Science

What exactly is a second?

Posted on by John

The previous post looked into the common definition of Unix time as “the number of seconds since January 1, 1970 GMT” and why it’s not exactly true. It was true for a couple years before we started inserting leap seconds. Strictly speaking, Unix time is the number of non-leap seconds since January 1, 1970.

This leads down the rabbit hole of how a second is defined. As long as a second is defined as 1/86400 th of a day, and a day is the time it takes for the earth to rotate once on its axis, there’s no cause for confusion. But when you measure the rotation of the earth precisely enough, you can detect that the rotation is slowing down.

Days are getting longer

The rotation of the earth has been slowing down for a long time. A day was about 23½ hours when dinosaurs roamed the earth, and it is believed a day was about 4 hours after the moon formed. For most practical purposes a day is simply 24 hours. But for precision science you can’t have the definition of a second changing as the ball we live on spins slower.

This lead to defining the second in terms of something more constant than the rotation of the earth, namely the oscillations of light waves, in 1967. And it lead to tinkering with the calendar by adding leap seconds starting in 1972.

Cesium

You’ll hear that the second is defined in terms of vibrations of a cesium atom. But what exactly does that mean? What about the atom is vibrating? The second is not defined in terms of motions inside an atom, but by the frequency of the radiation produced by changes in an atom. Specifically, a second has been defined since 1967 as

the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom.

Incidentally, “cesium” is the American English spelling of the name of atomic element 55, and “caesium” is the IUPAC spelling.

The definition of a second raises several questions. Why choose cesium? Why choose that number of periods? And what are hyperfine levels and all that? I’ll attempt to answer the first two questions and punt on the third.

OK, so why cesium? Do we use cesium because that’s what atomic clocks use? And if so, why do atomic clocks use cesium?

As I understand it, the first atomic clocks were based on cesium, though now some atomic clocks are based on hydrogen or rubidium. And one reason for using Cs 133 was that it was easy to isolate that particular isotope with high purity.

Backward compatibility

So why 9,192,631,770 periods? Surely if we’d started from this definition we’d go with something like 10,000,000,000 periods. Clearly the number was chosen for backward compatibility with the historical definition of a second, but that only approximately settles the question. The historical definition was fuzzy, which was the point of the new definition, so which variation on the historical definition was used for backward compatibility?

The time chosen for backward compatibility was basically the length of the year 1900. Technically, the number of periods was chosen so that a second would be

the fraction 1/31556925.9747 of the tropical year for 1900 January 0 at 12 hours ephemeris time.

Here “tropical year” means the time it took earth to orbit the sun from the perspective of the “fixed stars,” i.e. from a vantage point so far away that it doesn’t matter exactly how far away it is. The length of a year varies slightly, and that’s why they had to pick a particular one.

The astronomical definition was just motivation; it has been discarded now that 9,192,631,770 periods of a certain radiation is the definition. We would not change the definition of a second if an alien provided us some day a more accurate measurement of the tropical year 1900.

Unix Time and a Modest Proposal

Posted on by John

The time it takes earth to orbit the sun is not a simple multiple of the time it takes earth to rotate on its axis. The ratio isn’t even constant. The time it takes earth to circle the sun wobbles a bit, and the rotation of the earth is slowing down slightly.

The ratio is around 365.2422. Calling it 365 is too crude. Julius Caesar said we should call it 365 1/4, and that was good enough for a while. Then Pope Gregory said we really should use 365 97/400, and that’s basically good enough, but not quite. More on that here.

Leap seconds

In 1972 we started adding leap seconds in order to synchronize the day and the year more precisely. Unlike leap days, leap seconds don’t occur on a strict schedule. A leap second is inserted when a committee of astronomers decides one should be inserted, about every two years.

An international standards body has decided to stop adding leap seconds by 2035. They cause so confusion that it was decide that letting the year drift by a few seconds was preferable.

Unix time

Unix time is the number of seconds since the “epoch,” i.e. January 1, 1970, sorta.

