Transcript Spectroscopy – Lecture 1

Lecture 3
Radiative and Convective Energy Transport
I. Mechanisms of Energy Transport
II. Radiative Transfer Equation
III. Grey Atmospheres
IV. Convective Energy Transport
V. Mixing Length Theory
The Sun
I. Energy Transport
In the absence of sinks and sources of energy in the
stellar atmosphere, all the energy produced in the stellar
interior is transported through the atmosphere into outer
space. At any radius, r, in the atmosphere:
4pr2F(r) = constant = L
Such an energy transport is sustained by the temperature
gradient. The steepness of this gradient is in turn
dependent on the effectiveness of the energy transport
through the different layers
I. Mechanisms of Energy Transport
1. Radiation Frad (most important)
2. Convection: Fconv (important in cool stars like the Sun)
3. Heat production: e.g. in the transition between the solar
chromosphere and corona
4. Radial flow of matter: coronae and stellar winds
5. Sound waves: chromosphere and coronae
Since we are dealing with the photosphere we
are mostly concerned with 1) and 2)
I. Interaction between photons and matter
Absorption of radiation:
In + dIn
In
s
dIn = –knr In dx
tn =
kn : mass absorption coefficient
[kn ] = cm2 gm–1
s
∫ k r dx
n
Optical depth (dimensionless)
0
dtn = knrds
Convention: tn = 0 at the outer edge of the
atmosphere, increasing inwards
Optical Depth
In 0
In(s)
dIn = –In dt → In(s) = In0e–tn
The intensity decreases
exponentially with path length
If t = 1 → In = In ≈ 0.37 In0
e
Optically thick: t > 1
Optically thin: t < 1
We can see through the
atmosphere until tn ~1
Optical depth, t, of
ring material small
Optical path
length larger and
you see nebular
material as ring
Optical path
length small and
you see central
star and little
nebular material
Optical depth, t, of
ring material larger
Optical path
length roughly
the same one
sees disk and no
central star
Optical Depth
The quantity t = 1 has a geometrical interpretation in terms
of the mean free path of a photon:
t = 1 = ∫ krds = krs
s ≈ (kr)–1
s is the distance a photon will travel before it gets absorbed. In the
stellar atmosphere the abosrbed will get re-emitted and thus will
undergo a „random walk“. For a random walk the distance traveled
in 1-D is s√N where N is the number of encounters. In 3-D the
number of steps to go a distance R is 3R2/s2.
s
At half the solar radius kr ≈ 2.5 thus s ≈ 0.4cm so it
takes ≈ 30.000 years for a photon do diffuse
outward from the core of the sun.
Emission of Radiation
In + dIn
In
dx
jn
dIn = jnr In dx
jn is the emission coefficient/unit mass
[ ] = erg/(s rad2 Hz gm)
jn comes from real emission (photon created) or from
scattering of photons into the direction considered.
II. The Radiative Transfer Equation
Consider radiation traveling in a direction s. The change in the
specific intensity, In, over an increment of the path length, ds, is
just the sum of the losses (kn) and the gains (jn) of photons:
dIn = –knr In + jnr In ds
Dividing by knrds which is just dtn
dIn
= –In + jn/kn = –In + Sn
dtn
The Radiative Transfer Equation
tn appears alone in the previous equation therefore try solutions
of the form In(tn) = febtn. Differentiating this function:
dIn
df
bt
bt
n
n
= f be + e
dtn
dtn
= bIn + ebtn df
dtn
Substituting into the radiative transfer equation:
bIn + ebtn df = – In + Sn
dtn
The Radiative Transfer Equation
The first two terms on each side are equal if we set b = –1 and
equating the second term:
e–tn
df
dtn = Sn
tn
or
f = ∫ Snetndtn + c0
t is a dummy variable
0
tn
In(tn) = febtn =e–tn∫Sn(tn)etn dtn+ c0e–tn
0
Set tn = 0 → c0 = In(0)
bringing e–tn inside the integral:
tn
In(tn) =∫Sn(tn)e–(tn–tn) + In(0)e–tn
0
tn–tn
In(0)→
tn=0
tn
tn
At point tn the original intensity, In(0) suffers an exponential
extinction of e–tn
The intensity generated at tn, Sn(tn) undergoes an extinction of
e–(tn–tn) before being summed at point tn.
tn
In(tn) =∫Sn(tn)e–(tn–tn) + In(0)e–tn
0
This equation is the basic intergral form of the radiative transfer
equation. To perform the integration, Sn(tn), must be a known
function. In some situations this is a complicated function, other
times it is simple. In the case of thermodynamic equilibrium (LTE),
Sn(T) = Bn(T), the Planck function. Knowing T as a function of x or tn
amounts to a solution of the transfer equation.
