Positional Astronomy for Historical Studiesfverbunt/iac2011/hlpos.pdf · Positional Astronomy for...
Transcript of Positional Astronomy for Historical Studiesfverbunt/iac2011/hlpos.pdf · Positional Astronomy for...
Positional Astronomy for Historical StudiesA brief introduction
Frank Verbunt
Astronomical Institute, University of Utrecht, Netherlands
May 13, 2011
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 1 / 33
Outline
1 Coordinate systemsequatorial coordinatesecliptic coordinates
2 Coordinate transformationsequatorial to ecliptic and v.v.equatorial to horizontal and v.v.the effect of parallax
3 Daily motion of the starsat the pole, at the equatorat arbitrary geographical latitudedaily and yearly motion of the Sun
4 Precessionaccording to Hipparchos, Ptolemaiosmodern
5 Some consequences and example computations
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 2 / 33
Coordinate systems: equatorial coordinates
Coordinate systemsa coordinate system is definedby defining a large circle, and azero point on it
the line through the center ofthe circle and perpendicular to itcuts the sphere in two points:the poles
two angles determine theposition of a point S; to see thisdraw one large circle throughthe poles and the zero point,another through the poles andthe point S.
Equatorial coordinates
large circle: equator
right ascension α
declination δ
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 3 / 33
Coordinate systems: ecliptic coordinates
Coordinate systemsequatorial coordinates aresuitable to determineobservation times
ecliptic coordinates are suitableto describe the motion of theplanets
the zero point is the intersectionof the equator and eclipticwhere the Sun is at the springequinox
Ecliptic coordinates
large circle: orbit of Sun
longitude λ
latitude β
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 4 / 33
Coordinate systems: connection to Cartesian coordinates
Equatorial coordinates
xyz
=
cos δ cosαcos δ sinα
sin δ
(1)
Ecliptic coordinates
xyz
=
cos β cos λcos β sin λ
sin β
(2)
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 5 / 33
Coordinate transformations
Equatorial to ecliptic rotation ε around x-axis
x′
y′
z′
=
1 0 00 cos ε sin ε0 − sin ε cos ε
x
yz
(3)
substitute x′ = cos β cos λ, y′ =cos β sin λ, z′ = sin β and x =cos δ cosα, y = cos δ sinα, z = sin δ:
tan λ =sinα cos ε + tan δ sin ε
cosα(4)
and
sin β = sin δ cos ε − cos δ sinα sin ε (5)
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 6 / 33
Coordinate transformations
Ecliptic to equatorial rotation −ε around x-axisAnalogously: x
yz
=
1 0 00 cos ε − sin ε0 sin ε cos ε
x′
y′
z′
(6)
hence
tanα =sin λ cos ε − tan β sin ε
cos λ(7)
sin δ = sin β cos ε+cos β sin λ sin ε (8)
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 7 / 33
Equatorial to horizontal coordinates and v.v.
H is the hour angle of the star (anglebetween the plane through the starand the poles, and the plane throughthe poles and the south). H changeslinearly with time (15◦/hr)
A is the Azimuth measured from theSouth(Note that the Astronomical Almanactakes A = 0 in the North!)
h the altitude (height) in degreesabove the horizon
φ the latitude on earth of the observer
To convert the coordinatesrotate along the y-axis overangle π
2 − φ.
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 8 / 33
Equatorial to horizontal coordinates and v.v.
cos A cos hsin A cos h
sin h
=
cos(π2 − φ) 0 − sin(π2 − φ)0 1 0
sin(π2 − φ) 0 cos(π2 − φ)
cos δ cos H
cos δ sin Hsin δ
=
sin φ 0 − cos φ0 1 0
cos φ 0 sin φ
cos δ cos H
cos δ sin Hsin δ
(9)
which may be rewritten
tan A =sin H
cos H sin φ − tan δ cos φ; sin h = sin φ sin δ+cos φ cos δ cos H
(10)Analogously:
tan H =sin A
cos A sin φ + tan h cos φ; sin δ = sin φ sin h−cos φ cos h cos A
(11)Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 9 / 33
Complication: parallax
The effect of parallaxA plane parallel to the surfaceof the Earth cuts the sky in acircle which is smaller than agreat circle.
. . . ignored, except for Moonan object at finite distance is seenin different directions from differentlocations on Earth
the effect is negligible for mostobjects: we will locate the observerat the center of the Earth
exception: the Moon:maximum difference in angle: theratio of the radius of the Earth(6400 km) and the distance to theMoon (380 000 km), i.e. 1◦.
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Other complications
Other complicationsOther effects which should betaken into account are
the local horizon is not flat
close to the horizon, lightrays are refracted by the air;its effect depends on theweather conditions; at thehorizon, a typical value is35′.
the Earth is not a perfectsphere, but slightly flattened;therefore the lineperpendicular to a flathorizon doesn’t pass throughthe center of the Earth: themaximum difference occursat 45◦ and is 11.5′. The(geographic) latitude that onefinds in atlases is definedsuch that it equals φ in eq. 9.