If you were to naively calculate Unix time for this coming New Year’s Day, you’d get the right result.

New Year’s Day 2025

When New Year’s Day 2025 begins in England, the Unix time will be

(55 × 365 + 14) × 24 × 60 × 60 = 1735689600

This because there are 55 years between 1970 and 2025, 14 of which were leap years.

However, that moment will be 1735689627 seconds after the epoch.

Non-leap seconds

Unix time is the number of non-leap seconds since 1970-01-01 00:00:00 UTC. There have been 27 leap seconds since 1970, and so Unix time is 27 seconds behind elapsed time.

Leap year analogy

You could think of a second in Unix time as being 1/86400 th of a day. Every day has 86400 non-leap seconds, but some days have had 86401 seconds. A leap second could potentially be negative, though this has not happened yet. A day containing a negative leap second would have 86399 seconds.

The analogous situation for days would be to insist that every year have 365 days. February 28 would be the 59th day of the year, and March 1 would be the 60th day, even in a leap year.

International Atomic Time

What if you wanted a time system based on the actual number of elapsed seconds since the epoch? This is almost what International Atomic Time is.

International Atomic Time (abbreviated TAI, from the French temps atomique international) is ahead of UTC [1] by 37 seconds, not 27 seconds as you might expect. Although there have been 27 leap seconds since 1972, TAI dates back to 1958.

So New Year’s Day will start in England at 2025-01-01 00:00:37 TAI.

A Modest Proposal

It seems our choices are to add leap seconds and endure the resulting confusion, or not add leap seconds and allow the year to drift with respect to the day. There is a third way: adjust the position of the earth periodically to keep the solar year equal to an average Gregorian calendar day. I believe this modest proposal [2] deserves consideration.

Kepler’s law says the square of a planet’s orbital period is proportional to the cube of the semi-major axis of its orbit. This means that increasing earth’s orbital period by 1 second would only require moving earth 3.16 km further from the sun.

***

[1] UTC stands for Universal Coordinated Time. From an earlier post,

The abbreviation UTC is an odd compromise. The French wanted to use the abbreviation TUC (temps universel coordonné) and the English wanted to use CUT (coordinated universal time). The compromise was UTC, which doesn’t actually abbreviate anything.

[2] In case you’re not familiar with the term “modest proposal,” it comes from the title of a satirical essay by Jonathan Swift. A modest proposal has come to mean an absurdly simple solution to a complex problem presented satirically.

Starlink configurations

Posted on by John

My nephew recently told me about being on a camping trip and seeing a long line of lights in the sky. The lights turned out to be Starlink satellites. It’s fairly common for people report seeing lines of these satellites.

Four lights in the sky in a line

Why would the satellites be in a line? Wouldn’t it be much more efficient to spread them out? They do spread out, but they’re launched in groups. Satellites released into orbit at the same time initially orbit in a line close together.

It would seem the optimal strategy would be to spread communication satellites out evenly in a sphere. There are several reasons why that is neither desirable or possible. It is not desirable because human population is far from evenly distributed. It’s very nice to have some coverage over the least-populated places on earth, such as Antarctica, but there is far more demand for service over the middle latitudes.

It is not possible to evenly distribute more than 20 points on a sphere, and so it would not be possible to spread out thousands of satellites perfectly evenly. However there are ways to arbitrarily many points somewhat evenly, such as in a Fibonacci lattice.

It’s also not possible to distribute satellites in a static configuration. Unless a satellite is in geostationary orbit, it will constantly move relative to the earth. One problem with geostationary orbit is that it is at an altitude of 42,000 km. Starlink satellites are in low earth orbit (LEO) between 300 km and 600 km altitude. It is less expensive to put satellites into LEO and there is less latency bouncing signals off satellites closer to the ground.

Satellites orbit at different altitudes, and altitude and velocity are tightly linked. You want satellites orbiting at different altitudes to avoid collisions, they’re orbiting at different velocities. Even if you wanted all satellites to orbit at the same altitude, this would require constant maintenance due to various real-world departures from ideal Keplerian conditions. Satellites are going to move around relative to each other whether you want them to or not.

