What's purple and commutes? An Abelian grape.

—Nonsense mathematics riddle

LONG AGO, when I had just entered college, I took a course that was called Calculus H50A, but was in fact an amalgam of matrix algebra and differential equations. I hadn't taken a math course for two years (I'd finished what math there was early in my high school career) and didn't know a thing about matrices.

Matrices can be used for a great many things, such as precisely
specifying the rotation of the Earth or describing how the outputs of an
electrical circuit are related to its inputs, but they are a bit
counter-intuitive, if you've never seen them before. They are not numbers,
for one thing, but arrays or grids of numbers. You can add or subtract
them, if they are the same size, but not otherwise. You can square a
matrix if it is as tall as it is wide, so to speak, but not otherwise.
You can multiply two matrices if the one is as wide as the other is tall,
but not otherwise. You cannot divide one matrix by another at all.
Instead, many matrices (but not all) have what is called an
*inverse*, a sort of reciprocal matrix, so that if you multiply
them together, you get the matrix equivalent of the number 1. If you
think you want to divide matrix *A* by *B*, what you do
instead is multiply by *B*'s inverse.

As if that weren't enough, what was difficult for me to wrap around at first is that matrix multiplication is not commutative; that is, it is not always the case that

For whatever reason, I had great trouble internalizing this oddity. I could understand intellectually what it meant, and I could take care not to assume commutativity when multiplying two matrices, but because of my long experience with ordinary numbers, where multiplication is commutative, it took a long time for me to really "get it."

What finally did it was finding something else that wasn't commutative: three-dimensional rotations. For example, hold your left hand out in front of you, with your thumb and forefinger out so that they make an L. Your forefinger should be pointing upward, and your thumb to the right. Now, do the following rotations in the following order:

- Rotate your hand from right to left by 90 degrees. Now, your thumb should be pointing at you, but your forefinger should still be pointing up.
- Rotate your hand back away from you by 90 degrees. Now your thumb should be pointing up, and your forefinger pointing away from you.

However, if you do those rotations in the opposite order, you don't end up in the same final position. Instead, your thumb will be pointing at you, and your forefinger to the right. Rotations of your hand by 90 degrees don't commute.

As a matter of fact, since matrices are often used to model rotations in three dimensions (or more), I now had a handy visual concept for understanding many matrices; it wasn't just an analogy.

Here's another case where things don't work the way you think they should.

In ordinary circumstances, we're used to a very simple arithmetic for speeds. If I'm aboard a train travelling at 30 m/s, and I throw a tennis ball forward at 30 m/s, then it's intuitively obvious that the tennis ball is travelling at 60 m/s, with respect to the train tracks, since 30 plus 30 equals 60.

Except that that's not exactly right.

The special theory of relativity predicts that velocities don't add like ordinary numbers. Instead, because of time dilation and length contraction, the sum of two speeds is always less than you'd expect from just simple addition. For speeds as low as 30 m/s, the effect is tiny—only about 1 part in 100 trillion.

To get an idea of just *how* tiny that is, suppose that you
were to compute the ball's position using 60 m/s, instead of the correct
value. It would take over a thousand years before the error grew as
large as 1 cm. No wonder that this peculiar aspect of speed wasn't
even conceived of until the turn of the 20th century.

Here's the actual formula. If the train is travelling at
*f*_{1}*c*—that is, the speed of the train is
*f*_{1} times the speed of light—and the tennis ball
is moving forward, with respect to the train, at
*f*_{2}*c*, then the speed of the tennis ball
with respect to the train tracks is not
(*f*_{1} + *f*_{2}) *c*, but
instead

EQUATION 1

*v* =
^{
(f1 + f2) c
} /
_{
(1 + f1f2)
}

Let's try out a few numbers and see what happens. If
*f*_{1} is 1/4, and *f*_{2} is 1/4,
then Equation 1 gives us *v* = 8/17 *c*—not quite
half the speed of light, as you'd expect from adding a quarter to
a quarter, but close. If *f*_{1} is 1/2, and
*f*_{2} is 3/4, then *v* = 10/11 *c*.
We also get the same answer if *f*_{1} is 3/4, and
*f*_{2} is 1/2.

