View Full Version : On the Rotation of the Earth

Celestial Mechanic
2009-Aug-05, 05:03 AM
Celestial Mechanic Explains the Earth's Rotation--Part One: A Blast From the Past

My old friends from grad school and my newer friends from UW are here for a coffee and a bit of physics. Our cups refilled and more of Tensor's doughnuts in hand, we return.

CM: "I want to say a few things about the Earth's rotation. Over the years I've thought a bit about the Earth's rotation since we use it as a time-piece and as part of the definition of two different coordinate systems. I've come up with a different idea for handling the motion of the ecliptic and treating the precession and nutation by a different method."

DB: "Shouldn't this be in Against the Mainstream?"

CM: "No, because it uses the tested methods of celestial mechanics to arrive at its results. It is not that far from the mainstream's presentation, indeed it could become the mainstream view if it explains the mainstream results more simply and with less computation. This thread is entirely mainstream."

BH: "Can you give a brief summary without giving too much away?"

CM: "It doesn't matter if I give it away. In brief, I propose to treat the ecliptic and the equator equally, with both precessing around the pole of the Laplace or invariant plane of the Solar System.

CM: "To explain why I have chosen this, I'd like to tell about my early years of study in celestial mechanics."

Jimmy K.: "Get ready for a blast from the past!"

CM: "Oh, come on, it won't be that bad. I won't tell any stories about wild times at OU. In fact these recollections might be boring, unless of course you are interested in celestial mechanics.

CM: "As a child I was interested in astronomy, and I recall some of my early struggles with trying to understand such things as sidereal time, the magnitude system, and why the harvest moon rose fewer minutes later than other full moons. This is not easy stuff for a 10-year old! I would eventually teach myself logarithms at 11, trig (plane and spherical!) and a bit of differential calculus when I was 12, but it would take me until I was 15 to really master integral calculus. That's one reason why I have some sympathy for 17 and 18-year olds who get differential and integral calculus in consecutive semesters."

Virginia: "I had trouble with it in advanced placement calculus."

JK: "So did I."

CM: "Oddly enough, although I read about differential and partial differential equations in high school, I never tried my hand at celestial mechanics. My orientation was more towards geometry, especially in higher dimensions, and number theory.

CM: "I finished high school and arrived at the University of Oklahoma the day before my 17th birthday, almost 40 years ago. OU had (and I hope still has) a fabulous library, indeed it was called the 'Yale of the Southwest'. It had just acquired its one-millionth volume before I got there; it would acquire its two-millionth volume shortly before I left 18 years later."

JK: "Eighteen years!"

CM: "It's a long story."

V: "I'll bet!"

CM: "But I won't be telling it anytime soon. Anyway, during my freshman year I made the acquaintance of book called Theory of Eclipses by Buchanan. I wanted to try calculating eclipses, in particular lunar eclipses since I also ran across Jean Meeus' Canon of Solar Eclipses that year. The first textbook of celestial mechanics that I studied from was Celestial Mechanics by Brouwer and Clemence.

CM: "My first attempt during spring break of 1970 at predicting the lunar eclipse of 1971 Febrary 10 was a disaster, I was off by two hours and the magnitude off by 0.2. Of course I was not handling perturbations properly. Ultimately I would calculate lunar eclipses. One of the things that impressed me then were the then-current analytical theories of lunar and planetary motion. I had access to the Tables of the Inner Planets in Volume VI of Papers of the Astronomical Ephemeris and Nautical Almanac, which were used (with some modifications) through 1984, and Improved Lunar Ephemeris. One of the things that struck me was the difference in the ways the planetary and lunar theories were expressed.

CM: "The perturbations of the planetary theories were expressed as cosine series of the form K*cos(j*g+j'*g'+C), with j and j' as integers (possibly zero) and g and g' the mean anomalies of the planet and the perturbing planet respectively. The lunar theory had sine series for the longitude and latitude perturbations, a cosine series for the sine parallax (the inverse of the radius vector), and all arguments were linear combinations of four arguments, D, M, M' and F, without any phases, for example sin(M-2*F+2*D). What I wanted to do was develop the planetary theories in the same form as the lunar theory."

