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Where is the Sun in the Milk Way? • The distance of the Sun to the center of our Galaxy: 8 kpc • Why do we see the objects in different colors? – for instance, why does an object look red? Blackbody Radia8on • Why do we see the objects in different colors? – for instance, why does an object look red? – because it absorbs most colors except red, which is being reflected! • Blackbody is a physical enDty that absorbs all radiaDon that falls on it, at all wavelengths. • RadiaDon emanaDng from a blackbody is due uniquely to its thermal energy. • Planck func8on: – Max Planck showed that a blackbody with temperature T emits a conDnuous spectrum of radiaDon characterized by a funcDon Bν(T), called “Planck funcDon”. (erg s-‐1 cm-‐2 Hz-‐1 sr-‐1 in CGS or J s-‐1 m-‐2 Hz-‐1 sr-‐1 in SI), (J=Joules) – Planck funcDon depends only on T and ν and is given by the following expression 3 2h ν 1 Bν (T) = 2 hν c e kT −1 k is the Boltzmann constant, k = 1.380658 x 10−16 erg/K c is the speed of light, c = 2.99793458 x 1010 cm/s € ν is the frequency at which the spectral energy is emiced h is the Planck constant, h = 6.6260755 x 10−27 erg s • Planck func8on is isotropic and thus independent of the direc8on • Can also be wricen per unit wavelength (Bλ) and Bν dν = −Bλ dλ dν c 2hc 2 1 Bλ = −Bν = 2 Bν = 5 hc dλ λ λ e λkT −1 (erg s-‐1 cm-‐1 cm-‐2 sr-‐1 or erg s-‐1 Å-‐1 cm-‐2 sr-‐1) € Planck distribuDons (Bλ) as a funcDon of wavelength for T = 2000, 6000 and 12 000 K. λmax associated to each funcDon. Visible part of the spectrum is also shown. • The monochromaDc flux (Fν) is defined as the quanDty of energy in the spectral range between ν and ν+dν emiced per unit surface, per unit Dme in units of erg s-‐1 cm-‐2. • For a blackbody, Fν= πBν • The energy distribuDon emiced by a blackbody leads to two laws: – Stefan-‐Boltzmann law: gives the total power output per unit area, F, of a blackbody €with temperature T ∞ ∞ F = Fν dν = πBν dν = σT 4 0 0 ∫ ∫ σ is the Stefan-‐Boltzmann constant, σ = 5.67051 x 10−5 erg cm-‐2 K-‐4 s-‐1 € – Wien’s law: the wavelength λmax, at which the Bλ is at its maximum, varies inversely with temperature λmax 0.2898 K cm = T – It explains why hocer blackbodies (or stars) are blue and cooler ones are red € – Exercise: Calculate the €λmax for the Sun. • The surface temperature of the Sun, T=5800 K λmax 0.2898 K cm 0.2898 = = ≅ 5 ×10 −5 cm = ? Å T 5800 Luminosity, Effec8ve Temperature, Flux and Magnitudes • What is Luminosity (L) of a star? – The luminosity of a star is defined as the radiaDve power output emanaDng from its surface, in other words, the total energy output from a star’s (or an object’s) surface and given in units of “erg s-‐1” – It is independent of distance – and important to understand the energy producDon of a star • To obtain the luminosity (L), the radiaDon field emiced over the enDre electromagneDc spectrum and over the enDre surface of the star must be integrated! Teff : effecDve temperature, is the temperature of a blackbody which has the same radius R* and luminosity L* of a star. F : Flux energy emiced from an object’s (star) surface per unit area at per unit Dme € L∗ = 4πR∗2σTeff4 ⎛ L ⎞1/ 4 Teff = ⎜ ∗ 2 ⎟ ⎝ 4πR∗ σ ⎠ L∗ 4 F= = σ T eff 2 4 π R ∗ • At a distance d larger than R✽ from the center of the star ⎛ R ⎞2 L∗ 4 ∗ F(d) = σTeff ⎜ ⎟ = 2 d 4 π d ⎝ ⎠ d € L • This equaDon shows the effect of the geometrical diluDon of the flux as a funcDon of distance from a star. • It’s also called the “Inverse-‐Square Law” • Magnitude is a relaDve scale that measures the logarithmic value of the radiaDve flux. • DefiniDon of magnitude is given by #F & m2 − m1 = −2.5 log% 2 ( $ F1 ' m1 and m2 are the apparent magnitudes of two stars F1 and € F2 are their observed fluxes • Absolute magnitude, M: the magnitude of a star at a distance of 10 pc #L & M 2 − M 1 = −2.5 log% 2 ( $ L1 ' • The difference between the apparent and the absolute magnitude of a star is related to its distance d (in parsecs) to the observer via the equaDon ≡ !!m − M = −5 + 5log d • The value m – M is oqen called the “distance modulus”. € – Exercise: Knowing that the apparent visual magnitude of the Sun is m = -‐26.73, calculate its absolute magnitude, M? • The Sun is by definiDon at a distance of one astronomical unit (AU) from the Earth. • Since 1 AU = 1.496 x 1013 cm = 4.848 x 10−6 pc, the distance modulus equaDon Mv = 4m.84