Radiative Transfer Equation for Spherical Geometry
After all, stars are spheres!
z
To observer
dIn
knrdz = –In + Sn
dz
rdq
q
r
dr
dIn =
dz
x
y
∂In dr
∂In dq
∂z dz + ∂q dz
Radiative Transfer Equation for Spherical Geometry
Assume In has no f dependence and
dr = cos q dz
r dq = –sin q
∂In cos q – ∂In sin q = –I + S
n
n
k
rdr
k
rr
∂q
∂r
n
n
This form of the equation is used in stellar interiors and the
calculation of very thick stellar atmospheres such as
supergiants. In many stars (sun) the photosphere is thin thus
we can use the plane parallel approximation
Plane Parallel Approximation
To observer
q
ds
The increment of path
length along the line of
sight is ds = dx sec q
To center of star
q does not depend on z so there is no second term
dIn
= –In + Sn
cos q
knrdr
Custom to adopt a new depth variable x defined by dx = –dr. Writing dtn for knrdx:
dIn
cos q
= In – Sn
dtn
The optical depth is measured along x and not along the line of
sight which is at some angle q. Need to replace tn by –tnsec q. The
negative sign arises from choosing dx = –dr
Therefore the radiative transfer equation becomes:
tn
In(tn) =–∫Sn(tn)e–(tn–tn)sec q sec q dtn
c
The integration limit c replaces In(0) integration constant because the
boundary conditions are different for radiation going in (q > 90o) and
coming out (q < 90o).
In the first case we start at the boundary where tn=0 and work
inwards. So when In = Inin, c=0.
In the second case we consider radiation at the depth tn and deeper
until no more radiation can be seen coming out. When In=Inout, c=∞
Therefore the full intensity at the position tn on the line of sight
through the photosphere is:
In(tn) = Inout(tn) + Inin(tn) =
∞
= ∫Sne–(tn–tn)sec q sec q dtn
tn
tn
– ∫Sne–(tn–tn)sec q sec q dtn
0
Note that one must require that Sne–tn goes to zero as
tn goes to infinity. Stars obviously can do this!
An important case of this equation occurs at
the stellar surface:
Inin(0) = 0
∞
Inout(0) = ∫Sn(tn)e–tnsec q sec q dtn
0
Assumption: Ignore radiation from the rest of the universe (other
stars, galaxies, etc.)
This is what you need to compute a spectrum. For the sun which
is resolved, intensity measurements can be made as a function
of q. For stars we must integrate In over the disk since we
observe the flux.
The Flux Integral
Fn =
∫
In cos q dw
p
Fn = 2p
∫I
n
cos q sin q dq
0
p/2
= 2p
∫
0
Inout cos q sin q dq + 2p
p
∫
Inin cos q sin q dq
p/2
Assuming no azimuthal (f) dependence
The Flux Integral
Using previous definitions of Inin and Inout
p/2 ∞
Fn = 2p
∫∫
0
p
– 2p
tn
tn
∫∫
p/2 0
Sne–(tn–tn)sec q sin q dtn dq
Sne–(tn–tn)sec q sin q dtn dq
The Flux Integral
If Sn is isotropic
∞ p/2
∫ ∫
Fn = 2p Sn
0
tn
tn
p
∫ ∫
– 2p Sn
0
e–(tn–tn)sec q sin q dtn dq
e–(tn–tn)sec q sin q dtn dq
p/2
Let w = sec q and x = tn – tn
p/2
∫
0
∞
e–(tn–tn)sec q sin q dtn dq =
∫
1
e–xw
w2
dw
The Flux Integral
Exponential Integrals
∞
En(x) =
∫
1
e–xw
wn
dw
tn
∞
∫
Fn(tn) = 2p Sn E2(tn – tn)dtn – 2p
tn
∫ S E (t
n
2
n
– tn)dtn
0
In the second integral w = –sec q and x = tn – tn . The limit
as q goes from p/2 to p is approached with negative values of
cos q so w goes to ∞ not –∞
The Flux Integral
The theoretical spectrum is Fn at tn = 0:
∞
Fn(0) = 2p
∫ S (t ) E (t )dt
n n
2 n
n
Fn is defined per
unit area
0
In deriving this it was assumed that Sn is isotropic. It most
instances this is a reasonable assumption. However, in stars
there are Doppler shifts due to photospheric velocities, stellar
rotation, etc. Isotropy no longer holds so you need to do an
explicit disk integration over the stellar surface, i.e. treat Fn
locally and add up all contributions.
