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

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

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

Coordinate systems: connection to Cartesian coordinates

Equatorial coordinates

cos δ cosαcos δ sinα

sin δ

Ecliptic coordinates

cos β cos λcos β sin λ

sin β

Coordinate transformations

Equatorial to ecliptic rotation ε around x-axis

1 0 00 cos ε sin ε0 − sin ε cos ε

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)

sin β = sin δ cos ε − cos δ sinα sin ε (5)

Coordinate transformations

Ecliptic to equatorial rotation −ε around x-axisAnalogously: x

1 0 00 cos ε − sin ε0 sin ε cos ε

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)

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.

Equatorial to horizontal coordinates and v.v.

cos A cos hsin A cos 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

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◦.

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.

Daily motion of the stars

At the pole the horizon is theequator

Stars move parallel to horizon

Daily motion of the stars

At the equator the poles are atthe horizon

Stars move perpendicular tohorizon

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

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

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

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

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

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

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)

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

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

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

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)

above: for 15 brightest stars, below, for 49 brightest stars

Stars cross/do not cross horizon (Mauna Kea)

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.

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.

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◦

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.

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◦

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◦

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◦

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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