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Ohio University - Lancaster Campus slide 1 of 47 Spring 2009 PSC 100. A star’s color, temperature, size, brightness and distance are all related!. Ohio University - Lancaster Campus slide 2 of 47 Spring 2009 PSC 100. The Beginnings.

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the beginnings

Ohio University - Lancaster Campus slide 2 of 47Spring 2009 PSC 100

The Beginnings
  • Late 1800’s, early 1900’s – how light is produced by atoms is being intensely studied by…
    • Gustav Kirchoff & Robert Bunsen
    • Max Planck…Josef Stefan...
    • Ludwig Boltzmann…Albert Einstein
black bodies

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Black Bodies
  • In 1862, Kirchoff coins the phrase “black body” to describe an imaginary object that would perfectly absorb any light (of any wavelength) that hit it.
    • No light transmitted through, no light reflected off, just totally absorbed.
slide4

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  • a perfect absorber of light would also be a perfect emitter
  • amount of light energy given off each second (its brightness or luminosity) and the color of its light are related to the object’s temperature.
slide5

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  • Molten lava and hot iron are two good examples of black bodies, but…
  • a star is an excellent black body emitter.
slide6

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  • Max Planck, a German physicist, was able to make theoretical predictions of how much light of each color or wavelength would be given off by a perfect black body at any given temperature.
  • These predictions or models are today called Planck Curves.
slide8

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  • What 2 characteristics of the curves change as the temperature increases?
  • The size of the curve increases.

(2) The peak of the curves shift to theleft, to shorter wavelengths & higher

energies.

can we draw some conclusions

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Can we draw some conclusions?
  • Hotter stars should be brighter than cooler stars.
  • Hotter stars should emit more of their light at shorter wavelengths (bluer light)
  • Cooler stars should emit more of their light at longer wavelengths (redder light).
  • All stars emit some energy at all wavelengths!
slide10

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  • In 1879, Josef Stefan discovered that the luminosity of a star was proportional to the temperature raised to the 4th power.
  • In 1884, Stefan’s observations were confirmed when Ludwig Boltzmann derived Stefan’s equation from simpler thermodynamic equations.
stefan boltzmann law

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Stefan-Boltzmann Law
  • Today, we honor both scientists by naming the equation after them…the Stefan-Boltzmann Law:
  • At the surface of the star, the energy that’s given off per square meter (Watts / m2) called the luminous flux is...

W / m2 = 5.67 x 10-8 T4

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  • At 100 K (cold enough to freeze you solid in just seconds), a black body would emit only 5.67 W/m2.
  • At 10x hotter, 1000 K, the same black body would emit 104 times as much light energy, or 56,700 W/m2.
slide13

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  • If the temperature of a star were to suddenly double, how much brighter would the star become?
  • If the temperature of a star somehow fell to 1/3 of what it was, how much fainter would the star become?

24 = 16 times brighter(1/3)4 = 1/81, or 81 times dimmer

slide14

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  • In 1893, Wilhelm Wien (pronounce “vine”) discovered by experiment the relationship between the “main” color of light given off by a hot object and its temperature.
  • This “main” color is the peak wavelength, called λmax , at the top of the Planck Curve.
slide15

For each curve, the

top of the curve is the

peak wavelength.

wien s law

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Wien’s Law
  • Wien’s Law says that the peak wavelength is proportional to the inverse of the temperature:

λmax = 2.9 x 106 T = 2.9 x 106

T λmax

  • T must be in Kelvin, and λmax in nanometers.
slide17

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  • What is the peak wavelength of our sun, with a T = 5750 K?
  • What is the peak wavelength of a star with a surface temperature of 3500 K?

2.9 x 106 = 504 nm (yellowish-green)

5750 K

2.9 x 106 = 829 nm (this star emits the

3500 K majority of its light as

infrared, IR).

slide18

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  • A reddish star has a peak wavelength of 650 nm. What is the star’s temperature?

A star has a peak wavelength in the ultra-violet of 300 nm. What is the star’s temperature?

