After all the abstract modeling in my recent posts, I’m going to now switch over to a couple of practical problems (textbook exercises) on the subject of black-body radiation.
These
exercises come from Chapter 7 of Lasers, a textbook by Peter Milonni and
Joseph Eberly, published in 1988. This is a good book for learning physical optics—the
theory of the interaction of visible and near-visible light with matter—at the
junior or senior level. These two problems are actually sort of 2nd semester physics-freshman-level problems, unlike most other problems in the book.
Problem
7.5 Calculate the total intensity, over all
wavelengths, of light from the sun (a) at the earth’s surface and (b) at the
surface of Pluto. Assume that the sun is a blackbody of temperature 5800 K and
that the orbital distances are RE = 150 x 106 km
and RP = 5900 x 106 km.
Problem
7.6 Estimate the temperature of a blacktop road on a sunny
day. Assume the asphalt is a perfect blackbody.
These
exercises are related, since you need to know the "total intensity" of the sun’s
radiation at Earth’s surface to calculate the temperature of the
idealized-perfect-absorber asphalt pavement. This "intensity," which is more properly called flux (power/area), is given on page 232
of M&E’s text as 1.4 kW/m2, but that’s actually the amount of
sunlight striking the top of the atmosphere. The answer to 7.5 (a) and 7.6 really should be based on the lesser amount of
radiation reaching the bottom of the atmosphere, which is what I'll use when I get to that calculation.
Long-Winded Solution to Problem 7.5
We
have in this problem the inverse of what we have in the usual discussion of
black-body radiation. The usual discussion refers to the energy density of
black-body radiation in a cavity, which we can think of as a spherical cavity
inside a solid block of a material like carbon (sorry, too general) graphite, maintained at temperature T. Any cavity shape gives the same black-body
spectrum--an interesting exercise in itself, but I'll have to come back to it. Also, of course, there's the issue of Kirchhoff's radiation law that says the material from which the cavity is made is immaterial. I've chosen a spherical cavity which can be thought of as hollowed out of a large cube of graphite (pencil "lead") because it’s easy to visualize.
In equilibrium, the
energy density (watts/meter3) of the radiation is the same
throughout the cavity, and if viewed from any point in the cavity (except a point on the
wall) the radiation intensity doesn’t depend on direction. In other words,
black-body cavity radiation is homogeneous and isotropic.
In
this problem, however, a spherical cavity is replaced by a spherical object
with empty space surrounding it. The black-body problem has been inverted!
We don’t have a homogeneous radiation field, because energy density (well,
power density in this case) decreases with distance from the surface of the
sun. We still have isotropy, because radiant energy flow occurs at the same
rate in every direction away from the sun's surface. But, before solving this problem, it’s instructive to
consider a case where radiation is not emitted equally in all directions.
Imagine
we have instead of the sun a spherical piece of charcoal that isn’t yet fully
heated—it doesn’t yet have a uniform red glow, because different parts of it
are at different temperatures. We can’t use the Stefan-Boltzmann law,
Itotal
= σT4,
to
calculate the total intensity of the piece of charcoal, because we don't know the temperatures of the different parts. But there nevertheless is a total intensity, and this could
be calculated if you knew how the partial intensities depended on direction—if
the intensity as a function of “solid angle” was known. This in turn could be
measured with a bunch of photocells photometers of known surface area arranged on a
spherical surface surrounding the piece of charcoal. The total intensity is
then found by the integral of I(ν, θ, φ), where θ and φ
are the usual spherical polar coordinates, which when taken together constitute
a solid angle measured in steradians (one steradian = the area on a sphere's surface equal to the radius of the sphere squared).
So
now I’ve made my point—almost. The intensity of radiation emitted from a heated
surface of nonuniform temperature depends on direction and also will have a different
frequency dependence in different directions, meaning the peak of radiation
intensity will be at different frequencies for different temperatures (Wien
displacement law). The main point I want to make here is that the so-called “total
intensity” Itotal in the S-B law is an
example of radiant flux rather than intensity. Flux has units of energy per second per unit
area, while intensity is energy per second, per unit area, per unit frequency,
per unit solid angle in a given direction. (Okay, I got this definition from an astronomy text used for a class I took in 1977. Maybe the technical definitions are different now. I'll get back to you on that.) (November 25, I'm back: Still the same.)
In
this problem, since the sun has the idealized uniform surface temperature of
5800 K, we only have to integrate the
intensity as a function of frequency, I(ν).
But how is this related to the constant-temperature Planck spectral
energy density function, ρ(ν), which is our starting
point in this problem? Well, our starting point is really the S-B law. We
aren’t asked to integrate the Planck function and get the S-B law ourselves.
