Theoretical Astrophysics Flashcards
1
Q
spectrum
A
shows emission as a function of wavelength or frequency
2
Q
a hot, dense object produces
A
continuous spectrum (blackbody)
3
Q
a hot, tenuous gas produces
A
emission line spectrum
4
Q
a cooler, tenuous gas overlying a hotter dense object produces
A
an absorption line spectrum
5
Q
tenuous
A
not dense
6
Q
example of continuous spectra
A
cosmic microwave background spectrum
7
Q
example of emission line spectra
A
solar emission spectra
8
Q
example of absorption line spectrum
A
solar spectrum and Fraunhofer lines
cooler gas in the upper photosphere and lower chromosphere overlying hotter gas in the lower photosphere
9
Q
a spectral line results from the
A
transition of an electron between two discrete energy states in an atom
10
Q
chemical composition of an object can be identified from
A
the presence of spectral lines
11
Q
the shape/profile of spectral lines tell us
A
the properties of the atom
the properties of the emitting or absorbing material
eg: temperature, pressure, speed and magnetic field
12
Q
an emission line is characterised by:
A
its total (integrated) intensity
its central wavelength
its width (full width at half max/equivalent width)
its shape/profile
13
Q
examples of emission line shapes
A
gaussian
lorentzian
elliptical
14
Q
I(λ) describes
A
the intensity of the radiation at wavelength λ
15
Q
I(λ) d(λ) is
A
the amount of radiation in infinitesimal wavelength range λ to λ+dλ
16
Q
the total emission in the line is found by
A
integrating
I total = ∫ I(λ) dλ
17
Q
absorption lines are characterised by
A
central wavelength and shape
line width and depth is combined into ‘equivalent width’
18
Q
equivalent width of absorption line
A
draw rectangle extending from the continuum to zero intensity, with area equal to the total area of the absorption line
width of rectangle = equivalent width
19
Q
changing I(λ) to I(v)
A
I(λ)dλ = I(v)dv
I(λ)|dλ/dv| = I(v)
using λ=c/v
dλ/dv = -c/v^2 = -λ^2/c
I(v)=λ^2/c I(λ)
20
Q
units of I(λ)
A
wm^-2sr^-1m^-1
change m^-1 to Hz^-1 for I(v)
21
Q
why do we take absolute value of dλ/dv
A
as wavelength increases, frequency decreases
absolute value to ensure energy is positive
22
Q
a spectral line corresponds to
A
the transition of an electron in the atoms/ions of a gas between two energy levels/states
23
Q
if energy level is well-defined, we might expect the line to
A
have a very well defined frequency and therefore be infinitely narrow
i.e. photons emitted/absorbed at a single wavelength
24
Q
what are natural and collisional broadening due to?
A
finite lifetime of electrons in atomic states
25
what are thermal and rotational broadening due to?
motion of atoms, on microscopic and macroscopic scale
26
heisenberg's uncertainty principle
the time an electron typically stays in a state and the uncertainty in its energy linked by Heisenberg's uncertainty principle
deltaEdeltat > or = h bar /2
27
𝜏rad
natural or radiative lifetime which depends on the atomic structure and the quantum mechanical selection rules
average time between emission of photons by a single atom
27
natural broadening
from heisenberg approx
deltaE delta t = h bar
sub in E=hv
28
why is energy uncertainty of lower level neglected
lower level always more stable than the upper level
uncertainty much smaller so neglected in comparison
28
typically an electron sits in an upper state for time 𝜏rad before...
making a spontaneous transition downwards, radiating a photon
28
natural broadening
intrinsic property of an atomic transition line
always there but very small in comparison to other broadening effects.