BH: "What are D, M, M' and F?"

CM: "Sorry. D is the mean elongation, equal to the difference between the mean longitude of the Moon, L, and the mean longitude of the Sun, L'. M and M' are the mean anomalies of the Sun and Moon, respectively. And F is the argument of latitude, equal to the difference between the mean longitude L and the longitude of the Moon's orbit on the ecliptic, Omega."

JK: "Why is there no F'?"

V: "Because the Sun is in the ecliptic, it has no inclination and no node. Also, don't forget that the mean anomalies M and M' are the differences between the mean longitudes and longitudes of pericenters for the Moon and Sun respectively."

CM: "Thank you, Virginia. Anyway, I dropped out of school but continued to live in Norman, OK and work for the university. In 1973 I purchased the Dover reprint of Ernest W. Brown's An Introductory Treatise on the Lunar Theory, probably the best two dollars I ever spent."

JK: "Two dollars! You've got to be putting us on!"

CM: "No, I'm not. Let's refill our cups and I'll get my copy of the book and show you."

To be continued...

Celestial Mechanic
2009-Aug-05, 05:06 AM
Celestial Mechanic Explains the Earth's Rotation--Part Two: One Plane, Invariable

My old friends from grad school and my newer friends from UW are here for a coffee and a bit of physics. Our cups refilled and my very dog-eared copy of Ernest W. Brown's An Introductory Treatise on the Lunar Theory in hand, we return.

Celestial Mechanic: {Presents the book with a flourish.} "Here it is."

Virginia: "Oh my God! And to think of what I spent on textbooks last year!"

Jimmy K.: "I'd rather not."

CM: "I recently heard of a textbook costing $250. Now I'm willing to drop $40 now and then on a computer programming book, preferably one with a CD-ROM in it, though not for much longer. And I know that most of the more technical physics books that I can no longer browse at the UW bookstore are going to be $60 or $80, but I really can't imagine what textbook could possibly cost $250. I honestly don't know how college students are able to make it these days."

DB: "I don't know either."

BH: "I looked at the tuition at OU and the tuition is simply obscene."

CM: "It's obscene everywhere, and getting more obscene by 6, 8, 10 percent a year."

JK: "My summer jobs haven't been going up by 10 percent a year."

CM: "Nor has my income, so don't feel too bad. But let's get back to more pleasant things, such as the rotation of the Earth.

CM: "In 1977 I started work on a planetary theory. One of the first things I realized was that if the orbital planes of the planets are precessing the plane of the Earth's orbit must also precess. This means that the ecliptic coordinate system is not suitable as a coordinate system because it is not inertial. I then remembered that there is a plane that is invariable, or nearly so.

CM: "A small digression, let me note that starting with the 17th, each century has had a comprehensive work summarizing that century's progress in celestial mechanics. Newton's Principia was the work for the 17th century; Pierre Simon de Laplace's Mechanique Celeste which gave celestial mechanics its name summarized the 18th, although it was published in 1809. Tisserand's Mechanique Celeste of 1894 summarized the 19th, and the first two volumes of Celestial Mechanics by Hagihara summarized the 20th."

V: "Wasn't Hagihara's book supposed to be five volumes but he died after completing volumes 2a and 2b?"

CM: "I believe that is correct, but I think his students completed the remaining volumes. I've at least looked at these works, most of them at the OU library, but the last three volumes of Hagihara are at the math library at UW Madison and I have seen them there.

CM: "But back to Laplace. At OU they had all five volumes of his work, translated into English and copiously footnoted by Bowditch. And I do mean copiously footnoted! In one of the volumes Laplace presented perturbation theory and gave the larger perturbations for all the planets out to Uranus, which had been discovered only a few years before. It's been a long time since I've seen this book, so I don't remember whether he used a coordinate system based on the invariable plane or the standard ecliptic coordinates, but I'm willing to bet on the invariable plane because he was willing to use newer methods and definitions. For example, he used centesimal seconds for time, that is, he divided the day into 100,000 seconds, 10 hours with 100 minutes and 100 seconds each."