Notice that it doesn't matter if we switch the speed of the
train and the speed of the tennis ball (with respect to the train)—we
get the same answer in either case, since both the sum in the numerator
and the product in the denominator can be reversed without affecting
the result. That suggests that even if these speeds don't add like
ordinary numbers—in that 1/2 *c* and 1/2 *c* don't make
*c*—they might share some characteristics with ordinary
numbers. [1]

For example, if we add 1/2 *c* and 1/2 *c* according
to Equation 1, we don't get *c*, we get 4/5 *c*. If we
add another 1/2 *c* to that, we get 13/14 *c*. And if we
add another 1/2 *c* to *that*, we get 40/41 *c*.
We can keep on adding increments of 1/2 *c*, and always end up
with another speed that's still under *c*, although these speeds
keep getting closer and closer to *c*. (Don't take my word for
it—figure it out for yourself!)

Now, to make it easier to refer to these numbers, let's give them names. Let's call

.

.

.

In addition, let's give *w*_{0} the value of 0
*c*—that is, a speed of zero. From the way we constructed this
series of numbers, we can say that *w*_{0} plus
*w*_{1} equals *w*_{1},
*w*_{1} plus *w*_{1} equals
*w*_{2}, *w*_{2} plus *w*_{1}
equals *w*_{3}, *w*_{3} plus
*w*_{1} equals *w*_{4}, and so
forth—assuming that we define "plus" to operate on speeds as
described in Equation 1.

What if we don't always add increments of *w*_{1}
(that is, 1/2 *c*), but try instead to add other numbers in the
series? For example, what do you get if you add *w*_{2}
to itself? Equation 1 tells us that the result is 40/41 *c*,
which turns out to be none other than *w*_{4}. Similarly,
if you add *w*_{2} to *w*_{3}, you get
121/122 *c*—that is, *w*_{5}. In fact, for any
two numbers in the series, *w _{m}* and

So far, so good. Now, is there any way to characterize the *w*
series without resorting to Equation 1? As listed above, the fractions
have a pleasing regularity to them—the numerator is always 1 less than
the denominator. (This is even true of *w*_{0} if you
allow the 0 to be rewritten as 0/1.) The size of the numerator and
denominator also seem to increase smoothly and with accelerating pace
as the series proceeds. That suggests that there might very well be
a simple rule for generating the *w* sequence.

Things become more suggestive still when we note that if we add the
numerator and denominator of each fraction, and if we treat 0 as 0/1,
we get the simple series 1, 3, 9, 27, 81, 243, … which are just the
powers of 3. Then simple algebra suffices to give us the following
expression for the *w* series (omitting the *c* for now):

EQUATION 2

*w*_{n} = (3^{n}–1) /
(3^{n}+1)

(Try it!) We can also turn things around; if we're given a
velocity *w*, we can find out what order it comes in, in the
sequence. We can use algebra to get the following from Equation 2:

EQUATION 3

3^{n} = (1+*w*) / (1-*w*)

3

For example, we saw that *w _{4}* = 40/41. Equation 3
tells us that

3^{n} = (81/41) / (1/41) = 81

and sure enough, 3^{4} = 81. Another way of saying that is
that the log of 81, to the base 3, is equal to 4. In standard notation,
that would be written

log_{3} 81 = 4

In general, then, we can take the log base 3 of both sides of Equation 3 to give us

EQUATION 4

*n* = log_{3} [(1+*w*) / (1-*w*)]

None of this, incidentally, depends at all on *n* being an
integer. We could set *n* equal to 1.572, say, and Equations 2
and 4 would tell us how to get the velocity corresponding to
*w*_{1.572}, and how to get 1.572 back from that
velocity. What's more, we can say with confidence that
*w*_{1.572} + *w*_{2.164} =
*w*_{4.736}, since 1.572 + 2.164 = 4.736. All these
properties are simply the consequences of the special theory of
relativity.