28
lorentz profile equation
A is einstein coefficient
vo is line-centre frequency
29
what does 𝜏rad determine
whether the transition is forbidden or allowed
29
shape of a naturally broadened line
lorentz profile
29
lorentzian intensity diagram
longer wings
pointier peak
30
the average number of spontaneous transitions per second per atom from level j (higher) to level i (lower)
Aji = 1/𝜏rad
units of per second
Aji is the einstein A coefficient
31
allowed transition
transition with 𝜏rad< or = 10^-8s
32
forbidden transition
transition with 𝜏rad>10^-8s
33
why are allowed reasonance lines strong?
many photons emitted per second as the electrons are not in the upper level for long
34
why are forbidden transitions allowed to happen?
other factors come into play that can lead to strong emission of forbidden lines
in particular collisional de-excitation
35
well-known astrophysical allowed transitions
h alpha, Ly alpha
transitions of neutral H
36
well-known astrophysical forbidden transitions
S II, O III
37
A_ji can be used to calculate
power radiated in a spectral line
38
a downwards transition from level j to level i produces
a photon of energy E_ji
39
there are ____ transitions per second per atom
A _ji
40
energy emitted per second per atom
=A_jiE_ji
(where E_ji=hv_ji)
41
power radiated in spectral line by N_j atoms in state j is
P_ji=N_jA_jiE_ji
**important that it is not just N**
42
there are continuous excitations between all energy levels by
collisions with electrons and the absorption of photons
43
In a gas of N atoms total, not all atoms can be in an excited state unless
artificially kept there, ie in a laser
**N_j
44
the upper states of forbidden spectral lines are long-lived but can be
perturbed by encounters with nearby particles, and de-excite
45
collisions reduce the lifetime of
the upper state compared to its natural lifetime
so the line is broadened
46
collisional lifetime,
average time between collisions that lead to de-excitation
47
collisional lifetime depends on
speed of the perturbing particles
the density of particles
the interaction cross-section
48
coll lifetime - speed of particles
higher speed means more collisions per second
49
coll lifetime - density
denser gas means a higher chance of encountering another particle
50
coll lifetime - interaction cross-section
larger cross section means a larger effective area presented by the particle
51
increased temp - coll lifetime
decreased collisional lifetime
higher T, higher energy of particles, more collisions, less time between
52
average time between collisions - single particle - visual
particle moving in space
area of cross-section is sigma
length of cylinder = vt so in 1 second =v
volume v sigma
number density n so one second collides with nV=nvsigma other particles
53
to get actual collisional timescale, have to take into account
particle is moving randomly among all surrounding particles which are also moving
mean speed
54
collisional broadening is an example of
pressure broadening
55
pressure broadening
general term given to line broadening resulting from any kind of interaction between particles
56
if a radiating particle is moving with some velocity v along the line-of-sight then the observer sees
a red-shifted or blue-shifted photon
57
wavelength or frequency shift is given by
delta lambda / lambda0 = deltav/v0=v/c
this is non-relativistic Doppler shift
where v=observed frequency , v0=rest frequency (same for wavelength)
58
if there is a large number of radiating particles, all moving in different directions at different speeds, then
each photon emitted appears in a different part of the line profile
59
the total line profile is obtained by
summing over emission from all contributing particles
60
thermal broadening summary
ensemble of emitting particles with random directions/speeds --> total radiation is a combination of emission from all atoms --> observer sees broadened line
61
thermal broadening
if the emitting gas is hot and the particles have a random, small-scale motion the line will be broadened
62
thermal broadening - line width depends on
average speeds of the particles
63
doppler broadening
thermal and rotational
64
assumption made in rotational broadening
omega same at equator and poles i.e. solid body rotation
omega=constant
65
z-axis rotational broadening
rotation axis
66
y-axis rotational broadening
line of sight direction
67
the velocity at any point on the surface of the star at a distance r from its rotation axis is
v= omega x r
gets (-omegay, omegax,0)
68
line of sight speed at projected distance x
vlos = omega x
69
strips of the stellar disk at the same x
all have the same vlos
70
vlos along central meridian
0
71
doppler shift from all points at projected distance x from the rotation axis of the star
deltavx/v0 = vx-v0/v0 = vlos/c = omegax/c
72
maximum doppler shift from each end of the equator
delta v max/v0 = veq/c = omega Rstar /c
73
the amount of emission at each frequency v0+deltavx is proportional to
the length of the chord at x from the rotation axis
74
if the star is spherical then
there is the same amount of emission at v0-deltavx and v0+deltavx
75
calculating maximum rotational line broadening
need 2delta lambda to account for redshift and blueshift
76
the line profile, I(v) has the shape of a
half-ellipse
77
