ASPH305 Part 1 - Intro & Angular Measurements - PDF

ASTROPHYSICS 305 PART 1 Introduction & Angular Measurements Chapters 1, 2.1, 2.2 These slides are on D2L in Content: Lecture Notes 1 1 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 A SCI...

ASTROPHYSICS 305 PART 1 Introduction & Angular Measurements Chapters 1, 2.1, 2.2 These slides are on D2L in Content: Lecture Notes 1 1 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 A SCIENCE OF SUPERLATIVES The BIGGEST & smallest Hottest & Coldest Fastest & Slowest Furthest NOTE - Astronomy uses certain units for distance depending on what we’re looking at: Astronomical Units (AU) for solar system distances 1 AU = 1.496x1011 m Parsecs for stellar/Galactic/extragalactic distances 1 pc = 3.086x1016 m We also usually calculate masses in units of Solar Masses (M⊙) 1 M⊙ = the mass of the Sun = 1.99 x1030 kg 2 2 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  Space is REALLY BIG!!!! With lots of different things 100s of Mpc to try and understand......from a long ways away! Note the difference between: - a star and a planet: ~1 Mpc - R⊙ = 6.96 × 105 km, R⊕ = 6.4 × 103 km - ∴ R⊙ ∼ 100R⊕ 30,000 pc 10s of pc Betelgeuse Mars - a solar system and a galaxy - Rgal ∼ 5 × 1017 km (50,000 lyrs) ∼ 16,000 pc 40 AU - RSS ∼ 6 × 109 km (40 AU ) - ∴ Rgal ∼ 8 × 107 RSS 12,000 km M101 MWC758 3 A galaxy A “Solar “system 3 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 WE’RE NOT AS CONNECTED TO THE NIGHT SKY AS OUR ANCESTORS WERE. The night sky over Calgary The night sky in a dark location 4 4 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 CITY LIGHTS GLOBALLY 5 5 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  So, since your great-great-great grandparents probably knew more about the night sky than you do let’s start by laying the foundations of modern observational astronomy which starts with some ancient ideas…. 6 6 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  THE CELESTIAL SPHERE Ancients believed that all heavenly objects were located on crystal spheres that encircled the Earth. The Celestial Sphere rotates about the Earth which causes the daily rise & set of the stars. This is wrong, of course, but still provides a convenient tool to map the sky And to measure the motions of the stars, planets, etc astronomers still use this concept today Ignore the distance to the stars and simply describe their position on the Celestial sphere assume all stars are on the surface of a “celestial sphere” at in nite distance 7 7 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 fi  The Earth is rotating but, from our perspective, it seems like we are standing still and the Universe is rotating around us! So, it LOOKS to us that the Celestial Sphere is rotating around some rotation axis (that we call the Celestial Poles) But it is NOT https://www.youtube.com/watch?v=aFIR7hbqed4 8 8 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  CELESTIAL SPHERE You can imagine that all the stars in space are actually located on the surface of one gigantic sphere centred on the Earth. The Celestial Poles are the intersections of the Earth’s rotation axis with the celestial sphere The Celestial