Sunlight can heat asphalt to what temperature?

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 R_E = 150 x 10⁶ km and R_P = 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,

 

I_total = 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” I_total 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 I_total. This is what M&E show in their equation 7.7.15, page 231,

 

I_total = ∫ 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.  I_total 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 I_total 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 I_total 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 I_total 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, R_E, but the power-per-unit-area is reduced by the factor (R_S /R_E)².

So first multiply I_total by the surface area of the sun, which is 4π(R_S)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 I_earth, 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

 

I_earth = I_total 4π(R_S)² / 4π(R_E)² =  σT⁴(R_S /R_E)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 I_total, then

 

I_total = σT⁴

T⁴ = I_total / 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.

Indistinguishable vs distinguishable

This post relates to the two preceding posts concerning counting the number of microstates available to two quantum harmonic oscillators.

The total energy is shared by the oscillators, but the problem doesn’t say what the mechanism of energy transfer between oscillators is, or what kind of energy it is (acoustic = phonons, or electromagnetic = photons).  As stated, the problem is not physical. It’s merely a counting problem based on having two oscillators (boxes), one with i units of energy and the other with j energy units, and n’ total energy units (balls) to distribute in the oscillators. Plus there’s always the ZPE of one-half-a-ball in each of the two boxes. This ½-ball of ZPE in each box sums up to be one ball that can’t be counted when we are asked how many microstates there are, as shown by the energy constraint equation

i  +  j = n′ – 1

There’s also the undiscussed issue of the quantum energy elements being identical and indistinguishable. Using the example of n′ = 5, we have n′ – 1 = 4 indistinguishable balls to be distributed, which I did using a table in my post of Callen's problem 15.1-1. Now I’m going to use a different method of counting the number of microstates for indistinguishable balls and then use the same method to count states for distinguishable balls.

If we had four distinguishable balls we could label them ABCD, but in the quantum oscillator problem with n′ – 1 = 4, we have indistinguishable balls, which we can label AAAA. We can call them letters now instead of balls and use a divider | to show the number of ways these letters can be distributed in the two boxes (two oscillators), starting with zero letters in the left box and ending with zero letters in the right box:

 

|AAAA           A|AAA          AA|AA          AAA|A          AAAA|

 

It’s easy to see from this partitioning that there are only five microstates, because permutations of the letters are not distinguishable. This indistinguishability applies to energy elements in the two quantum oscillator problems, Callen's 15.1-1 and 15.1-2.

For a (supposedly) classical physics comparison, we can do a similar partitioning with four distinguishable letters ABCD, starting with the five arrangements as above

 

|ABCD           A|BCD          AB|CD          ABC|D          ABCD|.

 

But now we must also count the distinguishable permutations of the letters among the two boxes.

First we can visually count the number of permutations in each of the above arrangements. Only permutations or exchanges of letters between the boxes (across the partition) are counted. For multiple letters inside a box, their permutations among themselves are not counted as part of the total number of microstates. So we DO have indistinguishability at this single-box level in classical physics.

The first and last arrangements have zero balls in one box, so no exchanges between boxes are possible, and each of these arrangements represents ONE microstate. The second and fourth arrangements each have FOUR microstates, because each of the four letters can be permuted into the single-letter box. The middle arrangement’s permutations can be visually counted by showing all the possible distinguishable arrangements between the two boxes. There are SIX such arrangements:

 

AB|CD        AC|BD        AD|BC        CB|AD        DB|AC        CD|AB .

 

The total number of microstates is thus 2 + 8 + 6 = 16 when we consider the energy elements to be distinguishable.

Since 16 equals 2⁴, and we have two boxes and four distinguishable letters, this would seem to indicate a general way of calculating the number of microstates, namely mⁿ, where n is the number of distinguishable energy elements to be distributed in m oscillators. Nice formula! Where does it come?

Going back to four “balls” and two “boxes”, we know we have to put all four balls into the two boxes (because each arrangement must have the same total energy). How many ways are there to do this? There's no preference for putting any ball in either box, so there are two ways of placing the first ball, two ways of placing the second ball (independent of where the first ball is placed), and so on, giving 2x2x2x2 = 2⁴ ways of putting the four distinguishable balls in the two boxes. This is due to the randomness of the selection of the boxes. For each ball we can flip a coin to determine which box it goes in.