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Daily motion of the stars
At the pole the horizon is theequator
Stars move parallel to horizon
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 12 / 33
Daily motion of the stars
At the equator the poles are atthe horizon
Stars move perpendicular tohorizon
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 13 / 33
Equatorial coordinates to horizontal coordinates
Arbitrary geographic latitude illustrated for φ = 53o
star with δ > 90◦ − φ always above horizon, with δ < φ − 90◦ alwaysbelow horizon
with φ − 90◦ < δ < 90◦ − φ crosses horizon
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 14 / 33
Equatorial coordinates to horizontal coordinates
Cygnus for φ = 53o, horizon at 1 hr intervals
δ (αCyg, δCyg) = 45◦, δ (γCyg) = 40◦: always above horizon
δ (ηCyg) = 35◦, δ (βCyg) = 27◦: cross horizon
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 15 / 33
Daily and yearly motion of the Sun: β ' 0, λ nonlinear in t
3rd law of Kepler for ellipticorbit: λ changes non-linearlywith time
in equatorial coordinates:−ε ≤ δ ≤ ε, α(t) non-linear
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The Sun through the year 1 = Jan 1, 2= Feb 1, . . .
Setting of starsaround March 1: βCyg sets with Sunmid September: βCyg sets atsunrise
Rising of starsmid November: βCyg rises with
Sunearly May: βCyg rises at sunset
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Precession according to Hipparchos, Ptolemaios
Basic motionin most modern (!) textbooksprecession is described as themotion of the North (equatorial)Pole in a circle around the NorthEcliptic Pole
this is the precession asunderstood until Laplace
consequence: the zero point ofλ moves linearly with time⇒λ = λo + Kt , β =constant
Ptolemaios: K = 1◦/centurymodern: K ' 1◦/(72 year)
Rotation Earth axis aroundNEP
obliquity ε assumed constant
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 18 / 33
Precession according to modern understanding
Motion of Equatorial poles among stars: yyyy indicates year
Two small corrections: due to influence gravity of other planets, ε changesslowly, and orientation of ecliptic changes slowly
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 19 / 33
Modern precession: change in obliquity ε
Equation for ε(t)12h UTC on 1 Jan 2000 isJDo ≡JD 2451545.0
at that timeε ≡ εo = 23◦26′21.448′′
if t is number of Julian centuriessince JDo :t = (JD − JDo)/36525 then
∆ε(′) = −46.815t − 0.00059t2
+ 0.001813t3 (12)
ε(t) = εo + ∆ε (13)
Historic variation ε (theory)
(dashed line: theory of Laskar)
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 20 / 33
Modern precession: full equations
Equation for three anglesTo precess αo , δo at J2000.0 to α, δat another epoch, we first determinet as above. Then compute threeangles:
ζ(′′) = 2306.2181t + 0.30188t2
+ 0.017998t3
z(′′) = 2306.2181t + 1.09468t2
+ 0.018203t3
θ(′′) = 2004.3109t − 0.42665t2
− 0.041833t3
(14)
Precessionthen:
A = cos δo sin(αo + ζ)
B = cos θ cos δo cos(αo + ζ)
− sin θ sin δo
C = sin θ cos δo cos(αo + ζ)
+ cos θ sin δo
(15)
and finally:
tan(α − z) =AB
; sin δ = C
(16)Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 21 / 33
Precession and the rising/setting of the Sun and stars
The Sunthe most northern/southernsetting depends only onobliquity ε and geographiclatitude φ
to first order, precession hasconstant ε hence sunset/riseat solstitia was at the sameazimuth 5000 yrs ago as now
actually, small changes arecaused by the slow changein ε
Starsprecession changes theposition of the pole amongthe stars
hence the distance of a starto the pole, and itscomplement δ change
hence the azimuth where thestar rises/sets changes withtime
to know whether analignment pointed to a star,one must know itsconstruction date
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Probability on aligment to a star at fixed epoch
15 brightest stars5 are always below thehorizon in Utrecht (Achernar,Canopus, αCrucis, α and βCentauri)
2 always above horizon(Capella, Vega)
⇒ 8 stars rise and set;accept ±2◦ → 32◦ at easthorizon i.e. 17%.
49 brightest starslimiting magnitude is V = 2.0
24 stars rise and set;probability of hit ±2◦, inarbitrary direction is 38%.(Due to overlap in ranges)
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 23 / 33
above: for 15 brightest stars, below, for 49 brightest stars
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Stars cross/do not cross horizon (Mauna Kea)
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 25 / 33
Range of sunrises (Greece)
This picture and the previous one are from the website AstrophysicalPicture of the Day, antwrp.gsfc.nasa.gov for 20 and 21 December2005, respectively. They are made by Antony Ayiomamitis and PeterMichaud, respectively.
