# Symetry, time dilation, twin paradox and all that stuff

1. Dec 3, 2006

### resurgance2001

I spent several months last year trying to understand the twins paradox and didn't get anywhere. The standard answer given is that symetry is broken a/ becuase one of the twins has to accelerate and b/ becuase on the return journey she is actually in a different inertial frame to the outward journey.

My question is this; what is the actual maths that satisfactorally predicts then the outcome of the situation? I have heard and read the same answer about symetry being broken several times but never been able to find someone who can explain the underlying maths

Another (possibly) simpler example I found it the 1971 experiment done with two atomic clocks. One website author showed mathematics where the centre of the Earth was taken as an approximate intertial frame and the relative times found for someone on the surface and then at flying at altitude. The maths and the predictions were very close to the outcome. But my question(s) are these? What is proper time? Why can the centre of the Earth being considered in someway as a preferential frame of reference to consider as having a proper time? That sounds suspiciously like some kind of universal time. - Confusing!

Cheers

Peter

2. Dec 3, 2006

### robphy

"Proper time" is conceptually "wristwatch time".
Every massive object [idealized as a "point particle"] has its own "wristwatch time"... which for simplicity is identical in construction to every other object's wristwatch.

In special- and general-relativity, two objects that meet at two different events will generally disagree with the elapsed wristwatch time between those events. The difference in their wristwatch elapsed times is relatively small for "everyday motions"... however, accurate-enough watches will reveal the difference. In Galilean- and Newtonian-relativity, the difference in their wristwatch elapsed times is zero... i.e. In Galilean- and Newtonian-relativity, there is a [spacetime] path-independence in elapsed time between two events.

Geometrically, proper-time corresponds to the spacetime arc-length of a timelike curve [worldline].

Last edited: Dec 3, 2006
3. Dec 3, 2006

### Staff: Mentor

Here's a worked-out example that uses the full Lorentz transformation equations, not just the length-contraction and time-dilation equations, which are incomplete because they don't include relativity of simultaneity. I worked this out some time ago as a Usenet posting, with the equations in the crude pseudo-LaTex that people usually use on Usenet. I've converted the equations into real LaTex, and may have introduced some mistakes along the way, which I'll fix when I spot them.

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The scenario: You stay behind on Earth while your twin goes on a round-trip space journey to Star Base Alpha, which is 4 light-years away. His ship can travel at a speed of 0.8c, so from your point of view he needs 5 years for the outbound trip and another 5 years for the return trip. The total trip duration is 10 years by your reckoning.

Relativity predicts that your twin experiences less elapsed time because of time dilation:

$$5 \sqrt {1 - {0.8}^2} = 3$$

years for each leg of the trip, and from his point of view the round trip lasts only 6 years.

The relativistic time dilation equation predicts that each twin's clocks "run slower" in the other twin's reference frame. So why can't your twin conclude that the trip must be shorter for you, than it is for him? The answer lies in the fact that your experiences are not symmetrical. Your twin is at rest in two different inertial reference frames, one during the outbound trip and another one during the inbound trip. You remain at rest in a single inertial reference frame during the entire journey. Your twin has to fire his spaceship's engines at the turnaround point. You do nothing.

To convince ourselves that both you and your twin do in fact agree that the elapsed time for the trip is longer on your clock than on his, we need to look carefully at how the two clocks behave in your twin's reference frames (note the plural). In order to do this properly, we need to use the full Lorentz transformation equations, and not just the length-contraction and time-dilation equations, which don't contain the "full story" of relativistic kinematics.

Let's call your reference frame S, in which an event has coordinates (x,t); let's call your twin's reference frame during the outbound trip S', in which an event has coordinates (x',t'); and finally, let's call your twin's reference frame during the return trip S", in which an event has coordinates (x",t").

To simplify the analysis, let's assume that your twin's ship can accelerate so rapidly that it can effectively change velocity instantaneously.

If S and S' are set up so that at the instant when your twin departs (moving in the +x direction with speed v), both you and he are at x = x' =0, and your clock and your twin's clock are synchronized so that they both read zero, we can use the usual textbook version of the Lorentz transformation equations

$$x^\prime = \gamma (x - v t)$$

$$t^\prime = \gamma (t - vx/c^2)$$

or inversely

$$x = \gamma (x^\prime + v t^\prime)$$

$$t = \gamma (t^\prime + v x^\prime / c^2)$$

where $\gamma = 1 / \sqrt {1 - v^2 / c^2}$. In our example, v = 0.8 light-year per year, and c = 1 light-year per year, so $\gamma$ = 1.6667 = 5/3.

