The Stefan-Boltz law as a thermo equation of state

In this discussion, the Stefan-Boltzmann law will be written in terms of the energy contained in a "trapped" electromagnetic field, rather than being written in terms of the power of an "escaping" electromagnetic field (as I discussed for sunlight in a recent post).  An energy-versus-temperature equation is one example of a thermodynamic equation of state. The well-known energy-versus-temp relation for a monatomic ideal gas, for instance, is U=3NkT/2. I'll look at many more examples of equations of state, and at how they are related to the "fundamental relations" of thermodynamics in a future post.

Our main man Herb Callen has a good discussion (Section 3-6 in his 2nd edition text) on the thermodynamics of radiation in an empty box: 

If the walls of any 'empty' vessel are maintained at a temperature T, it is found that the vessel is, in fact, a repository of electromagnetic energy. The quantum theorist might consider the vessel as containing photons, the engineer might view the vessel as a resonant cavity supporting electromagnetic modes, whereas the classical thermodynamicist might eschew any such mechanistic models.

The empty vessel could be almost anything. A metal breadbox, a wooden breadbox, a closed cardboard box, a spherical Christmas tree ornament, or Santa Claus ornament, as long as it's hollow. Yes, it's generally going to have air in it, which is a point that doesn't get discussed much. That's not to say it will be inflated with air like a tennis ball, football or basketball.  But it could be, and there's an electromagnetic field in there with that compressed air. We can just let these "vessels" be at room temperature. Otherwise we have the problem of keeping the walls uniformly at the same high or low temperature. 

Even if a box has walls that are at different temperatures, there will be some sort of radiation spectrum in the box.  Let's say one wall is heated so it's emitting most of the radiation in the cavity, presumably with a different emission spectrum near its surface than the spectra at the surfaces of the other walls.  But is this even possible? Can the spectrum of the radiation in the box be different on different sides of the box?

Well, what we would have is the usual linear superposition (addition) of the amplitudes of the all the waves/photons in the box.,* resulting in constructive and destructive interference, which results in an equilibrium energy-versus-frequency spectrum.  If we have a rectangular or square box, we can ask what the role of the radiation is in changing the temperature distributions in the five walls of our box that are not directly heated. This changing of the temps is a temporary, or transient, behavior. An equilibrium temperature distribution will come into being after the transient behavior is over, due to heat conduction in the walls and radiation from the heated wall of the box to the other walls of the box.

But the  radiation in the box won't have a black-body spectrum unless the walls of the empty vessel are maintained, as Callen says, at the same temperature, which gives equality of absorption and emission rates for every part of the walls. At least this is the desired black-body cavity experimental set-up. The other thing we need for the usual, real, set-up is a little peek-a-boo hole in the box to observe the spectrum of the radiation. 

But we aren't going to try to observe it here. We just want to ask, for a given temperature how much electromagnetic energy is in the vessel? The answer is given by the Stefan-Boltzmann law in yet another form, different from the three ways I wrote it in my February post. Instead of flux (power per unit area) or intensity, the total energy of the radiation is given as proportional to Kelvin-temperature-to-the-fourth-power. Callen writes this as

U = bVT⁴ 

where V is the volume of the empty vessel and b = 4σ/c = 7.56x10^-16 J/m³K⁴, where σ is the Stefan-Boltzmann constant and c is the speed of light. (See my 12 November 2020 post, where the integral of the Planck spectral energy density function at temperature T is shown symbolically and the factor of  c/4 is used.  Here, we're getting rid of that factor. If you divide both sides of the above equation by V, the result is energy density.) Sooooooo, the above equation really comes from integrating the Planck spectral energy density formula over all frequencies while holding temperature constant.  

What, then, is the difference between the views of the quantum theorist, the engineer, and the thermodynamicist in Callen's description above? The photons of the quantum theorist are after all normal modes of the electromagnetic field (see page 97 of these notes on the EM field), and a normal mode is a standing wave that has one of the resonant frequencies of the box. The best brief reference on this subject is Rodney Loudon's two-page discussion, "Introduction: The Photon," at the beginning of his book The Quantum Theory of Light, 3rd ed., and the five references therein.

