Question about Special Relativity

I’m currently reading a science book called “Relativity Visualized”. It’s not a bad pop-science summation of Special Relativity. I’ve hit a stumbling block trying to understand a certain concept.

The book gives an example of identical twins. One is bundled in to a spacecraft and zoomed around at close to the speed of light. The other twin stays put on earth.

When the spacecraft returns, it turns out the that the twin who stayed on earth has aged at a faster rate than his space-zooming brother.

But why?

If all speed and motion is relative, then the spacecraft surely didn’t do anything that the planet earth wasn’t also doing, relative to each other?

If you leave and come back, then you have not been in an inertial reference frame, and that symmetry doesn’t apply. http://en.wikipedia.org/wiki/Twin_paradox

An inertial frame of reference relative to what?

The spaceship accelerated. That’s the difference. I suggest a book called “About Time” by Paul Davies for the best description of this, with some rudimentary math, I’ve ever read.

Thanks for the book tip, but in the meantime, why isn’t the earth also accelerating, relative to the spaceship?

Acceleration is caused by a force acting on an object. Only the spaceship has the force of the engines acting on it (and the traveling twin).

It’s not the acceleration directly, it’s how far down the “gravity” well you are. With one twin on Earth, and the other in a spaceship at 1 g constant acceleration, both will experience the same acceleration of 1 g. From the point of view of the twin on Earth, the difference in the passage of time due to Earth’s gravity well is very small.

From the point of view (frame of reference) of the traveling twin, he’s stationary in a constant gravitational field*. The Earth twin is either very far down the gravity well, with time passing slowly (when the traveling twin is accelerating away from the Earth), or very far up the gravity well, with time passing much faster (when the traveling twin is accelerating towards the Earth). The two effects only partially cancel, and when the traveling twin returns, he’s the one who has aged less.

In short, to look at it from the traveling twin’s point of view, you need to use general relativity, not just special relativity. If you do it correctly, you get the same answer.

ETA: * as in constant throughout space (and obviously this is in addition to all the real gravity due to matter).

This is the key point: Constant-speed movement is relative, but acceleration isn’t. If there’s only two things in the universe and they’re moving at a constant speed, you can’t say which one is moving. But you can see which one is accelerating.

And, of course, Dr. Einstein showed us that there is no way to tell the difference between those two.

I’m not sure what a “gravity well” is, but I think I understand John Mace’s explanation that the spaceship has a force acting upon it that the earth does not. I’m not sure why having a force acted upon you = time dilation, though.

While we are speaking of relativity and time dilation, I’ve got a question I’m too lazy to do the math on. In the new movie Avatar, humans travel to the Alpha Centuri system ( ~4.5 ly ) in about 6 years. I guess that works out to something like .8 c. If some humans make the return trip to earth, how much more time has passed for the earth than for the travelers?

Gravity Wells

Time passes at 60% of the normal rate for the travelers to Alpha Centauri. 12 years would have passed on Earth, 7.2 years for the travelers, ignoring acceleration time and all that. I’m not going to do that math.

The speed of the ship is about 4.5 c-yr / 6 yr = 0.75 c. Note that these numbers are in the Earth’s frame of reference. So when the travelers return to Earth, 12 years will have passed.

The speed gives a gamma factor of 1 / sqrt(1 - 0.75^2) = 1.5. That means from the ship’s point of view, the distance to Alpha Centauri is only 4.5 c-yr / 1.5 = 3 c-yr. And their destination is moving toward them at 0.75 c, so the travelers experience 3 c-yr / 0.75 c = 4 yr in transit. So, including the return trip, the travelers will have advanced 8 yr total.

From the Earth’s point of view, the travelers are time-dilated by a factor of 1.5. So the Earth’s 12 years, becomes 12 yr / 1.5 = 8 yr that will have passed on the traveling ship.

A force acting upon you does not = time dilation. In the scenario in my last post, both twins are experiencing the same force, but the traveling twin still ages less than the Earth-bound twin.

The comic linked to by UncleRojelio is actually a pretty good visual for gravity wells. Differences in height correspond to time dilation. Note that all the wells have a bottom. If you were floating out in space, away from planets and stars, for the comic you’d just have a flat, level surface. If you were accelerating, and took that as your frame of reference, you’d have a sloped surface. Something very far away could then be very high or very low relative to you, hence have a very different flow of time (again, relative to you).