Your twin travels at a speed of 0.8c, so in your frame, S, he takes 5 years to cover the 4 light-years to Star Base Alpha, where he turns around (instantaneously). In frame S, his turnaround takes place at x = 4 light-years and t = 5 years.

We calculate the position and time of the turnaround in his frame, S', as follows:

$$x^\prime = 1.6667 [4 - (0.8)(5)] = 0$$

which figures, because he's stationary in frame S', and

$$t^\prime = 1.6667 [5 - (0.8)(4)] = 3$$

So, during the twin's outbound trip (5 years according to you), 3 years elapse on his clock. That is, in your frame, S, his clock runs slower than yours, by a factor of $1 / \gamma = 0.6$.

Now, how does your clock behave in his frame, S'?

First, let's calculate your position in S', at t' = 3 years. In S', you are moving "backwards" (-x' direction) at speed 0.8 light-years per year. So at t' = 3 years, you are located at x' = -(0.8)(3) = -2.4 light-years.

We can now calculate what your clock (which shows "S time") reads at this point in time in your twin's frame, S', using the inverse Lorentz transformation:

$$t = 1.6667 [3 + (0.8)(-2.4)] = 1.8$$

years. So, during your twin's outbound trip (3 years according to him), 1.8 years elapse on your clock, in his frame. That is, in his frame, S', your clock runs slower than his, by a factor of $1 / \gamma = 0.6$.

Now, your twin fires his rocket engines to turn around very quickly so that he is now moving towards you with speed 0.8 light-years per year. He is now in a new reference frame, call it S".

We need a Lorentz transformation between S and S". We can't simply change primes to double primes in the equations that we used before, because S and S" do not coincide at x = x" = 0 and t = t" = 0. We have to use a more general version of the Lorentz transformation:

$$x^{\prime \prime} - x_0^{\prime \prime} = \gamma [(x - x_0) - v (t - t_0)]$$

$$t^{\prime \prime} - t_0^{\prime \prime} = \gamma [(t - t_0) - v (x - x_0) / c^2]$$

Inverse:

$$x - x_0 = \gamma [(x^{\prime \prime} - x_0^{\prime \prime}) + v (t^{\prime \prime} - t_0^{\prime \prime})]$$

$$t - t_0 = \gamma [(t^{\prime \prime} - t_0^{\prime \prime}) + v (x^{\prime \prime} - x_0^{\prime \prime}) / c^2)]$$

where some event (call it a "reference event") has coordinates $x_0$ and $t_0$ in frame S, and coordinates $x_0^{\prime \prime}$ and $t_0^{\prime \prime}$ in frame S". Note that if we set $x_0 = x_0^{\prime \prime} = 0$ and $t_0 = t_0^{\prime \prime} = 0$, we get back the original textbook version of the Lorentz transformation.

For this case, let the turnaround of your twin's ship be the reference event. Then $x_0 = 4$ light-years, $t_0 = 5$ years, $x_0^{\prime \prime} = 0$, $t_0^{\prime \prime} = 3$ years, and v = -0.8c (because now the ship is moving in the -x direction in S). Substituting these into the generalized Lorentz transformation gives

$$x^{\prime \prime} = 1.6667 [(x - 4) - (-0.8)(t - 5)]$$

$$t^{\prime \prime} - 3 = 1.6667 [(t - 5) - (-0.8)(x - 4)]$$

To get the reading on your clock in your twin's frame, just after the turnaound, we substitute x = 0 (you're at the origin of your own frame, S) and t" = 3 (the turnaround time in S"). This gives two equations in two unknowns, t (which is what we really want) and x" (which we get as a bonus):

$$x^{\prime \prime} = 1.6667 [-4 - (-0.8)(t - 5)]$$

$$0 = 1.6667 [(t - 5) - (-0.8)(-4)]$$

which we can solve to get x" = -2.4 and t = 8.2.

Therefore, in your twin's first reference frame, S', just before the turnaround, you are 2.4 light-years behind him and your clock reads 1.8 years; and in his second reference frame, just after the turnaround, you are 2.4 light years in front of him (remember, he's turned around!) and your clock reads 8.2 years. Apparently your position has shifted by 5.8 light-years and your clock has gained 6.4 years during the turnaround! But this has no physical significance as far as you yourself are concerned. These changes are simply because your twin is measuring space-time using a different coordinate system than before.

As an analogy, imagine an object resting on a sheet of paper that has a two-dimensional coordinate system (x,y) drawn on it. Rotate the the paper quickly that the object remains stationary because of inertia (like the old trick of yanking a tablecloth out from underneath the plates and silverware). The object's coordinates change suddenly, but nothing actually happens to the object itself.

