BBuild Your Own Analemma
Charles H. Holbrow
Colgate University & MIT (Dated: February 5, 2013)Earth’s analemma is the lopsided figure eight marked out over a year by the position of the Sunin the sky observed at the same clock time each day. It shows how solar time deviates from clocktime. The analemma’s shape results from the tilt of Earth’s axis of rotation relative to the planeof its orbit around the Sun and from the elliptical shape of that orbit. This tutorial paper usesvector analysis of the Earth-Sun geometry and a numerically generated quantitative description ofEarth’s motion around the Sun to construct Earth’s analemma. To visualize the geometry and themotion that give rise to the analemma is challenging, but this construction is a project within thecapability of physics students who have had basic undergraduate mechanics.
I. INTRODUCTION
FIG. 1. Earth’s analemma as a composite of pho-tographs of the Sun taken at the same location andtime every day for a year.
Gary Cooper’s shootout in the movie “HighNoon” almost surely was not at high noon. Hewas using clock time as established by the rail-roads. Even if he happened to be at the lon-gitude of his time zone corresponding to solartime, there are only four days in the year whenthe Sun crosses the meridian – its daily highestpoint in the sky – at 12:00:00 clock time.During a year solar noon varies from 16 min33 s earlier to 14 min 6 s later than clock time. This variation is apparent in Fig. 1 which isa composite of pictures taken in Cascade, Col-orado at the same clock time each day over thecourse of a year. The curve outlined by these365 images is Earth’s analemma, and the pur-pose of this paper is to show how to construct itfrom basic principles of geometry and physics.The collection of images of the Sun in Fig. 1is low in the western sky because the pictureswere taken in the afternoon. The belt of the fig-ure eight, i.e., the points slightly below wherethe lines of the eight cross, moves along an arcthat is the Earth’s equator projected onto thesky, i.e., the celestial equator. You can imaginethe figure eight of the analemma rising some-what tilted in the east – upper side first; thenthe figure eight, long axis perpendicular to thecelestial equator, moves across the sky and setsin the west – upper side last. At noon the longaxis of the figure eight lies along the meridian,half way between the eastern and western hori-zons.The analemma is a record of the Sun’s posi-tion in the sky as observed from the same pointon Earth at time intervals of exactly one day.The analemma displays the variation of the an-gle between the Sun’s path in the sky and thecelestial equator. Angle measured perpendicu-lar from the celestial equator is called declina-tion. The seasonal variation of the Sun’s decli-nation from − . ◦ to +23 . ◦ and back againgives the analemma its long dimension. A smallperiodic variation in the time of day when theSun crosses the meridian gives the analemma its a r X i v : . [ phy s i c s . pop - ph ] F e b width of a few degrees. This paper shows how the relative motion ofSun and Earth results in the two angular varia-tions that produce Earth’s analemma. The pa-per provides an analytic explanation in terms oftwo useful vectors, one that describes the mo-tion of the actual Sun, and the other that de-scribes a fictitious average or “mean Sun” thatcorresponds to clock time. These vectors aredefined in section II.Section III uses diagrams to show how geome-try results in the small periodic variation in thedaily time at which the Sun crosses a merid-ian. These diagrams show how tilt produces ananalemma even if the orbit is a perfect circle;they also show how to extract useful informa-tion from the two vectors. The vectors are thenused to produce the analemma for the specialcase of a circular orbit. The result is a sym-metric figure-eight analemma with its mid-pointexactly on the celestial equator.Section IV extends the vector approach togenerate the analemma arising from an ellipticalorbit. This section uses a numerical calculationof Earth’s orbit and shows how its ellipticitygives the analemma the different sized loops andslightly lopsided shape you see in Fig 1. The ex-cellent agreement of the calculated analemmawith the observed one is confirmed by layingone over the other.A higher level of vector analysis is exhibitedin section V by using vector algebra to con-struct the analemma. The outcome is elegantand compact.In sections VI and VII are a few fun factsfor fostering facultative physics fora, some sug-gestions for student projects that could use ex-tensions of the ideas and techniques developedhere, and some general conclusions.
