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The Twins Problem, Linear Version --
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So-called "time dilation" confuses the daylights out of most people starting out in SR, and leads one directly into the so-called "paradoxes". I really think it might have been better if no one had ever coined the term "time dilation". Similarly, "gamma", with its simplistic interpretation as a dilation/contraction factor, is a rich source of confusion.
So in this version, we dispense with Lorentz, we ignore Gamma, we don't use the word "dilation", and just do the problem "the easy way", by computing proper distances.
A Brief Explanation of the Technique
The "metric tensor" or "metric 2-form" defines an "inner product" on the 4-dimensional space in which relativity is normally studied. A general "inner product" is just like the "dot product" in ordinary Cartesian space, except that the formula may be more general than the simple sum of the products of the coordinates which defines ordinary dot product.Dot product:
Lorentz metric inner product, in an inertial frame:
The Lorentz metric tensor in an inertial frame is given as:
and we can apply it to two column vectors, using ordinary matrix multiplication, like this:
where the transpose of a column vector is a row vector. If we take the inner product of a vector with itself, we get the square of its length, just as we do in ordinary Cartesian space (save that the squared length may be negative).
In an inertial frame, in flat space, for a particle which is at rest in the frame, the elapsed time between two events on the particle's world line (or two events in a person's life) can be found by taking the square root of the absolute value of the inner product of the coordinate distance between the points (remember, the particle is at rest so only its time coordinate is actually changing):
Now, here's the first tricky part: As long as we get the same answer, it doesn't matter how we evaluate the "inner product" distance between the two events.
And the second tricky part: When we change frames of reference (i.e., change coordinate systems), we transform the metric so that the inner product of any two vectors doesn't change. So, it doesn't matter what coordinates we use to find the elapsed time between two events -- it will always come out the same.
And the third tricky bit: Lorentz transforms preserve the form of the metric. Thus, when we use a Lorentz transform to move from one coordinate system to another, we don't need to explicitly transform the metric tensor -- we can ignore the whole issue, because it doesn't change. If we use any other coordinates, however, we need to explicitly transform the metric tensor, and when we compute "proper distances" we need to use the transformed metric tensor to do it. But as long as we do that, we'll still get the same answer.
Finally, the biggest issue in the "Twins" problem is how old each twin is at the end of the experiment. In other words, what is the total elapsed time each twin experiences? To compute that, all we need to do is find the "proper distance" between each pair of events in each twin's life, and sum the distances -- those distances are exactly the elapsed times experience by the twins. So, there's no need to transform into the traveler's coordinate system; we can just evaluate the distances in the Earth coordinate system, and we'll get the correct answer. What I'm going to do on this page is find the answer using Earth coordinates, and then find the answer again using coordinates which are more closely related to the traveling twin.
But Wait -- What's Really Happening?
"Really" is a slippery word. But we can certainly ask "What's happening with the math that makes the moving twin age less?" and that has a couple of sensible answers -- but a simplistic notion of "time dilation" isn't part of the explanation. See the discussion of the metric in Against Gamma for more information.The Problem
Velocity = the ever-popular 0.866cDistance = 4 LY, which is about how far away Alpha Centaurus is.
Time is measured in years, distance measured in LY; C is 1 LY/Y.
Stationary frame = Earth frame = an inertial frame in which the metric is η = diag(-1,1), as shown above in eq. (2) and (3) (but restricted to 2 dimensions, 1 of time and 1 of space).
Coordinates in that frame are (t, x).
Twins are named "H" and "T" (Home and Traveler).
In special relativity, handling acceleration is awkward at best, so we just treat the problem as four events labeled A through D, separated by periods of linear motion, each event given with its Earth coordinates:
A: at (0,0) Twin T leaves Earth at V=0.866, t = x = 0.
B: at (4.619, 4) Twin T arrives at Alpha Centaurus and reverses course
C: at (4.619, 0) Twin H looks at calendar, says "T's at Alpha Centaurus!"
