Leap Seconds

News Flash

: There's a proposal afoot to eliminate leap seconds, making TAI - UTC = constant.

That proposal is effectively dead. See <http://www.ien.it/luc/cesio/itu/gambis_leap.pdf>.

Because of the tidal pull of the moon, the earth's spin is slowing down. In order to compensate for this, various standards organization agree to insert a leap second every once in a while. These leap seconds have been inserted on the last day of June or December, and are officially denoted as 23:59:60 on "clocks". -- BillTrost

Conservation laws tell us that something else must be speeding up. What is it? Deadlines?

The moon's tidal pull is causing the rotation of the Earth to slow down, but angular momentum must be conserved. I recall from college days that the moon is actually being boosted into a higher orbit by the transfer of angular momentum. I think that eventually the moon will be thrown entirely out of orbit (its orbital velocity will exceed escape velocity). I refuse to do integral calculus any more, so I can't work out when that will happen.

By the way, the moon doesn't really orbit the Earth. Both the Earth and the moon orbit the center of mass of the Earth-moon system. This point is roughly several hundred miles beneath the Earth's surface. If the moon were just slightly more massive or farther from the Earth, that point would be above the Earth's surface! Since the moon's orbital radius is increasing, eventually the center of rotation of the Earth-moon system will ascend to sea level. I guess it would be a good problem for a college physics text to work out the date at which that would occur. :-)

My information is based on a book The Nick Of Time by George Alec Effinger, rather than physics, so it would be prudent to take it with a grain of salt. According to GAE, tidal lock is causing the Earth's rotation around its axis to slow, and causing the moon to get farther from the Earth.

Eventually (I imagine, but this is not explicitly stated, that this will happen when the Earth stops rotating), this process will reverse: the Earth will revolve faster and faster, and "reel in" the moon. At some point, the moon will get too close (Roche's limit, IIRC, specifies how close) and will explode.

Except that, in The Nick Of Time, the moon doesn't explode because it's been wrapped in duct tape.

The moon is moving farther out, and it won't be coming back, as tidal friction can't act to speed up our rotation. You can compare the situation with the moon's rotation, and nearly all the other moons in the solar system, all of which present a constant face to their planet and haven't started spinning again. However, if I recall correctly, there won't be time for the moon to escape while the Sun still lives. Incidentally, the decline in our rotation - or rather, the decreased number of days per year - is supposedly visible in the fossil record, though I'm not sure how you can get the daily growth rate of an Ordovician clam. I know it varies enough that calculations done with ancient Chinese astronomical records, which are usually accurate past an hour, can pick it up. 1-3 milliseconds per day per century.

I understand that you can count days pre year in some fossils. There are diurnal growth processes that leave traces like growth rings in trees. Together with the yearly variation you can derive days per year and thus the length of a days (because the year is fixed).

A very practical puzzle for programmers: LeapSeconds can cause a day to be more or less than 24*60*60 seconds long. What else can? (See the LeapSecondsPuzzleAnswer) -- PaulChisholm

http://www.niceties.com/leapsec.html -- leap seconds observed
http://www.boulder.nist.gov/timefreq/pubs/bulletin/leapsecond.htm

Thus far, all deviations from 86400 second days have been for extra seconds. Given the physics of the rotational slowdown, that is to be expected.

So far, I haven't seen anything in this discussion that accounts for the elliptical aspects of the orbits (consider what this does to the environment of earth, for example.)

Aren't LeapSeconds sort of the NoOp for clocks?

The elliptical aspect (eccentricity) of the Moon orbit is irrelevant because it does not influence the total orbital energy of the Moon in any way.

What happens is the following: The Earth spins quicker around its axis than the Moon orbits around the Earth, while the viscosity of the Earth limits the rate at which the Earth can bulge / debulge under the gravity of the Moon. This means that there is an angle between the peak of the tidal bulge on the Earth, and the Moon's location relative to the Earth. The Earth therefore looks asymmetrical as seen from the Moon, and the Moon's gravity applies a non-zero nett torque on the Earth. This torque decreases the Earth's rotational rate, while, thanks to Newton's "action = -reaction" it drags the Moon forward in its orbit, accelerating it into a higher orbit.

