Prof. Jeff Kenney Class 4 May 31, 2018

Transcription

1 Prof. Jeff Kenney Class 4 May 31, 2018

2 Which stellar property can you estimate simply by looking at a star on a clear night? A. distance B. diameter C. luminosity D. surface temperature E. mass

3 you can estimate both the surface temperature and the brightness simply by looking at the star surface temperature from color since stars radiate approximately like blackbodies brightness or apparent brightness the intensity of light detected related to luminosity of star (how much energy it puts out) and distance

4 Blackbody an idealized opaque source at a uniform temperature which reflects no light, and emits photons with a spectral distribution of wavelengths given by Planck s Law

5 Blackbody an idealized opaque source at a uniform temperature which reflects no light, and emits photons with a spectral distribution of wavelengths given by Planck s Law

6 Planck s Law says: Intensity of radiation (I) depends only on temperature (T) and wavelength (λ) [h,c,k are all fundamental constants, not physical variables] Intensity energy per time per area per wavelength interval in a given direction (per solid angle) Units: Joule second -1 meter -2 Hertz -1 steradian -1 Units: J s -1 m -2 Hz -1 sr -1

7 Emitting per area, per solid angle

8 Solid angle

9 Planck s Law I If you specify T, then the spectrum (i.e. intensity I vs. wavelength λ) is fully determined. λ

10 Blackbody is not a very good name! Doesn t need to appear BLACK (& not all things which look black are blackbodies) Doesn t need to be BODY (can be gas rather than solid) Not necessarily dense or solid

11 Stars are not perfect blackbodies, but they are close to blackbodies Spectra of the Sun & T=5777K blackbody

12 Stars are not perfect blackbodies, but they are close to blackbodies Spectra of the Sun & T=5777K blackbody Blackbody is an idealized concept Some objects are well-approximated by idealized BB; others are not No real thing is a perfect blackbody

13 What is Temperature? Directly related to speed or kinetic energy (KE=½mv 2 ) of atoms or molecules or other particles that make up substance Motions of gas atoms Internal motion of protein molecule

14 What is Temperature? Directly related to speed or kinetic energy (KE=½mv 2 ) of atoms or molecules or other particles that make up substance T ~ v avg 2 energy kt = ½mv 2 Temperature is proportional to the average squared speed of the particles

15 What is Temperature? Absolute zero particles have NO motion

16 What is Temperature? Absolute zero particles have NO motion Kelvin temperature scale based on absolute zero = 0 K = -273 C = -460 F

17 What is Temperature? Absolute zero particles have NO motion Kelvin temperature scale based on absolute zero = 0 K = -273 C = -460 F T K = T C + 273

18 What is Temperature? Absolute zero particles have NO motion Kelvin temperature scale based on absolute zero = 0 K = -273 C = -460 F T K = T C Water freezes 32 F = 0 C = 273 K

19 T=5800 K

20 UV IR blackbody spectra I 12,000 K 6000 K 3000 K λ

21 The wavelength at which the intensity of a BB spectrum peaks is given by Wien s Law: λ max = K m T if T in K, lambda in m

22 UV IR blackbody spectra I 12,000 K, peaks at 0.25µm 6000 K, peaks at 0.5µm 3000 K, peaks at 1.0µm λ

23 blackbody spectra: shown on logarithmic scale 0.01 µm 0.1 µm 1 µm 10 µm 100 µm

24 The blackbody emission of the Earth is shown as curve A. If we move the Earth closer to the Sun, it will heat up. Which curve could represent its new blackbody emission? A B C D E--None of above

25 Flux if you add up radiation (integrate Planck s law) over all directions (solid angles) and all wavelengths, then you find the (energy) flux F emitted from the surface of a blackbody source flux is energy emitted per second per area units: Joule second -1 meter -2 F = σt 4 Stephan-Boltzmann law σ= 5.67x10-8 J s -1 m -2 K -4 Stephan-Boltzmann constant *note strong T dependence! If T doubles, F increases by 16x!!

26 A particle of interplanetary material is heated by friction from 400 K to 4000 K as it falls into Earth s atmosphere, producing a meteor or a shooting star. If it behaves like a perfect blackbody, how does its emitted radiation change as it is heated? 1. Its flux is 10x greater, while its peak wavelength is 10x shorter (IR->VIS) 2. Its flux is 100x greater while the peak wavelength is 100x shorter (IR->UV) 3. Its flux is 10,000x greater, while its peak wavelength is 10x shorter (IR->VIS) 4. Its flux is 10,000x greater while its peak wavelength is 10x longer (VIS -> IR)

27 Luminosities (of stars & other things) sum up (integrate) flux over entire surface area of emitting source Luminosity = flux x area L= F A (energy per time) = (energy per area per time) x (area) luminosity is energy emitted per second units: Joule second -1

28 luminosity of stars (~round blackbodies) L= F A For blackbody: F =σt 4 For sphere A = 4πR 2 (surface area) à L = 4πσR 2 T 4 luminosity of round blackbody For sun: R = 1 R sun = 6.96x10 8 m T = 1 T sun = 5800 K L = 1 L sun = 3.9x10 26 J s -1 ***show sun-earth to same scale 1 solar luminosity is standard unit of luminosity in astronomy

29 Some stars are 10x hotter than the sun and are 10x bigger. How much more luminous are they?

30 Some stars are 10x hotter than the sun and are 10x bigger. How much more luminous are they? L sun = 4πσR 2 sun T 4 sun L BH = 4πσR 2 BH T 4 BH = 4πσ (10R sun ) 2 (10T sun ) 4

31 Some stars are 10x hotter than the sun and are 10x bigger. How much more luminous are they? L sun = 4πσR 2 sun T 4 sun L BH = 4πσR 2 BH T 4 BH = 4πσ (10R sun ) 2 (10T sun ) 4 Take ratio of equations: L BH 4πσ (10R sun ) 2 (10T sun ) (4πσR sun2 T sun4 ) = = L sun 4πσR sun2 T sun 4 4πσR sun2 T sun 4

32 Some stars are 10x hotter than the sun and are 10x bigger. How much more luminous are they? L sun = 4πσR 2 sun T 4 sun L BH = 4πσR 2 BH T 4 BH = 4πσ (10R sun ) 2 (10T sun ) 4 Take ratio of equations: L BH 4πσ (10R sun ) 2 (10T sun ) (4πσR sun2 T sun4 ) = = L sun 4πσR sun2 T sun 4 4πσR sun2 T sun 4 L BH /L sun = 10 6!!!