One can resolve the apparent paradox without having to appeal to General Relativity, by realizing that we’re not using two (in which case there would be equivalence), but actually at least three different inertial frames (i.e. frames moving at a constant speed). One, the rest frame of the Earth. Two, the rest frame of the spaceship travelling away from earth. Three, the rest frame of the space ship travelling towards earth.

So, then, let’s take a look at what happens to our two twins: One, Trippy, embarks on a journey to Planet X ten light years away, while the other, Homey, stays on earth. Let’s say at the same time (in the Earth’s rest frame), on a planet ten light years further away than Planet X (so twenty light years away from Earth in the direction of Planet X) another spaceship embarks on a journey in the opposite direction at the same speed as Trippy’s ship (0.866c for convenience, because that gives a Lorentz factor of 2) (let’s call Trippy’s ship Ship 1, and the other one Ship B).

From Earth, the whole thing is simple enough: Trippy hops onto the ship going at 0.866c, goes to Planet X – which takes him 11.55 years --, then he hops onto the ship going the opposite direction, rides that for another 11.55 years, and is back home after 23.1 years – i.e., when Trippy comes back, Homey is 23.1 years older.

Now, to Trippy, things look slightly different – in the rest frame of Ship 1, the distance between Earth and Planet X is length contracted, half as long in fact, since we have a Lorentz factor of 2. Cruising at 0.866c, it thus takes him a cool 5.78 years to get there. And once he’s there, he decides – in no time at all! – that Planet X is really a bit of a shithole, hops onto Ship B, and makes back – again, covering five light years in 5.78 years. Thus, to him, the whole trek took a mere 11.55 years, half of what it seemed like to his brother Homey.

Of course, that’s exactly what Homey expected – Trippy moved at 0.866c for 23.1 years, meaning his clock ticked half as fast as the one on Homey’s wall, registering, in toto, 11.55 years.

However, depending on how well versed Trippy is in Special Relativity, he might be in for a surprise – it sounds entirely reasonable for him to expect that, since he was in an inertial frame of reference all the time and thus can view the rest frame of the Earth as moving, it ought to be Homey that has only experienced half the time passing – to wit, since Trippy’s trip took him 11.55 years, he might expect that Homey ought only have aged 5.78 years!

And the thing is, for each of the two legs of his journey, Trippy is entirely correct in his reasoning – Homey ages 2.89 years during the 5.78 years it takes Trippy to get to Planet X, and again 2.89 years during the journey back. Nevertheless, once Trippy arrives back home, he will find his brother having aged 23.1 years since his departure.

So when does he suddenly accumulate those missing 17.32 years?

The answer to that lies in the relativistic concept of simultaneity. Two events (x[sub]1[/sub], y[sub]1[/sub], z[sub]1[/sub], t[sub]1[/sub]) and (x[sub]2[/sub], y[sub]2[/sub], z[sub]2[/sub], t[sub]2[/sub]) are simultaneous, almost trivially, when t[sub]1[/sub] = t[sub]2[/sub]. Now, the rub is that if two events are simultaneous in one inertial system, they need not be in another that is moving relatively to the first one. That’s because the math you use to ‘translate’ coordinates in one inertial system into coordinates in another inertial system is that of Lorentz transformations, which mix up time and space – crucially, the time coordinate in the system you translate into depends on both time and location in the system you translate from; if we denote with a ’ the coordinates in an inertial frame moving at speed v in x-direction, the relation for the transformed time coordinate is t’ = γ(t - vx/c[sup]2[/sup]), with γ being the Lorentz factor. Hence, even though t[sub]1[/sub] = t[sub]2[/sub], it’s generally not the case that t’[sub]1[/sub] = t’[sub]2[/sub] (because x[sub]1[/sub] ≠ x[sub]2[/sub]), and two events that are simultaneous in one frame of reference may occur at different times in another, relatively moving, one.

Thus, the events on Earth a traveller on Ship 1 would consider simultaneous with himself generally will differ from the events on Earth a passenger on Ship B considers simultaneous with himself; and, when changing over from one ship to the other, the notion of simultaneity changes accordingly.