II. TWO VECTORS FORCONSTRUCTING AN ANALEMMA
It is the variation between clock time and Suntime that gives rise to the analemma. Conse-quently, to construct the analemma it is usefulto define two vectors. One, (cid:126)r S ( t ), correspond- ing to the Sun’s observed motion; the other, (cid:126)r m ( t ), corresponding to the motion of the fic-tional mean Sun and thus to the passage of clocktime t .The vector (cid:126)r S ( t ) points at the real Sun as itmakes its passage across the sky. The vectorvaries seasonally both in position and rate asit tracks the apparent geocentric motion of theSun.The vector (cid:126)r m ( t ) points at the mean Sun. Be-cause the mean Sun defines the rate at whichclocks run, (cid:126)r m ( t ) is in effect a clock vector.Thus, if (cid:126)r m ( t ) is set up on the equator at the lon-gitude used to define a time zone, it will pointat the mean Sun’s location directly overhead atexactly 12:00:00 noon in that time zone. Forsomeone living in the U.S. Mountain Time Zone(UTC-07) that longitude is 105 ◦ W. Set on theequator at that longitude, the vector (cid:126)r m ( t ) willpoint directly overhead at the fictitious meanSun as it crosses the meridian of people observ-ing from 105 ◦ W longitude — for example, fromdowntown Denver, CO.As you will see, the analemma’s shape is im-plicit in the difference between the two vectors: (cid:126) A ( t ) = (cid:126)r S ( t ) − (cid:126)r m ( t ) . (1)Figure 2 shows the positions of the Sun andthe mean Sun around a planet at successivetimes. (To make the tilt more evident, it hasbeen drawn as 45 ◦ . ) The dots are locations ofthe tips of the vectors (cid:126)r S ( t ) and (cid:126)r m ( t ) at 14-dayintervals. Thus, the figure maps two spatial andone time dimensions into three spatial dimen-sions. The connected dots correspond to theangular position (two spatial dimensions) of themean Sun exactly over a particular meridian —the meridian of the observer — day after day,and the arc length between dots of the meanSun corresponds to an exactly integer numberof days (the time dimension). The dots of theactual Sun are spaced by the same time intervalas those of the mean Sun, but, as you will see,except in four instances the actual Sun can notcross the observer’s meridian at the same timeas the mean Sun. The dots of the real Sun moveback and forth relative to the dots of the meanSun. FIG. 2. Intersecting circular paths traced out bythe tip of the mean Sun (clock) vector — shownby connected dots — and the tip of the Sun vectorfor a circular orbit. The points of intersection ofthe two circles correspond to Earth-Sun equinoxes.The axes are oriented so the x -axis lies on the lineof intersection (the line of nodes) of the two circles;the z -axis is perpendicular to the plane of the meanSun and, therefore, parallel to the planet’s axis ofrotation. For visual impact the plane of (cid:126)r S is heretilted 45 ◦ to the equatorial plane swept out by (cid:126)r m ;the dots are 14 days apart if the orbital period is ayear. To extract the analemma from (cid:126)r S ( t ) and (cid:126)r m ( t ) you need to represent them in a coordi-nate system. It is customary to use the pro-jection of Earth’s system of spherical coordi-nates (latitude and longitude) onto the sky. Forexample, the circular arc formed by projectingEarth’s equator onto the sky defines the celes-tial equator.Other projected lines of Earth’s latitude de-fine corresponding lines of celestial latitude, i.e.,declination. Lines of celestial longitude are de-fined similarly by projecting Earth’s lines of lon-gitude onto the sky.Because of Earth’s rotation, the lines of celes-tial longitude rise in the east, sweep overhead,and set in the west. By convention the line ofzero celestial longitude lies in the plane defined by the Earth’s axis and the line of nodes.To represent (cid:126)r S ( t ) and (cid:126)r m ( t ) it is convenientto use the celestial spherical polar coordinatesin a Cartesian reference frame with its origin atEarth, its z -axis along Earth’s axis of rotation,and its x - and y -axes in the equatorial plane.Referring to Fig. 2, you see that the x - and y -axes lie in the plane of the circle of connecteddots traced out by (cid:126)r m ( t ).By convention the x -axis is set to lie alongthe line of intersection (called the line of nodes)of the two circles and to point toward the pointwhere the rising (cid:126)r S ( t ) of the tilted circle crossesthe horizontal circle. In the geometry of Earthand Sun this crossing occurs around March 20— the spring equinox of the Northern Hemi-sphere.In this frame of reference (cid:126)r m ( t ) tracing out acircle in the x - y plane (Fig. 2) can be written (cid:126)r m ( t ) = cos ωt ˆ ı + sin ωt ˆ , (2)where ˆ ı and ˆ are unit vectors along the x - and y -axes respectively, t is time, and ω is the an-gular velocity of Earth on its axis. The timecoordinate is measured from t = 0 at the Marchequinox.The time dependence of (cid:126)r m ( t ) is made to besuch that after exactly one day, it again pointsdirectly overhead at the mean Sun as it crossesthe observer’s meridian. This time interval be-tween two successive occurrences of the meanSun crossing any chosen meridian is defined tobe 24 h, the internationally agreed upon clocklength of an Earth day. Clocks are built to tickoff exactly 24 h from one meridian crossing ofthe mean Sun to the next.It is important to understand that becauseEarth is traveling around the Sun, (cid:126)r m ( t ) rotatesthrough more than 2 π radians in a day. Thisextra rotation occurs because even if Earth werenot rotating on its axis, over the course of a year(365.25 d) the Sun and (cid:126)r m ( t ) would still makea rotation of 2 π radians around Earth.To see the origin of this extra rotation, imag-ine Earth did not rotate on its axis at all. Thenif on March 20 (for example) (cid:126)r m ( t ) points at theSun, six months later it will point directly awayfrom the Sun, and after a year, it will once againpoint at the Sun. Because of this extra rotation,in a year of 365.25 d an observer on Earth seesthe Sun circle the Earth 366.25 times. Becauseof this extra rotation, the celestial longitude ofthe mean Sun observed from Earth advances by ω = 2 π/ .
25 = 0 . (cid:126)r m ( t ) must rotate through an angle of 2 π +2 π/ .
25 radian each day. Then after n daysof rotation (cid:126)r m ( t ) will have rotated through anangle of n π + nω . If the tip of (cid:126)r m ( t ) made adot every 14 days, the dots would form a circlelike the connected dots in Fig. 2.If Earth’s orbit around the Sun were circularand if Earth’s axis were not tilted relative tothe plane of the orbit, the Sun would move inthe sky exactly like the mean Sun; (cid:126)r S ( t ) wouldtrace out a circle exactly like that traced outby (cid:126)r m ( t ). To see that this is so, relate (cid:126)r S ( t ) to (cid:126)r m ( t ) by expressing the time t as the number ofdays n . Then for no tilt and a circular orbit,the Sun vector is (cid:126)r S ( n ) = cos nω ˆ ı + sin nω ˆ . (3)The clock vector rotates faster, but, as discussedabove, every 24 h it points at the mean Suncrossing the same meridian that it crossed a dayearlier. It is the same meridian, but that merid-ian is aligned with an arc of celestial longitude ω = . (cid:126)r m ( n ) = cos( n π + nω ) ˆ ı + sin( n π + nω ) ˆ . (4)Because cos( n π + nα ) = cos( nα ) (and similarlyfor the sine), it follows that every 24 h (cid:126)r m ( n ) = (cid:126)r S ( n ), i.e., (cid:126)r m ( n ) = cos nω ˆ ı + sin nω ˆ (5)= (cid:126)r S ( n ) . Thus, for an untilted circular orbit theanalemma would be a single image of the Sunalways at the same place in the sky at the sametime each day.