------------

This section was rewritten on 24 Oct 2020. In fact, this method will work for indistinguishable balls also as long as we have them lined up, as we must if we want to write them down:  AAAA.  They are distinguishable by their positions, even though they aren’t labeled. We could say they are localized by the somewhat unavoidable process of writing or printing them. We’re now talking about an ordered process.

Before beginning the process, the four A’s can be rearranged and these arrangements are indistinguishable.  But once we begin the process of putting the A’s in boxes, we can observe which A is put in which box: Pull out the right-most A and put it in a box, pull out the next one, the next and then the left-most A. This selection and placement process is reversible, because the information of where each A has been placed is necessarily known. 

Well, actually, there are two cases to consider as the A's are being placed inside the boxes. Normally, there is no order within the box, so when more than one A is placed in a box, there's no way to know if the same A is chosen when the process is reversed. To make the process reversible, we have to violate this usual condition by, for instance, stacking the letters on top of each other, a "first in, last out" arrangement. If we do this, the process is reversible, the number of unique ways of placing the four A's can be counted. Therefore, in this ordered process the four otherwise indistinguishable A's have 2⁴ possible arrangements just like the distinguishable letters ABCD.

This is similar to labeling the corners of a square with the letters AAAA versus labeling them with ABCD. Normally, rotations by 90 degrees of the AAAA square about its center are "changes that don't make a difference,"*  and these preserve the symmetry of the square. But if you actually perform the rotations (this is the ordered process), and count the number of rotations, then the process can be reversed, because the recorded information can be used to distinguish the otherwise indistinguishable positions of the square. 

Even with the corners labeled ABCD, there is no way to tell whether the rotation was 90 degrees, for instance, or (360 + 90) degrees if the rotations haven't been observed. 

But for the moment I will consider selections to be random and not ordered, so that the AAAA arrangements are indistinguishable except for the five partitions

|AAAA           A|AAA          AA|AA          AAA|A          AAAA|,

as discussed at the beginning of this now-about-to-end screed.

-----------

Being “identical” or not identical is a property we think of as independent of observation. Being indistinguishable or distinguishable is more of a result of an observation, or an interaction, a result dependent on the method of observation and more applicable to arrangements than to individual particles.  

However, the concepts of identity and distinguishability both have limits of observation that prevent us from knowing if we’re missing something when we make our judgement.  Identity—the property of two or more things being identical or not—belongs more to classical mechanics, and indistinguishability—the concept of two or more things not being distinguishable from each other—belongs more to quantum mechanics. But does either of the concepts belong totally to classical mechanics or totally to quantum mechanics? There’s ambiguity involved that maybe depends on the method of measurement, sort of like the ambiguity involved in the wave-particle duality.

*This is Sean Carroll's definition of symmetry, expressed in his early books and DVDs. More recently he has said on his website, "I think of symmetries as transformations that don't change anything physical.".

Callen problem 15.1-2

 

Callen problem 15.1-2:  “A system is composed of two harmonic oscillators of natural frequencies ω_o and 2ω_o . If the system has total energy E = (n + ½)ħω_o, where n is an odd integer, what is the entropy of the system?”

As in Problem 15.1-1, the total energy constraint is used to count the total number of microstates. The energies possessed by both oscillators must equal the total energy

(i+½)ħω_o  +  (j+½)2ħω_o  = (n + ½ )ħω_o,

giving the constraint relating the integers i, j, and n

(i+½)  +  (j+½)2  = n + ½

i+½  +  2j+1  = n + ½

i  +  2j  = n – 1

2j  = n – 1 – i

2j = (n – i) – 1.

 

Since n is odd, i must be even (or zero), so that (n – i) – 1 is even and j is an integer. This relationship is shown in a table:

 

 

i	2j	j
0	n – 1	(n  – 1)/2
2	n – 3	(n – 3)/2
…
n – 3	2	1
n – 1	0	0

 

Comparing this to the first table in Problem 15.1-1, there would be (n – 1) + 1 = n states if we were counting with n = all integers.  But we’re only counting with n = odd integers. So do we just divide n by two? That doesn’t work, since n is odd. But I didn’t see how to avoid that at first, until I realized that counting from zero to n – 1 is one too many states to divide by two to get rid of the n = even numbers. We need instead to count from i = 1 to i = n – 1  and divide that by two, then add one state for the i = 0:  

 

(n – 1) / 2  + 1 = (n – 1 + 2) / 2  = (n + 1) / 2

 

This is the number of microstates of the system. The entropy is

 

S = k log [(n+1)/2]

= k log(n+1) – k log 2

= k log(n+1) – 0.693k.

