1. Thermal Radiation Homework 1: Due Classtime on Tuesday, Feb. 4 Reading for today: Chapters 2.3 - 2.5 Reading for next lecture: Chapters 2.6 - 2.7 Topics for Today: Blackbodies/Planck's Law Wien's Law Stellar Colors/Stellar Temperatures Luminosity of a Blackbody Radius of Star

2. Temperature Scales (Reminder) Temperature on the Kelvin scale (°K) is a measure of the amount of thermal energy or kinetic energy in the motions of the atoms or molecules. 0 K means no microscopic classical motion ! For an ideal gas, the thermal energy per particle is 3/2 kT, where k is the Boltzmann constant, k = 1.38x10 -16 ergs K -1.

3. Thermal Radiation: Radiation produced by the thermal motions of charged particles in matter. Objects with T > 0 °K produce light. Empirically found emission was governed by temperature and had a continuous spectrum. An idealized perfect radiator is called a blackbody (also a perfect absorber). The energy radiated comes from the objects thermal energy (motions of atoms and molecules).

4. Perfect Blackbody The distribution of light energy as a function of wavelength was measured in the 1800s, however classical physics could not reproduce what was observed. Planck made an ad hoc assumption (quantized the energy of the oscillators) and produced a function (Planck's Law) that could reproduce what was observed. The intensity of light from a blackbody is given by the Planck's Law (can defined either per unit wavelength or per unit frequency) which I will call either B λ or B ν :

5. Where h is Planck's constant (6.63x10 -27 ergs Hz -1 or ergs sec) and k is the Boltzmann constant. The intensity specifies amount of light energy per second, per unit area, per unit wavelength (or frequency) and per unit solid angle. The units of intensity are the following: B ν (ν, T): ergs s -1 cm -2 Hz -1 sr -1 B λ (λ, T): ergs s -1 cm -3 sr -1 What is a solid angle ???????

6. Angular Measurements (Refresher) We often measure the angular size of objects or the angular distances between objects in the sky. We can define an angle by the following: Angle θ is defined by the ratio of the arc length s and the radius R. θ = S/R The units of θ are radians. Note that if the angle was a complete circle, then s = 2 π r (the circumference of a circle), thus there are 2 π radians in a circle.

7. Note that if s = r, then the angle θ is one radian. Since there are 360° (degrees) in a complete circle, one radian is equal to: 360°/2 π ~ 57.3 degrees. Degrees are further divided into minutes of arc and seconds of arc: 1° (degree) = 60' (arcminutes) 1' = 60" (arcseconds) Thus, there are 3600" in 1°

8. In astronomy we often deal with small angles when measuring the angular size of an object (such as a star). As angle θ becomes small, the arc length s and the straight line approximation (r) become nearly the same. Note that tan θ = r/R, therefore for small angles can approximate: tan θ = θ = r/R. So if r is the radius of a star and R is the star's distance, θ is the angular radius of the star.

9. Solid Angle A solid angle is the opening angle of a cone (consider this a 2-dimensional angle). Have sphere of radius R. Define a cone originating at the center of the sphere with solid angle Ω. The cone defines an area A on the sphere. The defininition of solid angle is that: Ω = A/R 2. The units of solid angle are steradians. The surface area of a sphere is 4 π R 2, so there are 4π steradians in a sphere

10. Solid Angle For small solid angles Ω, one can do a similar approximation as for a simple angle. If solid angle Ω is small, then area, A, can be replaced by area of a flat disk. Most astronomical objects have small Ω. So a star with a radius of r at a distance of R, subtends a solid angle Ω ~ πr 2 /R 2. But the angular radius of the star is θ = r/R, so we can also write the solid angle as π θ 2. Note that the solid angle is the square of a simple angle.

11. Getting back to our blackbody..... So, for instance, B λ (λ, T) tells us the intensity of light produced by a blackbody or how much light energy is emitted per second, per unit area, per unit wavelength interval, and per unit solid angle – note that the units are ergs s -1 cm - 2 cm -1 sr -1 or ergs s -1 cm -3 sr -1.

12. Planck's Law: We can examine the dependence on wavelength and temperature. To the right, the intensity per unit area versus wavelength is shown. Note: Wien's Law 1. Peak intensity shifts to shorter wavelengths for higher temperatures 2. Hotter objects produce more total light power per unit area

13. Wien's Law: The peak is where the derivative is zero. Take the derivative of the Planck Function with respect to wavelength (note that the peak of B λ and B ν are different). Solve dB λ (T)/dλ = 0: How would we find the wavelength at which the Planck Function has its maximum intensity ? Substituting x = hc/λkT: B λ (T) = 2k 5 T 5 /(h 4 c 3 ) x 5 /(e x -1) dB λ (T)/dλ = 2k 5 T 5 /(h 4 C 3 ) d/dλ [x 5 /(e x -1)] Remember: d/dλ f(x) = d/dx f(x) dx/dλ First we find that: dx/dλ = -hc/(kTλ 2 )

14. With dx/dλ = -hc/(kTλ 2 ), we can rewrite as: dB λ (T)/dλ = -2k 5 T 5 /(h 3 c 2 λ 2 ) d/dx [x 5 /(e x - 1)] Differentiate {remember d/dx f(x)g(x) = g(x) df(x)/dx + f(x) dg(x)/dx and also d/dx [f(x)] n = n f(x) n-1 d/dx f(x)} to get: dB λ /dλ = -2k 5 T 5 /(h 3 c 2 λ 2 ) [5x 4 /(e x – 1) – x 5 e x /(e x – 1) 2 ] Set equal to 0 and solve. Thus: 5x 4 /(e x – 1) – x 5 e x /(e x – 1) 2 = 0 Divide by x 4 and multiple by (e x – 1), get: 5 = xe x /(e x – 1) Not an analytic function, solve by iteration to find x = 4.965

15. λ(max) = 0.290/5800 = 5x10 -5 cm (500 nm or 5000 Å) in the middle of the visible spectrum)‏ Remember that x = hc/λkT or λT = hc/xk. The maximum intensity occurs when x = 4.965. Thus: λ max T = hc/4.965 k Wien's law is therefore: λ max T = hc/4.965 k = 0.290 cm °K If the Sun has a surface temperature of 5800 °K and if it radiated like a perfect blackbody, at what wavelength would it produce its maximum intensity ?

