Assignment 3 Due April 8
1. Classical theory tells us that each degree of freedom associated with an oscillator in a blackbody should have energy E = kT . Quantum theory tells us this can be wrong if it requires
that kT be smaller than the universal quantum of energy E = h⌫. Classical theory also tells
us that the energy emitted by a blackbody increases exponentially with frequency according
to the Rayleigh-Jeans law. Instead we measure, and can derive from quantum theory, the
Planck function: the Planck function has the long tail of the Rayleigh-Jeans law at low frequencies, but unlike the Rayleigh-Jeans law, a sudden drop-off at high frequencies. Show
that the wavelength where this drop-off starts to occur (i.e. near where there is a maximum in
the Planck function) is roughly where classical theory would (impossibly) require kT ⌧ h⌫.
It may help to consider Wien’s law and look at a Planck function curve for guidance. Consider two cases, the temperature of the sun (T = 6000 K), and the temperature of the earth
(T = 300 K). Keep in mind you aren’t required to explicitly solve the Planck function. Your
answer shouldn’t be very long.
What you need to find here is the wavelength that satisfies h⌫ = hc/ < kT for the earth and
the sun, and relate this to the wavelength using the exact expression from Wien’s Law max =
2897/T . You will find that that > max but is roughly the same order of magnitude. The
Planck equation is different than the Rayleigh-Jeans law precisely because kT cannot be less
than h⌫.
2. In parts of the atmosphere where LTE does not hold, would you expect the temperature
associated with molecular translational motions (our normal concept of temperature) to be
warmer or colder than the temperature associated with the internal energy transitions within
the molecules? Why?
For Kirchoff’s law to apply (that is if the absortivity is unity, so is the emissivity) the precondition is that the atmosphere is in LTE. Non-equilibrium conditions hold in regions of
the atmosphere that are too rarified to redistribute absorbed energy to kinetic molecular
motion. Such conditions can occur in the upper atmosphere, and in situations of extreme
energy inputs, can be manifest as the aurora. The physical effect is that the atmosphere is
colder than what one would expect from molecular absorption alone. Emission is larger
than absorption, so the radiative loss of “heat” energy is particularly high.
3. Show that the equilibrium temperature of the surface of the moon is 273 K assuming it has
an albedo of 0.08.
2
2
4⇡Rm
T 4 = ⇡Rm
(1
A)S
The value of S is the same for the moon as the earth, give or take. The answer should be 273K
4. Much of what we see in the atmosphere is the consequence of multiple scattering. In other
words, light gets scattered by atmospheric scatterers many times before it reaches the observer. A good example of this is that the blue sky looks white (or whitish) if looking toward
the horizon. Thinking about the effect of multiple scattering, try to explain why by drawing
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a picture. You should draw a diagram with the sun, air molecules, and you, showing the impact of multiple scattering on the color of light that reaches your eye. Start by representing
light from the sun as being two thick rays of equal intensity (thickness) of blue and red, and
track their trajectories for a scenario where the rays encounter multiple scattering events, but
the blue is scattered most efficiently, and both final rays come straight down to your eyes.
Despite blue being preferentially scattered, the final intensities should be equal too.
Consider the image below. If light is directly scattered to your eye from a single scattering
event, the color will look predominantly blue. But if the density of particle is sufficiently high
that it is likely that there will be an additional scattering event, the blue light that would
have been scattered to your eye is then scattered away from the direction your eye is looking.
Thus, what your eye sees is a more uniform combination of red and blue – or white
Figure 1: Why the sky is white
5. The average distance between water molecules is about 3 ⇥ 10 10 m. Now, calculate the approximate radius, and number of molecules in that radius, of a water droplet that no longer
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behaves as a Rayleigh scatterer (i.e.
dependence) with respect to visible 0.5 µm wavelength light, because the dipoles no longer vibrate together in phase and there is destructive
interference in the backscatter direction (i.e. ✓ = ⇡). You should use the equation
=
2⇡r
(1
cos✓)
where ✓ is the angle from the forward scattering direction and
is the phase difference
between waves. You should find that the approximate number of molecules across is 1000
for a distance of about 0.1 µm.
A molecule is no longer a Rayleigh scatter when there is destructive interference in the
backscatter direction and
=⇡
2
and
= 2⇡r/ (1 cos ✓)
2⇡r
=
(1 cos (⇡))
=
4⇡r
Solving
⇡ =
r =
4⇡r
/4
for 0.5 µm light you will get the largest size that makes for a decent Rayleigh scatterer.
r = /4 ' 0.5 ⇥ 10 6 /4 ⇠ 1 ⇥ 10
7
m
Therefore a droplet that is no longer a Rayleigh scatterer is approximately 1000 molecules
across.
6. If the particle no longer acts as a Rayleigh scatterer because it is too large, would you expect
the value of the exponent for wavelength to be smaller or larger than four? Why? Consider
the role of interference.
A molecule that is no longer a Rayleigh scatter will have a smaller exponent because interference cancels the sensitivity of scattering to wavelength
7. Provide an explanation for why the sky is blue that would be suitable for a young child but
fairly accurately reflects the physics that when light encounters a dipole, the dipole preferentially radiates (or scatters) the highest energy or frequency of the incident light.
Could be something like this. The power of sunlight makes the molecules of air that we
breath jiggle very fast. When the molecules jiggle they make their own light, and it is this
light that we see with our eyes. If the air didn’t make its own light we would see the black
of outer space to either side of the sun. Faster jiggles are more powerful and brighter, and
what we call the color blue. Slower jiggles that we call red are less powerful. We see both
red and blue in the sky, but mostly blue, because it has more power and energy.
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