Since the number 3 figures prominently in Equations 2 and 4, you
might get the idea that it also figures prominently in special relativity.
But that turns out not to be the case; there's no real significance to
the number 3. Actually, it's just an artifact of our choice for
*w*_{1}. If we had chosen to set it equal to 2/3 instead
of 1/2, then our series of *w* numbers would have been
0/1, 2/3, 12/13, 62/63, 312/313, etc. Note that the numerator and
denominator of each fraction sum up to the series 1, 5, 25, 125, 625,
etc.—the powers of 5. So if we had chosen to set *w*_{1}
equal to 2/3, the number 5 would have taken on significance, and 3
wouldn't have appeared anywhere. Of course, the new *w* numbers
would no longer be the same as the old ones, but they would still
follow the same rules of addition, including the property that
*w _{m}* +

In fact, it makes some sense to choose, instead of 3, or 5, or any
other number, the base of the natural logarithms, *e* = 2.71828+.
The reason is that Equation 2 then can be written in terms of the
hyperbolic tangent function, tanh:

and in fact, some formulas relating to steady acceleration at faster-than-light speeds involve tanh and the other hyperbolic trigonometric functions. (See, for example, my article, "Relativistic Travel," in the February 2003 issue of Sky and Telescope.)

Incidentally, one thing that we've been dancing around without mentioning explicitly is that all of the fractions, no matter how we choose our magic numbers, are less than 1—that is, less than the speed of light. That's good, or else we'd have identified a way to accelerate gradually to the speed of light or beyond, and that is something that special relativity specifically forbids.

In my last essay, "The Unwinnable Race," I described how we discovered that the galaxies were each racing away from one another. Edwin Hubble showed that the speed with which the galaxies receded from us was proportional to its distance from us, so that if Galaxy A is three times as far from us as Galaxy B is, then Galaxy A is receding from us three times as fast as Galaxy B is. This property is now known as Hubble's Law, and the constant of proportionality—that is, the number that tells us how fast it's going for every unit of distance that separates it from us—is called Hubble's Constant.

Hubble's Constant is not a fundamental constant of nature. If the
universe's expansion were neither accelerating nor decelerating, the
recession velocity of M87—a galaxy in the Virgo cluster about 60
million light-years away—would have been the same five billion years ago
as it is now. But five billion years ago, M87 was considerably closer to
us than it is now. That means that Hubble's Constant must have been
larger than it is now. And if things continue as they are now, it will
be smaller in the future than it is now. (To be sure, there are recent
indications that the universe's expansion is *not* constant, but
has decelerated in the past, and is now accelerating. However, that only
further underscores that Hubble's Constant isn't really constant.)

But even if Hubble's Constant isn't really constant in time, it's still thought to be applicable everywhere in the observable universe. If it's 75 km/s/Mpc here, it should be 75 km/s/Mpc as measured in M87, or the most distant quasar we can see. If that's the case, then it appears there should be a limit to how far we can see. If an object is 4,000 Mpc away from us, and the Hubble Constant is 75 km/s/Mpc, then the recession speed of that object should be 300,000 km/s, which is the speed of light.