However!!! A higher Moon orbit leads to a decreased gravitational pull between the Moon and the Earth, which then decreases the amplitude of the tidal bulge, which decreases the exchanged energy. Similarly, the decreasing rotational rate of the Earth decreases the angle between the tidal bulge and the Moon, which decreases the torque, which decreases the exchanged energy. The entire process of the raising Moon orbit and the decreasing Earth rotation rate is therefore gradually grinding to a halt.

It can end only in one of three ways:

(1) The Earth's rotational rate becomes the same as the time that the Moon needs for (then) one orbit around the Earth. This has already happened with the (asymmetrical) Moon itself, which has a rotational rate that matches its orbital period around the Earth so that we only ever see one side of the Moon. In this scenario, the Moon becomes geosynchronous (... but will still be wobbling in North-South direction due to its orbital inclination) so that people from the opposite side of the Earth must travel if they ever want to see the Moon. Satellite TV will no longer be possible, because geostationary satellites can not be parked in their final orbit without hitting the enormous moon during their final non-geostationary orbits. In this situation, the tidal bulge of the Earth has become stationary as well, i.e. it forms a permanent deformation of the Earth due to the Moon's gravity, with a peak that is lined up exactly along the Moon direction. There is no more angular offset, therefore no more nett torque, so that the rotational rate of the Earth as well as the Moon's orbital height remain forever as they are.

(2) Before ever reaching the stationary situation above, the Moon is accelerated up to escape velocity and disappears from the gravitational pull of the Earth into deep space. However, as soon as it leaves "Earth orbit" it enters "Sun orbit" because its motion through the universe will now be dominated by the gravity field of the Sun. This means that the Moon effectively becomes another planet of our solar system (...it's bigger than Pluto). The Moon's orbit around the Sun will still be similar to that of the Earth, so that the two bodies will orbit closely together, one (the lower one) catching up very slowly with the other one. This is a disaster waiting to happen: under various other orbital perturbations (most notably, the gravity pull from Jupiter and other planets in the Solar System) two such closely orbiting planets will sooner or later "catch" each other, i.e. they either start orbiting around each other again, or, much more fun, they smash into each other and together crumble into a single, bigger planet. This is in fact how planets are formed, out of small lumps of mass that gradually crash together, and then stay together under internal gravity.

(3) The Moon orbit becomes so high that interactions with other bodies in the Solar System perturb its orbit to the extent that it escapes from the Earth's gravity field. The rest is like in (2), but the Moon might then also smash into Mars or Venus and not into the Earth.

With some calculations that are too clumsy to type in this text-only page, it can be shown that scenario 1 applies for the current values for Earth's rotation rate, viscosity, distance to the Moon, relative masses of Earth and Moon and universal gravitational constant. So, Effinger is wrong in two ways: first, the Moon will not come back to the Earth by reversal of the current process (he overlooks the second order effect of the reducing tidal bulge), and the Moon will not smash into the Earth under option (2) either.

The conclusion is that in the very distant future days and months will have the same length, so that the concept of a week may have to be defined more precisely, either as a quarter of a Moon orbit (= 6 hours) or as seven days (= seven months). Technically speaking, it will be Monday forever - nothing to look out for.

Now, back to leapseconds. The rate of UTC time is by definition the same as the rate of TAI time, and is defined unambiguously through the S.I. second as N oscilllations of some atom X. However, the rate of UT1 time is defined by the Earth's rotational rate, in such a way that there are always 86400 UT1 seconds in a single revolution of the Earth. Leap seconds are merely introduced in UTC to keep the calendar difference between UTC and UT1 within 1.5 TAI seconds of each other, for practical purposes. This means that by the time of reaching scenario (1) it will be inevitable to introduce thousands of leap seconds for every normal UTC second.

It is therefore incorrect to claim that the proposal to abandon leap seconds could ever be dead. On the contrary: it is only a matter of time (hoho) before the need for more than 1.5 leap seconds at every June 30th or December 31st will render the current definition of the UTC time scale unusable - and with it, the concept of leap seconds.

Wow. That is some fascinating reading. Where did you get all that from? -- JamesHollidge