4. THERMAL RADIATION


  4.1 Temperatures on Icarus, Moon and Mars
      Stefan-Boltzmann law

The problem: First, derive an equation for the surface temperature on the Sun-facing point of an asteroid, as a function of the distance, r (in A.U.), from the Sun. Assume as given that the energy flux from the Sun at 1 A.U. is F, in w/m2, that the surface reflects a fraction A of the incident energy, that a thermal surface emits sigmaT4w/m2, and that the Sun-facing point faces the Sun forever.

   Second, evaluate your equation for the temperature (in oK) on the asteroid Icarus (A = 0.1) when it is 0.2 A.U. from the Sun, on the Moon (also A = 0.1) at 1 A.U. from the Sun, and on Mars (A = 0.16) at 1.6 A.U. from the Sun. (Assume negligible atmosphere on Mars). Parameters needed: the energy flux at 1 A.U. is F = 1367 w/m2, and sigma = 5.67x10-8 w/m2/K4. Speculate on a design of a space suit or heat shield that might allow a human to survive in sunshine on Icarus when Icarus is 0.2 A.U. from the Sun.

   Third, suppose the asteroid rotates so rapidly that the temperature is the same all over the asteroid, so that the energy arriving from the Sun is re-radiated uniformly in all directions. By what factor is the temperature lower than your first computation? According to what (qualitative) criterion would you choose which method of estimating the temperature is better?

The solution: The energy flux received at distance r is F/ r2 w/m2 where r is in A.U. and F is given in the problem for r = 1. A fraction (1-A) of this energy is absorbed. Since the point of interest continually faces the Sun, this amount of energy must also be radiated away, at the rate sigmaT4w/m2. Therefore, T4 = (1-A)F/sigmar2. Given the numerical values of F and sigma, the result is T = 395 (1-A) 1/4r-1/2oK . The surface of Icarus (see problem 2.1), under the given conditions, has T = 863oK. Perhaps an astronaut could survive if the space suit or heat shield were extremely reflective and if the heat absorbed at the Sun-facing side were to be conducted to a much larger shaded area to be radiated away. On the Moon, with r = 1, T = 385 oK. For Mars, the temperature is 300oK.

   For a rapidly rotating body with uniform temperature, energy is absorbed over an area piR2 but radiated from an area 4 piR2. Therefore, T4 is lowered by a factor of 4, and T = 279 (1-A) 1/4r-1/2oK. This value requires that the period of rotation is much less than the time it takes for the surface to cool off.

Interpretation: For the Moon, the temperature estimated for the Sun-facing point is quite realistic. The space suits of the Apollo astronauts on the Moon had to be cooled when the men were in sunshine. But if the astronauts stepped into the shadow of a big rock, they radiated away enough heat so that their space suits needed heating. Away from the Sun, the lunar surface cools to 110oK.

   On Mars, T = 300oK = 27oC is really a theoretical maximum if there is no atmosphere, but the surface becomes this warm occasionally because of a slight greenhouse effect. At night the surface temperatures may drop to 130oK. Where the Mars Pathfinder landed in July 1997, the surface temperatures were much less extreme, 197oK to 263oK. The atmosphere there is much cooler than the surface : if you were standing on Mars, your nose would be 20oC cooler than your feet. Humans on Mars will need space suits in any case because the atmosphere has very low pressure and lacks oxygen.

   When a part of the Moon is eclipsed by the Sun, the surface cools off significantly in about two hours. Assuming that the surfaces on the Moon and asteroids are similar, an asteroid surface needs about two hours without sunshine to cool off. Therefore, an asteroid rotating with a period less than about one hour (and tumbling so that all parts are heated) more nearly satisfies the second version of the problem. The surface on Icarus, with a rotation period of 2.3 hours, more nearly satisfies the first version.

Didactics: The solar energy flux at the Earth must be measured from satellites above the Earth's atmosphere. Space experiments are usually very hard to calibrate accurately. It was a major technical achievement to build an instrument that could measure the solar energy flux accurate to about 0.1% . The measured value permits us not only to determine the solar luminosity but to detect that the solar luminosity in fact varies by roughly 0.1% over the years. (See problem 6.5.)