The quantum theorist is not so much interested anymore in cavity radiation by itself as he or she is in cavity quantum electrodynamics, when there are a controlled number of atoms in a very small cavity (and no air, unless the "atoms" in the cavity are a few oxygen and nitrogen molecules) . An electrical engineer who wants to use the "vessel" as a resonant chamber for electromagnetic radiation is just like an acoustical engineer who wants to use an acoustic chamber to enhance certain frequencies and suppress others. (Here we can think about the usual "box of gas" as having acoustical modes of vibration.) Microwave ovens possess electromagnetic resonant cavities to enhance the microwave cooking frequencies, while musical instruments, for example, enhance the desired harmonics of sound waves.

But I'm getting off the subject here. The subject, which I haven't fully explained yet, is the equations of state of blackbody radiation, and how they can be used to find the entropy of blackbody radiation in terms of U and V. This is a "fundamental relation" in thermodynamics, and would be something the thermodynamicist is interested in, particularly the fact that N--whether he or she thinks of it as the number of photons in the box or number of normal modes--isn't an independent variable along with U and V. (See the quote from Callen below, and note he uses an exclamation mark.)

Besides the above equation of state for the total energy of trapped blackbody radiation in terms of temperature, there's a well-known relation between radiation pressure P, total energy U, and volume V,

P = U/3V

which you can copy and paste into an internet search box to find out more about. "It will be noted," says Callen, "that these empirical equations of state are functions of U and V, but not of N. This observation calls our attention to the fact that in the 'empty' cavity there exist no conserved particles to be counted by a parameter N. The electromagnetic radiation within the cavity is governed by a fundamental equation of the form S = S(U, V) in which there are only two rather than three independent extensive parameters!"

The relevant equation for finding the entropic fundamental relation is 

ST  =  U  +  PV ,

or 

S  =  U / T  +  PV / T.

From the two equations of state, we have 

1/T   = b^1/4 (V/U)^1/4

and 

P/T   = (1/3) b^1/4 (V/U)^1/4

so that the entropy fundamental equation is 

S(U, V)   = (4/3) (bV)^1/4 U^3/4

One of the things I'm interested in is how this expression for the entropy of black-body radiation in terms of U and V compares with the logarithmic expression found by Planck, but I'll come back to that later. First, I'll solve the three problems in this section of Callen's book (Section 3.6), then I'll post some detailed typed handouts from a Statistical Physics class I took in 1995 at Texas State University that used Callen's book as the text (thank you,  Mark Loewe!).

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*Professor Hazeltine said in a junior-level electrodynamics class I took at the UofTexas at Austin 30 years ago that "photons don't interact." I raised my hand and said, "What about interference and diffraction?" Hazeltine explained that "interacting" has to do with things that exert forces on each other, and photons don't exert forces on each other. I asked, "Could we still say they interact linearly then?" I don't recall his response, but he stuck with the "photons don't interact" paradigm. It's not a good paradigm, because photons, being bosons, are gregarious and will indeed interact in a certain way (a sort of attraction) that fermions don't (they sort of repulse each other). We don't call mutual boson interactions and mutual fermion interactions by the name "forces," but maybe we should, or maybe we should modify our idea of forces.

Trinity bomb test described as ammo dump explosion

 (From p.1 of the El Paso Herald-Post, Monday afternoon, July 16, 1945)

Army Ammunition Explosion Rocks Southwest Area

Great Blast Near San Marcial Lights El Paso

            An ammunition magazine containing a considerable amount of high explosives and pyrotechnics exploded at 5:30 a.m. in the New Mexico desert near San Marcial on a remote section of the Alamogordo Air Base reservation.

            No one was hurt.

            The blast was seen and felt throughout an area extending from El Paso to Silver City to Gallup, Socorro and Albuquerque.

            Many persons saw a flash light up the sky, like daylight, and felt earth tremors.  They thought an earthquake had struck.

