and later flies by Altair at the same speed.

This is not correct. If you think it is, please show your work; please demonstrate, mathematically, how you would justify this claim.

Why not? You already said GR can be used to get the correct solution.

17. May 25, 2014

### Staff: Mentor

You are forgetting relativity of simultaneity here; that phrase "when the spaceship arrives at Altair" means different things to the different observers.

For the earthbound observer, "when the spaceship arrives at Altair" happens at the same time that the earthbound clock has accumulated a greater time than the onboard clock.

For the traveling observer, "when I and the spaceship arrive at Altair" happens at the same time that the earthbound clock has accumulated a smaller time than the onboard clock.

These results will hold whether you consider the acceleration or not; and there is no inconsistency in the results. In particular:
We have two equally valid inertial frames correctly and consistently computing the time dilation in the other frame. The "net difference" to which you refer is an illusion resulting from forgetting the relativity of simultaneity (to see this clearly, specify exactly which values you are subtracting to arrive at this "net difference").

18. May 25, 2014

### yogi

[R] No experiment in SR has ever been verified to yield a net time difference unless acceleration is inferred or included someplace during the experiment. Many notable authors have taken the position that SR is not applicable to the twin paradox. IMO the introduction of a pseudo G force at turnaround is also misleading. Time dilation in SR is illusory if all inertial frames are equal...its real in the sense of a mensuration distortion, but no accumulation of time on one clock relative to another clock in uniform motion can be greater than the other - for all parts of the trip to Altair that are non accelerating, I do not think any conclusions can be drawn since there is no physical consequence to uniform motion.

If you think the time difference will continue to accumulate after the acceleration phase ends - tell me the physics of how that is possible.

19. May 25, 2014

### PAllen

Give me a break. In an obtuse triangle, which part of the path along the short sides the extra length? The extra length does not have location. Similarly, for differential aging there is no time or spacetime location for where age difference 'happens'.

20. May 25, 2014

### stevendaryl

Staff Emeritus
Those notable authors are wrong.

That is just not correct. I really do think that you should look at the analogy with Euclidean planar geometry. You pick a direction, and consider that the x-axis. Relative to that axis, you can compute the length of a path in terms of the slope of the path at each point:

$L = \int \sqrt{1+s^2} dx$

where the slope $s$ is defined by: $s=\dfrac{\delta y}{\delta x}$

It's clear from this formula that for two paths connecting the same two points, the one with $s=0$ will have the smallest length. But how is that possible? Slope is relative to a choice of x-axis; given any two straight lines, we can pick an axis so that one has slope zero, or we can choose a different axis so that the other has slope zero. But length is not relative. It's absolute (in Euclidean geometry, anyway).

The fact is that even though $\sqrt{1+s^2}$ is relative, and $dx$ is relative, the combination $\sqrt{1+s^2} dx$ is absolute. We can use the formula for length to deduce that a path that has constant slope (that is, a straight line) connecting two points will have a smaller length than a path with variable slope connecting the same two points.

It's very similar with SR. The proper time formula is

$\tau = \int \sqrt{1-(v/c)^2} dt$

Although the quantity $\sqrt{1-(v/c)^2}$ is relative, and $dt$ is relative, the combination $\sqrt{1-(v/c)^2} dt$ is absolute. We can use the formula to deduce that a path that has constant velocity (that is, an inertial path) connecting two spacetime points will have a longer proper time than a path with variable velocity connecting the same two points.