Equator is the projection of the Earth’s Equatorial plane onto the Celestial Sphere 9 9 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  Constellations The night sky/celestial sphere is populated with stars the sky is divided up into 88 distinct regions - called constellations completely arbitrary - like countries on a world map 10 10 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 CONSTELLATIONS All stars within the arbitrary boundaries belong to the constellation 11 11 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 CONSTELLATIONS random groupings of stars can look like objects (if we use our imagination) The pattern stars that make up the “picture” are called the asterism 12 12 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  Sometimes there can be more than one asterism Ursa Major The Big Dipper (The Great Bear) 13 13 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  CONSTELLATIONS Stars in a constellation/asterism can be at different distances So although we SEE the sky as a 2 dimensional object, we have to remember that it is really a 3D object But, given what we see, how do we measure sizes of celestial bodies and distances between objects on the sky in 2 dimensions? This is a fundamental problem in astronomy that brings us to…….. 14 14 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  This is a fundamental problem in astronomy that brings us to…….. ….SPHERICAL TRIGONOMETRY Since, we can really ONLY measure sizes of objects (or distances between objects) as ANGLES A couple of important concepts from spherical trig are: Great Circle passes through any 2 points and plane bisects sphere plane passes through centre of sphere small circle pass through any 2 points but does NOT bisect sphere plane does not pass through centre of sphere 15 15 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  SPHERICAL TRIGONOMETRY lines of latitude are small circles Since, we can really ONLY (except for the equator) measure sizes of objects (or distances between objects) as ANGLES A couple of important concepts from spherical trig are: Great Circle passes through any 2 points and plane bisects sphere plane passes through centre of sphere small circle pass through any 2 points but does NOT bisect sphere plane does not pass through centre of sphere lines of longitude are great circles 16 16 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  SPHERICAL TRIGONOMETRY Great Circle shortest distance between any 2 points on the surface of a sphere which is why airplanes y apparently strange routes when we see them plotted on a at map 17 17 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 fl fl  SPHERICAL TRIGONOMETRY Great circle arc between 2 points r′ r′ A B R R Small circle arc between 2 points 18   18 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  c is the angle subtended by arc AB in radians SPHERICAL there are 2π radians in a 360˚ circle TRIGONOMETRY so to convert from degrees to radians multiply by π/180 length of arc AB (in physical units like m or km or r AU etc.) is: | AB | = rc r where r is the distance from r centre of circle to arc r = radius of the sphere ONLY if circle is a great r = radius = distance