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 26 / 33
Final remarks, Literature
Some remarksOur treatment is not complete.Things we have left out include
nutation
aberration
difference between atomictime and UT
The angles ζ, z and θ in Eq. 14have a physical interpretation.See Seidelman, whereexpressions for precession froman arbitrary date are also given.
LiteratureP.K. Seidelmann, 1992,Explanatory supplement to theastronomical almanac: Theofficial book, with goodexplanations.
J. Meeus, 1991, Astronomicalalgorithms: More compact, withworked examples (very useful fortesting code!), but littleexplanation.
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 27 / 33
Worked example 1:
Problem 1observatory Sonnenborgh inUtrecht has coordinates05◦07′46′′E 52◦05′12′′N
on February 4, 2003, theSun is at α(2000) =21h08m49.69s , δ(2000) =−16◦24′43.7′′.
Compute the azimuth where theSun sets, for an assumed flathorizon. How long was the Sunabove the horizon of Utrecht thatday?
Answersα(Sun) and l(Utrecht) don’t enter!All angles in decimal degrees.
δ = −16.41214
φ = 52.08667
h = 0⇒ sin h = 0. eq. 10⇒ cos H =− sin φ sin δ/(cos φ cos δ) =0.37818, sin H = 0.92573
eq. ??⇒ tan A =0.92573/0.47935,A = 62.6245◦
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 28 / 33
Worked example 1: Worked example 2a:
Answers 1 (c’td)H is the hour angle of the Sunwhen it sets, and expressed inhours gives the time between theSun in the South, astronomicalmid-day, and sunset. 2H is thusthe time that the Sun is up:H = 67.779◦ = 4.518 hr, and theSun was above the horizon for9.04 hr. (We ignore the change inδ in these 9 hrs.)
Problem 2aThe position of Vega incoordinates of 2000.0 isα(2000) = 18.615649h ,δ(2000) = 38.783692◦. Its propermotion, in ′′/yr, is µα∗ = 0.20103,µδ = 0.28747. Compute theposition of Vega, in eclipticcoordinates, on 1 Jan 1601, usingprecession with constant ε.
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 29 / 33
Worked example 2a:
Problem 2a (c’td)1 compute α(2000), δ(2000) of
Vega in 16012 convert to ecliptic
coordinatesλ(2000), β(2000) of Vega in1601
3 add 1◦/72 yrs to computeλ(1601)
Answers 2a: proper motiontime interval is dt = −399 yr
NOTE: to convert µα∗ in ′′/yrto µα in s /yr, we divide by15cos δ. Also 1s=1h /3600and 1′′=1◦/3600
in −399 yr, we havedα = −0.001905h anddδ = −0.031860◦
hence Vega in 1601 inJ2000.0 coordinates hasα = 18.613744h ,δ = 38.751832◦
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 30 / 33
Worked example 2a: Worked example 2b:
Answers 2a: conversion to eclipticcoordinates
ε ≡ εo = 23.439291◦
λ(2000) = 285.260751◦
β(2000) = 61.704294 = β(1601)
λ(1601) = 279.719227◦
Problem 2b: modern precession1 first step as 2a)2 precess to α(1601), δ(1601)
3 convert to ecliptic coordinatesλ(1601), β(1601)
Answers 2b: modernprecessionsee Eqs. 14,15,16
t = −3.99 century
ζ = −2.554395◦
z = −2.550915◦
θ = −2.222044◦
A = −0.774615
B = 0.114537
α(1601) =18.390669h
δ(1601) = 38.460520◦
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 31 / 33
Worked example 2b: comparison with 2a and Brahe
Answers 2b: conversion toecliptic coordinates
ε = 23.491142◦
λ(1601) = 279.725437◦
β(1601) = 61.753557
Difference with 2a for 1601:
βb − βa = 0.05◦ = 3′
λb − λa = 0.0062h
(×60 cos β): = 0.18′
Difference with Brahe:
βb − βB = 0.038◦ = 2.3′
λb − λB = 0.0087h = 0.25′
Vega in Brahe CatalogueTycho Brahe’s catalogue isdated Annum Completum1600, i.e. 1 Jan 1601
for Vega, Brahe givesλ = Capricornus 09◦43′
Capricornus is the tenthconstellation in the ecliptic,hence Brahe hasλ = 9 ∗ 30 + 9 + 43/60 =279.7167◦
β = 61◦47.5′ B(orealis):North, hence β = 61.7917◦
Frank Verbunt (Astronomical Institute Utrecht) Positional Astronomy for Historical Studies May 13, 2011 32 / 33
Exam problem: Vega in Hevelius’ catalogue
Front page and Vega entry Exam Problem 2Determine the date ofthe catalogue byHevelius and his λ, βfor Vega
Use precession atconstant ε to computeλ, β for Vega at thatdate from J2000.0values
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