1 Photons Observational Astronomy 2019 Part 1 Prof. S.C. Trager 2 Outline Wavelengths, frequencies, and energies of photons The EM spectrum Fluxes, filters, magnitudes & colors 3 Wavelengths, frequencies, and energies of photons Recall that λν=c, where λ is the wavelength of a photon, ν is its frequency, and c is the speed of light in a vacuum, c=2.997925×1010 cm s–1 The human eye is sensitive to wavelengths from ~3900 Å (1 Å=0.1 nm=10–8 cm=10–10 m) – blue light – to ~7200 Å – red light 4 “Optical” astronomy runs from ~3100 Å (the atmospheric cutoff) to ~1 µm (=1000 nm=10000 Å) Optical astronomers often refer to λ>8000 Å as “near- infrared” (NIR) – because it’s beyond the wavelength sensitivity of most people’s eyes – although NIR typically refers to the wavelength range ~1 µm to ~2.5 µm We’ll come back to this in a minute! 5 The energy of a photon is E=hν, where h=6.626×10–27 erg s is Planck’s constant High-energy (extreme UV, X-ray, γ-ray) astronomers often use eV (electron volt) as an energy unit, where 1 eV=1.602176×10–12 erg 6 Some useful relations: E (erg) 14 ⌫ (Hz) = 1 =2.418 10 E (eV) h (erg s− ) ⇥ ⇥ c hc 1 1 1 λ (A)˚ = = = 12398.4 E− (eV− ) ⌫ ⌫ E (eV) ⇥ Therefore a photon with a wavelength of 10 Å has an energy of ≈1.24 keV 7 If a photon was emitted from a blackbody of temperature T, then the average photon energy is Eav~kT, where k = 1.381×10–16 erg K–1 = 8.617×10–5 eV K–1 is Boltzmann’s constant. It is sometimes useful to know what frequency corresponds to the average photon energy: h⌫ kT ⇡ ⌫ (Hz) = 2.08 1010T (K) or ⇥ λT =1.44 cm K 8 The diference between the AVERAGE and PEAK wavelengths of the blackbody curve comes about because of the “heavy” long wavelength “tail” Note that this wavelength isn’t the peak of the blackbody curve. Consider the blackbody function of the curve. 2hc 1 We see the Sun as white, or, at sunrise/sunset, as yellow because of B (T )= λ λ3 exp(hc/λkT) 1 − scattering (which we’ll come back to in a later lecture). and assume that λ<hc/kT. Then setting dB λ ( T ) /d λ =0 we find λpT=0.290 cm K for the peak of the blackbody curve. For the Sun, whose surface temperature is T=5777 K, this implies λp≈5000 Å, or roughly a green color. 9 10 The relation between energy kT in eV and temperature T in K is particularly useful in high-energy astronomy: 5 kT (eV) = 8.617 10− T (K) ⇥ T (K) = 1.161 104 kT (eV) ⇥ Therefore X-rays with a wavelength of 10 Å and an energy of 1.24 keV may have been emitted from a blackbody with a temperature of ~1.4×106 K! 11-1 Note that the visible/optical range covers <1 decade in wavelength! The electromagnetic Note also that the names and boundaries of the bands are historical and spectrum poorly defined... 11-2 The electromagnetic spectrum 11-3 The electromagnetic spectrum 12 Approximate EM bands in astronomy Band λstart λend Telescopes Ground Radio ~1 cm WSRT, LOFAR @ ~2m Millimeter 1 mm 10 mm ALMA, JVLA Submillimeter 0.2 mm 1 mm ALMA Infrared 1 µm 0.2 mm near-infrared (NIR) 1 µm 2.5 µm ground-based Ground Space! mid-infrared (MIR) 2.5 µm 25 µm Spitzer, JWST far-infrared (FIR) 25 µm 200 µm (0.2 mm) Herschel Ground Optical 3100 Å 1 µm ground-based, HST visible ~4000 Å ~8000 Å eye Ultraviolet (UV) ~500 Å 3100 Å near-ultraviolet (NUV) 2000 Å 3000–3500 Å GALEX, HST Space far-ultraviolet (FUV) 900 Å 2000 Å GALEX, HST, FUSE extreme-ultraviolet (EUV) 500 Å 1000 Å EUVE X-ray 0.1 keV (100 Å) 200 kev (0.06 Å) XMM, Chandra γ-ray ~200 keV (0.06 Å) Fermi, INTEGRAL 13 Fluxes, filters, magnitudes, and colors For a point source – like an unresolved star – we can define the spectral flux density S(ν) as the energy deposited per unit time per unit area per unit frequency therefore S(ν) has units of erg s–1 cm–2 Hz–1 The actual energy received by a telescope per second in a frequency band Δν (the bandwidth) is P=Sav(ν)AeffΔν, where Aeff is the effective area of the telescope – which includes effects like telescope obscuration, detector efficiency, atmospheric absorption, etc. – and Sav(ν) is the average spectral flux density over the bandwidth 14 An example: bright radio sources have fluxes of 1.0 Jy (Jansky) at ν=1400 MHz near the 21 cm line of H. Then S(ν)=1×10–23 erg s–1 cm–2 Hz–1 (=1.0 Jy) If we observe a 1 Jy source