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 a galaxy 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 were "flat" -- that is, neither accelerating nor decelerating -- then 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 isn't exactly flat in that sense, 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
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. Our equations suggested that some photons were affected by an infinite red shift. 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. And yet we do. How can this be?
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 my head around at first is that matrix multiplication is not commutative; that is, it is not always the case that
A · B = B · A
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
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
To get an idea of just how tiny that is, suppose that you were to compute the ball's position using
Here's the actual formula. If the train is travelling at f1c -- that is, the speed of the train is f1 times the speed of light -- and the tennis ball is moving forward, with respect to the train, at f2c, then the speed of the tennis ball with respect to the train tracks is not
v = (f1 + f2) c / (1 + f1f2)
Let's try out a few numbers and see what happens. If f1 is
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
For example, if we add
Now, to make it easier to refer to these numbers, let's give them names. Let's call
w1 = 1/2 c
w2 = 4/5 c
w3 = 13/14 c
w4 = 40/41 c
w5 = 121/122 c
In addition, let's give w0 the value of 0 c -- that is, a speed of zero. From the way we constructed this series of numbers, we can say that w0 plus w1 equals w1, w1 plus w1 equals w2, w2 plus w1 equals w3, w3 plus w1 equals w4, 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 w1 (that is,
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 w0 if you allow the 0 to be rewritten as
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):
wn = (3n - 1) / (3n + 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:
3n = (1 + w) / (1 - w)
For example, we saw that
3n = (81/41) / (1/41) = 81
and sure enough,
log3 81 = 4
In general, then, we can take the log base 3 of both sides of Equation 3 to give us
n = log3 [(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 w1.572, and how to get 1.572 back from that velocity. What's more, we can say with confidence that
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 w1. If we had chosen to set it equal to
In fact, it makes some sense to choose, instead of 3, or 5, or any other number, the base of the natural logarithms,
wn = tanh (n/2)
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.
All of this is important because it relates to red shifting. Photons which move away from one another have a longer wavelength and lower energy. If the object's recession velocity is v, then its red shift is equal to
1 + z = sqrt[(1 + v/c) / (1 - v/c)]
(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
(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
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
Next, consider another galaxy, Galaxy Y, which is
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
And if Galaxy Z lies another
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
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.
Copyright © 2003 Brian Tung
Brian Tung is a computer scientist by day and avid amateur astronomer by night. He is an active member of the Los Angeles Astronomical Society and runs his own astronomy Web site. His previous publications in Strange Horizons can be found in our Archive.
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?)