Week 8--- Radiative Transfer and the T of Planets
I. Blackbody emission and absorption
An important service provided by the Sun is warming the planetary surfaces.
If all planets were perfect blackbodies, the only
complications for computing
their surface T would relate to whether
it is day or night, what is the angle of the sunlight,
and what are the heat transport timescales. Let us
begin with the simplest case-- a perfect blackbody with
no heat transport from place to place on the surface,
but rapid local attainment of radiative equilibrium.
In this situation, the surface T at any place would be
determined by an energy balance between the Stefan-Boltzmann
flux per unit area emitted based on the surface T, and the
incident flux per unit area from the Sun at whatever angle
in the sky it is seen from.
Right away we can conclude that the night side of a planet would
be incredibly cold, because the only incident light would be
from the distant stars and the CMB.
This is unrealistic, since it would need time to reach such cold
T, and internal heat transport would keep it from going down
that far, so we will not bother to calculate that situation.
Suffice it to say that the surface of the Moon at lunar night
has its T constantly dropping and gets very cold before the
Sun rises and warms it again.
However, we can reasonably calculate the surface T on the lunar
day side. First we find the flux per unit area from the Sun,
which depends on the flux per unit area perpendicular
to the ray to the Sun, times the foreshortening fraction at
the point of interest.
The foreshortening fraction takes into account that sunlight
coming from a low angle is spread over the surface of the
planet, so this multiplies the flux per area perpendicular to
the ray by the trigonometric factor cos(theta), where
theta is the angle of the Sun relative to the vertical (zenith).
Usually cos(theta) is renamed the "direction cosine" mu,
and that is what features in the "solid angle" dphi*dmu.
The flux per unit area at distance D from the Sun is
sigma*T_S^4*R^2 / D^2 where T_S is the surface T of the Sun,
and R is the radius of the Sun,
so after applying the foreshortening factor mu, this is
mu*sigma*T_S^4*R^2 / D^2, which we must equate to sigma*T^4 to
get the local T of the planet surface at angle theta to
the Sun. That solves to T = T_S * (R/D)^(1/2) / mu^(1/4).
Note this falls off relatively slowly with D, and has no dependence
on the size of the planetary surface, because it can be calculated
at any point on that surface independently of the planet size.
(This is why the Moon is a similar T to the Earth, and the common
idea that "space is cold" is untrue in the solar system.)
Of course the above result does not share any heat from place to
place, so is not appropriate to planets with atmospheres, or to
gas giants, both of which allow gases to carry heat from the warm
side to the cool side. Even Earth has enough atmosphere to experience
only minor day/night T variations, compared to the Moon, and Venus
shows almost no variation. So in these cases, it makes more sense
to compute the total rate that the planet absorbs sunlight, which
is sigma*T_S^4*R^2/D^2 * pi*r^2, where r is the radius of the planet
and we have used that pi*r^2 is the cross section the planet presents
to the Sun. Then the total rate the planet emits infrared blackbody
radiation is sigma*T^4*4pi*r^2, and equating those yields
T = T_S*(R/2D)^(1/2). Considering that T_S = 5800 K and for
the Earth, R/2D = 1/400,
gives for the Earth T = 290 K, which is pretty close to reality.
That it is so close is kind of a coincidence of the fact that the
two corrections to the blackbody case, the scattering albedo of the
Earth and the greenhouse effect, essentially cancel each other.
But given the current importance of the latter effect, let us
consider those corrections in more detail.
II. Scattering Albedo and the compensating greenhouse effect
The scattering albedo is the fraction of incident light that bounces
off the Earth rather than being absorbed by it. You know this isn't
zero if you've ever seen a photograph of the Earth from space, and
you can see other planets in the night sky. An important fact is that
the Earth scatters visible light much better than infrared light, so
the scattering albedo bounces some of the sunlight off, while
absorbing all the infrared. This contrast cools the Earth, because it
means the Earth is a good blackbody in the infrared where the
Stefan-Boltzmann law is used to calculate its emission rate, but it
does not absorb all the light from the Sun, it only absorbs a fraction
1-A where A is the scattering albedo. Including this correction in
the previous result gives T = T_S*(R/2D)^(1/2)*(1-A)^(1/4).
That would seem to make the Earth colder than it is, but it is compensated
by another contrast between the visible and infrared-- while the Earth's
surface is better at bouncing off visible light (which then does not go
into warming the surface), Earth's atmosphere (via greenhouse gases like
H2O and CO2) is better at absorbing infrared light than visible. So the
infrared blackbody emission from the surface is partially absorbed and
re-radiated back down, such that the Earth's surface receives light
from both the Sun and its own infrared-emitting air. If the optical
depth in the visible is included in A, then the optical depth in the
infrared can be called tau and included as an augmentation of the
incident light. Hence the full correction is more like (1-A+tau)^(1/4),
where tau is not exactly the optical depth of air in the infrared but
involves geometric corrections that are not significant and I will
leave them out because we don't have a clear definition of what tau
means anyway (it ends up depending a lot on wavelength in ways that
require a much more complete calculation to include correctly).