III. EFFECT OF AXIS TILT ON THEANALEMMA
Because Earth’s axis of rotation is tilted23.44 ◦ relative to the plane of its orbit aroundthe Sun, the equality of Eq. 5 does not hold forthe real Sun. (cid:126)r S ( t ) traces out a path at an an-gle to the path traced out by (cid:126)r m ( t ). The tips ofthe two vectors trace out paths similar to thoseshown in Fig. 2.The analemma is shaped by the tilt of Earth’saxis of rotation and by the ellipticity and ori-entation of its orbit. The major effect comesfrom tilt. This is apparent when you constructan analemma for an orbit that is a circle. Atilted, circular orbit is also easier to deal withthan an elliptical orbit, so it’s a good procedureto first try out (cid:126)r S ( t ) and (cid:126)r m ( t ) in this simplercontext and then modify them to take into ac-count the actual orientation and elliptical shapeof Earth’s orbit.Because of the tilt of Earth’s axis, the Sun’sangle relative to the celestial equator (its decli-nation) changes over the course of a year. Thus,for Earth, the points of (cid:126)r S ( t ) on the tilted cir-cle trace out the seasonal movement of the Sunas it goes from a declination of δ = 0 ◦ atthe March equinox to δ = 23 . ◦ (the NorthernHemisphere’s summer solstice) to 0 ◦ (autumnalequinox) to − . ◦ (winter solstice) and so on.Figure 2 illustrates the variation in declinationbut with the tilt set to 45 ◦ to make the effectmore apparent.It is because its path is at an angle to theequatorial plane that the real Sun arrives overa given meridian at a different clock time eachday. That is to say, when the mean Sun crossesa meridian, the real Sun will be over a slightlydifferent meridian. Because Earth rotates at15 ◦ per hour, the angular difference betweenthe longitudes of these two meridians corre-sponds to a time difference. For example, on aday when it is exactly clock noon at your merid-ian and the real Sun is overhead a meridian 1 ◦ to the east, it will cross your meridian 4 minlater at 12:04 pm. A difference of 1 ◦ in longi-tude corresponds to 4 min of time, and 1 arc-min corresponds to 4 s of time. FIG. 3. This is the appearance of the two intersect-ing circles shown in Fig. 2 as seen by a viewer look-ing directly down the z axis onto the plane of themean Sun, the planet’s equatorial plane — shownhere by the connected dots.FIG. 4. The mean Sun (o) moves at a steady ratearound the celestial equator. The tilted orbit’s pro-jection onto the plane of the mean Sun (Earth’sequatorial plane) is shown by the +’s. The (cid:5) ’s arethe projection of the +’s onto the celestial equator,and they move back and forth relative to the meanSun, i.e., relative to the regular time of a clock.(The tilt in this figure is 45 ◦ .) Figures 3 and 4 show how the tilt causes thereal Sun to vary in celestial longitude relativeto the mean Sun. It is worth examining thesefigures closely both to understand the geometricorigins of the effect and because they clarify howto extract a quantitative account of the effectfrom the two vectors (cid:126)r S ( t ) and (cid:126)r m ( t ).This difference in the longitudes of the twomeridians can be represented by a length of arcon the equatorial path of the mean Sun. To getthis arc length, project (cid:126)r S ( t ) onto the equatorialplane, and then extend the projection to inter-sect the circle of the mean Sun. Figures 3 and4 illustrate the construction of the arc length,and Fig. 4 shows how the movement of the Suncauses this arc length to vary as the real Sunmoves back and forth relative to the meridianof the mean Sun. Combined with the seasonalchange in declination this variation in arc lengthproduces an analemma that is a figure eight.Why a figure eight? The geometry of Figs. 3and 4 answers the question. These figures showthe relation between the mean Sun and the ac-tual Sun over a succession of days. On theequinox an observer sees them both at the samemeridian at the same time. But Fig. 3 showsthat this alignment is soon lost. The figure is aview along the z axis perpendicular to the planeof the mean Sun, i.e., the equatorial plane. Seehow the tilt foreshortens the projection. Thisprojection is shown by the +’s in Fig. 4, i.e.,the +’s are the projection of the tip of (cid:126)r S ontothe plane of (cid:126)r m ( t ). An observer will see an an-gular separation that corresponds to the arc be-tween the extension (the (cid:5) ’s) of the projectionof (cid:126)r S onto the arc along which the mean Sun isdisplayed (the o’s).In Fig. 4 the time interval between successiveo’s, +’s, and (cid:5) ’s is the same ( ∼
13 d), yet, as thefigure shows, the (cid:5) ’s move back and forth rel-ative to their corresponding o’s. To check thisstatement place a straight edge on the box ( (cid:3) )in Fig. 4 that marks the center of the graph, andnotice that a line from the box to a + connectsdirectly to a (cid:5) . This (cid:5) is the projection of the+ onto the equatorial circle. If you now followthe behavior point-by-point, you can see that atthe equinox the (cid:5) lines up with its correspond-ing o — the mean Sun and the actual Sun crossthe meridian at the same time. But then, atfirst, the Sun lags behind the mean Sun. Theselags accumulate until the mean Sun is half wayto the solstice (45 ◦ on the diagram); after thatpoint the (cid:5) moves faster than its correspond-ing o, and catches up to it so they both reach90 ◦ together. The (cid:5) passes its o, but after 135 ◦ the (cid:5) slows down and they both reach 180 ◦ to-gether.The cycle repeats over the second half of theyear, so the back-and-forth motion occurs twiceduring the time that the up-down motion — thechange in declination — goes through its cyclecorresponding to one period of the orbit. Back-and-forth twice for up-and-down once makes afigure eight.The preceding argument and conclusions areimplicit in Eq. 1. The declination angle δ isthe angle between (cid:126) A and its projection onto theequatorial plane. If ˆ k is a unit vector in thedirection of the Earth’s axis of rotation, thencos (cid:16) π − δ (cid:17) = ˆ k · (cid:126)r S sin δ = ˆ k · (cid:126)r S , (6)where ˆ k and (cid:126)r S are unit vectors.To extract a numerical result from Eq. 6, youneed representations of (cid:126)r S and (cid:126)r m in the sameCartesian coordinate system. For a frame withˆ k parallel to Earth’s rotation axis and ˆ ı and ˆ in the equatorial plane, (cid:126)r m is as given in Eq. 5.For (cid:126)r S , however, Eq. 3 is correct only for theframe designated by primes in Fig. 5, in which (cid:126)r S lies in the ˆ ı (cid:48) – ˆ (cid:48) plane which is perpendicularto the z (cid:48) axis, and, therefore, perpendicular toˆ k (cid:48) .The angle θ between ˆ k and ˆ k (cid:48) is the planet’stilt; θ = 23 . ◦ for Earth. For the case whereone plane is tilted by rotating it an angle θ around the x -axis, the connection between theprimed and unprimed unit vectors is, as you cansee from Fig. 5,ˆ ı (cid:48) = ˆ ı ˆ (cid:48) = cos θ ˆ + sin θ ˆ k ˆ k (cid:48) = − sin θ ˆ + cos θ ˆ k. (7) FIG. 5. The connection between the unit vectors ˆ ı ,ˆ , ˆ k , and the primed unit vectors of a frame rotatedan angle θ about their shared x -axis Using Eqs. 7 to express (cid:126)r S ( t ) = cos ωt ˆ ı (cid:48) +sin ωt ˆ (cid:48) in terms of the unprimed referenceframe, you will get (cid:126)r S ( t ) = cos ωt ˆ ı + sin ωt cos θ ˆ + sin ωt sin θ ˆ k. (8)Now you can evaluate Eq. 6 and find the dec-lination as a function of time:sin δ = sin ωt sin θ = sin nω sin θδ = sin − (sin nω sin θ ) . (9)Notice that, as expected, over the course of ayear, i.e., from n = 1 to n = 365, δ varies from+ θ to − θ and back again.What about the variation of the differencealong the equator? To find this follow the stepsused to obtain Fig. 4. First, project (cid:126)r S ( t ) ontothe plane of the mean Sun. The projection isthe ˆ ı and ˆ components of Eq. 8, so you justomit the ˆ k component and get the projectedvector (cid:126)r Sxy where (cid:126)r
Sxy ( t ) = cos ωt ˆ ı + sin ωt cos θ ˆ (10)This vector locates the +’s in Fig. 4. But youwant to compare to the o’s not the +’s but the (cid:5) ’s, i.e., the points located by the extension of (cid:126)r Sxy , to the unit circle defined by (cid:126)r m . To makethis extension, scale (cid:126)r Sxy to make it be the samelength as (cid:126)r m . Because (cid:126)r m is of unit length, youscale (cid:126)r Sxy by dividing by its magnitude | (cid:126)r Sxy | = (cid:112) (cid:126)r Sxy · (cid:126)r Sxy , which from Eq. 10 is | (cid:126)r S xy | = (cid:112) cos ωt + sin ωt cos θ, (11)so thatˆ r Sxy = cos ωt ˆ ı + sin ωt cos θ ˆ (cid:112) cos ωt + sin ωt cos θ . (12)Equation 12 describes a unit vector ˆ r Sxy thatis the same length as the unit vector (cid:126)r m . InFig. 4 the tip of ˆ r Sxy constructed in this waylies on the arc defined by (cid:126)r m ( t ).The arc length on the unit circle equals (inradians) the angle, call it α , between (cid:126)r m andˆ r Sxy . You can find cos α from the scalar productof the two unit vectorscos α = (cid:126)r m · (cid:126)r Sxy | (cid:126)r Sxy | = cos ωt + sin ωt cos θ (cid:112) cos ωt + sin ωt cos θ . (13)Alternatively, you can use the vector crossproduct to find sin α .sin α = ˆ k · ( (cid:126)r m × (cid:126)r Sxy )= sin ωt cos ωt (cos θ − (cid:112) cos ωt + sin ωt cos θ = − sin(2 ωt ) sin ( θ ) (cid:112) cos ωt + sin ωt cos θ . (14)Using the cross product has some benefits.One is that when (cid:126)r m and (cid:126)r S are nearly equal, α is small enough so that sin α ≈ α . Another isthat application of the double-angle trig iden-tity makes it evident that the abscissa varieswith twice the frequency of the ordinate.Equations 9 and 14 are parametric equa-tions of the analemma, and they produce theanalemma curve shown in Fig. 6. This plot of FIG. 6. Earth’s analemma if Earth’s orbit were acircle. Each dot is 1 d. declination δ vs. α , the difference between thelongitude of a Sun with a circular path and thelongitude of the mean Sun, was made using Ex-cel. As foreseen, it is a symmetric figure eight.As is customary, the units of the abscissa arein minutes of time rather than in radians ordegrees. IV. EFFECT OF ORBITALELLIPTICITY ON THE ANALEMMA