from centre circle of circle to surface 19 19 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  r = radius of the sphere ONLY if SPHERICAL circle is a great circle TRIGONOMETRY Length of arc ab: | ab | = r′θ Where r is the radius of a r′ θ the SMALL circle (NOT THE RADIUS OF THE b SPHERE) and θ is in radians Θ C R Length of arc CD: D | CD | = RΘ Where R is the radius of the GREAT circle (i.e. the sphere) and Θ is in radians 20   20 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  SPHERICAL TRIGONOMETRY A Note that this is a bit different r than our usual right-angle (Euclidian) trigonometry | AB | = rc The length of arc: | AB | = rc is c NOT the same as the length of r B the side of a right-angle triangle: | AB | = r tan c A However, when angle c is small, they are essentially equivalent We’ll look at this in a bit | AB | = r tan c c B r 21 21 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  SPHERICAL TRIGONOMETRY A A Note that this is a bit different than our usual right-angle trigonometry The length of arc: | AB | = rc is c NOT the same as the length of r BB the side of a right-angle triangle: | AB | = r tan c Nevertheless, we can use this trigonometry to However, when angle c is small, calculate the true size ( | AB | in units like (say) they are essentially equivalent metres) of any object if we can measure its We’ll look at this in a bit angular size (c) and we know its distance (r) Conversely, we can measure the distance to any object (r) if we can measure its angular size (c) and we know it’s true size ( | AB | in units like (say) metres) 22 22 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  C ANGULAR C SIZES Or angular separation between 2 points r D |AB| c r |AB| = rc so if you want the linear size/ diameter (D) of an object with an angular size/diameter of c, or the distance between 2 points (D) separated by an angular distance c D = |AB| = rc D=d↵ for small angles 23 23 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  SPHERICAL TRIGONOMETRY You are 1.3km away from the Calgary Tower and you measure its angular height/size to be 8.37˚. How tall is the Calgary Tower? Imagine you’re in the centre of a circle… h=? θ = 8.37∘ d = 1.3 km = 1300 m 24 24 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  SPHERICAL TRIGONOMETRY You are 1.3km away from the Calgary Tower and you measure its angular height/size to be 8.37˚. How tall is the Calgary Tower? π θ = 8.37∘ × = 0.14608 rads 180∘ h = | AB | = dθ = 1300m × 0.14608 = 189.9m A h=? θ = 8.37∘ d = 1.3 km = 1300 m B 25 25 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  SPHERICAL TRIGONOMETRY You are 1.3km away from the Calgary Tower and you measure its angular height/size to be 8.37˚. How tall is the Calgary Tower? Note, if we did this as a right angle triangle: h tan θ = d h = | AB | = d tan θ = 1300m tan(8.37∘) = 191.3m Similar…but not identical A You need to decide how much precision is needed to get the “correct” answer h=? θ = 8.37∘ d = 1.3 km = 1300 m B 26 26 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  ANGULAR SIZES 1∘ = 60′ (arcminutes) 1′ = 60′′ (arcseconds) So 1∘ = 3600′′ So a position written as (say): 16∘ 12′ 37′′ Can be converted to