All that’s left to do, therefore, is to calculate the difference in time between a event on Earth that Trippy on Ship 1 considers simultaneous with himself (t[sub]S[sub]1[/sub][/sub]), and a point in time that Trippy on Ship B considers simultaneous with himself (t[sub]S[sub]B[/sub][/sub]). With the transformation given above, that’s:

t[sub]S[sub]B[/sub][/sub] - t[sub]S[sub]1[/sub][/sub] = γ(t + vx/c[sup]2[/sup]) - γ(t - vx/c[sup]2[/sup]) (with the ‘+’ sign in front of the first v coming from the fact that the inertial frame in which Ship B is at rest moves at v in negative x direction in Earth’s rest frame)
= 2γvx/c[sup]2[/sup]
= 220.866*5 (because c is 1 in lightyears per year, and x is equal to five light years – the distance between Earth and Planet X from the rest frame of either ship)
= 17.32

So, when Trippy makes the jump from one ship to the other, the age of Homey, as calculated by Trippy, ‘jumps’ 17.32 years – there’s no paradox at all!

Of course, this isn’t actually physically possible – generally, you’d have to decelerate the ship to a relative full stop, then re-accelerate it, and decelerate it again at arriving on Earth, making the whole thing quite a bit more complicated; but, since acceleration can be modelled as a series of infinitesimal jumps from inertial systems of infinitesimally different speeds, the reasoning is essentially the same, although in that case, the ageing will occur smoothly, not in a sudden jump, which is much more in line with what one might expect (though I sometimes feel like having taken a great leap in age, particularly on weekend mornings when I’ve been out the night before).

I got different numbers because I went with your .8c speed.

Looks like I killed another one…

Sorry if my post was a bit muddled, I’m not the greatest explainer in the world. However, the point I made isn’t really a very complicated one – so just on the off chance anyone’s still interested, I’ll try to sum up (and if that doesn’t work, either, just disregard it entirely; ZenBeam’s explanation, for instance, is wholly sufficient to eliminate the paradox).

Anyway, the thing is this – it’s absolutely correct (necessary, in fact) for Homey to apply time dilatation to Trippy’s ageing process, concluding that he’ll have aged half the time Homey himself has when he returns (using the figures I gave above). Similarly, it’s perfectly correct for Trippy to view the Earth as the moving frame during each leg of his trip, and conclude that Homey will have aged half the time Trippy himself has on the trip to Planet X, and again half the time on the trip back to Earth.

Nevertheless, there’s no paradox, and Trippy will find Homey having aged twice as much as he has when he returns. The reason for this is that Trippy’s notion of the present moment, his now changes when Trippy changes reference frames, i.e. ‘turns around’ to go back to Earth. No such turnaround exists for Homey – he stays in the same inertial system all the time.

That the present moment – the set of points in space with the same time coordinate – is dependent on relative motion is perhaps one of the most counter-intuitive concepts in special relativity. To make this a little more plausible, consider a train car racing along the tracks at some significant fraction of the speed of light. As the train speeds past a person standing beside the tracks, a flash of light is given off right in the middle of the train car. To a person inside the train car, the distance between the origin of the light and the front and back walls is the same; he will thus observe the light hitting both walls at the same time – simultaneously. However, the speed of light is finite, and the same in all frames of reference. So, a stationary observer will see the light propagate inside the traincar at exactly c; but because the train is moving very fast, during its flight time, the back wall will approach the origin of the flash, and the front wall will recede from it, with the result that the outside observer sees the light reaching the back wall before it reaches the front wall – what’s simultaneous in one frame of reference isn’t in another.

Now, pretty much the same thing happens in the ‘paradox’ – the two frames of reference Trippy occupies have accordingly two notions of present, with the now of the return frame corresponding on Earth to a time just the right amount of years later than the now of the arriving frame to make it all work out; when Trippy ‘turns around’, the set of points simultaneous with him – his hyperplane of simultaneity – shifts or tilts into one in which Homey is the requisite number of years older to reach the right age during Trippy’s journey home so that both of their pictures ultimately agree.