BH: "That was during the French Revolution, wasn't it?"

CM: "Yes, but remember that it was ultimately published in 1809, well into the reign of Napoleon."

JK: "But what is this invariable plane?"

CM: "Thank you for steering us back onto topic. The continuous symmetries of a physical theory give us conservation laws, as proven by Emmy Noether in 1918. Newtonian gravity is invariant under translations in space which gives us conservation of linear momentum. Invariance under time translations gives us conservation of energy. And finally, invariance under rotations gives us conservation of angular momentum, and that is the key.

CM: "In the two-body problem the orbital angular momentum is a constant. The plane of the orbit does not move, so that neither its inclination to the plane of reference changes, nor do the nodes, the intersections of the plane of reference with the plane of the orbit. Also, the magnitude of the vector does not change. But when other planets are present, the individual orbits are perturbed. But the total angular momentum, the sum of all the angular momenta of the planets is conserved and is constant.

CM: "So, as Laplace did, let us take that vector as the Z-axis of our coordinate system. The vector does not move and neither does the plane perpendicular to it, which is our invariable plane."

DB: "How are the X- and Y-axes chosen?"

CM: "As always, the choice of an origin for the longitude is arbitrary. Not having seen Laplace's book in a long time, I don't know what choice he made. The choice that I would make for my coordinates is to have the X-axis be the position of the ascending node of the Earth's orbit at J2000.0. In other words, Omega3 equals zero at 2000 January 01d 12h UTC."

BH: "That's all fine and dandy, but how are you going to convert it into real coordinates so astronomers can use it? I mean, I've never seen any star atlases using the invariable plane in their coordinate system. I've never even seen one that shows the invariable plane!"

CM: "Very true. While these coordinates make it easy (well, somewhat easy) for me to express and calculate theories of planetary motion, it is not immediately useful for comparisons with star charts. I've got to bring my system 'down to Earth' by converting my coordinates to the traditional Earth-based coordinate systems. If you are content with astrometric coordinates using a fixed equator and ecliptic for the epoch, namely J2000.0, two triples of unit vectors, X, Y, and Z, for ecliptic and equatorial coordinates respectively are available. My theories give cartesian coordinates in my coordinate system, so you are only three dot-products away from having cartesian coordinates in the system of your choice.

CM: "But if you want coordinates in the ecliptic and equinox of date, then I'm going to have to roll up my sleeves and somehow calculate the unit vectors that I call C and Q. C, of course, is just the unit vector for the Earth's orbital angular momentum, the planetary theory took care of that. But Q is the rotational vector of the Earth, the topic of this thread. To calculate it I need both the theory of the Sun (which is really the Earth's motion) and the theory of the Moon. Well, there is a theory of the Moon, but most theories of the Moon use ecliptic coordinates, not invariable plane coordinates such as I am using.

CM: "It was then that I realized that maybe the lunar theory should also be redone using the invariable plane. Then the lunar theory does have an F', equal to the difference between the mean longitude of the Sun and the longitude of the Earth's orbit on the invariable plane. I started out trying to recast the planetary theory to be consistent with the lunar theory only to wind up recasting the lunar theory as well. Someday."

DB: "What do you mean someday?"

CM: "Well, so far I have not calculated more than a few terms here and there, as feasibility studies. I'm reasonably sure I can do the numerical work on the planetary theories fairly quickly. But the lunar theory is not going to be easy.

CM: "But anyway, with the new planetary and lunar theories in hand, calculation of the precession and nutation will follow."

JK: "Excuse me, my cup is empty. I'm just going for a refill."

CM: "Sounds like a good idea. When we return I will fill in more detail on my treatment of the Earth's rotation."

To be continued...

Celestial Mechanic
2009-Aug-06, 04:28 AM
Celestial Mechanic Explains the Earth's Rotation--Part Three: The Grand Nutation

My old friends from grad school and my newer friends from UW are here for a coffee and a bit of physics..