Related posts

GPS satellite orbits

Posted on by John

GPS satellites all orbit at the same altitude. According to the FAA,

GPS satellites fly in circular orbits at an altitude of 10,900 nautical miles (20,200 km) and with a period of 12 hours.

Why were these orbits chosen?

You can determine your position using satellites that are not in circular orbits, but with circular orbits all the satellites are on the surface of a sphere, and this insures that certain difficulties don’t occur [1]. More on that in the next post.

To maintain a circular orbit, the velocity is determined by the altitude, and this in turn determines the period. The period T is given by

T = 2\pi \sqrt{\frac{r^3}{\mu}}

where μ is the “standard gravitational parameter” for Earth, which equals the mass of the earth times the gravitational constant G.

The weakest link in calculating of T is r. The FAA site says the altitude is 20,200 km, but has that been rounded? Also, we need the distance to the center of the earth, not the altitude above the surface, so we need to add the radius of the earth. But the radius of the earth varies. Using the average radius of the earth I get T = 43,105 seconds.

Note that 12 hours is 43,200 seconds, so the period I calculated is 95 seconds short of 12 hours. Some of the difference is due to calculation inaccuracy, but most of it is real: the orbital period of GPS satellites is less than 12 hours. According to this source, the orbital period is almost precisely 11 hours 58 minutes.

The significance of 11 hours and 58 minutes is that it is half a sidereal day, not half a solar day. I wrote about the difference between a sidereal day and a solar day here. That means each GPS satellite returns to almost the same position twice a day, as seen from the perspective of an observer on the earth. GPS satellites are in a 2:1 resonance with the earth’s rotation.

(But doesn’t the earth rotate on its axis every 24 hours? No, every 23 hours 56 minutes. Those missing four minutes come from the fact that the earth has to rotate a bit more than one rotation on its axis to return to the same position relative to the sun. More on that here.)

Update: See the next post on the mathematics of GPS.

[1] Mireille Boutin, Gregor Kemperc. Global positioning: The uniqueness question and a new solution method. Advances in Applied Mathematics 160 (2024)

Maybe Copernicus isn’t coming

Posted on by John

Before Copernicus promoted the heliocentric model of the solar system, astronomers added epicycle on top of epicycle, creating ever more complex models of the solar system. The term epicycle is often used derisively to mean something ad hoc and unnecessarily complex.

Copernicus’ model was simpler, but it was less accurate. The increasingly complex models before Copernicus were refinements. They were not ad hoc, nor were they unnecessarily complex, if you must center your coordinate system on Earth.

It’s easy to draw the wrong conclusion from Copernicus, and from any number of other scientists who were able to greatly simplify a previous model. One could be led to believe that whenever something is too complicated, there must be a simpler approach. Sometimes there is, and sometimes there isn’t.

If there isn’t a simpler model, the time spent searching for one is wasted. If there is a simpler model, the time searching for one might still be wasted. Pursuing brute force progress might lead to a simpler model faster than pursuing a simpler model directly.

It all depends. Of course it’s wise to spend at least some time looking for a simple solution. But I think we’re fed too many stories in which the hero comes up with a simpler solution by stepping back from the problem.

Most progress comes from the kind of incremental grind that doesn’t make an inspiring story for children. And when there is a drastic simplification, that simplification usually comes after grinding on a problem, not instead of grinding on it.

3Blue1Brown touches on this in this video. The video follows two hypothetical problem solvers, Alice and Bob, who attack the same problem. Alice is the clever thinker and Bob is the calculating drudge. Alice’s solution of the original problem is certainly more elegant, and more likely to be taught in a classroom. But Bob’s approach generalizes in a way that Alice’s approach, as far as we know, does not.

Related posts

Why does FM sound better than AM?

Posted on by John

The original form of radio broadcast was amplitude modulation (AM). With AM radio, the changes in the amplitude of the carrier wave carries the signal you want to broadcast.