16. Stellar Surface Temperatures A star’s photosphere (surface) radiates approximately like a blackbody, thus stars follow Wien’s law. Above star has its peak intensity at a wavelength of 460 nm or (4.6x10 -5 cm), therefore it surface temperature is: T = 0.290/λ(max) = 0.290 cm °K/4.6x10 -5 cm = 6300 °K Peak

17. Stellar Surface Temperatures The star above has its peak intensity at around a wavelength of 800 nm or (8x10 -5 cm). Peak wavelength much longer than previous stars, so star is much cooler.

18. Stellar Surface Temperatures The color of a star is related to its surface temperature. The “color” of a star can be measured by determining its flux density at two different wavelengths Since stars are producing a broad range of wavelengths, the colors are not that pronounced.

19. Red, cool star Blue, hot star

20. The Constellation of Orion Betelgeusea cool, red star Rigel, a hot, blue star Bellatrix, a hot, blue star

21. One can determine uniquely the surface temperature of a star by measuring the star's flux density at two different wavelengths. Example: a blue filter, B, and a yellow filter, V “visual”. Form the flux density ratio: F V /F B

22. The color of a star can be defined by the ratio of the flux density of a star at two wavelengths. However, traditionally astronomers have expressed this using apparent magnitude. Remember apparent magnitude in the V band (Vega system): m V = - 2.5 log [F* ν,λ (V)/F Vega ν,λ (V)] Measure apparent magnitude of a star in the V and B filters, then a star's “color” is defined as following: (B-V) ≡ m B – m V (B-V) = -2.5 log [F* ν,λ (B)/F vega ν,λ (B)] ] + 2.5log[F* ν,λ (V)/F Vega ν,λ (V)] Rearanging: (B-V) = 2.5 log [F* ν,λ (V)/F* ν,λ (B)] – 2.5 log [F Vega ν,λ (V)/F Vega ν,λ (B)] The difference in magnitudes is the log of the flux density ratio. A Star’s Color

23. So the (B-V) color of a star provides a measure of the flux density ratio F* ν,λ (V)/F* ν,λ (B) compared to that of Vega. Vega is a hot star with a surface temperature of about 10,000 K. By definition, (B-V) for Vega is 0. For a hot star (hotter than Vega), the flux density ratio F* ν,λ (V)/F* ν,λ (B) will be smaller for the star than ifor Vega, thus: (B-V) < 0 (negative) For a relatively cool star (cooler than Vega), the flux density ratio F* ν,λ (V)/F* ν,λ (B) will be larger for the star than in Vega, thus: (B-V) > 0 (positive)

24. There is a simple relation between the surface temperature of a star and the observed B-V color Sun: B-V = +0.66 Bellatrix: B-V = -0.21 Betelgeuse: B-V = +1.85

25. We need to sum up the emission from the star over solid angle, over wavelength or frequency and over surface area. First consider solid angle. You might think that each unit surface would radiate equally in all directions, and the solid angle would be one hemisphere (solid angle of 2π steradians). Luminosity of a Star Not the case due to projection effects. The area is most effective radiating normal to its surface. The amount of solid angle is just π steradians. The amount of energy radiated per second, per unit area, per unit wavelength (or frequency) interval from star's surface is: π B λ (T) or π B ν (T) Assume star emits light like a blackbody. So how can we compute it luminosity ?

26. Integrate π B λ (T) or π B ν (T) over all λ or ν: Note that: ∫ π B λ (T) dλ = ∫ π B ν (T) dν As before, we assume x = hc/λkT, thus dλ = -hc/kT x -2 dx, and the integral becomes: ∫ π B λ (T) dλ = - π 2k 4 T 4 /(h 3 c 2 ) ∫x 3 /(e x – 1) dx The definite integral ∫ ∞ 0 x 3 /(e x – 1) dx = - π 4 /15, thus: ∫ π B λ (T) dλ = σ T 4, where σ = 2π 5 k 4 /(15h 3 c 2 ) = 5.67x10 -5 erg cm -2 sec -1 K -4 (σ is called the Stefan-Boltzmann constant) So we have summed over solid angle, how do we we sum over wavelength or frequency to get the energy radiated per second per unit area ???

27. For a Blackbody, the power per unit area: Power per unit area =  T 4  = 5.67x10 -5 ergs cm -2 s -1 °K -4 So how much more power is produced per unit area from a blackbody at 8000°K compared with one at 4000°K ? Answer: 16 times more

28. Luminosity = 4 π R 2  σ T 4 surface area of star  power per unit area Luminosity of a Star The total luminosity of a star is then given by:

29. Relation between luminosity (L), the radius of a star (R) and its surface temperature (T): L = 4 π R 2 σ T 4 If you know any two parameters, you can find the third. Radius of a Star It is easiest to measure the surface temperature and luminosity of a star, so this relation can be used to determine the radius of the star. R = [L/(4 π σ T 4 )] 1/2.