decimal degrees by: 12 37 16∘ + + = 16.21027778∘ 60 3600 Many calculators have a built in function to do this conversion 27          27 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  ANGULAR why is spherical trig only SIZES equal to Euclidian trig for |AB| = r2✓ small angles? A because for large angles R |AB| ≠ 2R (the diameter) θ 2θ r D θ R You can see that for large angles, there is a large discrepancy between D and |AB| B e.g. θ = 30˚ = 0.5236 rads R = r tan θ tan θ = 0.5774 D = 2R = 2r tan ✓ tan θ ≠ θ 28 28 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  ANGULAR SIZES for large angles |AB| ≠ 2R But for small angles |AB| ~ 2R |AB| = r2✓ A r R θ B D = 2R = 2r tan ✓ e.g. θ = 0.5˚ = 0.008726646 rads for small angles: tan θ = 0.008726868 tan θ ∼ θ tan ✓ ⇡ ✓ QED D ⇡ 2r✓ = |AB| AAACA3icbVDJSgNBEO1xjXEb9aaXxiB4CjNR1IsQl4PHCGaBzBB6Oj1Jk56F7hoxTAJe/BUvHhTx6k9482/sJHPQxAcFj/eqqKrnxYIrsKxvY25+YXFpObeSX11b39g0t7ZrKkokZVUaiUg2PKKY4CGrAgfBGrFkJPAEq3u9q5Ffv2dS8Si8g37M3IB0Qu5zSkBLLXP3GjskjmX0gEsSO9BlQPA5HlxcDlpmwSpaY+BZYmekgDJUWuaX045oErAQqCBKNW0rBjclEjgVbJh3EsViQnukw5qahiRgyk3HPwzxgVba2I+krhDwWP09kZJAqX7g6c6AQFdNeyPxP6+ZgH/mpjyME2AhnSzyE4EhwqNAcJtLRkH0NSFUcn0rpl0iCQUdW16HYE+/PEtqpaJ9VCzdHhfKJ1kcObSH9tEhstEpKqMbVEFVRNEjekav6M14Ml6Md+Nj0jpnZDM76A+Mzx/QXZZP 29 29 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  When you have an object with physical dimensions, you can calculate its area i.e. the area of a right-angle triangle But what if you don’t know the distance to the object (I.e. the radius of the sphere)? Since we can measure the angular lengths of arcs |AB|, |AC|, & |BC (I.e. angles a, b, & c) we can also calculate an angular area encompassed by a spherical triangle A spherical Triangle is not made of just angular area on the sky (dΩ) is called the any 3 points on solid angle surface of sphere. Its sides must be Solid angle is in units of steradians arcs of great circles. (abbreviated to sr) which is: radians × radians = radians2 dΩ We’ll see how to do this calculation in general in a bit But let’s look at a simpli ed scenario rst 30 30 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 fi fi  If the region on the sky is circular in shape (a common scenario in astronomy), the dΩ solid angle is given by: dΩ = 2π(1 − cos θ) θ for small angles: dΩ 2 ✓ E θ cos ✓ ⇡ 1 AAACE3icbVBNS8NAEN34WetX1aOXxSKIYEmiqMeCF48V7Ac0sWy2m3bpJht2J2IJ/Q9e/CtePCji1Ys3/43bNgdtfTDweG+GmXlBIrgG2/62FhaXlldWC2vF9Y3Nre3Szm5Dy1RRVqdSSNUKiGaCx6wOHARrJYqRKBCsGQyuxn7zninNZXwLw4T5EenFPOSUgJE6pWOPSo096DMg2CNJouQDdvAJ9kJFaDY17txR5o46pbJdsSfA88TJSRnlqHVKX15X0jRiMVBBtG47dgJ+RhRwKtio6KWaJYQOSI+1DY1JxLSfTX4a4UOjdHEolakY8ET9PZGRSOthFJjOiEBfz3pj8T+vnUJ46Wc8TlJgMZ0uClOBQeJxQLjLFaMghoYQqri5FdM+MWGAibFoQnBmX54nDbfinFbcm7Ny9TyPo4D20QE6Qg66QFV0jWqojih6RM/oFb1ZT9aL9W59TFsXrHxmD/2B9fkDvdKdag== 2 so for small angles this becomes: dΩ ≈ πθ 2 i.e. imagine you’re looking at a circle on the sky whose radius is there are 4π steradians in the θ radians instead of x meters. θ entire surface of a sphere* You measure a solid angle dΩ Z (angular area) of dΩ = πθ 2 sr instead of a “physical” area of dA = πx m 2 2 ⌦= d⌦ = 4⇡ 31 *we’ll see why in a bit 31 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  