            William O. Eareckson, commanding officer of the Alamogordo Air Base, released the following statement:

            “Several inquiries have been received concerning a heavy explosion which occurred on the Alamogordo Air Base reservation this morning.

            “A remotely located ammunition magazine containing a considerable amount of high explosives and pyrotechnics exploded.

            “There was no loss of life or injury to anyone, and the property damage outside the explosives magazine itself was negligible.

            “Weather conditions affecting the content of gas shells exploded by the blast may make it desirable for the Army to evacuate temporarily a few civilians from their homes.”

            El Paso men going to work at 5:30 a.m. said the flash illuminated Mt. Franklin.  E.R. Carpenter, Louie Ratliff and Jack Coulehan, riding down Alabama avenue, said the whole sky was ablaze with light.  Mr. Carpenter, mechanical superintendent of the Newspaper Printing Corporation, said many persons called the newspaper’s composing room to report seeing the flash and hearing the explosion.  Callers asked if a meteor had fallen.

Big Light in Sky

            L.R. Lessell, Gila forest headquarters supervisor, said rangers reported the shock was felt throughout the Mogollon mountains.

            Rangers at Chloride reported the blast lighted the sky brightly in the area of San Marcial.  The blast was followed by a terrific explosion, like a detonation, rangers said.

            Forest rangers, believing that an earthquake had struck, checked with  Alfred E. Moore at the Smithsonian Observatory on Burro mountain.  The observatory is near Tyrone and 800 feet high.

            Mr. Moore confirmed the flash, saying it was distinctly visible at the observatory, but the shock was unlike earthquake vibrations.  The observatory head has experienced earthquakes in South America and Mexico.

Silver City Shaken

Silver City residents reported three distinct blasts were felt there.  The shock cracked plate glass windows in downtown buildings.

            “The blasts sounded like heavy claps of thunder,” Don Lusk of Silver City, said.  “Houses shook.  People were roused out of sleep from the noise and tremors.” 

            Mrs. H.E. Wieselman of 901 North Ochoa street saw the explosion as she crossed the Arizona-New Mexico state line.  She was enroute to El Paso from California.

            “We had just left Safford, and it was still dark,” Mrs. Wieselman said.  “Suddenly the tops of high mountains by which we were passing were lighted up by a reddish, orange light.

            “The surrounding countryside was illuminated like daylight for about three seconds. 

            “Then it was dark again.

            “The experience scared me.  It was just like the sun had come up and gone down again.”

Front Seat at Sky Show

            Ed Lane, Santa Fe railroad engineer, was at Belen, N.M., when the blast occurred.

            He said he had a front seat to the greatest fireworks show he had ever seen.  The blast was in the direction of San Marcial and seemed to be only a few miles from Belen.
            “I was coming to El Paso,” Mr. Lane said.  “My engine was standing still.  All at once it seemed as if the sun had suddenly appeared in the sky out of darkness.  There was a tremendous white flash.  This was followed by a great red glare and high in the sky were three tremendous smoke rings.  The highest was many hundreds of feet high.  They swirled and twisted as if being agitated by a great force.  The glare lasted about three minutes and then everything was dark again with dawn breaking in the east.”

I copied this article sometime in the early 1990s by hand (pencil and paper) from the original July 16, 1945 El Paso Herald-Post in the bound newspaper archives of the Barker Center for Texas History at the University of Texas at Austin.  In May1994, I used it as an appendix to a paper I wrote ("Zero Patience for Zero Hour") for a History of the Atomic Bomb class taught by Bruce Hunt. Here are some other front-page Herald-Post headlines from July 16, 1945:

450 Super-Forts Heap 2500 Tons of Fire on Japan

Woman Happy as She Gets Back Lost $1480

Ice Cream Firm Owner Freezes to Death in Plant

Capitalizing on Prejudice, Nationalist Party Drives for Control of Government

B-29 Crew Throws Man to Safety from 9000 Feet

Temperature, thermal conductivity and heat capacity

In regard to my previous post, I'll remind you of this: an object's temperature doesn't by itself determine how hot or cold it feels when you touch it. For a given temperature, when the object (or surface) is in contact with your skin, how fast how well it conducts heat is the primary property that determines how hot or cold it feels. This property is known as thermal conductivity. 