with a single Westerbork telescope – diameter 25 m, efficiency ≈0.5 at this frequency – with a bandwidth of Δν=1.25 MHz, and assuming Sav(ν)= S(ν) over this bandwidth, the telescope will receive 23 1 2 1 P =1 10− erg s− cm− Hz− ⇥ 0.5 ⇡(12500 cm)2 1.25 106 Hz ⇥ 9⇥ 1 ⇥16 ⇥ 3 10− erg s− =3 10− W ⇡ ⇥ ⇥ 15 This is a tiny amount of power! It would take ~80% of the age of the Universe to collect enough energy to power a 100W lightbulb for 1 second! In reality, S(ν) and Aeff will (likely) not be constant over the bandwidth Δν, so we should really write ⌫2 P = S(⌫)Ae↵ d⌫ Z⌫1 16 The total power flowing across an area is called the flux density F, ⌫2 F = S(⌫)d⌫ Z⌫1 This is the “Poynting flux” in E&M It has units of erg s–1 cm–2 17 To find the luminosity, we multiply the flux density over the area of a sphere with a radius equal to the distance between the observer (us!) and the emitting object: r so that L=4πr2F over some bandwidth Δν=ν2–ν1. 18 The luminosity is therefore the total power of an object in some frequency range Δν. Note that we often use the term luminosity to mean the bolometric luminosity, the total power integrated over all frequencies. 19 This definition of luminosity assumes 1. the emission is isotropic – that is, the same in all directions 2. an average spectral flux density over the bandwidth If (2) is incorrect, we should write ⌫2 L =4⇡r2F =4⇡r2 S(⌫)d⌫ Z⌫1 20 Optical and near-infrared astronomers use magnitudes to describe the intensities of astronomical objects. To define magnitudes, it’s useful to know that NUV– optical–NIR detectors (usually) have a response proportional to the number of photons collected in a given time. 21 We can define a photon spectral flux density Sγ(ν), which is the number of photons (γ) per unit frequency per unit time per unit area. It is simply S(⌫) S (⌫)= γ h⌫ and has the units s–1 cm–2 Hz–1 22 The number of photons per unit time and unit area detected is then the photon spectral flux density times an efficiency factor that depends on frequency, integrated over all frequencies: 1 Fγ = Sγ (⌫)✏(⌫)d⌫ Z0 Here ε(ν) is the efficiency which includes all effects like the filter curve, detector efficiency, absorption and scattering of the telescope, instrument, and atmosphere, etc. 23 note: from here on, I mean log-base-10 for “log” and log-base-e for “ln” Consider two stars with fluxes Fγ(1) and Fγ(2) Then the magnitude difference between these stars is F (2) m m = 2.5 log γ 2 − 1 − 10 F (1) ✓ γ ◆ We use logarithms because human perception of intensity tends to be in logarithmic increments We’ll come back to the zeropoint of this scale shortly! Note that this definition defines the apparent magnitude, the magnitude seen by the detector 24 The coefficient of “–2.5” is important. It says that a ratio of 100 in fluxes (received number of photons) corresponds to a magnitude difference of 5 magnitudes If star 2 is 100 times brighter than star 1, it is 5 magnitudes “brighter” but actually 5 magnitudes less. Confusing, eh? 25 This means that a 1st magnitude (m=1) star is brighter than a 2nd magnitude star (m=2). By how much? Invert our equation for magnitudes: Fγ (2) 0.4(m m ) = 10− 2− 1 Fγ (1) So if m2–m1=1, then Fγ(2)/Fγ(1)=1/2.512... — a factor of ~2.51 in flux. 26 Some useful properties and “factoids” about magnitudes... The magnitude system is roughly based on natural logarithms: m m =0.921 ln(f /f ) 1 − 2 1 2 If ∆ f 1 , then ∆ m = m m 1.086∆f 2 − 1 ≈ so the magnitude difference between two objects of nearly-equal brightness is equal to the fractional difference in their brightnesses – i.e., a difference of 0.1 magnitudes is ~10% in brightness A factor of 2 difference in brightness is a difference of 0.75 magnitudes 27 Let’s return to our efficiency term ε(ν): we can write this as ✏(⌫)=f⌫ R⌫ T⌫ where f ν is the transmission of any filter used to isolate the (frequency) region of interest R ν is the transmission of the telescope, optics, and detector T ν is the transmission of the atmosphere (if any) 28 Let’s consider the filter term fν: the transmission of the filter can be chosen as desired (assuming the right materials can be found) so that a specific bandpass can be observed There are many filter systems (see next slide)..