The Mean Intensity and K Integrals
tn
∞
∫
∫
Jn(tn) = 1/2 Sn E1(tn – tn)dtn + 1/2 Sn E1(tn – tn)dtn
0
tn
tn
∞
∫
∫
Kn(tn) = 1/2 Sn E3(tn – tn)dtn + 1/2 Sn E3(tn – tn)dtn
tn
0
The Exponential Integrals
Exponential Integrals
∞
En(x) =
∫
1
∞
En(0) =
∫
1
e–xw
wn
dw
wn
dw
1
=
1–n
∞
dEn
dx =
∫
1
1 d e–xw
dw = –
wn dx
∞
∫
1
dEn
dx = –En–1
e–xw
wn–1
dw
The Exponential Integrals
Recurrence formula
En+1(x) = e–x –xEn(x)
The Exponential Integrals
For computer calculations:
–xEn(x)
From Abramowitz and Stegun (1964)
E1(x) = –ln x – 0.57721566 + 0.99999193x – 0.24991055x2 +
0.05519968 x3 – 0.00976004x4 + 0.00107857x5 for
E1(x) =
x4 + a3x3 + a2x2 + a1x + a0
x4 + b3x3 + b2x 2 + b1x + b0
a3 = 8.5733287401
b3= 9.5733223454
a2= 18.0590169730
b2= 25.6329561486
a1= 8.6347608925
b1= 21.0996530827
a0= 0.2677737343
b0= 3.9584969228
1
xex
x≤1
x >1
Assymptotic Limit:
1
n
En(x) = xex 1 – x
Polynomials fit E1 to an error less than 2 x 10–7
≈
1
xex
[
+ n(n+1)
– …
2
x
[
E1(x) =
e–x
Radiative Equilibrium
• Radiative equilibrium is an expression of conservation of energy
• In computing theoretical models it must be enforced
• Conservation of energy applies to the flow of energy through the
atmosphere. If there are no sources or sinks of energy in the
atmosphere the energy generated in the core flows to the outer
boundary
• No sources or sinks in the atmosphere implies that the divergence
of the flux is zero everywhere in the photosphere.
In plane parallel geometry:
d
dx F(x) = 0
∞
or F(x) = F0
A constant
F(x) = ∫F0 (tn)dn =F0 for flux carried by radiation
0
Radiative Equilibrium
tn
∞
∫ [ ∫ S E (t
n
0
2 n
– tn)dtn –
tn
∫ S E (t
n
2
n
– tn)dtn
0
[
∞
dn = F0
2p
This is Milne‘s second equation
It says that in the case of radiative equilibrium the solution of
the radiative transfer equation is found when Sn is known that
satisfies this equation.
Radiative Equilibrium
Other two radiative equilibrium conditions come from
the transfer equation :
dIn
cos q
= knrIn – knrSn
dx
Integrate over solid angle
d
dx
∫I
n
cos q dw = knr
∫ I dw – k r ∫ S dw
n
n
n
Substitue the definitions of flux and mean intensity in the first and
second integrals
dFn
= 4pknrJn – 4pknrSn
dx
Radiative Equilibrium
Integrating over frequency
∞
d
dx
∫
∞
Fn dn = 4pr
0
∫
∞
knJn dn – 4pr
0
0
But in radiative equilibrium the left side is zero!
∞
∞
∫ k J dn = ∫ k S dn
n n
n n
0
Note: the value of the flux
constant does not appear
∫
0
knSn dn
Radiative Equilibrium
Using expression for Jn
tn
∞
∫ S E (t
n
1 n
– tn)dtn + ½
n
0
n
1 n
1
n
– tn)dtn
tn
∞
∫ k [½ ∫ S E (t
n
0
tn
∞
∫ S E (t
∫
[
Jn(tn) = ½
– tn)dtn + ½ Sn E1(tn – tn)dtn dn = 0
tn
First Milne equation
0
Radiative Equilibrium
If you multiply radiative equation by cos q you get the K-integral
∫
cos2
dIn
q
dw =
dx
∫ k rI
n
n
cos q– knrSn cos q dw
2nd moment
1st moment
∞
∞
∫
0
dKn
F0
dtn dn = 4p
∞
∫ [ ∫
0
tn
∫
[
And integrate over
frequency
d ½ S E (t – t )dt + ½
F0
dn
SnE3(tn – tn)dtn
=
n 3 n
n
n
dtn
4p
0
tn
Third Milne Equation
Radiative Equilibrium
The Milne equations are not independent. Sn that is a solution for
one is a solution for all three
The flux constant F0 is often expressed in terms of an effective
temperature, F0 = sT4. The effective temperature is a fundamental
parameter characterizing the model.
