This page was originally done as
something of a joke. After all,
how helpful can it be to solve the Twins Paradox without ever using the
Lorentz transforms? It seems like all the useful content would be
left
out of the solution! Since putting it up, however, I've cleaned
up
the
formula formatting a bit (which was rather rough) and added a
spacetime diagram here,
and all in all I think reading and understanding this particular
approach could be quite helpful to a general understanding of the twins
"paradox". 

Socalled
"time dilation" confuses the daylights out of most people
starting out in SR, and leads one directly into the socalled
"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 2form" defines an "inner product" on
the 4dimensional 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 everpopular
0.866c
Distance = 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
(lightyears), 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
spacetime 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: Spacetime 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 spacetime
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
spacetime 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’
worldlines 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
“Etransformed” 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 “Etransformed
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.