if we integrate up all of the intensity in the line profile this gives
the line flux
Ftot
78
integrating intensity is equivalent to
finding area under the hlaf-ellipse
area of half ellipse is piab/2
79
rotational broadening is also detectable in
spectral lines from distant galaxies
however profile is not usually simple ellipse due to complicated distribution of material in galaxies
more typical shape has 'horns'
80
angular speed of material travelling in own galaxy depends on
the distance from the galactic centre
(galaxy does not rotate as a solid body)
81
at the sun, distance R0 from galactic centre, angular speed is
Ω(R0)
82
at point P, at a distance R, angular speed is
Ω(R)
83
velocity of sun at radius R0
Vs=Ω(R0)R0
84
component of Vs along line of sight
Vs sin gamma
85
LOS speed depends on
difference in angular speeds
distance from sun
observing position
86
if galaxy was a solid body, would have Ω(R)-Ω(R0)=
0
and vp=0
87
if spins in H atom are both in same direction, the energy is
slightly higher than if they are anti-parallel
88
hyperfine splitting
the two possible energy levels for the ground state of neutral hydrogen
89
transition between the two states in ground state of neutral H
spin flip transition
90
why do we see loads of neutral H even though forbidden
very abundant in the galaxy so 21cm line is very strong
91
doppler shift and rotational broadening of the 21cm line can be measured easily because
radiative lifetime is very long so natural broadening is very small
in ISM neutral H is cold so thermal broadening negligible
densities are low in ISM so collisional broadening negligible
interstellar dust does not absorb 21cm radiation (optically thin)
92
a system
a fixed group of atoms or molecules under study
this can be a single particle or many particles
it can be closed or open
93
a state
a condition of the system at a particular time that can be described by a parameter or set of parameters
94
degenerate states
distinct states (ie different quantum numbers) which have the same energy
95
thermodynamic equilibrium
the condition of thermal, radiative, mechanical and chemical balance
no net flows of energy, matter or net changes of phase
(ie no large scale changes with time)
96
the probability, p, of a system in thermodynamic equilibrium at temp T and in a state with energy E is proportional to
the boltzmann factor
p prop to e^-E/kT
97
if the system can be in one of n different states with equal probability p, then the sum of the probabilities
must equal 1
ie is normalised
98
degeneracy is the number of
distinct ways that a system can occupy a state of given energy
99
the probability of a system being in state i is directly proportional to
the number of ways state i, with energy Ei can be occupied
100
if there are 3 ways to arrange g=
3
101
pi=
a gi e^-Ei/kT
a=constant of proportionality
gi=degeneracy
e^-Ei/kT = boltzmann factor
102
evaluating a and pi for a simple 2 level system
degeneracies g1=g2=1
plug these into p equation
p1+p2=1 and degeneracies same so
a[e^-E1/kT + e^-E2/kT]=1
sub in a into p1, set E2-E1=deltaE and divide top and bottom by exp(-E1/kT)
same for state 2
103
evaluating a and pi for a simple 2 level system
if deltaE>>kT then
exp(-deltaE/kT) approx = 0
so p1 approx =1, p2 approx=0
(at low temp, system is in its lowest energy state)
104
evaluating a and pi for a simple 2 level system
if deltaE<
exp(-deltaE/kT) approx = exp(0) =1
so p1=1/2 and p2=1/2
(at high temps, both states are equally populated)
105
evaluating a and pi for a degenerate 2 level system
three ways to arrive at total (p_e) spin of stste E2
g2=3 and g1=1
sub in
summing p1+p2 and rearrange for a
sub a back in for p1 and p2
106
evaluating a and pi for a degenerate 2 level system - at temps characteristic of H I clouds, we have delta E << kT so
exp(-deltaE/kT) approx =1
so p1 = 1/4 and p2=3/4
107
evaluating a and pi for a degenerate multi-level system
pi has sum on denominator which is the normalisation constant, summed over all possible levels
108
ratio of probabilities
ratio of numbers of atoms
makes maths easier as constants cancel
109
the emission from an astrophysical abject is produced by
particles - electrons, ions and neutrals
110
way to quantify number of particles at a given position or velocity
using a particle distribution function
111
reading particle distribution functions
y-axis value is prop to number of particles in the state given on x axis
112
probability density function f(x) is
the probability that some property of a particle takes value x
(x could be position, velocity etc)
113
f(x) must be
normalised
such that a particle lies somewhere in the range of possible values xmin to xmax
integral between xmin and xmax of f(x)dx =1
114
if f(x) is normalised, the fraction of particles between x1 and x2 is
integral between x1 and x2 of f(x)dx
115
the mean value of a property q(x) is calculated from
the integral between x min and x max of q(x)f(x)dx
if f(x) is normalised
116
if all particles have equal mass, the mean energy of the particles is
117
mean square speed normalisation
integral between + and - infinity f(vx)dvx
=c integral between + and - infinity of e^-mvx^2/2kT dvx
(note g=1 so not included)
c=normalisation constant
118
to evaluate mean square speed need to set a to
m/2kT
119
particle has energy per degree of freedom of
1/2kT
120
the more particles in dvx...