But
it’s helpful to write down the conversion of the integral of ρ(ν)
into the integral of I(ν) which is needed to get Itotal. This is what M&E show in their
equation 7.7.15, page 231,
Itotal
=
∫ I(ν)dν =
(c/4) ∫ ρ(ν)dν = σT4,
where
ρ(ν) is the Planck spectral energy density function at a single temperature, given by a particular curve of the several curves usually shown in
a black-body radiation graph. Itotal is just the area under this curve. So
you can now see where the S-B law comes from, except I haven’t shown the
calculation of the integral of ρ(ν). “Left to the reader,” but not a freshman physics/calculus calculation. See the lower part of the box discussing the Planck Radiation Formula at this Hyperphysics website to see where the above factor of c/4 comes from.
Our
given Itotal is the radiation flux at the surface of the sun,
the energy-per-second emitted by unit area of the sun’s photosphere. Once we find this number, how do we calculate the much smaller radiation flux at Earth’s
surface?
Well,
doesn’t
radiant "intensity" decrease as the inverse square of distance? Can’t we just divide Itotal by
the surface area of a sphere having radius equal to Earth’s distance from the
sun? Or by the Earth-Sun distance itself squared?
If you try to divide Itotal
by an area, or by a radius-squared, the units are wrong, for one thing. They
are power per meter-to-the-fourth-power. We need to find power per
meter-squared at Earth’s distance from the sun.
So maybe you’ve figured out that we first have to find the power, in watts, of solar radiation at the surface of the sun. This same amount of energy per second is flowing through the surface of a sphere with radius equal to the Earth-Sun distance, RE, but the power-per-unit-area is reduced by the factor (RS /RE)2.
So first multiply Itotal by the surface area of the sun, which is 4π(RS)2 = 4π(6.957x108) m2, to get the sun’s power output. Then divide this power output by the surface area of a sphere with radius equal to the Earth-Sun distance to get Iearth, i.e., the solar radiation flux hitting the top of Earth's atmosphere facing the sun at any time. Doing these two things at one time gives
Iearth = Itotal 4π(RS)2 / 4π(RE)2 = σT4(RS /RE)2
= (5.67 x 10-8watt/m2/K4)(5.8
x 103K)4 (6.957x108m / 1.5x1011m)2
= (5.67)(5.8)4 x
104 (4.64x10-3)2 watt/m2 =
(5,67)(1132)(21.5) x 10-2 watt/m2
= 1.38 x 103 watt/m2
This is the power-per-square-meter anywhere on the sphere with radius equal to the Earth-Sun distance, including any square meter at the top of Earth’s atmosphere. This is the expected answer as far as the textbook is concerned. But it's not hard to find the data on absorption and reflection of Earth's atmosphere.
According to a January 2009 posting by NASA (earthobservatory.nasa.gov),
At Earth's average distance from the Sun (about 150 million kilometers), the average intensity of solar energy reaching the top of the atmosphere directly facing the Sun is about 1,360 watts per square meter, according to measurements made by the most recent NASA satellite missions. … About 29 percent of the solar energy that arrives at the top of the atmosphere is reflected back to space by clouds, atmospheric particles, or bright ground surfaces like sea ice and snow. This energy plays no role in Earth’s climate system. About 23 percent of incoming solar energy is absorbed in the atmosphere by water vapor, dust, and ozone, and 48 percent passes through the atmosphere and is absorbed by the surface. Thus, about 71 percent of the total incoming solar energy is absorbed by the Earth system.
How do we use this information to calculate an
approximate asphalt surface temperature? We don’t want to include sea ice and
snow reflection, which is averaged into the overall reflected energy of 29%. I’ll
make the unwarranted assumption that sea ice and snow—of which there’s less now
than in 2009—constitute about 6% of that 29% reflection coefficient, and
subtract that 6%, and see what happens.
Thus we assume 23% reflection and 23%
atmospheric absorption, and that leaves 54% of the sun’s radiation flux at the top of
the atmosphere reaching our asphalt surface. Rounding up my 1.38 and NASA’s 1.36
to 1.4, we have
1.4 x 103
W/m2 x 0.54 = 756 W/m2
I’ll leave the Pluto calculation to the intrepid reader. Yeh, it’s not even considered a planet anymore, so who cares? Sunbathing Plutonians maybe.
Problem 7.6
Based on the above radiation flux at Earth's surface being our Itotal,
then
Itotal
= σT4
T4
= Itotal / σ
T4
= (756 W/m2) / (5.67x10-8 W/ m2K4
)
T4 = 1.33 x
1010 K4
T = 340 K
= 67°
C
= 153°
F.
Not hot enough to safely cook a hamburger (160° F), but hot enough, as you probably know, to burn your skin if you have prolonged contact with it. Asphalt isn’t actually a black-body, because it does reflect some radiation, so the temp should be lower than this. (How much lower?) If the reflection factor is less than the 23% I estimated, then the asphalt temp would be higher than 153° F. Maybe later I'll do a more realistic calculation based on these considerations.
Oh, yeah, there's also the effect of the tilt of Earth's axis of rotation relative to Earth's orbit around the sun that causes summers to be hot and winters to be cold. The calculation done here assumes the sun's rays hit the asphalt surface perpendicularly, as when the sun is directly overhead. That makes it a summertime calculation, and only applicable to the mid-latitudes.