 

When n = 1, the entropy is zero, as expected, since there’s only one microstate. As n increases, there’s always the constant 0.693k to be subtracted from the increasing entropy.

 

Callen problem 15.1-1, two quantum oscillators

Callen problem 15.1–1.  (part one) "A system is composed of two harmonic oscillators, each of natural frequency ω_o and each having permissible energies (n +½) ħω_o, with n being any non–negative integer. Given that the total energy of the system is E′ = n'ħω_o, where n′ can be any positive integer, calculate the number of microstates available to the system, and the entropy of the system." (See Chapter 15 in Thermodynamics and an Intro to Thermostatistics (1985) for background reading that leads up to this problem.)

The first thing to note is these aren't coupled oscillators.  Thus this is the simplest possible example of a Planckian oscillator system with zero-point energy. The oscillators can share the total energy available to them, but otherwise don't interact. Yeh, like photons themselves. As Peter Milonni (The Quantum Vacuum, chapter 2) says, an electromagnetic field mode is a (quantum) harmonic oscillator. Also, to be more precise, "mode" should be "normal mode".

The other thing to consider is indistinguishability versus distinguishability. Callen says nothing that I see in Chapter 15 or earlier about this, so we just rather automatically accept that the energy elements are indistinguishable. This makes it easier to count the microstates because permutations of the energy elements for a particular microstate don't need to be counted.

You can write the energy constraint of the two oscillators as

(i+½) ħω_o  +  (j+½) ħω_o  = n′ ħω_o ,

which gives

(i+½)   +  (j+½)   = n′ ,

i  +  j = n′ – 1,

j = n′ – i – 1

From this expression, you can just build a table of all possible i and j values, using the fact that the lowest “non-negative integer” is 0, not 1:

i	j
0	n′ – 1
1	n′ – 2
2	n′ – 3
…	…
n′ – 3	2
n′ – 2	1
n′ – 1	0

For any given positive integer n'  this table gives all the ways the energy can be shared between the two oscillators. So, just counting the number of rows in the table gives the number of microstates. Since we have to start counting from zero, the number of microstates is 

(n′ – 1) + 1 = n′ .

The “–1” comes from having to account for zero-point energy (ZPE) in the energy constraint, and the “+1” from starting the count at zero. So there are n′ microstates. Yeah, sort of an anticlimactic result—the number of microstates is just equal to the number of "energy elements" the system has.  This is because, with only two oscillators, whatever part of the total energy isn't in one is in the other. When there are three or more oscillators, permutations of the total number of oscillators among the energy states must be counted. (But--not to be dealt with in this problem--is this equal to the number of permutations of the total number of energy elements among the oscillators? That is, we can count oscillators per energy level, or energy elements per oscillator. Is the number of microstates the same for both methods of counting?)

Also, in this two-oscillator case, the equality of # of microstates and # of energy elements is only true when both oscillators have a zero-point energy, as you can see if you leave out the ZPE in the energy constraint equation above.

How many microstates does a single oscillator have? Remember, the given energy of the system is not the highest energy available because this is the thermally-isolated-system case, the microcanonical case, and the given energy is the total energy. A single oscillator with constant total energy (with or without ZPE) has one microstate.

As an example for our two oscillators, let n′ = 5.  Then i + j = n′ – 1 = 4, and there are five microstates: 

i	j
0	4
1	3
2	2
3	1
4	0

Solution to part one:  Since there are n′ microstates, the entropy is

S₁ =k log (n′) = k log (E′/ħω_o).

where log means natural log. I’ll come back later to this part of the problem to examine what kind of physical system this two–oscillator model might represent.

(What if the problem involved N oscillators without ZPE, a la Dr. Planckenstein in 1900?  I’ll look at that case below.)

-----------------------------

Now for part two of the problem:

"A second system is composed of two harmonic oscillators, each of natural frequency 2ω_o.  The total energy of this system is E″ = n″ħω_o, where n″ is an even integer.  How many microstates are available to this system, and what is its entropy?"