the more emission in dv
121
gaussian line profile varies like
exp(-delta v)^2
122
FWHM is calculated by
evaluating (v-v0) where I(v)=I0/2
123
FWHM is also called
the thermal width of the line
124
FWHM of line from 1D distribution of atoms with boltzmann distribution of energies is obtained by
I(v)=I0/2
which is true if exp(-mc^2/2kt (v-vo)^2/v0^2)=1/2
take logs of both sides and rearrange for (v-v0)
thermal width is 2(v-v0)
125
to determine the distribution function of v we need
to find the degeneracy of states with speed v
then normalise
126
degeneracy of velocity states
count how many allowed combinations of (vx,vy,vz) correspond to the same total speed v
127
for particle confined in box length L
allowed states of particle correspond to
allowed wavelengths that will fit in the box (resonances)
lambda=2L/n correspond to de broglie h/p
128
derivation of vn=n pi hbar/m L
pn=h/lambdan = nh/2L
v=p/m so vn=pn/m = nh/2mL
sub in h bar = h/2pi
129
energy of state n
sub in vn to 1/2mv^2
can be used to evaluate the boltzmann factor
130
if the 1D box gets bigger such that L tends to infinity
spacing in energy between states decreases, tending towards a continuum of states
all values of 1D speed are allowed and each corresponds to a single classical state
131
in 3D velocity space, the velocity states form
a regular lattice with a uniform density of states
132
a thin shell in velocity space of speed v, thickness dv has volume
dV where
dV=4piv^2 dv
133
the density of states is uniform so the number of states in range v, v+dv is
proportional to v^2 dv
134
the number of states with the same speed v(ie same total energy) is
the degeneracy of the state of speed v
this is prop to v^2
135
the fraction of particles in each state is prop to
the boltzmann factor of that state
136
the number of states in range v to v+dv is prop to
v^2 dv
137
the maxwell boltzmann distribution
area1=area2 for normalised distribution
very high and very low speeds are improbable
most particles cluster around some characteristic speed
distribution is not symmetric - it has a more extended high energy tail
138
the MB distribution can be written as
f(v)dv = normalisation constant x degeneracy x BF
degeneracy=v^2
139
approach for normalising the MB distribution
1. obtain the reduction formula that links In and In-2
2. evaluate I0 and I1
140
normalising the MB distribution
integral is of standard form x^ne^-ax^2
integrate by parts
bit in brackets =0 because at 0 first part=0, at infinity second part=0
plug back into find C
141
most of the particles are at a speed shown by
the peak in the distribution function
142
how to evaluate most probable speed
finding turning point of MB function = where gradient=0
143
trivial solutions to turning point for most probable speed
v=0 and v=infinity
these correspond to the extreme ends of the MB distribution, we want peak in middle
144
the FWHM of a thermally-broadened line is related to the
speed that most of the particles have in thermal equilibrium
145
the mean speed is found by evaluating
integral between 0 and infinity of v f(v) dv
sub in MB equation
using reduction formula, evaluate I3
sun back in
146
calculating mean square speed
similar to mean speed but using I4 in reduction formula
already have I0 from before so sub in
147
rms speed
the sqaure root of the mean square speed
148
mean energy
149
working out fraction of particles in tail only works if
speed is significantly greater than the most probable speed
150
calculating fraction of particles in tail
integral between ve and infinity of v^2exp(-mv^2/2kT)dv
change variable to x=mv^2/2kT
x^1/2varies slowly compared to e^-x for large x so take out as constant
151
lower limit xe=mve^2/2kT is
the ratio of the kinetic energy of the particle to a typical thermal energy
152
to use equation for fraction of particles in tail, need to
check that we are in the limit of large x
153
a blackbody is an object that
absorbs all radiation incident upon it and also re-radiates it all
the matter and radiation are then in thermodynamic equilibrium
154
classical theory of cavity radiation - Rayleigh-Jeans approach
cavity with walls at T and small aperture where radiation can enter/exit
radiation inside is emitted/absorbed by walls and has blackbody spectrum
aperture samples the radiation inside cavity and emergent radiation also bbody
155
radiation in cavity vs astrophysical object - walls of cavity
lots of particles in thermodynamic equilibrium at T, ie in a MB distribution
156
radiation in cavity vs astrophysical object = heating of the cavity
there is a source of radiation at the centre of the object
157
