Lecture 3: Astronomical Magnitudes

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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 p d2 , where the 1/d2 term comes
  from the area of a sphere

• ? Flux from an object decreases as the inverse
  square of its distance

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• e.g. Suns luminosity LO 3.8 x 1026 W and it is
  1AU (1.5 x 1011 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 19th 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 m1 m2 -2.5 log10
  (F1/F2)
• (check by putting F1 100 F2)
• N.B. Faint stars have large magnitudes

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Apparent and Absolute Magnitudes

• Magnitudes measured relative to Vega defined to
  have m 0
• ?6th 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 Pogsons equation m1 m2 -2.5
  log(F1/F2)
• ? m M -2.5 log (10 pc / d)2 or m M 5
  log (d / 10 pc)

F1 L / 4 p d2
F2 L / 4 p (10 pc)2
i.e. m M 5 log d - 5
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• 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 11th mag.