In "The Unwinnable Race," I described how photons from such an object
could get to us, even though its recession velocity was the speed of
light (or even if it was greater than that). But a new problem arose.
Those photons don't have the same energy that they did when they were
first emitted by the object. Instead, they have been red-shifted, so
that they have a longer wavelength and lower energy. If the object's
recession velocity is *v*, then its red shift is equal to

EQUATION 5

1 +*z* = sqrt[(1+*v*/*c*) /
(1-*v*/*c*)]

1 +

(Ordinarily, the red shift is defined as just *z* itself,
but I find that inelegant for reasons that should soon become clear.)
Therefore, if *v* is 3/5 of the speed of light, then the number
under the square root is 8/5 divided by 2/5, or 4, and then the red
shift is equal to 2. That is, the arriving photons have 2 times the
wavelength and 1/2 the energy as the photons that originally left the
object. If *v* is 4/5 of the speed of light, then the red shift is
3, and the arriving photons have 3 times the wavelength and 1/3 the
energy.

(Now you see why I don't like the definition of red shift to be
*z* itself. Historically, it referred to the amount by which
the wavelength shifted, not the factor by which it was multiplied.
If you add the original wavelength to the shift in wavelength, then
you get the resulting wavelength; that's why the extra 1 is in there.
But in order to figure out what the resulting wavelength, you might as
well just add the 1 back in and multiply by the original wavelength.)

You might expect that if *v* is 5/5 of the speed of light—that
is, the speed of light itself—the trend would continue and the red shift
would be 4, but that's not true. Instead, Equation 5 tells us that the
red shift will be infinite. That is, the arriving light would have its
wavelength stretched infinitely long.

Now, an infinitely red-shifted photon has no energy. It cannot stimulate any detector, whether it's our eye at the telescope, or any other imaging device, like a camera. We should therefore not be able to see it, even if the photons are able to reach us.

But look at the expanding universe in a new light: step by step. Instead of using Hubble's Law directly to the furthest objects, let's get there, one galaxy at a time. If Galaxy X is 10 Mpc away, then assuming a Hubble's Constant of 75 km/s/Mpc (as we'll do from here on out), it must be receding from us at 750 km/s, about 1/400 the speed of light. Observers on Galaxy X would say the same thing about our Milky Way Galaxy.

Next, consider another galaxy, Galaxy Y, which is 10 Mpc away from Galaxy X, but in the opposite direction from us. If Hubble's Constant holds everywhere in the observable universe—and there's no reason to think it shouldn't—then Galaxy Y will be receding from Galaxy X at 750 km/s.

How fast is Galaxy Y receding from us? We could use Hubble's Law directly on it, but instead, let's just add the velocities. Since Galaxy X is receding from us at 750 km/s, and Galaxy Y is racing ahead of it at another 750 km/s, it would seem that Galaxy Y should be going away from us at 1,500 km/s—except that as we've already seen, velocities don't add that way. We should, instead, get something smaller than 1,500 km/s. Not much smaller—the difference is about 1 part in 16,000—but smaller.

And if Galaxy Z lies another 10 Mpc beyond Galaxy Y, *its*
recession velocity will be measured, not at 2,250 km/s, which is what
you'd expect, but again, something a little less. What's more, the
difference this time is even greater, about 1 part in 5,000.

If we were to imagine a chain of galaxies, extending to any distance we like, we end up with a recession velocity that is the "sum" of lots of terms of 750 km/s, but this "sum" is computed according to the equations above. And since this sum is always less than the speed of light, the red shift never does reach infinity, and we can always see them. It does become harder and harder to detect them, but at least it isn't fundamentally impossible.

You may be wondering, by the way, what this means for Hubble's Law. Since the sum of two velocities is always less than you'd expect, does that mean that the relationship of distance to velocity isn't quite as linear as Hubble's Law suggests?

The answer to that is yes, but it does confuse people because there's more than one way to measure distance and velocity. We can measure them as some omniscient observer from outside the universe would, or we can measure them the only way we really can—from within the universe. Hubble's Law holds as long as you measure distance from the same perspective that you measure velocity.

But that's an essay for another time.

[1] So the speeds of the train and the tennis
ball *do* commute (with respect to addition), even if they don't add
like ordinary numbers. (That makes the train a commuter train. Isn't
that funny?)

Copyright (c) 2003 Brian Tung