The energy constraint is now

(i+½) 2ħω_o  +  (j+½)2ħω_o  = n″ħω_o ,

(i+½)  +  (j +½)   = n″/2 ,

i  +  j = n″/2 – 1,

j = n″/2 – i – 1

Why does Callen give us this second part of the problem?  What does it tell us that’s not told in the first part of the problem?  This is the same relation between i and j as in part one, but with n″/2 instead of n′.  And n″/2 still gives us consecutive positive integers as possible values, since we are told n″ is an even integer. The table of available microstates is

 

i	j
0	n″/2 – 1
1	n″/2 – 2
2	n″/2 – 3
…	…
n″/2 – 3	2
n″/2 – 2	1
n″/2 – 1	0

 

There are (n″/2 – 1) + 1  =  n″/2 microstates.  The entropy is

S₂ = k log (n″/2).

This part of the problem definitely has some Planck-original-quantum-paper similarity to it, although it's entirely superficial: Planck used different numbers of "primes" on his variables (E, E′, E″, etc., for energy, for instance)  to indicate different frequencies  ω, ω', ω'', etc. And that's all we have here in part two, a different frequency.  But, the big thing with Planck's theory was and is that the energy of a quantum mechanical oscillator is determined by its single, perfectly monochromatic frequency (amazing!) and by integer multiples of that frequency.  I think this is the point of Callen's problem.  Is the number of microstates going to change just because we have a higher frequency, and thus a higher energy?  What's really different about this and part one? 

Consider this:  

E″ = n″ħω_o =  (n″/2) 2ħω_o, 

where the consecutive energy levels for n″/2 are 1, 2, 3, ... , since n″ is even. But this is just like counting the number of energy levels for part one.  What we have is n″/2  = n′,  and that means 

E″ =  (n″/2) 2ħω_o=  n′ 2ħω_o

Now the energy constraint is

(i+½) 2ħω_o  +  (j+½)2ħω_o  = n′ 2ħω_o ,

(i+½)  +  (j +½)   = n′  ,

i  +  j = n′ – 1,

j =n′ – i – 1

This is the same relation between i and j as in part one. The number of microstates is the same, n′ , so the entropy is the same.

Now for part three of the problem:  What is the entropy of the system composed of the two preceding subsystems if they are separated and enclosed by a totally restrictive wall? 

Since the subsystems are noninteracting in the most restrictive way possible--they can't exchange energy or particles (oscillators) with each other--their entropies just add together to give

S₁ + S₂  =  k log (n′) + k log(n″/2)

=  k log (E′/ħω_o) + k log(E″/2ħω_o)

=  k log (E′ E″/2ħ²ω_o²  ).

This is the answer Callen gives in the text. But based on my answer to part two above, this is equivalent to 

S₁ + S₂  =  k log (n′) + k log(n″/2)

 = k log (n′) + k log(n′)

= 2k log(n′)

=2k log(E′/ħω_o).

Now that the problem is solved,  I'll just relate it to the general problem of N oscillators by using Planck's (and before him, Boltzmann's) combinatorial formula for the total number of permutations of a system's "complexions" . This is the total number of microstates, and is the formula Planck used to calculate the entropy of his discrete-energy-level oscillators, with P being the number of energy elements and N being the number of oscillators:

(P+N–1)! / P!(N–1)!  

In part one of this problem we have N = 2 oscillators and P = n′ energy elements, giving the number of complexions as

(n′  + 2 – 1)!/ n'!1!  =  (n′ + 1)! / n′!  = n′ + 1.

We don't have the 1/2 + 1/2 ZPE terms here. With the zero-point energy included, the minus one as shown above in part one of the problem would cancel the plus one here, and we'd again have n′ microstates.

Now let N=3 oscillators and keep P=n′, no ZPE: 

(n′  + 3 – 1)!/ n'!2!  =  (n′ + 2)! / 2n′!  = (n′ + 2)(n′ + 1)/2.

Now let N=4 oscillators:

(n′  + 4 – 1)!/ n'!3!  =  (n′ + 3)! / 6n′!  = (n′ + 3)(n′ + 2)(n′ + 1)/6.  

(To be continued in a later post using Planck’s example with ZPE from his Theory of Heat Radiation, 2^nd edition, 1912.)

Planck included or postulated or uncovered the zero-point energy in a paper published in 1911, and included this derivation in his 1912 book The Theory of Heat Radiation (the 2nd edition, which is very different from the 1st edition of 1906, with ZPE being one of the big differences). Einstein and Stern in their 1913 paper discussed some consequences of ZPE, although its physical meaning caused confusion theoretically until the 1920s when quantum non-commutativity of certain operators representing physical quantities was discovered.