SOLID ANGLE TO PHYSICAL AREA Sinceyou know that the full surface area of a sphere is: A = 4πr 2 dA dΩ we know that: Ω = 4π E θ And Youcan see that the full area of a sphere is equivalent to: A = Ωr 2 Soif you wanted to nd the physical area of a region smaller than the entire surface (i.e dA) that has a solid angle dΩ you can see that: 2 dA = d⌦r AAAB+HicbVBNS8NAEJ34WetHox69LBbBU0mqoBeh6sWbFewHtLFsNpt26WYTdjdCDf0lXjwo4tWf4s1/47bNQVsfDDzem2Fmnp9wprTjfFtLyyura+uFjeLm1vZOyd7da6o4lYQ2SMxj2faxopwJ2tBMc9pOJMWRz2nLH15P/NYjlYrF4l6PEupFuC9YyAjWRurZpeASXQTd24j2MZIP1Z5ddirOFGiRuDkpQ456z/7qBjFJIyo04Vipjusk2suw1IxwOi52U0UTTIa4TzuGChxR5WXTw8foyCgBCmNpSmg0VX9PZDhSahT5pjPCeqDmvYn4n9dJdXjuZUwkqaaCzBaFKUc6RpMUUMAkJZqPDMFEMnMrIgMsMdEmq6IJwZ1/eZE0qxX3pFK9Oy3XrvI4CnAAh3AMLpxBDW6gDg0gkMIzvMKb9WS9WO/Wx6x1ycpn9uEPrM8fFtaSEg== 32 32 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 fi  EXAMPLE As we’ll see later in the class, a telescope has a certain “resolution” which is the smallest angular scale it can detect on the sky (i.e. the smallest detail it can see - anything smaller than the telescope’s resolution would appear fuzzy or “unresolved”). This resolution is usually called the FWHM (Full Width at Half Maximum) and can be approximated by the angle 2θ in the gure to the right. If the FWHM of a the human eye is 1’ (effectively a small telescope) calculate the solid angle of the eye’s resolution (in sr). Using a distance = 384,400 km, calculate the linear size (in km) and linear area (in km2) of the smallest object that could be clearly seen on the Moon. ( 60 ) ( 60 ) 180 ∘ ∘ FWHM 0.5 0.5 π θ= = 0.5′ = = × = 1.454441043 × 10−4 rads 2 dΩ = 2π(1 − cos θ) ≈ πθ 2 = π(1.454441043 × 10−4 rads)2 = 6.645721168 × 10−8 sr For small angles like this one the linear size/diameter and θ linear area can be calculated from the small angle formulae : dΩ D = r × 2θ = 384,400 ⋅ 2(1.454441043 × 10−4) = 111.8174 km E θ (2) 2 2 D So: dA = πR = π = π55.90872 = 9820 km 2 Or dA = r 2dΩ = (384,400)2(6.645721168 × 10−8) = 9820 km 2 33  33 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 fi  SPHERICAL TRIGONOMETRY We are used to Cartesian coordinates X, Y, Z Butsince in astronomy we deal with the Earth and the sky above it both of which can be approximated as spheres itmakes more sense to use Note that in geography and probably high spherical coordinates school math, θ is usually de ned as this angle (e.g. latitude on the earth) r, θ, φ 34 34 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 fi  SPHERICAL We can convert TRIGONOMETRY from cartesian coordinates to spherical r0 = rsin✓ coordinates z = rcos✓ adj x opp y cos = = sin = = hyp r sin ✓ hyp r sin ✓ x = r sin ✓ cos y = r sin ✓ sin x = r sin θ cos ϕ os y = r sin θ sin ϕ c r 0= n ✓ r si r si n✓ x= z = r cos θ y = rsin✓sin 35 35 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  BACK TO SOLID ANGLE r′ = r sin θ In the r, θ, ϕ coordinate system we de ned, what is the area (dA) of an arbitrary region on the surface of a sphere that has “width” r dϕ and “height” dθ? Note - r is the radius of the sphere And remember: | AB | = rc this distance is rsinθ (it sweeps out a small circle which is why the distance is NOT just