Basic Properties of Stars - 4 §3.4-3.6

1. Basic Properties of Stars - 4 §3.4-3.6 Colours of stars and blackbody radiation

2. Colours of stars Stars have colours. Why?

3. Colours of stars Stars have colours. Why? Its not due to their redshift!!

4. Betelgeuse: red Rigel: blue-white. Blackbody Radiation Surface Temp = 3600K Surface Temp = 13,000K Orion

5. Blackbody Radiation History In 1792, Thomas Wedgewood first observed that all his ovens glowed “red-hot” at the same T, regardless of size, shape or materials. All objects T > 0 K emit radiation. Below 800 K in the IR, 800 to 1000 K detected in optical By 3000 K white hot - as T goes up Spectrum shifts to shorter wavelengths & power increases Perfect emitter absorbs all light it receives and reradiates it - called a blackbody.

6. Blackbody Radiation History Perfect blackbody is an idealization, but closely approximated as below. Cavity constant Temp  J. Stefan 1879 found relation between total power emitted and T P=T4 =5.67 x 10-8 W m-2 K-4 Small hole

7. Blackbody Radiation Blackbody spectrum depends only on T of source.

8. Blackbody Radiation Blackbody spectrum depends only on T of source. As T increases, max decreases and B()d increases maxT = d (Wien’s Law) and d = 2.898 x 10-3 m K.

9. Blackbody Radiation and Quanta By 1900, Max Planck found empirical formula for blackbody curve. B(T) = (a-5) / e b/T -1 and he tried to derive the constants a and b. Problem is shown on following slides

10. Standing waves

11. Standing waves

14. Blackbody radiation To circumvent this problem, Planck assumed that a standing E-M wave could not acquire any arbitrary amount of energy, but only allowed values that were multiples of a minimum wave energy. This quantum is given by h (or hc/), where h is constant = 6.63 x 10-34 J sec (Planck’s constant). Higher  (shorter ) of wave, greater minimum energy. Short , high  waves cannot contain even 1 quantum!

15. The Planck Function B(T) = energy emitted per second, per unit wavelength interval d at wavelength , per unit area into a unit solid angle by a blackbody of temperature T (whew!) B(T) = (2hc2/5)(1/ehc/kt - 1) w m-2 m-1 sterad-1 Where c = speed light = 3 x 108 m s-1 k = Boltzmann constant = 1.38 x 10-23 J K-1 h = Planck’s constant = 6.63 x 10-34 J s Only variable in the Planck function is T In terms of frequency B(T) = (2h3/c2)(1/eh/kt - 1) (require Bd =-Bd - as  decreases with increasing , d/d = -c/2, B =-Bd/d =Bc/2)

16. The Planck Function How does Planck function behave in the limits of very high and very low frequency? i.e h/kt >> 1 and << 1 Set h/kT = x  B(T) = (2h3/c2)/(ex - 1) For x >> 1, ex -1 = ex so B(T) = (2h3/c2) e-h/kT This called Wien approximation For x << 1, ex = 1 + x + x2/2 + x3/6 + ……… xn/n! = 1 + x Thus B(T) = (2h3/c2)(1/x) = 2kT2/c2 Thus log B(T) = 2 log  + log T + constant

17. Blackbody radiation

18. BB radiation - total intensity B(T) = (2h3/c2)(1/eh/kt - 1) The total intensity emitted by the BB is the integral B(T) = B(T)d = B(T)d = (2h3/c2)(1/eh/kt - 1)d where the integral goes from 0 to . Substitute x = h/kT, so that d = (kT/h)dx. Then B(T) = (2hk4T4/c2h4)(x3/ex -1)dx Integral just a real number so that B(T) = AT4 with A = 2k44/15c2h3 So B(T)  F  T4 (F = T4). This is the Stefan-Boltzmann Law  = 5.67 x 10-8 w m-2 K-4

19. Blackbody Radiation Relation between max and T is known as Wien’s Law maxT= 0.002897755 m K = 0.290 cm K. For a spherical source: F = L / 4R2, (R radius circle surrounding source) from S-B Law F = T4 so L = 4R2Te4 Te isthe effective T as stars are not perfect BB radiators - T of BB that puts out same energy as the star

20. Blackbody Radiation A simple problem: Lsun = 3.839 x 1026 W and its radius is Rsun = 6.955 x 108 m. (a) What is Te of Sun? (b) Where does Sun’s flux peak? (c) Any relation to the sensitivity of human eye?