Photons. Observational Astronomy 2018 Part 1 Prof. S.C. Trager

Transcription

1 Photons Observational Astronomy 2018 Part 1 Prof. S.C. Trager

2 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= 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

3 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 nearinfrared (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!

4 The energy of a photon is E=hν, where h= 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= erg

5 Some useful relations: (Hz) = E (erg) h (erg s 1 ) = E (ev) (Å) = c = hc 1 E (ev) = E 1 (ev 1 ) Therefore a photon with a wavelength of 10 Å has an energy of 1.24 kev

6 If a photon was emitted from a blackbody of temperature T, then the average photon energy is Eav~kT, where k = erg K 1 = ev K 1 is Boltzmann s constant. It is sometimes useful to know what frequency corresponds to the average photon energy: h kt (Hz) = T (K) or T = 1.44 cm K

7 Note that this wavelength isn t the peak of the blackbody curve. Consider the blackbody function B (T )= 2hc 3 and assume that λ<hc/kt. Then setting 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. 1 exp(hc/ kt) 1 db (T )/d =0

8 The relation between energy kt in ev and temperature T in K is particularly useful in high-energy astronomy: kt (ev) = T (K) T (K) = 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 ~ K!

9 The electromagnetic spectrum

10 The electromagnetic spectrum

11 The electromagnetic spectrum

12 Approximate EM bands in astronomy Band λstart λend Telescopes Radio ~1 cm WSRT, ~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 mid-infrared (MIR) 2.5 µm 25 µm Spitzer, JWST far-infrared (FIR) 25 µm 200 µm (0.2 mm) Herschel 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 Å GALEX, HST 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 Ground Space! Ground Space Ground

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)= 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 P = erg s 1 cm 2 Hz (12500 cm) Hz erg s 1 = 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 P = Z 2 1 S( )A e d

16 The total power flowing across an area is called the flux density F, Z 2 F = S( )d 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=4πr 2 F 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 L =4 r 2 F =4 r 2 Z 2 1 S( )d

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( ) 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: F = Z 1 0 S ( ) ( )d 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 Consider two stars with fluxes Fγ(1) and Fγ(2) Then the magnitude difference between these stars is F (2) m 2 m 1 = 2.5 log 10 F (1) 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 1 st magnitude (m=1) star is brighter than a 2 nd magnitude star (m=2). By how much? Invert our equation for magnitudes: So if m2 m1=1, then Fγ(2)/Fγ(1)=1/ a factor of ~2.51 in flux. F (2) F (1) = (m 2 m 1 )

26 Some useful properties and factoids about magnitudes... The magnitude system is roughly based on natural logarithms: m 1 m 2 =0.921 ln(f 1 /f 2 ) If, then f 1 m = m 2 m f 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 is the transmission of any filter used to isolate the (frequency) region of interest R is the transmission of the telescope, optics, and detector T 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)...

29 Two common filter systems

30 So the (apparent) magnitude difference between two objects is F,B (2) m B (2) m B (1) = B(2) B(1) = 2.5 log 10 F,B (1) where Z 1 Z 1 F,B = 0 S ( ) B ( )d = 0 S ( )f B R T d

31 We define the color of an object as the magnitude difference of the object in two different filters ( bandpasses ) if the filters are X and Y, then the color (X Y) is (X Y ) m X m Y = 2.5 log F,X F,Y

32 Most (but not all) magnitude systems are based on taking a magnitude with respect to a star with a known (or predefined) magnitude So to get a magnitude on system X, one observes stars with known magnitudes and calibrates the instrumental magnitudes onto the standard system We ll discuss this calibration process in great detail later in the course!

33 The Vega system defines a set of A0V stars as having apparent magnitude 0 in all bands of a system The Johnson-Cousins-Glass system is a Vega system, where the magnitudes of all bands in the system are set to 0 for an idealized A0V star at 8 pc Another common magnitude zeropoint system is the AB system, in which magnitudes are defined as m AB, = 2.5 log S( ) at a given frequency ν; see Fukugita et al. (1995) and Girardi et al. (2002) for more info.

34 Apparent magnitudes depend on the flux of photons received from a source; but this depends on the distance to the source! Remember that L=4πr 2 F, so for a given L, F r 2

35 To have a measurement of intrinsic luminosity, we must remove this distance dependence. We define the absolute magnitude M to do this: we choose a fiducial distance of 10 pc and define the distance modulus apple F (r) µ = m M = 2.5 log F (10 pc) r = 5 log 10 pc = 5 log r (pc) 5...ignoring absorption by dust and cosmological effects.

36 We define the absolute bolometric magnitude as the total power emitted over all frequencies expressed in magnitudes. We set the magnitude scale zeropoint to the (absolute) bolometric magnitude of the Sun, M =4.74 bol Thus M bol = 2.5 log L L where L = erg s 1 Solving for the luminosity of an object, then, we have L = M bol erg s 1 independent of the temperature (color) of the source.