In the theory of stellar atmospheres much of the technical effort
goes into iterative schemes using Milne‘s equations of radiative
equilibrium to find the source function, Sn(tn)
III. The Grey Atmosphere
The simplest solution to the radiative transfer equation is to assume that
kn is independent of frequency, hence the name „grey“. It occupies an
„historic“ place and is the starting point in many iterative calculations.
Electron scattering is the only opacity source relevant to stellar
atmospheres that is independent of frequency.
Integrate the basic transfer equation over frequency and denote:
∞
I =0∫Indn
∞
S =0 ∫Sndn
∞
J =0 ∫Jndn
dI
cos q
= –I + S
dt
∞
F =0 ∫Fndn
∞
K =0 ∫Kndn
Where dt = krdx, the „grey“
absorption coefficient
The Grey Atmosphere
This grey case simplifies the radiative equilibrium and Milne‘s equations:
F(x) = F0
J=S
dK
F0
=
dt
4p
The Eddington approximation, hemispherically isotropic outward and
inward specific intensity:
Iout(t) for 0 ≤ q ≤ p/2
I(t) =
Iin(t) for p/2 ≤ q ≤ p
The Grey Atmosphere
p/2
J=
1
4p
∫
I dw = 1 Iout
2
∫
sin q dq +
0
1
=
2
Iout (t) + Iin(t)
F(t) = p
Iout (t) – Iin(t)
K(t) =
1
6
p
1 in
I
2
∫ sin q dq
p/2
Iout (t) + Iin(t)
But since the mean intensity equals the source function in this case
S(t) = J(t) = 3K(t)
The Grey Atmosphere
We can now integrate the equation for K:
dK
F
= 0
dt
4p
K(t) =
F0
4p
t +
F0
6p
Where the constant is evaluated at t = 0. Since S = 3K we get Eddington‘s
solution for the grey case:
S(t) =
3F0
(t + ⅔)
4p
The source function varies linearly with optical depth. One gets a similar
result from more rigorous solutions
III. The Grey Atmosphere
Using the frequency integrated form of Planck‘s law: S(t) = (s/p)T4(t) and
F0 = sT4 the previous equation becomes
¼
T(t) = ¾ (t + ⅔) Teff
At t = ⅔ the temperature is equal to the effective temperature and T(t)
scales in proportion to the effective temperature.
Note that F(0) = pS(⅔), i.e. the surface flux is p times the
source function at an optical depth of ⅔.
III. The Grey Atmosphere
Chandrasekhar (1957) gave a complete and rigorous solution of the grey
case which is slightly different:
S(t) =
3F0
[t + q(t)]
4p
¼
T(t) = ¾ [t + q(t)] Teff
q(t) is a slowly varying function ranging from 0.577 at t = 0 and 0.710 at t = ∞
Eddington:
„This, however is a lazy way of handling the problem and it is not surprising
that the result fails to accord with observation. The proper course is to find
the spectral distribution of the emergent radiation by treating each
wavelength separately using its own proper values of j and k.“
IV. Convection
In a star the heat flux must be sufficiently great to transport all
the energy that is liberated. This requires a temperature
gradient. The higher the energy flux, the larger the temperature
gradient. But the temperature gradient cannot increase without
limits. At some point instability sets in and you get convetion
In hot stars (O,B,A) radiative transport is more efficient in the
atmosphere, but core is convective.
In cool stars (F and later) convective transport dominates in
atmosphere. Stars have an outer convection zone.
r2
IV. Convection
P2
r2 *
P2*
DT
Dr
Consider a parcel of gas that is perturbed
upwards. Before the perturbation r1* = r1
and P1* = P1
For adiabatic expansion:
Dr
P = rg
g = CP/CV = 5/3
Cp, Cv = specific heats at constant V, P
r1
P1
r1 *
P1 *
After the perturbation:
P2 = P2*
r2 * = r1
P2* 1/g
P1*
( )
IV. Convection
Stability Criterion:
Stable: r2* > r2 The parcel is denser than its surroundings and
gravity will move it back down.
Unstable: r2* < r2 the parcel is less dense than its surroundings and
the buoyancy force will cause it to rise higher.