Because Earth’s orbit is not a circle but anellipse with the Sun at one focus, the actualanalemma (shown in Figs. 1, 11, and 12) differsnoticeably from the circular orbit’s analemmashown in Fig. 6. The effects of the ellipticitycan be included by replacing (cid:126)r S ( t ) for a tiltedcircular orbit (Eq. 8) with (cid:126)r S ( t ) for the actualelliptic orbit.A good way to get the (cid:126)r S that corresponds tothe correct shape of Earth’s orbit is to numeri-cally integrate the equations of motion of Eartharound the Sun to obtain (cid:126)r E ( t ), a vector thattraces out Earth’s orbit, and then use the factthat (cid:126)r S ( t ) = − (cid:126)r E ( t ). FIG. 7. Step 1: With the orbit (diamonds) in theplane of the mean Sun (connected dots) set the co-ordinate axes to have their origin at the focus of theellipse and the y -axis along the ellipse’s major axis2 a . Perihelion is at y = − a (1 − (cid:15) ). To do this integration you need the right ini-tial conditions. A good way to find them pro-ceeds in three steps. Step one, find the initialconditions in a coordinate frame aligned withthe axes of the ellipse. This orientation of theaxes makes determining the initial conditionsparticularly simple.Step two, transform the initial conditions de-termined in step one into the frame of referencethat lines up its x -axis with the line of nodes— the line of intersection of the orbital planewith the plane of the circular orbit of the meanSun (see Fig. 2). For Earth, this corresponds toa rotation of the orbital plane 12 . ◦ about the z -axis.Step three, transform the initial conditions tocorrespond to the orbital plane tilted around itsline of nodes.Numerical integration with the initial condi-tions from step three gives a vector (cid:126)r E that ac-curately represents Earth’s orbit. The resulting FIG. 8. Step 2: Rotate the axes to align the x -axiswith the orbit’s line of nodes. In this picture, theaxes have been rotated around the z -axis by − ◦ so the major axis of the ellipse is now at 60 ◦ tothe y -axis. The line of nodes is the segment of the x -axis crossing the ellipse.FIG. 9. Step 3: Rotate the axes around the x -axisso that the orbital plane is tilted 23 . ◦ around theline of nodes (solid line). To make the tilt moreapparent it has been set to 40 ◦ in this picture. (cid:126)r S , which equals − (cid:126)r E , is no longer the unit vec-tor of Eq. 3 sweeping out a circle. Now it sweepsout an ellipse with eccentricity (cid:15) = 0 . θ =23 . ◦ relative to the plane of the mean Sun,speeding up at perihelion and slowing down ataphelion. This vector (cid:126)r S , the mean Sun vec-tor (cid:126)r m , and slightly modified versions of Eqs. 6and 14 yield a quantitatively accurate represen-tation of Earth’s analemma. FIG. 10. The sum of the distances from any pointon the curve of an ellipse to the two points labeled“focus” is a constant equal to the length of the ma-jor axis 2 a . The eccentricity (cid:15) is the distance fromthe center of the ellipse to either focus, measured inunits of the semi-major axis. Construction of this more elaborate (cid:126)r S vectorrequires some basic properties of ellipses sum-marized in Fig. 10. The two points labeled “fo-cus” are key. The curve is defined by the prop-erty that the distance from one focus to anypoint on the curve plus the distance from thatpoint to the other focus is constant. (Use a rulerto check that this statement is true for Fig. 10.)Another important feature is the eccentricity (cid:15) .It measures the distance of the focus from thecenter of the ellipse in units of the semi-majoraxis a . (Use a ruler to confirm that the eccen-tricity of the ellipse in Fig. 10 is (cid:15) = . b = a (cid:112) (1 − (cid:15) ) (15)area of ellipse: A = πab. (16)Newton’s law of universal gravitation statesthat the force (cid:126)F exerted on a spherical mass m by a spherical mass M is (cid:126)F = − GM m | (cid:126)r − (cid:126)r | ( (cid:126)r − (cid:126)r ) | (cid:126)r − (cid:126)r | (17)where (cid:126)r and (cid:126)r are the position vectors of thetwo masses, and G is the constant of universalgravitation. This equation is the inverse squarelaw of gravitational attraction multiplied by theunit vector ( (cid:126)r − (cid:126)r ) / ( | (cid:126)r − (cid:126)r | ), which pointsalong the line between the centers of the twomasses. The unit vector asserts that this is acentral force, and the minus sign at the begin-ning of the equation points the unit vector from m to M to reflect the fact that M is attracting m .For M >> m and with (cid:126)r = (cid:126)r − (cid:126)r , Newton’slaws of motion and Eq. 17 lead to a differentialequation for the orbit: (cid:126) ¨ r = − GM | (cid:126)r | (cid:126)r | (cid:126)r | . (18)The orbiting mass m has energy E where Em = 12 (cid:126)v · (cid:126)v − GM | (cid:126)r | , (19)and angular momentum (cid:126)L such that (cid:126)Lm = (cid:126)r × (cid:126)v = r (cid:126)ω, (20)where (cid:126)ω is the angular velocity of the planetaround the Sun. E and L are conserved (con-stant) quantities.For initial values of position and velocity suchthat the energy E <
0, the solutions to thedifferential equation are orbits that are ellipseswith semi-major axis a and the Sun at one fo-cus. The following equations connect the ge-ometry of the elliptical orbit to the underlying0 TABLE I. Some Parameters of Earth’s Orbit (cid:15)
Eccentricity 0.0167 a Semi-major axis 1 AU T Period 1 y E Energy per solar mass -19.739 AU / y L Angular momentumper solar mass 6.2823 AU /y φ Angle from line ofnodes to perihelion -78.2 deg physics: (cid:15) = (cid:18) EL G M (cid:19) (21) a = (cid:12)(cid:12)(cid:12)(cid:12) GM E (cid:12)(cid:12)(cid:12)(cid:12) (22) T = 4 π GM a (23)where T is the period of the planet’s orbit. No-tice that with a measured in AU and T in years,Eq. 23 becomes T = 1 = 4 π /GM for Earth,so that in these units GM = 4 π when M is themass of the Sun.These equations are interesting for what theysay about planetary motion. They are essentialfor checking the validity of orbits produced bynumerical integrations. They are also useful forobtaining initial conditions for orbits with pa-rameters different from those of Earth’s orbit.To find initial conditions that correspond toEarth’s orbit, that