AM signal

Frequency modulation (FM) came later. With FM radio, changes to the frequency of the carrier wave carry the signal.

I go into the mathematical details of AM radio here and of FM radio here.

Pinter [1] gives a clear explanation of why the inventor of FM radio, Edwin Howard Armstrong, correctly predicted that FM radio transmissions would be less affected by noise.

Armstrong reasoned that the effect of random noise is primarily to amplitude-modulate the carrier without consistently producing frequency derivations.

In other words, noise tends to be a an unwanted amplitude modulation, not a frequency modulation.

FM radio was able to achieve levels of noise reduction that people steeped in AM radio thought would be impossible. As J. R. Carson eloquently but incorrectly concluded

… as the essential nature of the problem is more clearly perceived, we are unavoidably forced to the conclusion that static, like the poor, will always be with us.

But as Pinter observes

The substantial reduction of noise in a FM receiver by use of a limiter was indeed a startling discovery, contrary to the behavior of AM systems, because experience with such systems had shown that the noise contribution to the modulation of the carrier could not be eliminated without partial elimination of the message.

Related posts

[1] Philip F. Pinter. Modulation, Noise, and Spectral Analysis. McGraw-Hill 1965.

Increasing speed due to friction

Posted on by John

Orbital mechanics is fascinating. I’ve learned a bit about it for fun, not for profit. I seriously doubt Elon Musk will ever call asking me to design an orbit for him. [1]

One of the things that makes orbital mechanics interesting is that it can be counter-intuitive. For example, atmospheric friction can make a satellite move faster. How can this be? Doesn’t friction always slow things down?

Friction does reduce a satellite’s tangential velocity, causing it to move into a lower orbit, which increases its velocity. It’s weird to think about, but the details are worked out in [2].

Note the date on the article: May 1958. The paper was written in response to Sputnik 1 which launched in October 1957. Parkyn’s described the phenomenon of acceleration due to friction in general, and how it applied to Sputnik in particular.

Related posts

[1] I had a lead on a project with NASA once, but it wasn’t orbital mechanics, and the lead didn’t materialize.

[2] D. G. Parkyn. The Effect of Friction on Elliptic Orbits. The Mathematical Gazette. Vol. 42, No. 340 (May, 1958), pp. 96-98

Calculating when a planet will appear to move backwards

Posted on by John Cook

When we say that the planets in our solar system orbit the sun, not the earth, we mean that the motions of the planets is much simpler to describe from the vantage point of the sun. The sun is no more the center of the universe than the earth is. Describing the motion of the planets from the perspective of our planet is not wrong, but it is inconvenient. (For some purposes. It’s quite convenient for others.)

The word planet means “wanderer.” This because the planets appear to wander in the night sky. Unlike stars that appear to orbit the earth in a circle as the earth orbits the sun, planets appear to occasionally reverse course. When planets appear to move backward this is called retrograde motion.

Apparent motion of Venus

Here’s what the motion of Venus would look like over a period of 8 years as explored here.

Venus from Earth

Venus completes 13 orbits around the sun in the time it takes Earth to complete 8 orbits. The ratio isn’t exactly 13 to 8, but it’s very close. Five times over the course of eight years Venus will appear to reverse course for a few days. How many days? We will get to that shortly.

When we speak of the motion of the planets through the night sky, we’re not talking about their rising and setting each day due to the rotation of the earth on its axis. We’re talking about their motion from night to night. The image above is how an observer far above the Earth and not rotating with the Earth would see the position of Venus over the course of eight years.

The orbit of Venus as seen from earth is beautiful but complicated. From the Copernican perspective, the orbits of Earth and Venus are simply concentric circles. You may bristle at my saying planets have circular rather than elliptical orbits [1]. The orbits are not exactly circles, but are so close to circular that you cannot see the difference. For the purposes of this post, we’ll assume planets orbit the sun in circles.

