1. Lecture 3PHYS1005 – 2003/4 Lecture 3: Astronomical Magnitudes Objectives: To define what is meant by brightness To justify the inverse square law To describe the astronomical system of magnitudes Brightness = Flux = Power/unit area observed from an object N.B. do not confuse with Luminosity, the total power output by the object Brightness and Flux vary with distance Luminosity is an intrinsic property of an object Object of luminosity L, at distance d, radiating equally in all directions  Flux F = L / 4 π d 2, where the 1/d 2 term comes from the area of a sphere  Flux from an object decreases as the inverse square of its distance

2. Lecture 3PHYS1005 – 2003/4 e.g. Sun’s luminosity L O = 3.8 x 10 26 W and it is 1AU (1.5 x 10 11 m) distant what is the flux from the Sun at Earth? Answer: 1,350 W m -2 (verify) what would the flux be at 10 pc? Answer: 3.18 x 10 -10 W m -2 (verify) Astronomical Magnitude System Hipparchus (120 BC) classified stars into 6 levels (1 – brightest, 6 – faintest) but measurements in 19 th century  eye response is logarithmic!  5 magnitudes ≈ x 100 Pogson (1856) defined 5 magnitudes exactly = factor 100 in brightness (unfortunately) he kept 1 as bright and 6 as faint! mathematical definition: m 1 – m 2 = -2.5 log 10 (F 1 /F 2 ) (check by putting F 1 = 100 F 2 ) N.B. Faint stars have large magnitudes

3. Lecture 3PHYS1005 – 2003/4 Apparent and Absolute Magnitudes Magnitudes measured relative to Vega – defined to have m = 0  6 th mag ≈ faintest you can see N.B. these are all apparent magnitudes Define absolute magnitude as that observed if object at d = 10 pc Conventional symbols used (problem-solving tip): –m – (lower case) apparent magnitude –M – (upper case) absolute magnitude Comparing true flux: With that at d = 10 pc: Then from Pogson’s equation: m 1 – m 2 = -2.5 log(F 1 /F 2 )  m – M = -2.5 log (10 pc / d) 2 or m – M = 5 log (d / 10 pc) F 1 = L / 4 π d 2 F 2 = L / 4 π (10 pc) 2 i.e. m – M = 5 log d - 5

4. Lecture 3PHYS1005 – 2003/4

5. Lecture 3PHYS1005 – 2003/4 e.g. 1) how far away could we see the Sun with the naked eye? Answer: Sun has M = +4.8, faintest we can see is m = 6 using m – M = 5 log d – 5  1.2 = 5 log d – 5  log d = 1.24  d = 17 pc e.g. 2) how faint could you see with a 50mm aperture telescope ? Answer: dark-adapted eye pupil has diameter ~ 5 mm  telescope increases collecting area by factor (50 / 5) 2 = 100 ≡ 5 mags  can see as faint as 6 + 5 = 11 th mag.