Theoretical Astrophysics Flashcards
spectrum
shows emission as a function of wavelength or frequency
a hot, dense object produces
continuous spectrum (blackbody)
a hot, tenuous gas produces
emission line spectrum
a cooler, tenuous gas overlying a hotter dense object produces
an absorption line spectrum
tenuous
not dense
example of continuous spectra
cosmic microwave background spectrum
example of emission line spectra
solar emission spectra
example of absorption line spectrum
solar spectrum and Fraunhofer lines
cooler gas in the upper photosphere and lower chromosphere overlying hotter gas in the lower photosphere
a spectral line results from the
transition of an electron between two discrete energy states in an atom
chemical composition of an object can be identified from
the presence of spectral lines
the shape/profile of spectral lines tell us
the properties of the atom
the properties of the emitting or absorbing material
eg: temperature, pressure, speed and magnetic field
an emission line is characterised by:
its total (integrated) intensity
its central wavelength
its width (full width at half max/equivalent width)
its shape/profile
examples of emission line shapes
gaussian
lorentzian
elliptical
I(λ) describes
the intensity of the radiation at wavelength λ
I(λ) d(λ) is
the amount of radiation in infinitesimal wavelength range λ to λ+dλ
the total emission in the line is found by
integrating
I total = ∫ I(λ) dλ
absorption lines are characterised by
central wavelength and shape
line width and depth is combined into ‘equivalent width’
equivalent width of absorption line
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
changing I(λ) to I(v)
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)
units of I(λ)
wm^-2sr^-1m^-1
change m^-1 to Hz^-1 for I(v)
why do we take absolute value of dλ/dv
as wavelength increases, frequency decreases
absolute value to ensure energy is positive
a spectral line corresponds to
the transition of an electron in the atoms/ions of a gas between two energy levels/states
if energy level is well-defined, we might expect the line to
have a very well defined frequency and therefore be infinitely narrow
i.e. photons emitted/absorbed at a single wavelength
what are natural and collisional broadening due to?
finite lifetime of electrons in atomic states
what are thermal and rotational broadening due to?
motion of atoms, on microscopic and macroscopic scale
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
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
natural broadening
from heisenberg approx
deltaE delta t = h bar
sub in E=hv
why is energy uncertainty of lower level neglected
lower level always more stable than the upper level
uncertainty much smaller so neglected in comparison
typically an electron sits in an upper state for time 𝜏rad before…
making a spontaneous transition downwards, radiating a photon
natural broadening
intrinsic property of an atomic transition line
always there but very small in comparison to other broadening effects.
lorentz profile equation
A is einstein coefficient
vo is line-centre frequency
what does 𝜏rad determine
whether the transition is forbidden or allowed
shape of a naturally broadened line
lorentz profile
lorentzian intensity diagram
longer wings
pointier peak
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
allowed transition
transition with 𝜏rad< or = 10^-8s
forbidden transition
transition with 𝜏rad>10^-8s
why are allowed reasonance lines strong?
