Review Lecture The Central Problem in astronomy is distance. What we see is basically a twodimensional picture of the sky. To interpret many pieces of information available to the astronomer we need to know how far away a star or galaxy is. Example: If you look at the sky, Sirius is brighter than etelgeuse. ut etelgeuse is actually 300 times more luminous or brighter than Sirius. That is, etelgeuse actually emits far more energy than Sirius. The reason is that etelgeuse is just much further away from us than Sirius; thus, apparently less bright or dimmer. To get at these quantities we need three definitions. Luminosity = L, measured in Watts, is the power output(at all wavelengths) of the star, L =σat in Watts W where σ= 67. x0 4 m K temperature measured in Kelvin. 4, A is the surface area of the star, and T is the This power is spread out into the area of a sphere, where spherical area =4 according to the Inverse Square Law. πd Apparent brightness and the Inverse Square Law, measured in Watts/meter, is what we appear to see. luminosity L brightness = b = = spherical area 4 π d () Thus, we can only find the luminosity of the star if we know the apparent brightness of the star and the distance to the star, d. The brightness at the surface of a star is called the flux 4 lu minosity L flux = σt = = surface area in W/m. 4πd [Note: This looks like the Stefan-oltzmann Law and is similar. If you will, flux is the brightness of a star measured at the surface of the star.]
rightness - unfortunately, Hipparchus (30.C.) measured brightness in terms of apparent magnitude, m, with a scale such that m = the brightest star in the sky visible to the eye m = 6 the dimmest m = represents 00 times brighter by instrument in a logarithmic fashion. Thus, if x represents the increase in brightness for m =, then x = 00 x =.. So, going from the brightest to the dimmest brightness magnitude (.) = 00 m = brightest (.) 4 = 39. m = (.) 3 =.7 m = 3 (.) = 6. m = 4. m = m = 6 dimmest Note: The eye has a different response to light than mechanical devices. Thus: m V = visual apparent magnitude m = bolometric apparent magnitude (magnitude including all wavelengths) Relationship between apparent magnitude and apparent brightness What this says is that if I have two stars with magnitudes m and m, where is brighter than, then the ratio of brightnesses is given by ( ) ( ) m m m b. =. = b This can be rewritten to solve for the difference in magnitudes by taking the log of both sides, i.e. m b log 0. = mlog 0. = 04. m = log0 or b (3) b m =.log0 b As an example, consider α Centauri(m = 0) and ernard s Star(m = 0), then ()
m = 0 and the ratio of brightnesses is (.) 0 = (.) (.) = 00 = 0,000. Thus Centauri has an apparent brightness 0,000 greater than ernard s Star. α The inverse square law of equation says that equation 3 can be rewritten as m = b d.log 0 =.log0 = log b d 0 d d (4) Absolute magnitude measures the apparent brightness of a star at 0 parsecs. y implication, a star at a distance of 0 parsecs would have the same apparent magnitude as absolute magnitude. Let the standard distance then be d standard = 0 parsecs = 3 x 0 7 m Example: The Sun m Sun = -6., the Sun is the brightest star in the sky d sun = AU =. x 0 m The Inverse Square Law says thatbrightness~ = distance d = = 7 b ( ) sun dstandard 3x0 m = x0 = 4x0 b d. x0 m standard sun 6 This says the ratio of brightness is inversely proportional to the squared distances. Thus, the Sun is 4 x 0 times brighter where it is, as seen from the Earth, than at 0 parsecs. What is the absolute magnitude of the Sun at 0 parsecs. Note: increased distance decreased intensity m positive. According to equation, we need to determine m such that Two methods: (. ) m = 4x0 m m + m ) Estimate 4 x 0 = 4 x (00) 6 m ( ) =. x. = (. ) (.) =. 4 is somewhere between. and 6.. Guess m =. (.) = 6. ( ) ( ) 6 6 30 00 =. =. m = 30 m= m + m =. + 30= + 3.
) Calculate by taking logarithms of both sides ( ) m. = 4x0 m ( 4x0 ) = ( ) = m ( ) log( 4x0 ). 6 log log. log. m = = = + 37. 3. log (. ). 4 estimate The Absolute Visual Magnitude of the Sun is M Sun = -6. + 3. = + The Sun would be a dim star just visible in Cortland at 0 parsecs. Distance Distance can only be determined directly from a local group of stars near enough to have their parallax measured. d = in par s p( in arcsec onds) ( sec ) From Earth the smallest parallax measurable is about.0 arcseconds corresponding to a distance of 00 pc = 36 light years. There are only about 000 stars within this range. Satellite measurements extend this distance to about 000 pc and 00,000 stars. In comparison to the Sun, equation () can be written as L d b = L d b Sun Sun Sun () Absolute Magnitude from distances Equation 3 can be rewritten as D m = M m = log = log D logd d = log0 logd = logd
This can be rearranged to say that the absolute magnitude can be determined from the apparent magnitude and the distance from the observer, i.e. (6) M = m+ log0 d This equation can be used to produce the log-log plot shown below. To use the plot you need two of the three quantities: relative magnitude, distance, or absolute magnitude. In the figure below, Sirius has a bolometric apparent magnitude of m = -.4 (the brightest star in the sky other than the Sun has a visual magnitude of ), and from parallax measurements d =.7 parsecs away. Finding this point on the plot and extrapolating over we find that Sirius has an absolute magnitude of M Sirius = + and a luminosity relative to the Sun of L Sirius = 3L sun. Conversely, the Sun has an absolute magnitude of +; thus, at 0 parsecs the relative magnitude would be +.
Distance (pc) 0.0 Sun Jupiter Name Part Apparent Magnitude A Stars Within 4 Parsecs -6.6 -.6 Absolute Magnitude 6 Spectral Type.3 Alpha Centauri A C 0 4 6 G K0 M G.0 0.00 Mass (M Sun ) Radius (R Sun ). ernard s Star 0 3 M.3 Wolf 39 4 7 M. D +36 o 47 0 M 0.3.7 L76- A 3.9 Sirius A - A A Wh Df.9 Ross 4 3 M 3. Ross 4 3. L79-6 3.3 ε Eridani 4 6 K 0.9 3.3 Ross 4 M 3.3 6 Cygni A 6 K K7 ε 3.4 Indi 7 K 3. Procyon A 0 3 3 F Wh Dwarf 3. Σ 39 A 9 0 3. D +43 o 44 A C τ 3.6 Ceti 4 6 G.0 3.6 CD -36 o 693 7 0 M 3.7 D + o 66 A 0 3.7 G- 7 3. L7-3 4 M 3. CD -39 o 49 7 9 M0 3.9 Kapteyn s Star 9 M0 4.0 Kruger 60 A 0 C 4.0 Ross 64 A 4.0 D - o 43 0 M 6 0 3 3 3 7 M4 M M K M4 M4 M. 0.9 0. 0. 0..3.0 0.63 0.6. 0.6 0.4 0.4 0.7 0.6 0.0 0.4 0.0.0 0.. 0.9. 0.0.7 0.0 0.