The bottom line is that the scattering albedo of the surface at visible
wavelengths is about the same as the optical depth of the atmosphere at
infrared wavelengths, to the two corrections largely cancel.
However, the rising optical depth due to human emissions of greenhouse
gases is creating global warming, a problem that will likely hound
younger generations for your entire lives, unfortunately.
The deeper problem is a general lack of use of critical thinking skills
in the population at large, which is not helped by the fact that
"radiative transfer" is actually a fairly tricky topic in astronomy.
III. Radiative transfer and the Brightness Theorem
Recall that for resolved sources, we define the "specific intensity"
I to be the energy per second per area per frequency per solid angle
along a ray. The "per frequency" is there simply because we have in
mind a spectrum, so of course doubling the frequency bin doubles the
energy flux. The important elements are the per area and per solid
angle, where the area is the area of a detector that must be arranged
perpendicular to the ray in question (if it is angled you must use
the foreshortened area, which is reduced by mu), and the solid angle
means you are breaking up the spatially resolved source into little
bins of solid angle and treating each with a different I. Of course
doubling the size of the detector, or doubling the solid angle bin
being detected, will double the energy flux, so that is why these
factors must be divided out to get a meaningful "brightness" I.
The first thing to recognize about I is that when you have a blackbody
emitter anchoring the end of some ray of interest,
the I along that ray (if nothing else happens to the light)
is equal to the
"Planck function" B_nu(T), so that is the connection between I and
the Planck function.
The second thing is that if the ray does not encounter any "opacity"
(meaning the light is not absorbed or scattered by any intervening
cross sections), then I stays the same-- this was the whole point
of defining I the way it is. This means that if you look at the Sun
from Mars, it has the same "brightness" I as from Earth, but it just
occupies a smaller solid angle in the sky. So Mars is colder not
because the Sun looks less bright in each of its resolved solid-angle
elements, but because it just looks smaller in the sky.
This rule also holds for lenses that refract and focus light, because
that only changes the angle of the rays, not the brightness along each
ray-- so when you look at the Sun through binoculars (bad idea), you
again see a Sun with the same I, but it looks larger (and hence you
get more energy focused into your eyes, which made it a bad idea).
But if you are orbiting Saturn, and look at the Sun through binoculars
that magnify it by a factor of 10, it will look just like the Sun does
from Earth. This is also true for mirrors that use "specular reflection",
since although it seems like a form of opacity, actually it again only
changes the angle of the rays, not the brightness I along each ray.
That only opacity can change I is called the brightness theorem, and is
the starting point of radiative transfer much like Galileo's law of
inertia is the starting point of mechanics (you first need to understand
what happens when nothing is happening before you can understand what
happens when something is happening).
IV. Radiative transfer with absorbing opacity
The "something" that can happen to a ray of light is it can be
absorbed and replaced by a different brightness. This happens little
by little along the ray, as the ray encounters opacity, so is described
by a differential equation for dI/dtau, where tau is called the optical
depth associated with the opacity.
When tau is small, we call it dtau, and it equals the small probability
dP that a photon will be absorbed as it crosses the distance dx associated
with dtau, and when tau is large, it equals the number of mean-free paths
the ray crosses.
The cross section per gram kappa is the fundamental property of
the intervening matter that controls this, leading to the expression
dtau = rho*kappa*dx. Since rho is mass density (and of course, the higher
the intervening mass density, the higher the optical depth), we can see
that tau is unitless, as required by the meaning given it.
When the ray crosses some optical depth tau that is comprised of purely
absorbing gas (like a blackbody at T=0, so no re-emission), it reduces
the brightness I according to the differential equation:
dI = - I*dtau , so dI/dtau = -I,
where dI is proportional to I because dtau is an absorption probability
per photon, so the more photons the more absorption.
The solution to this is I = I_o * e^(-tau), that is, purely absorbing
optical depth tau causes an exponential truncation of the initial
brightness I_o. So this is how the intensity of a laser beam would
drop off if you shined it through a fog.
If we allow
the absorbing opacity to function like a blackbody at a nonzero T,
then the opacity is essentially trying to replace I with its own
Planck function B, and its ability to make that replacement is in
proportion to the optical depth increment dtau, so we now have:
dI = - I*dtau + B*dtau = (B-I)*dtau , so dI/dtau = B-I.