The return trip is similar to the outgoing trip as far as elapsed time is concerned. In the classic words of textbook writers , "the details are left as an exercise for the reader." 5 years elapse in your frame, S, and the total elapsed time on your clock, according to you, is 5 years (time at turnaround) + 5 years = 10 years. Meanwhile, in your twin's frame, S", 1.8 years elapse on your clock, so according to your twin, the total elapsed time on your clock is 1.8 years (during his outbound trip) + 6.4 years (the "jump" when he turns around) + 1.8 years (during his inbound trip) = 10 years.

The two of you agree that 10 years have elapsed on your clock. Likewise, you both agree that 6 years have elapsed on his clock. The fact that each of your clocks "runs slow" compared to the other one, during each leg of the trip, does not lead to an actual physical contradiction when the clocks are re-united at the end of the trip.

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For another approach which gives the same result, see posting #3 in this thread:

Last edited: Jan 24, 2007
4. Dec 3, 2006

5. Dec 3, 2006

### resurgance2001

Thanks Jitbell. Now having read the rest of your message I think that I am starting to get it, though I will need to spend some time on the more general Lorentz transformations you give.

Can you say anthing about the case of the real experiments such as the one dones in 1971 and 1996 with ceasium clocks, and the relelvant calculations for those?

6. Dec 3, 2006

### Staff: Mentor

For the 1971 experiment, do a Google search on "Hafele and Keating". You may need to be careful in evaluating what you find, however. Anti-relativity crackpots love to bash that experiment. There are more recent experiments that do similar things, but I don't have any names at the top of my head. You'll probably find some leads here:

Note that this list was made in 2000, so it doesn't include any more recent experiments that might have been done.

Analyzing these experiments is not a simple SR calculation, in fact it's really a GR calculation because the gravity-dependent effects can be as important, or even more important, than the velocity-dependent effects.

As for your earlier question about the "earth center inertial" (ECI) reference frame, it's purely a matter of mathematical convenience. In principle, one can do the calculations using any reference frame, but it's easier in some frames than in others. For things like the H-K experiment (where the planes flew around the world) or analyzing the GPS satellites, it turns out that the ECI frame is mathematically the simplest one to work with.

7. Dec 4, 2006

8. Dec 4, 2006

### kesh

has anyone done a thought experiment of the twin 'paradox' in a closed universe? so one twin would return to the other by just continuing the outward journey, without any stopping and turning round, and no return trip would be necessary.

sorry if this is another idiot "i'm interested in gtr, but not interested enough to actually read the books" type question

9. Dec 4, 2006

### Staff: Mentor

10. Dec 4, 2006

### Severian

I think the key to understanding this is that you can't compare measurements made in different frames without applying a Lorentz transformation. So, in order for the twins to stand side by side to one another and really compare their ages, one of them has to change frame. It isthis act that breaks the symmetry. In other words, the stopping and turning around at Alpha-Centauri is a bit of a red herring. It is really the slowing down at the end which is the important part (i.e the turning around at AC breaks the symmetry, but isn't really needed because the slowing down at the end does too).

Funnily enough, I have to lecture this tomorrow morning.

11. Dec 4, 2006

### kesh

they could read each others clocks when their worldlines momentarily intersect

12. Dec 4, 2006

### Staff: Mentor

As is the speeding up at the the beginning of the trip. One doesn't even need to formulate the twin paradox in terms of twins and a round-trip. Just have the traveler make a one-way trip to Star Base Alpha, and stipulate that Earth and Alpha are at rest with respect to each other, and their clocks and calendars are synchronized. When the traveler arrives at Alpha and steps out of his spaceship, his clocks and calendars are behind those on Alpha, and he is younger than he "ought" to be.

13. Dec 7, 2006

### resurgance2001

In the 1971 experiment, why would the eastbound clock loose time, but the westbound clock gain time? This just seems to add further to the confusion!

14. Dec 7, 2006

### JesseM

Because one was moving along with the earth's rotation, the other against it. In every inertial frame, the average speed of the plane moving against the earth's rotation will be less than the average speed of the plane moving with it.

15. Dec 8, 2006

### resurgance2001

Right, so the clock that is goes west gains time because its relative motion is much less and the gravitional GR effects take over - ,right?

Any way here two pence worth which might get shouted down:

It seems to me that the issue of 'moving clocks goe slowly' is somewhat similar to the QM wavefunction. According to SR two intertial frames moving relative to one another are both entitled to say that the other person's clock is running slower. Who can know which one is running slower? Surely no one, not until one of them decelerates to meet the other and in doing so breaks the symetry. So isn't this (just a bit) like QM where no one knows what the state of the wavefunction is until an observation is made?