many photons emitted per second as the electrons are not in the upper level for long
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
well-known astrophysical allowed transitions
h alpha, Ly alpha
transitions of neutral H
well-known astrophysical forbidden transitions
S II, O III
A_ji can be used to calculate
power radiated in a spectral line
a downwards transition from level j to level i produces
a photon of energy E_ji
there are ____ transitions per second per atom
A _ji
energy emitted per second per atom
=A_jiE_ji
(where E_ji=hv_ji)
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
there are continuous excitations between all energy levels by
collisions with electrons and the absorption of photons
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<N for excited states j
the upper states of forbidden spectral lines are long-lived but can be
perturbed by encounters with nearby particles, and de-excite
collisions reduce the lifetime of
the upper state compared to its natural lifetime
so the line is broadened
collisional lifetime,
average time between collisions that lead to de-excitation
collisional lifetime depends on
speed of the perturbing particles
the density of particles
the interaction cross-section
coll lifetime - speed of particles
higher speed means more collisions per second
coll lifetime - density
denser gas means a higher chance of encountering another particle
coll lifetime - interaction cross-section
larger cross section means a larger effective area presented by the particle
increased temp - coll lifetime
decreased collisional lifetime
higher T, higher energy of particles, more collisions, less time between
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
to get actual collisional timescale, have to take into account
particle is moving randomly among all surrounding particles which are also moving
mean speed <v></v>
collisional broadening is an example of
pressure broadening
pressure broadening
general term given to line broadening resulting from any kind of interaction between particles
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
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)
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
the total line profile is obtained by
summing over emission from all contributing particles
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
thermal broadening
if the emitting gas is hot and the particles have a random, small-scale motion the line will be broadened
thermal broadening - line width depends on
average speeds of the particles
doppler broadening
thermal and rotational
assumption made in rotational broadening
omega same at equator and poles i.e. solid body rotation
omega=constant
z-axis rotational broadening
rotation axis
y-axis rotational broadening
line of sight direction
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)
line of sight speed at projected distance x
vlos = omega x
strips of the stellar disk at the same x
all have the same vlos
vlos along central meridian
0
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
maximum doppler shift from each end of the equator
delta v max/v0 = veq/c = omega Rstar /c
the amount of emission at each frequency v0+deltavx is proportional to
the length of the chord at x from the rotation axis
if the star is spherical then
there is the same amount of emission at v0-deltavx and v0+deltavx
calculating maximum rotational line broadening
need 2delta lambda to account for redshift and blueshift
the line profile, I(v) has the shape of a
half-ellipse
if we integrate up all of the intensity in the line profile this gives
the line flux
Ftot
integrating intensity is equivalent to
finding area under the hlaf-ellipse
area of half ellipse is piab/2
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’
angular speed of material travelling in own galaxy depends on
the distance from the galactic centre
(galaxy does not rotate as a solid body)
at the sun, distance R0 from galactic centre, angular speed is
Ω(R0)
at point P, at a distance R, angular speed is
Ω(R)
velocity of sun at radius R0
Vs=Ω(R0)R0
component of Vs along line of sight
Vs sin gamma
LOS speed depends on
difference in angular speeds
distance from sun
observing position
if galaxy was a solid body, would have Ω(R)-Ω(R0)=
0
and vp=0
if spins in H atom are both in same direction, the energy is
slightly higher than if they are anti-parallel
hyperfine splitting
the two possible energy levels for the ground state of neutral hydrogen
transition between the two states in ground state of neutral H
spin flip transition
why do we see loads of neutral H even though forbidden
very abundant in the galaxy so 21cm line is very strong
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)
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
a state
a condition of the system at a particular time that can be described by a parameter or set of parameters
degenerate states
distinct states (ie different quantum numbers) which have the same energy
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)
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
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
degeneracy is the number of
distinct ways that a system can occupy a state of given energy
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
if there are 3 ways to arrange g=
3
pi=
a gi e^-Ei/kT
a=constant of proportionality
gi=degeneracy
e^-Ei/kT = boltzmann factor
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
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)