is, to an orbit that has (cid:15) = . x -axis coinci-dent with the ellipse’s minor axis, and the y -axiswith its major axis, Earth’s position vector atperihelion is (cid:126)r E = 0 ˆ ı − a (1 − (cid:15) ) ˆ and its velocityis (cid:126)v E = v p ˆ ı + 0 ˆ .The quantity v p is the magnitude of Earth’svelocity at perihelion. You can determine v p using Kepler’s law of equal areas in equal times(conservation of angular momentum (cid:126)L ). A vector (cid:126)r from the Sun to a point movingwith velocity (cid:126)v sweeps out area at a rate d (cid:126)Adt = 12 (cid:126)r × (cid:126)v. (24)This is essentially the formula for the area ofa triangle — 1/2 the base times the altitudewhere the base is the arc length swept out in atime dt . Kepler and Newton tell us that for acentral force, the quantity d (cid:126)A/dt is constant inboth magnitude and direction. It follows thatfor Earth the value for that constant must bethe same as the total area swept out in a year,i.e., Eq. 16, divided by the time of one year: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d (cid:126)Adt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = πabT = πa √ − (cid:15) T = 3 .
141 AU /y . (25)But, as Eq. 20 states, (cid:126)r × (cid:126)v is also the angularmomentum (cid:126)L of a unit mass. (For any giveninitial conditions, the orbit is independent ofthe mass of the orbiting object, so set m = 1to simplify the discussion.) Lm ≡ L = 2 dAdtL Earth = 6 .
282 AU /y (26)And, because at perihelion (cid:126)r E and (cid:126)v E are ex-actly perpendicular, L Earth = a (1 − (cid:15) ) | (cid:126)v E | = (1 − . v p = . v p = 6 . , (27)and the magnitude of the velocity at perihelionis v p = 6 .
389 AU/y . (28)In the chosen reference frame the position ofEarth at perihelion is (cid:126)r E = − . AU, andits velocity is (cid:126)v E = 6 .
389 ˆ ı AU/y. It would beequally convenient to determine the initial con-ditions at aphelion.With these initial conditions an Euler integra-tion of the differential equation with an initial1half-step gives a good result. This is the methodthat Feynman used in his lectures. In the unitsused here a convenient time step for the inte-gration is ∆ t = 1 / .
25 = . (cid:126)r m given in Eq. 5. The integration can bedone with any mathematical software package.A spreadsheet works well, and the calculationused here is available with the online version ofthis paper. TABLE II. Initial values of (cid:126)r E in AU and (cid:126)v p inAU/y for calculating Earth’s orbit in three differentorientationsOrientation of el-lipse relative tocoordinate axes P e r i h e li o n o n − y a x i s O r b i t p l a n e r o t a t e d − . ◦ O r b i t p l a n e t il t e d . ◦ x y -0.9833 -0.9588 -0.8796 z v x / v y / v z / Integration with these initial conditions startsfrom perihelion (January 3) and gives the day-by-day position of Earth on its orbit with thespecified eccentricity. However, this orbit needsto be correctly oriented relative to the idealizedorbit of the mean Sun.Do this with steps two and three. Step two:make the line of nodes coincide with the x -axisby rotating the coordinate frame about its z -axis until Earth crosses the x -axis on March 20,the vernal equinox of the Northern Hemisphere.To find the angle of rotation use orbit dataproduced by numerical integration with a tilt TABLE III. Selected values of (cid:126)r E from the numer-ical integration of Earth’s orbit.Date x n y n z n r n angle but no rotation of the orbit plane. Thesedata show Earth crossing the x -axis 13 d laterthan the actual March 20 equinox. This re-sult means you must rotate the reference frameclockwise by − / . ×
360 = − . ◦ rel-ative to the orbit to place the x -axis so thatEarth crosses it on March 20. To rotate the coordinate frame, use the fol-lowing version of Eq. 7 modified to take intoaccount that this rotation is about the z -axis: x = x (cid:48) cos α + y (cid:48) sin αx = 0 + ( − . × − . . y = − x (cid:48) sin α + y (cid:48) cos αy = 0 + ( − . × . − . . These equations produce the second set of ini-tial conditions in Table II.Step three: tilt the orbital plane. Do this byapplying Eq. 7 to the set of orbital coordinateslabeled “rotated” in Table II; you will get theinitial conditions labeled “tilted” in Table II.Integration with these values gives Earth’s orbittilted relative to the equatorial plane.The result of the numerical integration is atable of values of x n , y n , and z n , the Cartesiancomponents of (cid:126)r E ( nω ), the vector that locatesEarth relative to the Sun on day n since peri-helion: (cid:126)r E ( nω ) = x n ˆ ı + y n ˆ + y n ˆ k (29)2Some results from the numerical integrationare given in Table III. These should make youconfident that the calculation is accurate. Be-tween March 20 and 21 y n and z n pass throughzero as they should at the vernal equinox of theNorthern Hemisphere. And between June 21and June 22, when the summer solstice occurs, x n changes sign as it should. The entry datedJuly 5 was selected because on that date thedistance r n has its largest value; in other wordsthe calculation gives the correct date for Earth’saphelion.But notice that the aphelion distance isslightly larger than a (1 + (cid:15) ); this discrepancyreflects the limitations of the method of numer-ical integration and the step size used here. It’sa warning that integrations over more than ayear may need a smaller step size or a more ac-curate method of integration.To construct the analemma, you