16. Dec 8, 2006

### JesseM

You don't need GR to analyze accelerated motion. In a given inertial frame, over a time-interval of t in that frame, a clock moving at constant speed v will be measured to only elapse a smaller interval of $$\sqrt{1 - v^2/c^2}*t$$ due to time dilation. So for a path which has a non-constant speed v(t) as measured in this inertial frame, the amount of time elapsed on the moving clock between two times $$t_0$$ and $$t_1$$ (in terms of the inertial frame's time-coordinates) can be calculated by doing this integral:

$$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$

In effect, you're just breaking the path up into a lot of little segments lasting an infinitesimal time dt where the velocity can be treated as constant during that segment, using the time dilation equation I mentioned earlier to say that the moving clock will elapse a time of $$\sqrt{1 - v^2/c^2} * dt$$ during this segment, and then summing over all these segments (since an integral is like a kind of infinite sum of a function's value over a bunch of infinitesimal increments).
Not really, it's not like there's any objective truth their disagreeing about, it's just a matter of using different coordinate systems to divide up space and time. In a similar way, in Newtonian physics different reference frames disagree about the velocity of a given object, but there's no objective answer. Another example is if you draw a line on a sheet of paper and then overlay an x and y axis on the paper which you can use to determine the slope of the line, the slope you get depends on how you orient the x and y axes, but this is just a matter of an arbitrary choice of coordinate system as well, not a disagreement over anything objective.

The geometric analogy is a pretty good way of thinking about things like the twin paradox, IMO. For instance, if you have two points on a piece of paper, and one path between them that's a straight line and another that curved, then you can use different coordinate systems to calculate the length of each path (you can do this using an equation that looks remarkably similar to the one for calculating the elapsed time on an accelerating clock above--I can give more details if you're interested), but no matter which coordinate system you use you'll always get the same answer for the length of each path, and no matter how you draw the curved path you'll always find that the straight line is the shortest distance between the points. Similarly, if you have two events in space and time (like the events of the two twins departing from each other and the later event of them reuniting), and two worldlines for the twins that cross both events, one of which is a "straight" path through spacetime (no acceleration) while the other isn't, then you can use different reference frames to calculate the time elapsed by a clock moving along each worldline, but all frames will agree on the time elapsed on each clock between the two events (known as the 'proper time' along a given worldline), and you'll always find that the time elapsed along a straight path is greater than the time elapsed along any non-straight path between the same two points in spacetime.

17. Dec 8, 2006

### resurgance2001

Thanks for the help Jesse - I am beginning to see this - BUT!! - I just don't see why then one would not be entitled to do the same summing over al the infinitessimal times from the spaceship (or aircraft's) point of view and find that the Earth bound (stationary) observer's clock is slow! I understand that in reality it doesn't happen that way - but what I don't understand is how the theory is predicting it.

18. Dec 8, 2006

### resurgance2001

Or are we saying that the time interval recorded on each person's clock (the proper time) is the same as would be found by an other inertial observer. In other words, if you take the centre of the Earth as an approximate intertial frame and calculate the expected interval of time for a person on the surface of the earth and the interval of time for a person who flies in a plane for x number of hours - that there will be a difference in these two and that this is the real difference observed on the clocks when the plane lands?

19. Dec 8, 2006

### JesseM

Because the equations of SR, such as the time dilation equation, only work in inertial reference frames. If you use a non-inertial coordinate systems, it will no longer be true that a clock with coordinate velocity v will elapse $$t*\sqrt{1 - v^2/c^2}$$ in coordinate time t, nor will other statements in SR hold like the one that tells you the coordinate velocity of light must always be c. And in flat spacetime there's a clear way to distinguish between inertial and non-inertial observers--if you're weightless you're moving inertially, if you experience G-forces you're moving non-inertially.
Exactly! And you'd get the same answer if you calculated the time elapsed on the clocks in the frame of any observer moving at constant velocity relative to this center-of-the-Earth frame.

I should mention that I'm idealizing to explain how the Hafele-Keating experiment would work in flat spacetime, in the real experiment the effects of gravitational time dilation were also taken into account, so the actual analysis of the time elapsed on each clock required GR. This page mentions that "In this experiment, both gravitational time dilation and kinematic time dilation are significant - and are in fact of comparable magnitude."

Last edited: Dec 8, 2006
20. Dec 8, 2006

### JesseM

To antonio carlos motta--did you read the IMPORTANT! Read before posting thread at the top of this forum? This forum is not a place to advertise your own personal theories, just to discuss the mainstream theories of special and general relativity.

Last edited: Dec 8, 2006