   Since F is known so accurately, the temperatures were evaluated accurate to three decimals, but variations in A (called the albedo), changes in r (the Moon and Earth together change solar distance by about 1% during a year) and variations in the direction of the surface relative to the Sun make even the second decimal inaccurate. The quoted observed temperatures, uncertain by several degrees, were determined from the infrared emissions of the lunar and Martian surfaces, observed by satellites orbiting the Moon and Mars, respectively.

Effect of the Earth's atmosphere: The formulae derived in this problem assume radiation directly from the surface into space, without interference by an atmosphere. For the Earth, A = 0.35, averaged over the Earth and over the year. 35% of the solar energy is reflected back into space, mostly by oceans, clouds, and ice near the poles. The visible light that does reach the Earth's surface is re-radiated in the infrared (according to Wien's law). The Earth's atmosphere (largely its water vapor and carbon dioxide) absorbs the infrared. The radiation is re-emitted and re-absorbed many times, but it gradually wanders upward to cooler layers of the atmosphere, from where it finally escapes into space. The more water vapor or carbon dioxide reside in the atmosphere, the warmer must be the surface so that the radiation migrates upwards and escapes as fast as solar energy is absorbed. This warming is known as the greenhouse effect.

   An often quoted measure of the natural greenhouse effect is the following: The Earth's average temperature at the surface is 15oC = 288oK. If we did not have the greenhouse effect, but still had enough atmosphere and oceans to makes the temperature uniform over the Earth, then the second version of the problem would apply and our temperature would be -20oC = 253oK. The Earth would probably be frozen and without life.

   The greenhouse effect is a completely natural phenomenon which has occurred for many millions of years. Since the amount of water vapor and carbon dioxide change in the course of millions of years, due to changes in Earth's volcanism and in the oceans, the Earth's surface temperature has also changed slowly. For instance, at the time of the dinosaurs, 100 million years ago, our atmosphere contained more carbon dioxide and the climate was warmer globally.

   Since the beginning of the industrial era, about 150 years ago, human activity has added carbon dioxide to the atmosphere, thus increasing the greenhouse effect, and forcing a gradual warming of the atmosphere . This process has accelerated in the last three decades. The global warming measured during the last two decades is probably due to the human-produced carbon dioxide. But there are alternative explanations for the measured global warming, such as a small increase in the average energy received from the Sun (see problem 6.5).


4.2 The radii of stars
      Stefan-Boltzmann law - selection effect in data interpretation

The problem: The diagram is a plot of the luminosity L and surface temperature T of the twenty nearest stars (symbols o) and of the twenty apparently brightest stars. Identify the largest and the smallest star shown on the diagram. Determine their radii (in units of the Sun's radius Ro), either from L = 4piR2sigmaT4 or by drawing lines of constant r/Ro = 10-2, 1, and 102 Ro into the diagram and interpolating.

   Which are larger, the stars on the upper or on the lower main sequence?

   On the basis of this diagram, is the following text written by a young student correct? "Astronomers observing the apparently brightest stars have learned that most stars are much larger than the Sun." Explain your answer.

The setting: On observing a star with a telescope, we can measure three quantities. One is the color of the star, which tells us its surface temperature, T. (Stellar radiation fits a thermal Planck "black body" radiation curve fairly accurately). A second quantity is the distance to the star. For a sufficiently nearby star, we measure its parallax, that is, its apparent shift relative to distant stars caused by the Earth's motion around the Sun. A third quantity is its apparent brightness (units w/m2). Distance and apparent brightness separately tell us nothing directly about the star, but the two can be combined to yield the stellar luminosity, L = (surface of sphere at Sun's distance from the star)x(apparent brightness). For ease of thinking, all values of L will be expressed in terms of the solar luminosity, Lo.

   On a photograph, the size of a star image is determined by our atmosphere (twinkling) and by the size of the telescope (diffraction). The size of the star image tells us nothing about the actual size of the star. The radius of nearly all stars is obtained theoretically from L = 4piR2sigmaT4 or, relative to the Sun, log(R/Ro) = 1/2 log(L/Lo) - 2 log(T/To).

   Do stars come in all sizes and temperatures? No, as shown by the adjacent "L-T" diagram. (Diagrams like this are often called an HR diagram after the astronomers who first recognized the usefulness of the diagram, Hertzsprung and Russell. For such historical reasons, T increases to the left in HR and L-T diagrams.) Most stars fall along a diagonal band called the "main sequence". But there are also other stars, notably the largest "supergiants" in the upper right and the smallest "white dwarfs" in the lower left. (The "white" in white dwarf is a misnomer, they tend to be blue.) The physical reason for the distribution of stars in the diagram is outlined in section V.