need (cid:126)r S thevector that points from Earth to Sun. It is (cid:126)r S ( nω ) = − (cid:126)r E ( nω )= − x n ˆ ı − y n ˆ − y n ˆ k (30)To extract α and δ , modify Eqs. 11, 12, and14 to be Eqs. 31, 32, and 34 and use them to get α and δ as functions of time, i.e., as functionsof n the day number. | (cid:126)r S xy ( nω ) | = (cid:112) x n + y n , (31)so that (cid:126)r Sxy ( nω ) = x n ˆ ı + y n ˆ (cid:112) x n + y n . (32)Then sin α = ˆ k · (cid:126)r m × (cid:126)r Sxy (33)so that α n = sin − (cid:32) − y n cos nω + x n sin nω (cid:112) x n + y n (cid:33) (34)and δ n = sin − (cid:32) z n (cid:112) x n + y n + z n (cid:33) . (35) FIG. 11. Here the calculated analemma is displayedoverlaid on the observed analemma of Fig. 1. Forthe best overlap the calculated analemma has beenrotated through 35.6 ◦ . This corresponds to about2:15 pm MST in reasonable agreement with the 2:28pm MST when the photographs of the observedanalemma were taken. Equation 34 is not entirely satisfactory for ex-tracting α . It is set up so that when n = 0, (cid:126)r S points 12 . ◦ counterclockwise from the nega-tive y-axis while (cid:126)r m is pointing along the x-axis.This makes the angle between them 90 − . . ◦ . In principle this large angle is not impor-tant; the variation in α is still the correct effect;it’s as though you were measuring the time atwhich the Sun is near its zenith with a clockin a time zone some seven or eight hours away.However, it is tidier to set (cid:126)r m so that it corre-sponds to the direction of (cid:126)r S . To do this, add aphase constant φ = − . ◦ so (cid:126)r m becomes (cid:126)r m ( n ) = cos( nω + φ ) ˆ ı + sin( nω + φ ) ˆ (36)and α ≈ x n sin( nω + φ ) − y n cos( nω + φ ) (cid:112) x n + y n (37)The analemmas shown in Figs. 11 and 12 areplots of δ vs. α obtained from Eqs. 35 and3 FIG. 12. This is the same curve as in Fig. 11 butoriented to be at noon and with the scale of theabscissa expanded by ×
6. Each dot is a differentday.
37 for n = 1 to n = 365 with φ = − . ◦ .In Fig. 11 the calculated analemma has beenlaid over the observed analemma; the agree-ment between the two is quite satisfactory. Theanalemma in Fig. 12 is the same as in Fig. 11but with the horizontal dimension enlarged by6 × . It shows the day-by-day variation of thearrival of the noontime Sun. V. VECTORIA’S SECRET
The construction of the analemma in the pre-ceding sections emphasizes visualization of howthe analemma results from the relative motionof Earth and Sun in three dimensions. The goalwas to show how the geometry connects withthe physics.But more formal, less visual approaches arepossible, and they have their merits. Done properly they can carry you past various diffi-culties of establishing relative phases, of gettingthe algebraic signs correct, and they can pro-vide an aid or an alternative for those who findit difficult to visualize three dimensions. Thefollowing paragraphs make a fuller use of vec-tor algebra to reprise the construction of theanalemma.The key question in constructing theanalemma is “What is the angle α between (cid:126)r m and the projection of (cid:126)r S onto the plane of (cid:126)r m ?”One way to find the answer is to use the factthat α is the dihedral angle between two planes,one defined by ˆ k and (cid:126)r m and the other by ˆ k and (cid:126)r S . Figure 13 shows the configuration. FIG. 13. The vectors ˆ k , (cid:126)r m , and (cid:126)r S define twoplanes. The variation between the positions of theSun and the idealized mean Sun causes α to varyfrom day to day. The angle α between the two planes is thesame as the angle between unit vectors drawnperpendicular to the two planes. This fact sug-gests the following procedure to find sin α : Con-struct unit vectors ˆ n S and ˆ n m normal to the twoplanes; take their cross product, which will givea vector in the ˆ k direction with magnitude ofsin α ; take the dot product of ˆ k with the crossproduct to find sin α (see Eq. 40).4Use cross products to construct ˆ n S ˆ n S = ˆ k × (cid:126)r S | ˆ k × (cid:126)r S | (38)and ˆ n m ˆ n m = ˆ k × (cid:126)r m | ˆ k × (cid:126)r m | = ˆ k × (cid:126)r m (39)where Eq. 39 follows from the fact that | ˆ k × (cid:126)r m | = 1 because ˆ k and (cid:126)r m are both unit mag-nitudes and perpendicular to each other.This is enough information to determine sin α by direct substitution intosin α = ˆ k · (ˆ n m × ˆ n S ) , (40)but using the vector identity (cid:126)A · ( (cid:126)B × (cid:126)C ) = ( (cid:126)A × (cid:126)B ) · (cid:126)C and the fact that (cid:126)r m is (cid:126)r m = − ˆ k × ˆ n m . (41)yields the more compact expression:sin α = ˆ k · (ˆ n m × ˆ n S )= ( ˆ k × ˆ n m ) · ˆ n S = − ˆ n S · (cid:126)r m . (42)Substituting Eqs. 36 and 38 into Eq. 42 givesEq. 37.The vector expression for the declination wasalready used to obtain the result given in Eq. 35. VI. FACTOIDS AND PROJECTS
The analemma occurs because solar days varyin length. A curious consequence is that thelongest solar day, about 24 h 30 s, occurs justafter the winter solstice, the day on which inthe Northern Hemisphere there are the fewesthours of daylight. Thus, “the shortest day ofthe year is also the longest day of the year” —or nearly. It is also interesting that for aboutthree weeks after the solstice sunrise continues to get later. Nevertheless, the amount of day-light increases in the Northern Hemisphere be-cause sunset is getting later by an amount thatmore than offsets the later sunrise.