D: at (9.238, 0) Twin T arrives home, H and T shake hands
In plain English, event A is at the origin, t = x = 0. Events B and C both take place at Earth time 4.619 (time measured in years), but at widely separated locations: Event B as at Alpha Centaurus, with Earth x coordinate +4 (light-years), while event C is on Earth with Earth x coordinate 0. Event D is back on Earth, with Earth x coordinate 0 again, but the Earth time at event D is 9.238.
That's the ground work. So let's solve it!
The Solution: First, in the frame of Earth, where everything's easy:
Twin H, while staying at home, "travels" along his world line from event A to event C, and then from event C to event D. He ages by his proper elapsed time, which is just the sum of those two proper distances: the distance from A to C and the distance from C to D. We’ll find these by applying eq (4).Just to avoid being too opaque, I'll write out the matrix multiplication in full, showing the metric, η, for the first case. Note that transpose of a row vector is a column vector, which is what those "T" superscripts are about: they’re just to make the matrix multiplications work out.
or more concisely, with the metric shown as η and all vectors written horizontally,
We also need the "coordinate distance" from D to C:
Then, finding the “proper distance” associated with that “coordinate” distance, we see:
so H ages by 4.619 + 4.619 = 9.238 years.
Twin T, on the other hand, travels from A to B, and then from B to D. He ages by his elapsed proper time, which is just the sum of those two distances – the distance from A to B and the distance from B to D:
Again, we need the coordinate distance between B and D:
and we see that the “proper” distance from B to D is:
so T ages by 2.310 + 2.310 = 4.620 years.
And we see that H aged 9.238/4.620 = 2.000 times as much as T.
(See the gamma? Pretty well hidden, isn't it? No gamma, no dilation, no apparent paradox. Just distances.)
Just a moment – could we see that in a picture, please?
This has been cute, but I’m afraid it hasn’t been terribly informative. Let’s draw a space-time diagram to try to get a better notion of what’s going on.To do this right, we need to know how fast the traveler’s clock appears to be ticking, as viewed by observers who are stationary in Earth’s frame of reference.
Since the traveler moved at constant velocity, we can find this just by dividing the amount of elapsed time his clock registered during the trip by the amount of time an Earth clock measured during the trip.
And that’s just,
where I’ve used τ for time as measured by the traveler.
With that in hand, we can draw the diagram. Drawn to scale, it’s rather bulky, which is unfortunate; you’ll probably have to scroll to see the whole thing; see figure 1.
Figure 1: Space-time Diagram in Earth's Frame of Reference:
This requires a bit of explanation, I think.
The X axis is horizontal, and the time runs on the vertical axis. The coordinates are those of Earth’s frame of reference. Earth remains at location x=0, and so just moves straight up the time axis. Alpha Centaurus remains at location x=4 and so also just moves straight up the time axis. The traveler, who follows the blue line, moves from Earth to Alpha Centaurus at 0.867C, and so moves diagonally across the diagram, from Earth to Alpha Centaurus. Upon arriving at Alpha Centaurus, the traveler immediately reverses direction and comes back, again moving diagonally across the diagram.
Time is marked off in years. Along the time axis, the values shown are time on Earth (since this is, after all, showing Earth’s frame of reference). Along the traveler’s line, there are tick marks showing time as it passes according to the traveler’s clock. These are determined from equation (S.1), above. There is one tick mark every six months of traveler time.
I’ve added something new, as well. Every six months, Earth sends a flash of light to the traveler. The paths followed by the flashes from Earth are shown in red. Every six months, by his own clock, the traveler sends a flash of light back to Earth. The paths of the traveler’s flashes are shown in green. This provides us, in a very graphic way, with the “porthole view” as seen by an Earth observer and as seen by the traveler: The number of flashes the traveler sees gives him a direct measure of how much time has passed on Earth. The number of the traveler’s flashes seen by the Earth observer gives that observer a direct measure of how much time has passed for the traveler.
The moments when the traveler receives a flash are marked with red dots on the traveler’s line. Note how few there are on the outbound path: In the entire trip from Earth to Alpha Centaurus, T sees just one flash from Earth! He sees H as aging very, very slowly during this period. But on the return path, T sees the flashes coming in very rapidly – they’re spaced much more closely; he sees H as aging very quickly on the way back. Counting the red dots, we see that T sees 17 flashes on the way home, while he saw just one on the entire trip out.