evaluating a and pi for a simple 2 level system
if deltaE«kT then
exp(-deltaE/kT) approx = exp(0) =1
so p1=1/2 and p2=1/2
(at high temps, both states are equally populated)
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
evaluating a and pi for a degenerate 2 level system - at temps characteristic of H I clouds, we have delta E «_space;kT so
exp(-deltaE/kT) approx =1
so p1 = 1/4 and p2=3/4
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
ratio of probabilities
ratio of numbers of atoms
makes maths easier as constants cancel
the emission from an astrophysical abject is produced by
particles - electrons, ions and neutrals
way to quantify number of particles at a given position or velocity
using a particle distribution function
reading particle distribution functions
y-axis value is prop to number of particles in the state given on x axis
probability density function f(x) is
the probability that some property of a particle takes value x
(x could be position, velocity etc)
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
if f(x) is normalised, the fraction of particles between x1 and x2 is
integral between x1 and x2 of f(x)dx
the mean value <q(x)> 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
if all particles have equal mass, the mean energy of the particles is
<E>=1/2 m <vx^2>
</E>
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
to evaluate mean square speed need to set a to
m/2kT
particle has energy per degree of freedom of
1/2kT
the more particles in dvx…
the more emission in dv
gaussian line profile varies like
exp(-delta v)^2
FWHM is calculated by
evaluating (v-v0) where I(v)=I0/2
FWHM is also called
the thermal width of the line
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)
to determine the distribution function of v we need
to find the degeneracy of states with speed v
then normalise
degeneracy of velocity states
count how many allowed combinations of (vx,vy,vz) correspond to the same total speed v
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
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
energy of state n
sub in vn to 1/2mv^2
can be used to evaluate the boltzmann factor
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
in 3D velocity space, the velocity states form
a regular lattice with a uniform density of states
a thin shell in velocity space of speed v, thickness dv has volume
dV where
dV=4piv^2 dv
the density of states is uniform so the number of states in range v, v+dv is
proportional to v^2 dv
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
the fraction of particles in each state is prop to
the boltzmann factor of that state
the number of states in range v to v+dv is prop to
v^2 dv
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
the MB distribution can be written as
f(v)dv = normalisation constant x degeneracy x BF
degeneracy=v^2
approach for normalising the MB distribution
- obtain the reduction formula that links In and In-2
- evaluate I0 and I1
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
most of the particles are at a speed shown by
the peak in the distribution function
how to evaluate most probable speed
finding turning point of MB function = where gradient=0
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
the FWHM of a thermally-broadened line is related to the
speed that most of the particles have in thermal equilibrium
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
calculating mean square speed
similar to mean speed but using I4 in reduction formula
already have I0 from before so sub in
rms speed
the sqaure root of the mean square speed
mean energy
<E>=1/2m<v^2> = 3/2kT
</E>
working out fraction of particles in tail only works if
speed is significantly greater than the most probable speed
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
lower limit xe=mve^2/2kT is
the ratio of the kinetic energy of the particle to a typical thermal energy
to use equation for fraction of particles in tail, need to
check that we are in the limit of large x
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
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
radiation in cavity vs astrophysical object - walls of cavity
lots of particles in thermodynamic equilibrium at T, ie in a MB distribution
radiation in cavity vs astrophysical object = heating of the cavity
there is a source of radiation at the centre of the object
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
radiation in cavity vs astrophysical object - radiation leaving aperture in the cavity
eventually the photons reach the ‘edge’ of the object and leave
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)
for reflection, E(x,t) must be
zero at the walls
ie sin(2pix/lambda)=0 at the walls
setting x=0 and x=a as each end of cavity. Then reflection means
2a/lambda = n
the integer mode number n can be obtained in terms of
frequency and the box size
each n corresponds to
a different allowed frequency mode
a small number dn of modes corresponds to
a small range of frequency dv
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
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
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
modes are separated by integers so density of modes =
1
the voume dV_n of the spherical shell, radius n is
4pi n^2 dn
where n=2av/c
there is an independent mode of propagation in which
E and B are rotated through 90 degrees
any polarisation of the wave can be formed by the
sum of these two independent modes of propagation
two independent waves per frequency so
for each frequency, g=2