Celestial Mechanic: "We're back and now the moment you've all been waiting for -- my theory of the Earth's rotation, or at least an outline of it.

CM: "First, imagine a unit sphere. We take the north invariable plane pole as our north pole, so the invariable plane intersects the unit sphere at the great circle of zero latitude in the invariable plane system of coordinates. Call the unit vector pointing to this north pole I. You'll have to imagine a 'hat' on the I and the other unit vectors I mention, I'm editing these proceedings in plain text with some BBCode in it.

CM: "Second, imagine a point at about 66.6 degrees latitude, that is, 23.4 degrees from I. The unit vector pointing to this point is Q, the north celestial pole. It precesses in the clockwise direction at a rate that is currently 5029 arcsec/Jcy. It does not describe a perfect circle as it is not always at this latitude. It just so happens that our current value of the obliquity of the ecliptic is at about the middle of its range, 23.4 degrees."

CM: "Lastly, imagine a point about 1.58 degrees away from I. The unit vector pointing in this direction is C, the north pole of the ecliptic. It also precesses in a clockwise direction, with a rate of about 1700 arcsec/Jcy as I will calculate later."

BH: "But doesn't the Earth's celestial pole circle the ecliptic? Here you have it circling the invariant pole. Are you sure of this?"

CM: "Well, the Earth's celestial pole doesn't really 'circle' the ecliptic pole either. The obliquity, which is the angle between the equator and ecliptic and also the distance between their poles varies. The value of the obliquity varies between 23.5 +/- 0.75 degrees, and as I said earlier we happen to be at the mid-point of this variation. Note that when C and Q are lined up at the same side of I at the same longitude the obliquity is at a minimum. When they are on opposite sides of I, that is, when their longitudes differ by 180 degrees, the obliquity is at a maximum."

V: "Just one thing: if C is at 1.58 degrees from I shouldn't the range of obliquities be plus-or-minus 1.58 degrees instead of 0.75 degrees?"

CM: "Good observation, Virginia. The Earth's rotational axis also describes a nodding motion in latitude and longitude called 'nutation'. The nutation and precession have the same cause, the only difference is that what we call the precession is the non-periodic part of the motion and the nutation is the periodic part.

CM: "The nutation has a very long term period in it whose argument happens to be the difference in the longitudes of C and Q. I predict it will have a value of roughly 0.83 degrees. When C and Q are on the same side of I, Q will be 0.83 degrees farther away from I, so that C and Q are only 0.75 degrees closer instead of 1.58. And when they are on opposite sides of I, Q will be 0.83 degrees closer to I so that C and Q are only 0.75 degrees farther apart."

DB: "That's a pretty tall order. Can you actually show this?"

CM: "I am going to gear up and give what I hope is a better calculation of the precession than I did in the original 'Precession Dialogues'. This time I will also calculate the well-known principal periodic terms of the nutation, that is the terms of period 18.6 years, and then apply the same technique for my much larger and longer-period term that I call the 'Grand Nutation'. But first let me wrap up one loose end, my claim that C precesses at 1700 arcsec/Jcy.

CM: "You see, the ecliptic is usually treated treated as rotating about some point on the ecliptic at a rate of 47.0029 arcsec/Jcy, according to Meeus in Astronomical Algorithms. I propose to treat its pole C as precessing about the invariable pole I at some rate. Let P and Q be the intersections of the ecliptic with the invariable plane at some time t and let P' and Q' be the intersections at some later time t+dt. The two ecliptics intersect at some point R. The angles PRP' and QRQ' are equal to a*dt, where a is 47.0029 arcsec/Jcy. Opposite this angle is the side PP' in the triangle PRP', call it A*dt, where A is the angular rate to be determined. If dt is made small enough the other two sides of the triangle are close enough to 90 degrees (PR is greater than 90, P'R is less than 90) that we can ignore the difference. The law of the sines then gives:

sin(a*dt)/sin(A*dt)=sin(1.58 deg)/sin(90),
(a*dt)/(A*dt) = a/A = sin(1.58 deg) (for dt small enough),
A = a/sin(1.58 deg) = 1707.7 arcsec/Jcy.
V: "Could we also do it the other way, treating the equator as rotating about one of its points?"