One of the reasons I wanted to write the previous post on the temperature of an asphalt road in the summertime sun is to compare how hot the asphalt feels with how hot an iron manhole cover feels. The question is, which surface will burn you faster?

Another factor that comes into play in this regard is heat capacity, also called specific heat capacity, or just specific heat, which is usually described in terms of how much the temperature of a given mass of a certain substance increases when a given amount of heat is supplied to it.  Turn that heat transfer process around and the question becomes how much heat (internal energy) is available to be transferred to a cooler object or surface (your skin, for instance) from the heated substance.

Now we have two questions concerning asphalt and iron: which will burn you faster, and which will give you a more serious burn due to the amount of heat it can transfer. These are related, but exactly how, I don't know.  There is (or was) a spirited debate about this relationship at the physics "stack exchange" website.

We're not considering the emission of energy by radiation now.  We're considering the transfer of energy by conduction, because we have two surfaces at different temperatures in contact with each other: the asphalt (or iron) and your skin. The role of the radiation is to bring the asphalt and the manhole cover to their Stefan-Boltzmann-law temperatures, as discussed in my previous post.

Are the pavement and the manhole cover even going to be at the same temperature in the direct sunlight?  If we treat them both as black-body surfaces we are assuming they reach the same temp once equilibrium between absorption and emission of radiation has been established in each material.

Even when we don't treat them as black-bodies, if they are both equally good absorbers-emitters of radiation, we expect them to reach the same temperature.  That's the subject of this post:  studying the Stefan-Boltz law as it applies to non-black-bodies.

I mentioned in my previous post that the two problems I solved from the Milonni/Eberly textbook Lasers are elementary problems (2nd semester calculus-based freshman physics level), and in fact I found good explanations of how to use the Stefan-Boltzmann law for non-black-bodies in two freshman-level textbooks: University Physics by Sears, Zemansky and Young (SZY), and Principles of Physics by Frank Blatt. 

The Stefan-Boltzmann law as Milonni/Eberly write it, 

 I_total = σT⁴,

is applicable only to idealized surfaces or objects that are assumed to be emitting and absorbing a Planck black-body spectrum. But the fact that real objects aren't perfect emitters and absorbers can be taken into account if their emissivity, ε, is included in S-B law, which SZY's text does, using R instead of I_total for the radiative flux:

R = εσT⁴ 

Prof. Blatt takes this a step further and explicitly shows how the surface area of an object is used to calculate the amount of energy per unit time it’s emitting:

P = AεσT⁴ 

where A is the surface area of the object and P  is the power (joules per second or watts in our MKS case). This is a helpful equation when it comes to finding the radiant energy emitted by an object (the sun for instance) as opposed to just a surface, where only the power per unit area (flux) can be calculated (the asphalt surface or manhole surface, for instance). 

Emissivity is defined as the "ratio of the energy radiated from a material's surface to that radiated from a perfect emitter, known as a blackbody, at the same temperature and wavelength and under the same viewing conditions. It is a dimensionless number between 0 (for a perfect reflector) and 1 (for a perfect emitter)". 

Based on "too-much-information" I found at various websites, I'll estimate the emissivities of asphalt and of oxidized cast iron to both have a value of about 0.9, so they will both be heated by the sunlight to about the same temperature. Given the 756 watts/meter² of solar radiation flux I used in my previous post, how much does a change from ε = 1 (black-body) to ε = 0.9 affect the temperature as calculated by the above equation?

Well, yeah, right, by a factor of the fourth root of 0.9, which is .974, which we might as well call 1, which means we might as well say the temp of both the asphalt and the manhole cover will be 153° F (67° C), as I previously calculated.

Now we need to know the thermal conductivities of asphalt and cast iron.  I'm going to use 1.3 watts per meter per kelvin (1.3 W/m/K) for asphalt and 56 W/m/K for cast iron. Based on these numbers, the manhole cover is capable of burning your skin 40 times faster than asphalt.