radiation in cavity vs astrophysical object - radiation reflecting from the walls of the cavity
photons emitted by the source are absorbed and re-emitted by the particles
158
radiation in cavity vs astrophysical object - radiation leaving aperture in the cavity
eventually the photons reach the 'edge' of the object and leave
159
equation for the electric field E(x,t) of a standing electromagnetic wave in 1D
E(x,t)=E0 sin(2pix/lambda)sin(2pimu t)
160
for reflection, E(x,t) must be
zero at the walls
ie sin(2pix/lambda)=0 at the walls
161
setting x=0 and x=a as each end of cavity. Then reflection means
2a/lambda = n
162
the integer mode number n can be obtained in terms of
frequency and the box size
163
each n corresponds to
a different allowed frequency mode
164
a small number dn of modes corresponds to
a small range of frequency dv
165
the number of modes in range v to v+dv is given by
the frequency range dv divided by the frequency spacing between allowed frequencies
since n is an integer, modes are separated by n=1
166
in 3D, nx,ny and nz define
a grid of points in n-space uniformly distributed at integer values
each point corresponds to an allowed 3D standing wave
167
the number of allowed frequencies in v to v+dv is equal to
the number of points between shells of radii n and n+dn in n-space
this has volume dV_n
168
modes are separated by integers so density of modes =
1
169
the voume dV_n of the spherical shell, radius n is
4pi n^2 dn
where n=2av/c
170
there is an independent mode of propagation in which
E and B are rotated through 90 degrees
171
any polarisation of the wave can be formed by the
sum of these two independent modes of propagation
172
two independent waves per frequency so
for each frequency, g=2
photon view: there are two independent photon spin states
173
to find the specific energy density
combine equation for number density of states with equation for equipartition of energy and divide by the volume
174
how to find the total energy density emitted by a blackbody
integrate specific energy density u(v) over all frequencies
175
ultraviolet catastrophe
integrating using the rayleigh jeans law gives infinite energy density
RJ has excellent agreement at low frequencies and high temps but fails towards higher frequencies
176
RJ limit
hv << kt
177
specific energy density units
Jm^-3 Hz^-1 (or m^-1 if in terms of wavelength)
178
Planck's theory of cavity radiation: the problem
assumption that classical equipartition of energy could describe the average energy for the allowed frequency modes
179
Planck's theory of cavity radiation: the solution
for a blackbody, the average energy of the standing waves is a function of the frequency
180
if the degeneracy in independent of E then the average energy is
integral E P dE / integral P dE
181
planck's idea
treat energy as a discrete quantised variable instead of a continuous variable
182
average energy of a mode with frequency v
sum of E.BF/sum of BF
insert E-nhv
set x=BF
expand sum
simplify using maclaurin series
183
in the quantum approach, the average energy is a function of
T and v instead of being a constant kT
184
can calculate the total energy density of the blackbody radiation using
planck's solution
change variables
use standard integral
185
regimes for planck function: Rayleigh jeans law
usually valid at radio an IR wavelengths
BF approx = 1+hv/kT
(set x=hv/kt and use maclaurin expansion for e^x)
186
regimes for planck function: Wien's law
valid at x-ray and gamma wavelengths
1/BF-1 approx = BF
187
if star's spectrum can be approximated by bb then power per unit area emitted by the star is
B=sigma T^4
sb law
188
total power radiated by surface of star is
L=4piR^2sigmaT^4
surface area x radiated flux
189
if planet has radius R then the power absorbed by it is
P=(1-a)FpiR^2
incoming flux x area
190
a
albedo
quantifies the power reflected by the planet
(1-a) gives fraction of absorbed power
191
assuming planet radiates as a BB in equilibrium so Pin=Pout then Pout=
4piR^2sigma T^4
surface area x radiated power per unit area
192
temperatures of planets in the solar system
venus much higher than expected - runaway greenhouse effect
the moon has big range due to no atmosphere
193
habitable zones
distance from the star where the planets could have liquid water
194
it is necessary to assume models for atmospheric composition to account for
albedo
greenhouse effect
195
refining the calculation to look at a small patch of the planet's surface
projected area as seen by the star is
dAp=dAcosB
196
local temp on planet surface - 3 assumptions
1. local value of planetary albedo is a so a fraction 1-a of the power arriving is absorbed