The solution for this is I = I_o * e(-tau) + B * [1 - e^(-tau)],
where tau is the total optical depth encountered from where the
initial intensity I_o is given,
assuming B is constant. (If B varies, one must solve the
differential equation in detail, often numerically.)
What can make this solution tricky is that often the temperature,
and hence B, are ruled by the radiative transfer, so one must
solve for I and B self-consistently, rather than taking B as given
in advance when you solve for I.
Another wrinkle is that often tau is used as a global coordinate,
rather than the optical depth from the point where the ray
is "anchored", so that means the tau in the above expression must
be replaced by tau_o - tau, where tau_o is the optical depth to
the point where I_o is given, and tau is the global coordinate in
optical depth units (so tau is zero at the observer, not at the
start of the ray where I_o is given).
V. Radiative transfer with scattering opacity
The other thing that makes radiative transfer tricky is that the
opacity is not generally acting like a perfect blackbody, often
it includes some scattering. Scattering does not remove or add
light, it takes the brightness already there and redistributes
it over angle, so it depends not on T but on the I in all other
directions. This means you cannot solve for I along each ray
independently, you must do a self-consistent calculation of all
the rays at the same time, similar to how B must be calculated
self-consistently with all the I.
The way we handle this is by defining the "source function" S
to be the intensity that the opacity is trying to replace I with
(even though we don't necessarily know what S is yet), and we
write the "radiative transfer equation" by replacing the blackbody
B with the more general S, giving
dI/dtau = S-I .
The above is generally not the form of the "radiative transfer
equation" that you will see, however, because it is convenient
to treat tau as a global coordinate that measures the optical
depth along a vertical ray that starts at the observer and
penetrates into the source region, rather than coming out
through it.
Also, any ray tilted from the vertical by a direction cosine mu
will encounter a longer path length, and more optical depth, than
would a vertical ray along mu=1, by the factor 1/mu.
Hence to use the radial optical depth as our global coordinate,
we must take the above equation and replace its meaning of tau
with (tau_o - tau)/mu.
Making that replacement gives the usual form of the radiative
transfer equation:
-mu*dI/dtau = S-I ,
where the -mu is there because of the new meaning of tau that
appears in (tau_o - tau)/mu.
It is simply there because from now on, tau does not mean the
optical depth along the ray from where I_o is specified, it
means the optical depth coordinate measured along a ray with
mu=1 starting at the observer and pointing backward,
sort of like the way surface T is backward in an H-R diagram.
VI. Scattering in 1D
Since S is not generally known prior to solving the radiative transfer
equation self-consistently between I and S, especially when scattering
is present, a
great deal of work goes into solving it
in 3D.
This is beyond our current scope so we will content ourselves
with the much simpler version in 1D.
The key application for a simple understanding is looking at the
sky on a cloudy day, where thick clouds will make it dimmer outside,
and you may or may not be able to see the outline of the Sun directly
through the clouds, all depending on their "optical depth."
The key point is that in 1D, the action of pure scattering (which
is how we are treating the clouds) is to mix the two opposite
directions together, meaning that the action of the scattering is
to try to replace the downward I_d and upward I_u along any
ray with the combination (I_d + I_u)/2. That means
S = (I_d + I_u)/2 goes into -dI/dtau = S-I, which gives
-dI_d / dtau = (I_u - I_d)/2 and dI_u / dtau = (I_d - I_u)/2,
where note we use mu=-1 for downward, and mu=+1 for upward (and
we could improve accuracy by tilting the rays so that |mu|=0.7 or
some such thing, but let us not worry about improving accuracy a
little). Then we only need a boundary
condition at the top of the clouds to solve for both I_d and I_u
simultaneously as a 2x2 matrix of solutions (the matrix is not
diagonal because of the coupling between I_d and I_u induced by
the scattering).
The physical meaning of I_d and I_u are like downward and upward
contributions to the total radiative flux, and we will find in the
solution that if tau is large, a fraction ~1/tau of the downward flux
that hits the top of the cloud layer penetrates through the bottom
of the cloud layer, to giving us a gray but not dark day below.
What this treatment cannot do, being 1D, is explain the difference
between a ray that points directly to the Sun, and one that does not,
which comes up when the clouds are thin enough that you can still
see the outline of the Sun directly through them. For a ray pointing
to the Sun that allows us to directly see the Sun, we can treat scattering
as if it merely removes light from that (bright) ray, and go back to the
B=0 blackbody treatment above, yielding I = I_o * e^(-tau),
where I_o is the surface brightness of the Sun, and tau is the optical
depth of the cloud along the ray to the Sun.
Together, the above two solutions explain the two things we wish to
understand-- why when tau gets large we cannot directly see the Sun any
more, but the day is not dark-- this is because e^(-tau) falls off much
more steeply than does 1/tau.