CM: "Of course we can. Given A=5029.0966 arcsec/Jcy and taking the obliquity as 23.439 degrees we have:"

a = A * sin(23.439) = 2000.4 arcsec/Jcy.
CM: "Compare that with the angle theta in Meeus's book above, formula (20.3) on page 126:"

theta = 2004.3109"*t - 0.42665"*t2 - 0.041833"*t3.

To be continued...

2009-Aug-10, 05:49 PM
Here here!

I love this dialogue format. Sorry to interrupt, but I stopped in for a cup and over-heard your discussion.

The nutations: Are you of the opinion these date back to 'initial conditions', or was/is there considerable bell ringing along the way? Is there any observable dampening of any of the basic parameters?

Celestial Mechanic
2009-Aug-11, 05:08 AM

The nutations: Are you of the opinion these date back to 'initial conditions', or was/is there considerable bell ringing along the way? Is there any observable dampening of any of the basic parameters?
The nutation that I will calculate here are forced nutations, the periods of the terms come from the torques of the Sun and Moon. There is also free nutation and precession but that appears to be heavily damped. You may want to look up "Chandler wobble".

The main parameters that I can see might cause dampening might be increase in the length of the day, increase in the Moon's mean distance, possible change to the moments of inertia as the Earth's rotation slows down.

Tobin Dax
2009-Aug-11, 06:08 AM
CM, these dialogue posts of yours have inspired me to steal the basic idea for use in my online classes. (If I get the chance to finish them in the next two weeks.) These are great posts, keep 'em coming.

Celestial Mechanic
2009-Aug-12, 04:01 AM
CM, these dialogue posts of yours have inspired me to steal the basic idea for use in my online classes. (If I get the chance to finish them in the next two weeks.) These are great posts, keep 'em coming.Thank you! Feel free, after all I stole it from Galileo who probably stole it from Socrates ... :D

Tobin Dax
2009-Aug-12, 06:41 PM
Thank you! Feel free, after all I stole it from Galileo who probably stole it from Socrates ... :D
The next time the dean talks about the grand academic tradition of the theft of good ideas, I might have to mention how far back it goes. :)

Celestial Mechanic
2009-Aug-15, 05:05 AM
Celestial Mechanic Explains the Earth's Rotation--Part Four: On Buzzard's Wings

My old friends from grad school and my newer friends from UW are here for a coffee and a bit of physics.

Celestial Mechanic: "The time has come for the actual calculations. I am going to calculate the following things: the main contributions from the Moon and the Sun on the precession of the Earth's rotational axis, as was done in the first set of precessional dialogues. I will do it in a more rigorous fashion so that we can also treat the main periodic terms of the nutation. In doing so we will uncover the correction to the precessional rate due to the eccentricities and inclinations. I have a new result, given below that carries us to within one arcsecond/Jcy of the adopted value. I will then calculate the principal terms of the nutation due to the lunar part and compare them with adopted values. Then I will do the same for the solar part, my so-called 'grand nutation' and prove the statements I made in the previous post.

Correction factor = (1-(3/2)*sin2I)*(1+(3/2)*e2)/(1-e2)3. (1)
Correction factor = (1-(3/2)*sin2I+(9/2)*e2). (2)
CM: "I'm going to go out on a limb and say that formula (1) is valid for all orders of eccentricity and inclination for Keplerian orbits. Formula (2) is valid to the second order in eccentricity and inclination."

DB: "You had (3/2) times the eccentricity squared when last you mentioned it, now you have 9/2 times. Where did the extra 3*e2 come from?"

CM: "I don't have my original derivation handy, so I don't remember what I did. I looked at a web site that I don't recall now and saw that they had virtually the same formula, (1-(3/2)*sin2I)/(1-e2)3/2, and they mentioned averaging over a single revolution, and that set off alarm bells. You'll see when we get to that part.