The NewYork Times finds these factoids fit to print. ? This tutorial can be the basis for furtherprojects. A project might be as modest asan exercise in vector manipulation. For exam-ple, you might replace ˆ n m and (cid:126)r S in Eq. 40with their cross product definitions from Eqs. 38and 39 and use the identity (cid:126)A × ( (cid:126)B × (cid:126)C ) = (cid:126)B ( (cid:126)A · (cid:126)C ) − (cid:126)C ( (cid:126)A · (cid:126)B ) to derive Eq. 42.Or you could measure the declination of theSun and observe the time at which it reaches itszenith and see if these agree with the predictionsof your analemma. You will need to understandhow to make (cid:126)r m correspond to your clock time– a worthwhile task.You can modify the input parameters for thenumerical calculation of Earth’s orbit and ex-amine how Earth’s analemma will change as theequinoxes precess. You can also replace Earth’sorbital and rotational parameters with those ofMars and find the analemma that an observeron Mars will see. By properly modifying theinput parameters of the numerical integration,you can find the analemma of any planet.The analemma is the result of the superposi-tion of two periodic motions at right angles toeach other. In effect it is an interference pat-tern, a Lissajou figure. The rather small eccen-tricity of Earth has a surprisingly large effect onthe analemma. Can you invert the argument,and find the eccentricity of Earth’s orbit fromits observed analemma?A nice project would be to calculate the tran-sits of Venus using this paper’s approach. Nu-merically calculate the position vector (cid:126)r V ofVenus relative to the Sun. Then construct apointer from Earth to Venus (cid:126)r EV = (cid:126)r V − (cid:126)r E .Look at the angular difference between (cid:126)r S (= − (cid:126)r E ) and (cid:126)r EV as a function of time. Find thetimes at which the angular difference is between − / ◦ and +1 / ◦ , i.e., approximately withinthe angular diameter of the Sun. You will needa method of numerical integration that is sta-ble over the long interval of time between theoccurrence of one pair of transits and the next5pair. The Euler method will not work, but afourth order Runge Kutta method available onmost packages of mathematics software may dothe job. VII. CONCLUSION
This tutorial shows how to build your ownanalemma using only the physics taught incalculus-level introductory physics. The resultis the construction of a quantitatively accurateanalemma for Earth. It is a concrete exampleof how successfully Newtonian mechanics de-scribes the solar system.We tell our students that Newtonian physicsis a triumph of human intellect; we describeinteresting motions of planets, moons, comets,etc. and then say that Newtonian mechanicsexplains them. It is important for physics stu- dents to go beyond praise for the achievementsof Newtonian physics. This tutorial enablesthem to recreate and experience some of thatachievement.
ACKNOWLEDGMENTS
I thank Howard Georgi (Harvard Univer-sity), Ed Bertschinger and Ramachandra Dasari(MIT) for their hospitality. It has greatly facil-itated my work. I thank Joe Amato (ColgateUniversity) for his close reading and thought-ful suggestions for improving this paper. Thegoals of this paper grew out of an exchange ofideas with Shane Larson (Utah State Univer-sity), and the decision to write it was inspiredby “Astronomy’s Discoveries and Physics Edu-cation,” a conference supported by NSF Grant1227800. Wojtek Rychlik, (2005),
Mechanics (Addison Wesley,Reading, MA, 1953) pp. xiii, 358 (see pp. 111- 117). Richard P. Feynman, Robert B. Leighton, andMatthew Sands,
Feynman Lectures on Physics ,Vol. 1 (Addison Wesley, Reading, MA, 1963) pp.9-6 – 9-9 You can also use the fact that perihelion forEarth does not occur on the day of the sol-stice, December 21, as would be the case ifyou did not rotate the axes. Perihelion oc-curs 13 days later on January 3, so the ro-tation of the axes that puts perihelion at theproper date is (as it should be) the same asthe rotation that places the March equinoxon the x -axis. Another alternative is to lookup the celestial coordinates of perihelion; see,for example,