2.9 x 106 = 4462 K (cooler than the sun)

650 nm

2.9 x 106 = 9667 K

300 nm

slide19

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  • We now have a “color thermometer” that we can use to determine the temperature of any astronomical object, just by examining the light the object gives off.
  • We know that different classes of objects are at different temperatures and give off different peak wavelengths.
what kinds of objects
Clouds of cold hydrogen gas (nebulae) emit radio wavesWhat kinds of objects?

http://www.narrowbandimaging.com/images/vdb142_small.jpg

protostars emit ir
Protostars emit IR.

http://www.antonine-education.co.uk/Physics_GCSE/Unit_3/Topic_10/protostar.jpg

sun like stars emit mostly visible light while hotter stars peak in the uv
Sun-like stars emit mostly visible light, while hotter stars peak in the UV.

http://www.nasa.gov/images/content/138952main_whywe16full.jpg

star cores emit gamma rays
Star cores emit gamma rays.

http://aspire.cosmic-ray.org/labs/star_life/images/star_pic.jpg

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  • Where would the peak wavelength be for
    • your body
    • a lightning bolt
    • the coals from a campfire
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  • A star’s spectrum is also influenced by
  • its temperature.
  • In 1872, Henry Draper obtained the first
  • spectrum of a star, Vega, in the
  • constellation Lyra.

photojournal.jpl.nasa.gov/jpeg/PIA04204.jpg

Credit: Lick Observatory Archives

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  • In 1885, Edward Pickering began a project at Harvard University to determine the
  • spectra of many stars. Draper’s widow
  • funded the work.
  • The first 10,000 spectra obtained were
  • classified by Williamnia Fleming, using the
  • letters A through Q.
slide29

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  • From 1901 to 1919, Pickering & his assistant
  • Annie Jump Cannon classified and published
  • the spectra of 225,000 stars (at the rate of
  • about 5000 per month!)
  • When Pickering died in 1919, Cannon
  • continued the work, eventually classifying
  • and publishing the spectra of 275,000 stars.

Credit: amazing-space.stsci.edu

slide30

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Hotter stars have

simpler spectra.

Cooler stars have

more complex

spectra, since most

atoms are not ionized.

slide31

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  • Class O >30,000 K bluish
  • He lines in spectrum.
  • (These stars are so hot that H is mostly ionized & doesn’t shows lines.) Pleiades
      • Class B 11,000-30,000 K bluish
      • He lines, weaker H lines
  • Rigel, Regulus, Spica
      • Class A 8,000-11,000 K blue-white H lines (Balmer Series)
  • Sirius, Vega
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      • Class F 6,000-8,000 K white
      • H, Ca lines, weaker H lines Procyon
      • Class G 5,000-6,000 K yellow
      • Ca, Na lines, + other metals
  • Sun, Capella, -Centauri
      • Class K 3,500-5,000 K orange
      • Ca & other metals
  • Arcturus, Aldebaran
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      • Class M <3,500 K red
      • metal oxides (TiO2), molecules
  • Betelgeuse, Antares
      • Oh, Be AFine Girl, Kiss Me!
slide34

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The stellar classes (OBAFGKM) are further

subdivided with a number 0 to 9 following the

letter.

Our sun, a G2 star, is slightly cooler than the

F range. A G9 star would be just a bit warmer

than the K range.

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  • 1910-1913, Henry Russell, a professor at
  • Princeton, and Ejnar Hertzsprung, an
  • astronomer at Leiden Observatory in the
  • Netherlands, used the data from the Draper
  • catalog to plot the temperature of the stars
  • vs. their brightness or luminosity.
  • What kind of result would you expect, a
  • random scatter, or a pattern?
slide37

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Betelgeuse and Antares show on the diagram

as being red stars, and red stars should be

faint.

Both stars are also hundreds of light

years distant, so why do they appear so

bright in our sky?

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slide39

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slide41

‘Red’

‘Red’

Red Dwarfs

slide42

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The H-R Diagram makes a lot more

sense when you realize that the

different regions don’t show different

kinds of stars…

…but stars at different stages

of their lives.

slide43

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  • Determining distance using the HR Diagram
  • From a star’s color-temperature, determine
  • its absolute magnitude (M).
  • Observe the star’s apparent magnitude (m)
  • through a telescope.
  • Use the distance modulus equation to
  • calculate the distance.
slide44

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How far away is an F1 star that has a surface

temperature of 8000 K, if its apparent

magnitude is +9.6?

slide46

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distance in parsecs =

10^[(9.6 - 3.0 +5)  5] =

10^[11.6  5] =

10^2.32 =

209 parsecs (or 681 light years)

slide47

Where might this method run into trouble?

Red & Orange star come in 2 varieties:

giants & dwarfs.

The spectrum of the star must be used to

determine if the star is large or small.

The presence of what element(s) in higher

than normal percentages might indicate

that the star is a giant, not a dwarf?

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