r) this arclength is rsinθdɸ (since arc length = distance A B (r sin θ) x angle (dϕ) i.e. | AB | = r′c = r′dϕ ) C this arclength is rdθ (because it’s part of a great circle r I.e. | BC | = rc = rdθ) r Thus the boxed area (dA) is: dA = | AB | × | BC | dA = r′dϕ × rdθ dA = r sin θdϕ × rdθ dA = r 2 sin θdθdϕ 36 And since the de nition of dA is: dA = r 2dΩ dΩ = sin θdθdϕ     36 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 fi fi  BACK TO dΩ = sin θdθdϕ SOLID ANGLE so…going back to real surface area…. Z Area = dA AAAB9XicbVDLSsNAFL2pr1pfVZduBovgqiRV1I3Q4sZlBfuANpbJZNIOnUzCzEQpof/hxoUibv0Xd/6NkzYLbT0wcDjnXO6d48WcKW3b31ZhZXVtfaO4Wdra3tndK+8ftFWUSEJbJOKR7HpYUc4EbWmmOe3GkuLQ47TjjW8yv/NIpWKRuNeTmLohHgoWMIK1kR4aJouuUZ8JjfzGoFyxq/YMaJk4OalAjuag/NX3I5KEVGjCsVI9x461m2KpGeF0WuonisaYjPGQ9gwVOKTKTWdXT9GJUXwURNI8s36m/p5IcajUJPRMMsR6pBa9TPzP6yU6uHJTJuJEU0Hmi4KEIx2hrALkM0mJ5hNDMJHM3IrICEtMtCmqZEpwFr+8TNq1qnNWrd2dV+oXeR1FOIJjOAUHLqEOt9CEFhCQ8Ayv8GY9WS/Wu/UxjxasfOYQ/sD6/AHP1pFg d A1 d A2 d A3 d A4 d A8 d A5 d A6 d A7 d A9 Etc… 37 37 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  BACK TO dΩ = sin θdθdϕ SOLID ANGLE so…going back to real surface area…. Z Area = dA dθ AAAB9XicbVDLSsNAFL2pr1pfVZduBovgqiRV1I3Q4sZlBfuANpbJZNIOnUzCzEQpof/hxoUibv0Xd/6NkzYLbT0wcDjnXO6d48WcKW3b31ZhZXVtfaO4Wdra3tndK+8ftFWUSEJbJOKR7HpYUc4EbWmmOe3GkuLQ47TjjW8yv/NIpWKRuNeTmLohHgoWMIK1kR4aJouuUZ8JjfzGoFyxq/YMaJk4OalAjuag/NX3I5KEVGjCsVI9x461m2KpGeF0WuonisaYjPGQ9gwVOKTKTWdXT9GJUXwURNI8s36m/p5IcajUJPRMMsR6pBa9TPzP6yU6uHJTJuJEU0Hmi4KEIx2hrALkM0mJ5hNDMJHM3IrICEtMtCmqZEpwFr+8TNq1qnNWrd2dV+oXeR1FOIJjOAUHLqEOt9CEFhCQ8Ayv8GY9WS/Wu/UxjxasfOYQ/sD6/AHP1pFg Z = r2 d⌦ dϕ AAAB/HicbVDLSgMxFM3UV62v0S7dBIvgqsxUUTdCwY07K9gHdMaSyWTa0CQzJBlhGOqvuHGhiFs/xJ1/Y9rOQlsPBA7n3Ms9OUHCqNKO822VVlbX1jfKm5Wt7Z3dPXv/oKPiVGLSxjGLZS9AijAqSFtTzUgvkQTxgJFuML6e+t1HIhWNxb3OEuJzNBQ0ohhpIw3sKryCHhUayocGDL1bToZoYNecujMDXCZuQWqgQGtgf3lhjFNOhMYMKdV3nUT7OZKaYkYmFS9VJEF4jIakb6hAnCg/n4WfwGOjhDCKpXkmx0z9vZEjrlTGAzPJkR6pRW8q/uf1Ux1d+jkVSaqJwPNDUcqgjuG0CRhSSbBmmSEIS2qyQjxCEmFt+qqYEtzFLy+TTqPuntYbd2e15nlRRxkcgiNwAlxwAZrgBrRAG2CQgWfwCt6sJ+vFerc+5qMlq9ipgj+wPn8AoMWTcA== dΩ1 dΩ2 dΩ3 dΩ4 Z Z dΩ8 = r2 sin ✓d✓d dΩ5 dΩ6 dΩ7 AAACE3icbVDLSgMxFM34rPU16tJNsAjiosxUUTdCwY3LCvYB7VgymUwbmskMyR2hDP0HN/6KGxeKuHXjzr8xbQfR1gNJDufcS+49fiK4Bsf5shYWl5ZXVgtrxfWNza1te2e3oeNUUVansYhVyyeaCS5ZHTgI1koUI5EvWNMfXI395j1TmsfyFoYJ8yLSkzzklICRuvYxvsTqroI7XEJ+aS5xB/oMCA5+3qTPu3bJKTsT4Hni5qSEctS69mcniGkaMQlUEK3brpOAlxEFnAo2KnZSzRJCB6TH2oZKEjHtZZOdRvjQKAEOY2WOmWqi/u7ISKT1MPJNZUSgr2e9sfif104hvPAyLpMUmKTTj8JUYIjxOCAccMUoiKEhhCpuZsW0TxShYGIsmhDc2ZXnSaNSdk/KlZvTUvUsj6OA9tEBOkIuOkdVdI1qqI4oekBP6AW9Wo/Ws/VmvU9LF6y8Zw/9gfXxDaQmnMM= dΩ9 Etc… Where the integration limits span the dθ & dϕ which dθ cover the area of concern, I.e. This area Or this area 38 dϕ 38 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  BACK TO SOLID ANGLE Z Z Any arbitrary area on the surface of a sphere is: Area = r2 sin ✓d✓d AAACE3icbVDLSgMxFM34rPU16tJNsAjiosxUUTdCwY3LCvYB7VgymUwbmskMyR2hDP0HN/6KGxeKuHXjzr8xbQfR1gNJDufcS+49fiK4Bsf5shYWl5ZXVgtrxfWNza1te2e3oeNUUVansYhVyyeaCS5ZHTgI1koUI5EvWNMfXI395j1TmsfyFoYJ8yLSkzzklICRuvYxvsTqroI7XEJ+aS5xB/oMCA5+3qTPu3bJKTsT4Hni5qSEctS69mcniGkaMQlUEK3brpOAlxEFnAo2KnZSzRJCB6TH2oZKEjHtZZOdRvjQKAEOY2WOmWqi/u7ISKT1MPJNZUSgr2e9sfif104hvPAyLpMUmKTTj8JUYIjxOCAccMUoiKEhhCpuZsW0TxShYGIsmhDc2ZXnSaNSdk/KlZvTUvUsj6OA9tEBOkIuOkdVdI1qqI4oekBP6AW9Wo/Ws/VmvU9LF6y8Zw/9gfXxDaQmnMM= What