CM: "So let's get started, shall we? First I have to say a few words about integration. We will do a lot of computing of torques acting on typical pieces of the Earth and then adding them all up at the end. When most people see a triple integral like IntegralV(A(x,y,z)dxdydz) they just sort of have a triple-integral brain-freeze and look away."

Jimmy K.: "I've had that happen a few times on tests."

Virginia: "Me too!"

CM: "But the thing is, we don't really do these integrals the way they are done in calculus class, in the sense of slogging through it one variable at a time remembering our integral formulas or looking them up in Gradshteyn and Ryzhik and hoping the math comes out right at the end. The thing is, for even the simplest one we would want to do, the volume integral of rho(x), where rho(x) is the density of the Earth at x we don't even know rho at enough points to perform the integral numerically. But we do know the answer, it is simply M, the mass of the Earth. My earlier set of dialogues used a sphere of uniform density and a ring of uniform density to emulate the gross details of the mass distribution of the Earth. I still had to perform integrals, but only over the longitude of the ring. I'm going to try a different approach here."

CM: "I am going to replace all those scary triple integrals with a friendlier notation:"

{A}=IntegralV(A(x)*rho(x)dxdydz). (3)
CM: "A pair of braces looks a lot friendlier than all that integral folderol, doesn't it? Remember what Dr. K used to call braces?"

BH: "I remember he used to call them 'buzzard's wings'."

CM: "Right! In the following development we will calculate various functions of the coordinates for typical locations within the Earth, and then apply the buzzard's wings in order to total it over the entire Earth. The simplest such formula is {1}=M, where M is the mass of the Earth. It is the first, or really the zeroth, in a sequence of functions called 'moments'. The mass is the zeroth moment, {x}, {y}, and {z} are the first moments, and {x2}, {y2}, {z2}, {xy}, {yz}, and {zx} are the second moments. There are 10 third moments, 15 fourth moments, and in general, (n+1)*(n+2)/2 n-th moments, but we will be using no higher than second moments."

BH: "Isn't this a bit elementary a place to be starting from?"

CM: "Perhaps, but I want to do this properly. Most people here probably have not seen this, though, and I want to show the separation into orbital and spin angular momentum."

BH: "OK, fire away!"

CM: "We have 10 moments of order two or less. We already know the zeroth-order moment, {1} = M. By the way, note that the buzzard's wings operation has the effect of multiplying the units of A by units of mass, so that our first-order moments have units of kg*m and the second-order ones have units of kg*m2. We are going to choose the origin of our coordinates so that the three first moments vanish, and we will choose the axes so that three of the six second moments vanish. Instead of 10 moments we only have four non-zero ones to worry about, and in the case of the Earth its axial symmetry causes two of them, {x2} and {y2}, to be equal. So that's three quantities to express the mass distribution of the Earth. It will be enough to explain the gross features of the Earth's rotation.

CM: "As I mentioned before, the three first moments have units of kg*m. They also form a vector, {x}, which suggests defining a coordinate vector R = {x}/{1}, or M*R = {x}. The vector R is the center of mass, of course. We can then define another vector by r=x-R, which represents the coordinate of some point of the Earth relative to the center of mass. The buzzard's wing of r has the following important property:"

{r} = {x-R} = {x}-{1}*R = M*R-M*R = 0. (4)
CM: "The buzzard's wings of r is zero, as are the buzzard's wings of all time derivatives of r. This has a wonderful effect, because whenever we take buzzard's wings of some expression the first powers of r and its derivatives drop out. I will give two examples:"

Kinetic Energy:
T = (1/2)*{dx/dt*dx/dt}
= (1/2)*{(dR/dt+dr/dt)*(dR/dt+dr/dt)}
= (1/2)*M*dR/dt*dR/dt + (1/2){dr/dt*dr/dt}. (5)

Angular Momentum:
h = {x X dx/dt}
= {(R+r)X(dR/dt+dr/dt)}
= M*RXdR/dt + {rXdr/dt}. (6)
CM: "Oh, one more important bit before we go. For each particle at x we have an acceleration F caused by distant objects, such as the Sun and the Moon. In other words, d2x/dt2=F and:"

{F} = {d2x/dt2}
= {d2R/dt2+d2r/dt2}
= M*d2R/dt2, (7)
CM: "Which, of course, is the equation of motion for the center of mass. Most of the time it is sufficient to replace {F} with M*F with F evaluated at the center of mass."

