“We demand rigidly defined areas of doubt and uncertainty “We demand rigidly defined areas of doubt and uncertainty.” Vroomfondel, the philosopher, The Hitchhiker's Guide to the Galaxy Sept. 30, 7:30-10:30pm, open house at Baker Observatory. If you can help, please contact Dr. Patterson

Photometry Photometry is converting counts (corrected, but not sky corrected) on CCD pixels into a brightness. Aperture: This puts a circle (or possibly an ellipse) around a star and sums all the counts on those pixels. Of course pixels are squares, so it uses fractional values.

The dark blue pixels are counted for the actual values, the light blue pixels are weighted by the fraction of area the circle encloses and the white pixels are not counted at all.

New terms: FWHM, HWHM

Photometry IRAF will give us magnitudes too. The goal of photometry is to determine flux for a given time. So you end up with the following: Time1 flux1star1 flux1star2 flux1star3 .. Time2 flux2star1 flux2star2 flux2star3 .. Time3 flux3star1 flux3star2 flux3star3 .. … .... IRAF will give us magnitudes too.

Definition of the magnitude scale

Magnitude nomenclature. Instrumental magnitudes: mV, mB.

Magnitude nomenclature. Standard magnitudes: V, B, R, etc.

Magnitude difference to brightness ratio conversion. A difference of 5 magnitudes is a brightness ratio of 100. For every 5 magnitudes multiply factors of 100. For every ratio of 100, add 5 magnitudes.

Distance in magnitudes Distance in magnitudes. Absolute magnitude (M) is the magnitude an object would be at 10pc.

How far away is an object with a distance modulus of 15?

Definitions: mV= instrumental V magnitude V = apparent (standard) V magnitude MV= absolute V magnitude (10pc=32ly) Dm of 5 is a brightness ratio of 100 and a distance ratio of 10.

Ultraviolet Blue Visual Red Infrared

ultraviolet green red infrared z = farther IR.

We will very briefly discuss how starlight is emitted. Colors: When we say a star is red, it means that its brightness in the R filter is larger than in the B filter. Why is that? We will very briefly discuss how starlight is emitted.

The power of light!

Spectrum Newton noticed that a prism spread light out into its constituent colors. This is what we call a spectrum

if l gets bigger, E gets smaller if l gets smaller, E gets bigger Wavelength and energy Since E = hc/l, if l gets bigger, E gets smaller if l gets smaller, E gets bigger

400 700 nm

Eletromagnetic spectrum

Wien's principles All bodies have a continuous (blackbody) spectrum. The peak of the continuous spectrum is determined by temperature only, not composition. T = 2.9x106/l for l in nm. The energy emitted by the continuous spectrum only depends on the temperature (in Kelvins).

Note on temperature scale: We use Kelvins because there are no negative values. For large Kelvin values, just double it to get Fahrenheit.

Shorter wavelengths = hotter object. Temperatures given in Kelvins T=2.9x106/max for  in nm. Shorter wavelengths = hotter object. Objects glowing blue are hotter then objects glowing red.

Stefan-Boltzmann Law Can determine energy per square meter from the temperature. E/m2 = T4 Where = 5.67x10-8 W/m2K4

Then the total energy emitted by a star, from its surface is: To get the total energy, multiply by the surface area of the star: A=4pR2 where R is the radius. Then the total energy emitted by a star, from its surface is: L=4pR2T4 This is called Luminosity

Light drops off as the square of the distance (1/d2) INVERSE SQUARE LAW Light drops off as the square of the distance (1/d2)

Put it all together: Apparent Luminosity Include distance into the calculation. Lap=R2T4/d2 This is how bright stars appear to us when we look up into the sky at night. Units: Temperature must be in Kelvin, size and distance must be in meters.

3 types of spectra Continuous spectra are made by objects under high pressure (like solids) Emission line spectra are made by (comparatively) hot, low pressure gases. Absorption line spectra are made by cool (comparatively) low pressure gases.

Below are the 3 kinds of spectra again. As you can see from the labels on the left, different elements produce different spectral lines.

The lines give us composition! So we can see what our Sun is made of ! A spectrum of our Sun: The lines give us composition! So we can see what our Sun is made of !

Now we can get the Sun's composition. By Mass 76% Hydrogen 22% Helium 0.8% Oxygen 0.4% Carbon 0.2% Neon 0.1% Iron and Nitrogen ~0.08% Silicon and Magnesium ~0.24% Everything else Astronomers call this stuff “metals”

Light We have to know how light is emitted (created) or absorbed (destroyed). The processes are: 1) bound-bound (atomic) transition 2) bound-free (ionization) 3) Free-free (Bremsstrahlung) absorption 4) Thomson (electron) scattering

Bound-Bound: Atomic Emission/Absorption This is when a bound electron changes orbitals.

Only orbits that have integer wavelengths are allowed. The Bohr model.

Light is emitted (or absorbed) when an electron changes orbital. But the energy of the light is quantized and can only have certain values.

DE=hc/l

The energy is determined by the exchange between electron 'orbitals.' For H, it looks like this.

Generally: E=-Z2*13.6eV (1/n2-1/n2o)

Transitions for hydrogen

Atomic Series n=1: Lyman n=2: Balmer n=3: Paschen n=4: Bracken Within each series, the lowest energy transition is labeled , the second lowest , and so on. So L is the n=2 to n=1 transition.

Atomic Series Because the Balmer series for hydrogen occurs in the optical, it is often written as H, H, etc. with it being understood that the H means the Balmer series for hydrogen.

Bound-free(photoionization): Ionization energy An atom can absorb any photons that have energy higher than the ionization energy (13.6eV for H). The extra energy goes into the momentum of the electron. This is also called Compton scattering. So gas is very opaque to photons with wavelengths shorter than the ionization energy.

hc/l=13.6eV using hc=1240 eV*nm Ionization energy Since stars are mostly H (92% by number), they become completely opaque to photons higher than the H ionization energy of 13.6eV. What wavelength does this correspond to? hc/l=13.6eV using hc=1240 eV*nm

Ionization energy hc/l=13.6eV, l=91.2nm. Since stars are mostly H (92% by number), they become completely opaque to photons higher than the H ionization energy of 13.6eV. What wavelength does this correspond to? hc/l=13.6eV, l=91.2nm.

Balmer Jump In fact, opacity greatly increases for electrons being stripped from the second orbital of H (the first excited state), which only requires 3.4eV. This corresponds to a wavelength of 364.7nm, which is also called the Balmer jump, since blueward of this wavelength, hot stars do NOT fit a blackbody spectrum.