if we want to calculate the surface area of the entire sphere? First integrate ɸ over a full circle (i.e 2π radians) π 2π π 2π π ɸ ∫0 ∫0 ∫0 ∫0 Ω= sin θdθ dϕ = sin θdθ ϕ = 2π sin θdθ 0 But why do we only integrate θ from 0 to π instead of 2π? Because we now have a full circle, if we rotate that circle by π in the θ direction the right half of the circle sweeps out the northern hemisphere and the left half of the circle sweeps out the southern hemisphere. So we have a full sphere! If we integrate over 2π we integrate over the surface TWICE ( 0) π Ω = 2π −cos θ Ω = 2π (−cos π − (−cos 0)) Ω = 2π (−(−1) − (−1)) = 2π(2) = 4π ∴ Asphere = Ω r 2 = 4πr 2 Q.E.D. 39 39 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  EXAMPLE Assuming the Earth is perfectly spherical with a radius of 6365 km and using the formulae from the notes calculate the solid angle (in sr) and geographical area (in km2) of a portion of the Earth’s surface with the coordinates of: θ1 = +25˚ 28’ 43.0’’ Close to the North pole θ = +25˚ 30’ 26.6’’ (i.e. ~ 64.5˚ N latitude) 2 ɸ1 = 0˚ 3’ 1.1’’ ɸ2 = 0˚ 11’ 40.2’’ Z ✓2 Z 2 ✓2 d⌦ = sin ✓d✓ d = cos ✓ ✓1 2 1 ✓1 1 d ⌦ = ( cos ✓2 + cos ✓1 ) · ( 2 1) Use θ1 = +25˚ 28’ 43.0’ (N) = 25.478611111˚ = 0.444685652 rads RADS θ2 = +25˚ 30’ 26.6’ (N) = 25.507388888˚ = 0.445187919 rads since dΩ ɸ1 = 0˚ 3’ 1.1’ (W) = 0.050305555˚ = 8.779975765x10-4 rads has units of sr ɸ2 = 0˚ 11’ 40.2’ (W) = 0.1954˚ = 3.394665396x10-3 rads dΩ = [−cos(0.44519) + cos(0.44468)] × [3.39 × 10−3 − 8.78 × 10−4] dΩ = (2.16176136 × 10−4) × (2.51666782 × 10−3) = 5.440435249 × 10−7 sr To nd the area: Both dA and dΩ are dA = r2 d⌦ always positive! dA = r 2dΩ = (6365km)2 × (5.440435249 × 10−7) = 22.0 km 2 40 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 fi  EXAMPLE Assuming the Earth is perfectly spherical with a radius of 6365 km and using the formulae from the notes calculate the solid angle (in sr) and geographical area (in km2) of a portion of the Earth’s surface with the coordinates of: θ1 = +25˚ 28’ 43.0’’ dA = r 2dΩ = (6365km)2 × (5.440435249 × 10−7) = 22.0 km 2 θ2 = +25˚ 30’ 26.6’’ θ2 ϕ2 ∫ ∫ ɸ1 = 0˚ 3’ 1.1’’ dΩ = sin θdθ dϕ ɸ2 = 0˚ 11’ 40.2’’ θ1 ϕ1 The integral accounts for the fact that this area is NOT a “square” but a “trapezium” If you just calculated dΩ = dθdɸ (i.e. approximated the area by a square) you’d get 1.264x10-6 sr θ1,ɸ1 θ1,ɸ2 which gives A = 51.2 km2 Lines of latitude have the same separation θ2,ɸ1 θ2,ɸ2 regardless of their longitude Lines of longitude get closer together at the poles (with changing In other words….. latitude). i.e. they have a smaller separation but have the SAME dφ 41 41 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  SPHERICAL COORDINATES The actual angular separation (arc length) between two points of longitude (when viewed from the centre of the Earth) is: ang sep = Δlongitude × cos( latitude) So between ϕ = 60∘ & 45∘, dϕ = 15∘ If I am at the centre of the Earth and I look at 2 points separated by 15˚ of longitude (dϕ = 