JK: "Won't the higher-order moments in the BW of F make a contribution?"

CM: "BW for buzzard's wings? I like that, I'll use it. The effects are very small. The Earth-Moon system can be treated as a point-mass located at the Earth-Moon barycenter. The difference between the one and two point-mass treatments amounts to an acceleration about one-thirteen-millionth of the main attraction due to the Sun. A high-precision theory of the Earth or the Moon will have to take this into account, but only eventually. It can be ignored in the first approximations.

CM: "So far I have given the dynamics of the center of mass, but now we have to look at the dynamics of the individual particles. Earlier I spoke about accelerations F being due to distant bodies, deliberately side-stepping the forces exerted by the particles of the Earth on one another. Fortunately we do not have to give a detailed accounting of these forces; the gravitational and contact forces constrain individual particles to describe, to a first approximation, circular paths around one or more axes.

CM: "First, let's consider objects for which radial motions are unimportant. In other words, a particle stays at a constant distance from the center of mass, or d(r2)/dt = r*dr/dt = 0. This rules out stars such as Cepheids and other pulsating variable stars. The Sun and other stars also have significant convection that transports heat from the core to the surface. Even the Earth has some convective motion in its core, but it is slow enough that we can ignore it. The simplest way to maintain a constant distance from the center of mass is to set dr/dt =w(r)Xr. By the way, that w should be a lower-case omega, but I write this off-line with a plaintext editor, blah-blah-blah."

JK: "Yes, we know the drill."

CM: "By allowing w to be a function of r we can treat bodies with differential rotation such as the Sun or Jupiter. But for bodies that are mostly solid such as the Earth, Moon, Mercury, Venus, Mars, etc. we can impose a stronger condition, namely rigidity: for all pairs of points ri and rj that d((ri-rj)2)/dt = 0. The easiest way to impose rigidity is to set w(r) = w, so that w is no longer a function of the coordinates although it can still be a function of time. We will call w the rotation vector. Now we can expand our rotational kinetic energy (5) and spin angular momentum (6):"

Trot = (1/2)*{(dr/dt)2}
= (1/2)*{(wXr)2}
= (1/2)*{r2w2-(r*w)2}
CM: "I'm going to write this out in component form, with indices i and j that take values of x, y, and z, so that wx is the x-component of w. Then:"

Trot = (1/2)*wi*{r2*dij-ri*rj}*wj
= (1/2)*wi*Iij*wj.

Iij = {r2*dij-ri*rj} (8)
CM: "Equation (8) is the definition of the moment of inertia tensor. It is a symmetric tensor, that is Iij=Iji and so on. Here I'm using dij as the Kronecker delta symbol, which is equal to one if i=j, zero otherwise. Also, I am using the Einstein summation convention, so my expression for Trot is actually a double summation. You can also think of it as a product of a row matrix (wi), a 3x3 matrix (Iij), and a column matrix (wj)."

BH: "But shouldn't we be working with angular momentum?"

CM: "Of course we will. And here it is:"

h = {rXdr/dt} = {rX(wXr)}
= {r2*w-(r*w)*r}, or

hi = Iij*wj. (9)
CM: "So we can think of the inertia tensor as operating on the rotation vector to produce another vector, the angular momentum. The two vectors do not have to be parallel.

CM: "I see that our cups are now empty, so we will pause for a refill. When we return we will discuss the properties of the inertia tensor, derive Euler's equations and derive our first result, the main part of the precession."

To be continued...