photon view: there are two independent photon spin states
to find the specific energy density
combine equation for number density of states with equation for equipartition of energy and divide by the volume
how to find the total energy density emitted by a blackbody
integrate specific energy density u(v) over all frequencies
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
RJ limit
hv «_space;kt
specific energy density units
Jm^-3 Hz^-1 (or m^-1 if in terms of wavelength)
Planck’s theory of cavity radiation: the problem
assumption that classical equipartition of energy could describe the average energy for the allowed frequency modes
Planck’s theory of cavity radiation: the solution
for a blackbody, the average energy of the standing waves is a function of the frequency
if the degeneracy in independent of E then the average energy is
integral E P dE / integral P dE
planck’s idea
treat energy as a discrete quantised variable instead of a continuous variable
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
in the quantum approach, the average energy is a function of
T and v instead of being a constant kT
can calculate the total energy density of the blackbody radiation using
planck’s solution
change variables
use standard integral
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)
regimes for planck function: Wien’s law
valid at x-ray and gamma wavelengths
1/BF-1 approx = BF
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
total power radiated by surface of star is
L=4piR^2sigmaT^4
surface area x radiated flux
if planet has radius R then the power absorbed by it is
P=(1-a)FpiR^2
incoming flux x area
a
albedo
quantifies the power reflected by the planet
(1-a) gives fraction of absorbed power
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
temperatures of planets in the solar system
venus much higher than expected - runaway greenhouse effect
the moon has big range due to no atmosphere
habitable zones
distance from the star where the planets could have liquid water
it is necessary to assume models for atmospheric composition to account for
albedo
greenhouse effect
refining the calculation to look at a small patch of the planet’s surface
projected area as seen by the star is
dAp=dAcosB
local temp on planet surface - 3 assumptions
- local value of planetary albedo is a so a fraction 1-a of the power arriving is absorbed
- heat is not conducted away, and the area re-radiates freely
- The area re-radiates as a blackbody.
local temp depends on
the inclination angle B of the location on the surface of the planet relative to the star
an atmosphere can be either
transparent (optically thin) or opaque (optically thick)
scattering
photons change direction
elastic: no change in photon energy
inelastic: change in photon energy
refraction
the direction of the propagation of the radiation changes
absorption
photons are absorbed by the medium and their energy is transferred to the medium
absorption process
any process that removes radiation from the line of sight of the observer
ie scattering
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
cross section usually depends on
v, particle properties and the process of interest
thomson scattering
scattering by free electrons
if hv«mc^2
cross section independent of frequency of photons
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
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
thomson scattering - oscillating electric field of incident photon causes
free charged particle to oscillate
thomson scattering - oscillating particle emits
radiation at the same frequency as incident wave so wave scattered
thomson cross section is the same for
all wavelengths
resonant scattering - photon interacts with bound electron and its frequency matches
the frequency of a transition
resonant scattering - cross section depends on
photon energy since this must match energy of transition
rayleigh scattering - scattering particle are much smaller than
wavelength of radiation
prop to lambda^-4
longer wavelengths are scattered
less than shorter
(red scattered less than blue)
ionisation potential
energy necessary to free an electron from the atom
kinetic energy of ejected electron
E-Ei
recombination
at later time, electron can be recaptured by another ion
an equilibrium is reached in the H cloud when
ionisation rate Ni equals recombination rate Nr
stromgren radius
radius of the sphere around a star inside which the material is kept ionised by the star’s radiation
stromgren radius assumptions
region in photon-ionisation equilibrium
H plasma fully ionsides
region spherical
units of specific intensity
Wm^-2Hz^-1sr^-1
in free space (ie no processes affect radiation) the specific intensity is
conserved along a ray
dI/dr=0
absorption coefficient
relates interaction cross section and the number density of particles in the medium
av=nsigmav
to evaluate the overall absorption coefficient, need to
consider all relevant processes
av = n1sigma1 +n2sigma2
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
optical depth tells us if
the medium is transparent or opaque
low optical depth = transparent
for a uniform medium the optical depth is
Tv=avL
L is distance along the ray path
source function
Sv=jv/av
observed specific intensity has three contributions
- radiation incident on medium and exponentially attenuated by it by the optical depth
- contribution by medium itself from source function
- exponential absorption of the medium’s own radiation
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
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