IV. Convection
Stability Criterion:
r2 *
r1 =
P2
P1
dP
1/g
( )
=
{
P1 + dr Dr
P1
r2 * = r1 +
1/g
}
1
g
(
1/g
{
r1 dP
P dr
) Dr
dr
r2 = r1 +
dr Dr
1
g
(
r1 dP
P dr
)
>
}
1 dP
= 1 + P dr Dr
dr
dr
This is the Schwarzschild criterion for stability
IV. Convection
By differentiating the logarithm of P =
k
m
rT we can relate
the density gradient to the pressure and temperature gradient:
dT
dr
( )star > (
1
1– g
)
T
P
dP
dr
( ) star
The left hand side is the absolute amount of the temperature gradient of the
star. Both gradients are negative, so this is the algebraic condition for stability.
The right hand side is the „adiabatic temperature gradient“. If the actual
temperature gradient exceeds the adiabatic temperature gradient the layer is
unstable and convection sets in:
dT
dr
dT
dr
( )star < ( ) ad
The difference in these
two gradients is often
referred to as the „Super
Adiabatic Temperature
Gradient“
The Convection Criterion is related to gravity
mode oscillations:
Brunt-Väisälä Frequency
The frequency at which a bubble of gas may oscillate vertically with
gravity the restoring force:
N2
=g
(
1
g
dlnr
dlnP
–
dr
dr
)
g is the ratio of specific heats = Cv/Cp
g is the gravity
Where does this come from?
r2
The Brunt-Väisälä Frequency
P2
r2 *
P2*
DT
Dr
r1
P1
r 2 = r1 +
dr
dr Dr
(
r1 dP
P dr
) Dr
Difference in density between inside the
parcel and outside the parcel:
Dr
r1 *
P1 *
r2 * = r1 +
1
g
1
Dr = g
=
=
1
g
r
(
r
dP
P dr
(
r dlnP
dr
(
1 dlnP
g dr
) Dr
–
dr
dr Dr
)
dlnr
r
Dr –
dr Dr
)
dlnr
r
Dr –
dr Dr
A
=
(
1 dlnP
g dr
)
–
dlnr
dr
Dr = rADr
Buoyancy force:
FB = –DrVg
V = volume
FB = –grAVDr
w2 = k/m
For a harmonic oscillator: FB = –kx
In our case x = Dr, k = grAV, m ~ rV
N2 = grAV/Vr
= gA = g
(
1 dlnP
g dr
–
dlnr
dr
)
The Brunt-Väisälä Frequency is the just the harmonic oscillator frequency of a
parcel of gas due to buoyancy
V. Mixing Length Theory
Convection is a difficult problem for which we still have no good theory. So
how do most atmospheric models handle this complicated problem? By
reducing it to a single free parameter whose value is left to guess work. This
is mixing length theory. Physically it is a piece of junk, but at this point there
is no alternative although progress has been made in hydrodynamic modeling.
Simple approach:
1. Suppose the atmosphere becomes unstable at r = r0 → mass
element rises for a characteristic distance L (mixing length) to r + L
2. Cell releases excess energy to the ambient medium
3. The cell cools, sinks back, absorbs energy and rises again
In this process the temperature gradient becomes shallower than in the purely
radiative case.
V. Mixing Length Theory
Recall the pressure scale height:
H = kT/mg
The mixing length parameter a is given by
L
a=
H
a = 0.5 – 1.5
V. Mixing Length Theory
The convective flux is given by:
F = rCpvDT
v is the convective
velocity
½rv2 = gDrL = gDraH
Dr = DT
r
T
→ v ≈ g½(aH)½(DT/T)½
gaH ½ 3/2
F = rCp T DT
Or in terms of the temperature gradients
F = rCp
½
g
2
(aH)
T
dT
dx
–
cell
3/2
dT
dx
average
How large are the convection cells? Use the scale height to
estimate this:
Scale height H = kT/mg
The Sun: H~ 200 km
A red giant: H ~ 108 km
Schwarzschild (1976)
Schwarzschild
argued that the size
of the convection
cell was roughly the
width of the
superadiabatic
temperature
gradient like seen in
the sun.
In red giants this has
a width of 107 km
Convective Cells on the Sun and Betelgeuse
~107 km
Betelgeuse = a Ori
From computer
hydrodynamic
simulations
1000 km
The Sun
Numerical Simulations of Convection in Stars:
http://www.astro.uu.se/~bf/movie/movie.html
http://www.aip.de/groups/sternphysik/stp/box_simulation.html
With no rotation:
With rotation:
• cells small
• large total number on surface
• motions and temperature differences
relatively small
• Time scales short (minutes)
• Relatively small effect on integrated
light and velocity measurements
• cells large
• small total number on surface
• motions and temperature
differences relatively large
• Time scales long (years)
• large effcect on integrated light and
velocity measurements
RV Measurements from
McDonald
AVVSO Light Curve