# Measurements of Luminosity For a refresher on Stefan’s Law, upon which this tutorial is based, please consult the Tutorial on Continuous Spectra.

## Presentation on theme: "Measurements of Luminosity For a refresher on Stefan’s Law, upon which this tutorial is based, please consult the Tutorial on Continuous Spectra."— Presentation transcript:

Measurements of Luminosity For a refresher on Stefan’s Law, upon which this tutorial is based, please consult the Tutorial on Continuous Spectra

Luminosity (brightness) of a Star Luminosity is the amount of energy per second (Watts) emitted by the star Recall: The luminosity of the sun is about 4 x 10 26 W Absolute Brightness: The luminosity per square meter emitted by the star at it’s surface. This is an intrinsic property of the star. Apparent Brightness: The power per square meter as measured at the location of the earth.

Luminosity (brightness) of a Star How it works: Photons flow radially outward from the center of the star. The luminosity is the energy output per second, that is, the power output from the star. The luminosity, therefore, is a constant for a given star. Direction of photon (energy) flow (radially outward)

Luminosity (brightness) of a Star However, the “brightness” of a star decreases as one moves farther and farther away. Reason: The energy/sec flows from the interior of the star through its surface. The surface of the star is a sphere, with a surface area given by 4 π R 2, where R is the radius of the star. Direction of photon (energy) flow (radially outward) R

Luminosity (brightness) of a Star However, the “brightness” of a star decreases as one moves farther and farther away. If a sphere of radius d is drawn around the star, it should be clear that the energy/sec through the surface of this sphere is the same as the energy/sec emitted through the surface of the star, since there is no mechanism to create or destroy photons in the space outside the star. The surface of the sphere has an area given by 4 πd 2, where d is the radius of the sphere. Direction of photon (energy) flow (radially outward) d

Luminosity (brightness) of a Star Absolute Brightness: The luminosity per square meter emitted by the star at it’s surface. This is an intrinsic property of the star. Apparent Brightness: The power per square meter as measured at a distance d from the star. If the observation of brightness of the star is made from the earth, the apparent brightness is the power per square meter as measured from the earth. d would then be the distance from the earth to the star. Ab. Bright = L / 4 π R 2 Where R is the radius of the star App. Bright = L / 4 π d 2 Where d is the distance to the observer d R Direction of photon (energy) flow (radially outward) L (the Luminosity) is the SAME in both expressions.

Luminosity (brightness) of a Star Prove that the luminosity of the sun is 4 x 10 26 W. From Appendix 2, Radius of earth’s orbit = 1.496 x 10 11 m Also, the magnitude of the sun (Pogson scale) is -26.7. Now, there is a footnote in the chapter (you would not need to know this) that a magnitude 1 star has an apparent brightness of 1.1 x 10 -8 W/m 2. Recall that each step in magnitude corresponds to a brightness 2.512 times greater than the lower magnitude. So, a magnitude 0 is 2.512 times greater than a magnitude 1 A magnitude -1 is (2.512) 2 times greater than a magnitude 1 A magnitude -2 is (2.512) 3 times greater than a magnitude 1 And continuing the logic, A magnitude -27 is (2.512) 28 times greater than a magnitude 1

Luminosity (brightness) of a Star Prove that the luminosity of the sun is 4 x 10 26 W. From Appendix 2, Radius of earth’s orbit = 1.496 x 10 11 m A magnitude -26.7 is a bit less than (2.512) 28 times greater than a magnitude 1. To make the calculation simpler, we will use -27 for the magnitude of the sun. Apparent Brightness of the sun = (2.512) 28 (1.1 x 10 -8 W/m 2 ) = (1.59 x 10 11 ) (1.1 x 10 -8 W/m 2 ) = 1749 W/m 2

Luminosity (brightness) of a Star Prove that the luminosity of the sun is 4 x 10 26 W. From Appendix 2, Radius of earth’s orbit = 1.496 x 10 11 m And finally Apparent Brightness = 1749 W/m 2 = L / 4πr 2 So, L = ( 1749 W/m 2 ) 4 π (1.496 x 10 11 m) 2 = 4.9 x 10 26 W Now, since the magnitude of the sun is actually -26.7, using a magnitude of 27 is an overestimate, and the calculated luminosity is a bit too high, as expected.

Luminosity (brightness) of a Star Two stars have the same apparent brightness as measured from the earth. Star A is 15 pc from the earth, star B is 30 pc from the earth. Which star has the greater luminosity, and how much greater is the luminosity of this star as compared to the lower luminosity star? Apparent Brightness = L / 4πd 2 Since the two stars have the same apparent brightness, the star that is further from the earth (star B) must have the higher luminosity. Since (App Bright) A = (App. Bright) B L A / 4πd A 2 = L B / 4πd B 2 L B = L A (d B / d A ) 2 = L A (30 pc / 15 pc) 2 = L A (2) 2 = 4 L A The Luminosity of Star B is 4 times greater than the Luminosity of Star A.

Download ppt "Measurements of Luminosity For a refresher on Stefan’s Law, upon which this tutorial is based, please consult the Tutorial on Continuous Spectra."

Similar presentations