The moments when H receives a flash from T are marked with green tick marks on Earth’s path (the time axis). Notice that, for the first eight and a half years, the flashes arrive very slowly: H sees just 4 flashes in that whole time. Then, around 8.6 years after T blasted off from Earth, H finally sees through his telescope that T has arrived at Alpha Centaurus and turned around. From that point on, the flashes come thick and fast: In the last few months before T’s arrival back home, H sees five more flashes.
There is quite a bit more information that can be wrung from the space-time diagram. One could read off the Doppler shift with little effort, for instance. Note, furthermore, that this diagram was constructed geometrically: I laid it out with a pencil and ruler on an old sheet of graph paper. What little calculation was required was done using a slide rule which happened to be nearby. (Here’s the draft of the diagram, in case anyone’s interested.) The rendered version looks reasonably slick but no computer time at all is needed to produce a useful ST diagram! In general, if you’re having a hard time understanding what happens in some problem in special relativity, it can help to draw a space-time diagram. As long as C is 1, all flashes of light travel on 45 degree lines, which is easy to do “by eye” when using graph paper! Many objects’ world-lines can be put in immediately, to good approximation, just by comparing the slope of the lines – their “velocities” -- with the slope of the line followed by a beam of light.
It would also be straightforward to construct a diagram in any other coordinates (the traveler’s, for example). One need merely transform the coordinates of the end points of the lines, and then draw in the connecting lines between them.
And now, digression over: Let’s get on with looking at the problem from the traveler’s point of view, as I said I would do back at the start.
Finally, in the frame of the traveler, which is a bit more interesting:
Well, we need some other coordinates. But I don't want to compute gamma or fiddle with the Lorentz transform, so let's just use the coordinates Einstein used in his famous 1905 paper. If they were good enough for Einstein they're good enough for me:
OK? Sound good? Then let's do it. Call these the “E coordinates”, and call this coordinate transform the "E" transform; in matrix form, it looks like this:
Here are the event coordinates in the “E frame”, for the same set of four events, found by using the E transform:
A: (0,0) -- The story begins, traveler time 0, traveler location 0
B: (4.619, 0) -- Twin T gets to AC and turns around
C: (4.619, -4) -- Twin H, still on Earth, looks at the calendar
Oh, oops, I guess we should use yet another coordinate system for T to come home, shouldn't we?
Nah, that's too much trouble. Let's just use the same coordinates for all the events:
D: (9.238, -8) -- Twin T arrives home. The coordinate system has continued on its merry way, though, so the space coordinate looks a little goofy: the twin is 8 lightyears the the “left” of the origin in the moving frame. Oh, well.
Now, the coordinate differences in the “E frame” look pretty simple:
B - A = (4.619, 0)
C - A = (4.619, -4)
D - C = (4.619, -4)
D - B = (4.619, -8)
Let's find the proper distances. No prob, right?
But wait, there's a slight hitch -- the metric's not quite right, because the "E" transform didn't preserve it. We need to adjust it a bit. To assure that distances measured by the metric are the same in both coordinate systems, we need to (implicitly) transform all vectors back to the original coordinate system before operating on them with the old η metric. We do that by multiplying on the left and right by the inverse transform, which will then automatically convert all “E-transformed” vectors back to “Earth frame” vectors when we apply the new metric. It looks like this:
E-1 is just E with the sign flipped on v (as we can check just by multiplying E and E-1 together and noting that the result is I) so this is
or
And we can now find the proper distances between events, using the equivalent to eq (5) in the new coordinates:
where "[delta]" is a row vector representing a coordinate distance in “E-transformed coordinates”,
Again, the twins age by their elapsed proper times, so
and
and the ratio of their ages is 9.238/4.620 = 2.000.
The distances had to work out the same way from the traveler's point of view, because we explicitly adjusted the metric to assure that they would.
Alternatively, if one uses the Lorentz transform, it preserves the metric and no adjustment is necessary. Since the metric is preserved in that case, the same answer is guaranteed to come out.