15∘): Same At the Equator where θ = 0∘ that difference in longitude would correspond to Δlongitude an angular size/separation of: s = 15∘ cos(0∘) = 15∘ Up towards the pole at (say) θ = 75∘ the angular separation between the same two lines of longitude would be: s = 15∘ cos(75∘) = 3.9∘ In terms of linear size of these arcs…at the Equator: different arc lengths ( 180∘ ) ∘ pi | AB | = rs = (6365 km) 15 × = 1666.4 km and so different linear At θ = 75∘ distances ( 180∘ ) ∘ pi | AB | = rs = (6365 km) 3.9 × = 433.3 km 42 42 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  SPHERICAL COORDINATES Let’s turn this around. If I am standing in the centre of the Earth and a see two objects that are 500km in size One is at the equator and one is at a latitude of 75˚ 1) What arc length does the object subtend? 2) How many degrees of longitude does it span? Same size 1) | AB | = rc → c = | AB | /r = 500/6365 = 7.85546 × 10−2 rads And it’s the same for BOTH locations 2) How many degrees of longitude does it span? Δlat = ang. sep. /cos(lat) Or Δϕ = c/cos θ = 7.85546 × 10−2 /cos θ At θ = 0∘, Δϕ = 7.85546 × 10−2 /cos 0 = 7.85546 × 10−2rads = 4.5∘ At θ = 75∘, Δϕ = 7.85546 × 10−2 /cos 75 = 0.3035116 rads = 17.4∘ 43 43 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  EXAMPLE Assuming the Earth is perfectly spherical with a radius of 6365 km and using the formulae from the notes calculate the solid angle (in sr) and geographical area (in km2) of a portion of the Earth’s surface with the coordinates of: θ1 = +25˚ 28’ 43.0’’ dA = r 2dΩ = (6365km)2 × (5.440435249 × 10−7) = 22.0 km 2 θ2 = +25˚ 30’ 26.6’’ ɸ1 = 0˚ 3’ 1.1’’ ɸ2 = 0˚ 11’ 40.2’’ By taking the integral we account for the fact that the arclength changes with latitude (θ) i.e. Total Area = the sum of the areas of the small orange rectangles θ1,ɸ1 θ1,ɸ2 So at high latitudes (small θ) the area is strongly trapezoidal θ2,ɸ1 θ2,ɸ2 but what if we went θ2 ϕ2 closer to the equator? ∫θ ∫ϕ dΩ = sin θdθ dϕ The area gets bigger 1 1 and is more square-like 44 44 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024  EXAMPLE Assuming the Earth is perfectly spherical with a radius of 6365 km and using the formulae from the notes calculate the solid angle (in sr) and geographical area (in km2) of a portion of the Earth’s surface with the coordinates of: θ1 = +89˚ 28’ 43.0’’ same CHANGE in latitude but close to θ2 = +89˚ 30’ 26.6’’ the equator ɸ1 = 0˚ 3’ 1.1’’ same longitudes as in ɸ2 = 0˚ 11’ 40.2’’ previous example Z ✓2 Z 2 ✓2 d⌦ = sin ✓d✓ d = cos ✓ ✓1 2 1 ✓1 1 d ⌦ = ( cos ✓2 + cos ✓1 ) · ( 2 1) θ1 = +89˚ 28’ 43.0’ (N) = 89.478611111˚ = 1.561696374 rads θ2 = +89˚ 30’ 26.6’ (N) = 89.507388888˚ = 1.56219864 rads ɸ1 = 0˚ 3’ 1.1’ (W) = 0.050305555˚ = 8.779975765x10-4 rads ɸ2 = 0˚ 11’ 40.2’ (W) = 0.1954˚ = 3.394665396x10-3 rads dΩ = [−cos(1.5622) + cos(1.5617)] × [3.39 × 10−3 − 8.78 × 10−4] dΩ = (5.022463307 × 10−4) × (2.51666782 × 10−3) = 1.263987178 × 10−6 To nd the area: dA = r2 d⌦ exactly what we got by approximating the dA = r2 ⌦ = (6365km)2 · 1.264 ⇥ 10 6 sr = 51.2 km2 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 45 area as a square! 45 ASPH305 - Part 1 - Intro & Angular Measurements - September 13, 2024 fi