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I crossed out the words "how fast" in the first paragraph above, but it's actually a good descriptor because time is implicitly present in the units for thermal conductivity: a watt is a joule per second. So how quickly (or slowly) something that's hot can burn you is indeed taken into account by its thermal conductivity.

If you reach into an oven where a loaf of bread has just finished baking, and accidentally let the metal oven rack touch your skin briefly, you'll get burned at least a little. You can touch the bread, however, without getting an immediate burn, and the air in the oven, which is at the same temp as the bread surface and the metal rack, would take even longer to burn you.  In our asphalt and iron manhole cover exercise, the manhole cover is like the oven rack--yeah, both are metal--and the asphalt is a little like the bread.  For much more about the thermodynamics of baking bread, see The Fresh Loaf  for a discussion on relevant heat transfer mechanisms. Radiation is one of them.

Now what about the heat capacity of asphalt versus the heat capacity of cast iron?  Heat capacity, as its name implies, is about the ability of a substance to store energy.  This is the currently missing middle component in the energy conservation process we're considering here:  Absorption of energy, storage of energy, and emission of energy. 

First we have solar radiation as the medium for the heat absorbed by the asphalt and the iron manhole cover. Second--since we assume no radiation is being reflected or transmitted--we have energy being stored in the material as kinetic energy of the atoms, molecules, and free electrons (if there are any) that make up the material. And third we have emission of heat from the surface of the material.  This was being done solely by radiation emission (so far we're ignoring the role of the heated air above the surfaces) in an equilibrium situation with absorption and storage of energy--until you came along and put one hand on the manhole cover and the other on the pavement to find out which feels hotter.  Or maybe it was me who did that. 

No matter who did it, that little one-second personal experiment would show the manhole cover feels hotter. You or I might not even be able to keep a hand on the manhole cover for the length of time it takes to say "Mississippi-one".

Why, then, when we use a thermometer to measure the temps of both the asphalt and the manhole cover do we not predict the manhole cover will be at a higher temp? It's a matter of perception, which is something discussed in a typical Coen-Brothers' way in A Serious Man.  Our un-serious man, Prof. Gopnick, remarks after smoking pot and drinking iced tea with his next-door neighbor Mrs. Samsky, that maybe everything is just a matter of perception as the junior rabbi, Rabbi Scott, had suggested to him.  But Rabbi Scott had said "perspective" not perception. 

Our perception of temperature depends on the rate at which energy is conducted to or away from our skin. Thermometers register this same effect by absorbing or losing energy (from a hotter object or to a cooler object, respectively) either rapidly or slowly while coming into thermal-conductance equilibrium with the object or surface they are in contact with. How quickly or slowly this equilibrium happens depends on how well the object or surface conducts heat, and also on how well the thermometer itself conducts heat.  I recall one of my favorite experiments in first-semester physics was called "The Time Constant of a Thermometer."

Now back to heat capacity. The heat capacity of asphalt is about twice that of cast iron:  900 J/kg/K for asphalt versus 460 J/kg/K cast iron. I merely leave this to the reader (and writer) as something to think about, that asphalt stores energy better than iron but conducts it about 40 times more slowly. One thing to think about in this regard is a block of iron in contact with a block of asphalt, where each is at a different temperature to begin with. Another thing is this: what is the heat capacity of a black-body? Einstein in 1907 used Planck's 1900 relation of discrete energy levels for continuously distributed frequencies to model the vibrations of solids, which are all about the heat capacity of solids, but heat capacity seems to be ignored when it comes to discussions of black-body radiation.

The "Pavements" website of the U.S. Federal Highway Administration is where I got the asphalt heat capacity number. The full name of this website is 

Pavement Thermal Performance and Contribution to Urban and Global Climate

The info there is relevant to most of this blog post of mine, and also mentions "albedo" in contrast to emissivity, although the relation between the two isn't simple. Well-written, check it out. I'm outta here.

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.

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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.

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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.".