Calculating retrograde periods

There is a surprisingly simple equation [2] for finding the points where a planet will appear to change course:

\cos(kt) = \frac{\sqrt{Rr}}{R + r - \sqrt{Rr}}

Here r is the radius of Earth’s orbit and R is the radius of the other planet’s orbit [3]. The constant k is the difference in angular velocities of the two planets. You can solve this equation for the times when the apparent motion changes.

Note that the equation is entirely symmetric in r and R. So Venusian observing Earth and an Earthling observing Venus would agree on the times when the apparent motions of the two planets reverse.

Example calculation

Let’s find when Venus enters and leaves retrograde motion. Here are the constants we need.

r = 1       # AU
R = 0.72332 # AU

venus_year = 224.70 # days
earth_year = 365.24 # days
k = 2*pi/venus_year - 2*pi/earth_year

c = sqrt(r*R) / (r + R - sqrt(r*R))

With these constants we can now plot cos(kt) and see when it equals c.

This shows there are five times over the course of eight years when Venus is in apparent retrograde motion.

If we set time t = 0 to be a time when Earth and Venus are aligned, we start in the middle of retrograde period. Venus enters prograde motion 21 days later, and the next retrograde period begins at day 563. So out of every 584 days, Venus spends 42 days in retrograde motion and 542 days in prograde motion.

Related posts

[1] Planets do not exactly orbit in circles. They don’t exactly orbit in ellipses either. Modeling orbits as ellipses is much more accurate than modeling orbits as circles, but not still not perfectly accurate.

[2] 100 Great Problems of Elementary Mathematics: Their History and Solution. By Heinrich Dörrie. Dover, 1965.

[3] There’s nothing unique about observing planets from Earth. Here “Earth” simply means the planet you’re viewing from.

What can JWST see?

Posted on by John

The other day I ran across this photo of Saturn’s moon Titan taken by the James Webb Space Telescope (JWST).

If JWST can see Titan with this kind of resolution, how well could it see Pluto or other planets? In this post I’ll do some back-of-the-envelope calculations, only considering the apparent size of objects, ignoring other factors such as how bright an object is.

The apparent size of an object of radius r at a distance d is proportional to (r/d)². [1]

Of course the JWST isn’t located on Earth, but JWST is close to Earth relative to the distance to Titan.

Titan is about 1.4 × 1012 meters from here, and has a radius of about 2.6 × 106 m. Pluto is about 6 × 1012 m away, and has a radius of 1.2 × 106 m. So the apparent size of Pluto would be around 90 times smaller than the apparent size of Titan. Based on only the factors considered here, JWST could take a decent photo of Pluto, but it would be lower resolution than the photo of Titan above, and far lower resolution than the photos taken by New Horizons.

Could JWST photograph an exoplanet? There are two confirmed exoplanets are orbiting Proxima Centauri 4 × 1016 m away. The larger of these has a mass slightly larger than Earth, and presumably has a radius about equal to that of Earth, 6 × 106 m. Its apparent size would be 150 million times smaller than Titan.

So no, it would not be possible for JWST to photograph an expolanet.

 

[1] In case the apparent size calculation feels like a rabbit out of a hat, here’s a quick derivation. Imagine looking out on sphere of radius d. This sphere has area 4πd². A distant planet takes up πr² area on this sphere, 0.25(r/d)² of your total field of vision.

Earth : Jupiter :: Jupiter : Sun

Posted on by John

The size of Jupiter is approximately the geometric mean of the sizes of Sun and Earth.

In terms of radii,

\frac{R_\Sun}{R_{\text{\Jupiter}}} \approx \frac{R_\Jupiter}{R_\Earth}

The ratio on the left equals 9.95 and the ratio on the left equals 10.98.

The subscripts are the astronomical symbols for the Sun (☉, U+2609), Jupiter (♃, U+2643), and Earth (, U+1F728). I produced them in LaTeX using the mathabx package and the commands \Sun, Jupiter, and Earth.

The mathabx symbol for Jupiter is a little unusual. It looks italicized, but that’s not because the symbol is being used in math mode. Notice that the vertical bar in the symbol for Earth is vertical, i.e. not italicized.