Luminosityandbrightness亮度和亮度.doc

Luminosity and brightnessLuminosityLuminosity is a measure of the total energy given output by a star at all wavelengths from gamma radiation to radio waves. For example the Sun gives out about 500 million million million MJ of energy every second so its luminosity is 500 million million million MJ.The luminosity depends on:(a) the size of the star(b) the temperature of the star(a) for a star with a certain surface temperature the bigger the star the more energy it gives out. A star with double the radius of another one will have an area four times as great and so have a luminosity four times greater than the first star(b) for a star of a certain size the hotter the star the more energy it gives out and so the greater its luminosity. A star with a temperature of double another one will have a luminosity sixteen times greater.BrightnessThe brightness is how bright a star appears when seen from the Earth. This depends on:(a) the actual luminosity of the star(b) the distance of the star from the observer on the EarthIf we have two stars of the same luminosity with one star double the distance of the other from the Earth the closer star will look four times brighter. It obeys the inverse square law.The photograph shows the Pleiades star cluster. The brighter stars look about the same brightness in fact they are. They are all part of the cluster and about the same distance from the Earth. However some of the background stars may be just as bright they dont look it because they are much further away.I am very grateful to Marcos Mataratzis and Vivek Hira for the permission to use their photograph. To see the full collection please visit: .br/arte/astrophotography/index.htmlLuminosity theoryThe luminosity depends on the surface area of the star. The surface area (A) of a star of radius R is 4pR2. A star with a radius of 2R will therefore have four times the luminosity of a star of radius R if the stars are at the same temperature.The luminosity of a star also depends on its temperature. If we consider a star to be a perfect black body then the radiation emitted per second (E) is given by the Stefan-Boltzmann law.Energy emitted per second (E) = sAT4 where s is Stefans constant (5.7x10-8 Wm-2K-4) and T is the absolute temperature of the star. (Strictly we should write the difference in the temperature between the star and its surroundings but since the temperature of a star is very large, as least a few thousand K and the temperature of space is very low about 3 K, this slight inaccuracy can be ignored).We can calculate this energy output for a star of the same size as our Sun. The radius of the Sun (R) = 6.96x108 m and its absolute temperature (T) = 6000K. Therefore E = 5.7x10-8x6.08x1018x60004 = 4.5x1026 J. A star with a temperature twice that of our Sun would have an energy output sixteen (24) times as great.Consider a star like Rigel in Orion (on the right in the photo). The surface temperature of Rigel is 11600 K (1.9 times that of the Sun) and its radius is seventy times that of the Sun. The energy output of Rigel per second is therefore 702x1.94 = over 68000 times that of the Sun. In other words an enormous 3.08x1031 J. Altair in Aquila at nearly 17 light years away has an output of just less than ten times that of the Sun.I am grateful to David Darling for permission to use this photo. To see his full collection see: /encyclopedia/ETEmain.htmlWiens displacement law and the temperature of a starIf we assume that stars are perfectly black bodies then we can use Wiens displacement to calculate their temperature. The law states that:Wavelength (lmaxA) X Temperature (TA) = Wavelength (lmaxB) X Temperature (TB)ElBAWavelength (l)lmaxAlmaxBwhere lmaxA ,TA and lmaxB ,TB are the peaks of the energy wavelength curve and the absolute temperature of two black bodies, in this case two stars. Consider two stars A and B as shown in the diagram.Star A has an energy peak nearer the blue end of the spectrum and is at a higher temperature than star B that has an energy peak towards the red end of the spectrum.Example problem A black body (the Sun) with a surface temperature of 6000 K emits radiation with max = 420 nm. Calculate the temperature of Sirius (the brightest star in the northern skies) if max for Sirius is 72 nm.max1T1 = max2T2 therefore 420x10-9x6000 = 72x10-9xT2Therefore: T2 = 35 000 KInverse square lawThe brightness of a star seen from the Earth depends on its intr