2. heat is not conducted away, and the area re-radiates freely
3. The area re-radiates as a blackbody.
197
local temp depends on
the inclination angle B of the location on the surface of the planet relative to the star
198
an atmosphere can be either
transparent (optically thin) or opaque (optically thick)
199
scattering
photons change direction
elastic: no change in photon energy
inelastic: change in photon energy
200
refraction
the direction of the propagation of the radiation changes
201
absorption
photons are absorbed by the medium and their energy is transferred to the medium
202
absorption process
any process that removes radiation from the line of sight of the observer
ie scattering
203
how well a target can scatter/absrob EM radiation is often described through its
cross-section which represents an effective area presented by the target particle to the radiation
204
cross section usually depends on
v, particle properties and the process of interest
205
thomson scattering
scattering by free electrons
if hv<
206
resonant scattering (spectral lines)
photon interacts with a bound electron exciting it to a high level
energy of transition matches the photon energy
excited electron de-excites, re-emitting photon with hv in a random direction
207
rayleigh scattering
photon interacts with bound electron but photon energy too small to excite
scattering is by particles smaller than wavelength of photon
causes interstellar reddening
208
thomson scattering - oscillating electric field of incident photon causes
free charged particle to oscillate
209
thomson scattering - oscillating particle emits
radiation at the same frequency as incident wave so wave scattered
210
thomson cross section is the same for
all wavelengths
211
resonant scattering - photon interacts with bound electron and its frequency matches
the frequency of a transition
212
resonant scattering - cross section depends on
photon energy since this must match energy of transition
213
rayleigh scattering - scattering particle are much smaller than
wavelength of radiation
prop to lambda^-4
214
longer wavelengths are scattered
less than shorter
(red scattered less than blue)
215
ionisation potential
energy necessary to free an electron from the atom
216
kinetic energy of ejected electron
E-Ei
217
recombination
at later time, electron can be recaptured by another ion
218
an equilibrium is reached in the H cloud when
ionisation rate Ni equals recombination rate Nr
219
stromgren radius
radius of the sphere around a star inside which the material is kept ionised by the star's radiation
220
stromgren radius assumptions
region in photon-ionisation equilibrium
H plasma fully ionsides
region spherical
221
units of specific intensity
Wm^-2Hz^-1sr^-1
222
in free space (ie no processes affect radiation) the specific intensity is
conserved along a ray
dI/dr=0
223
absorption coefficient
relates interaction cross section and the number density of particles in the medium
av=nsigmav
224
to evaluate the overall absorption coefficient, need to
consider all relevant processes
av = n1sigma1 +n2sigma2
225
the total optical depth of a gas measures
what fraction of the incoming radiation is removed from the line-of-sight by scattering or absorption
226
optical depth tells us if
the medium is transparent or opaque
low optical depth = transparent
227
for a uniform medium the optical depth is
Tv=avL
L is distance along the ray path
228
source function
Sv=jv/av
229
observed specific intensity has three contributions
1. radiation incident on medium and exponentially attenuated by it by the optical depth
2. contribution by medium itself from source function
3. exponential absorption of the medium's own radiation
230
optically thin - the probability that the photon is absorbed or scattered is
very low
typically the distance between scatterings is bigger than the size of the cloud
231
optically thick - the probability that the photon is absorbed or scattered is
very high
distance between scatterings much smaller than the size of the cloud
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