 


 
 


 

The solution: The largest stars are in the upper right of the diagram. Betelgeuse is some 800 times larger than the Sun. If Betelgeuse were in the place of the Sun, it would reach past Mars and almost to Jupiter. No wonder stars like Betelgeuse are called supergiants. The smallest stars are the white dwarfs, roughly 100 times smaller than the Sun, similar to the size of the Earth.

   Since the axes of the diagram are logarithmic, lines of constant radius are straight lines. Their slope of 1/4 is less than the slope of the main sequence. Thus blue main sequence stars are larger than the Sun, which is larger than red main sequence stars.

   The student's sentence is wrong. "Most stars" are more like the nearest stars. Each star in a given nearby volume has been listed. The apparently brightest stars are generally not nearby. Being very luminous, according to the diagram, they can appear bright even if seen at large distances. They must be spread over a very large volume. In any given volume, these luminous stars are only a small fraction of all the stars. The apparently brightest stars are not "most stars".

Didactics:

  1. This astronomical problem gives a good example of a selection effect, which affects many aspects of the physical sciences and of everyday life. An analogy in towns of developing countries might be: headlights from the infrequent cars are seen from far away; and they attract attention ; the much weaker lights and reflections from the bicycles are seen only if the bicycles are nearby ; yet the most common mode of transportation is the bicycle. An analogy for air travelers might be that distant bright lights from city shopping centers are much more visible from airplanes than are the ordinary lights on ordinary streets almost underneath the plane. Newspapers often report on "surveys" of opinions which show strong selection effects.

  2. Each student or group of students should receive a copy of the diagram so that they may read off L and T and/or draw constant-R lines.

  3. Sirius A is the only star (other than the Sun) that is in both groups, both nearby and very bright. Its companion, Sirius B, is one of the white dwarfs. Many luminosities are at least as uncertain as the sizes of the symbols in the diagram. Distances have recently been revised, for some stars significantly, by the observations of the European satellite HIPPARCOS. Temperatures of white dwarfs are also uncertain by more than the sizes of the symbols.




4.3 The temperature of a neutron star
      energy conservation - Wien's law

The problem: An isolated neutron star moves through the interstellar gases and gravitationally accretes on the order of dM/dt = 107kg/s of interstellar hydrogen. Assume that this material falls onto the surface of the neutron star, distributed uniformly over that surface. The gravitational energy released appears as thermal energy that must be radiated away. What is the surface temperature of the star? Assume a uniform density inside the star rho = 1018kg/m3.

   In what wavelength band (radio, infrared, visible, x-ray or gamma-ray) does the star radiate most intensely?

   Compare the total luminosity to that of the Sun. Use a neutron star radius R = 10 km .

The setting: Supernova explosions may leave behind a neutron star (introduced in problem 3.1). Some neutron stars remain observable for many thousands of years, either because they are pulsars (problem 3.4) or because they acquire gas from a companion star (problems 3.2, 3.3) But if the neutron star is old and isolated, the only source of energy may be the interstellar gas falling onto the star. Can we detect old lone neutron stars by the radiation from infalling interstellar gases?

The solution: Energy is released on the surface of the neutron star because matter falls from infinity to the surface. The gravitational energy released per unit of mass is GM/R. The rate at which mass is accreted is dM/dt. Therefore, the total rate of energy release or luminosity is L = GM (dM/dt)/R . The matter is accreted over the entire surface and thermalized, so we have for the total luminosity L = 4piR2sigmaT4, where sigma = 5.7x10-8 w/m2/K4. Equating the luminosities yields sigmaT4 = GM (dM/dt)/(4piR3) = 1/3 Grho dM/dt . The result is T = 2.5x105oK. By Wien's law, lambda(max) = 2.9x10-3/T = 10-8m . This is in the far-ultraviolet and "soft" x-ray range. The total luminosity is about 0.7x10-3Lo.

Interpretation: The rate of infall, dM/dt, is extremely uncertain because it is proportional to the interstellar gas density and is very sensitive to the velocity of the star relative to the gas. Furthermore, if the star is strongly magnetized and rotating rapidly, the centrifugal forces may give much smaller infall rates. The assumption that the gases fall onto all parts of the star is also debatable. If the star's magnetic field is still strong, the gas may be funneled onto the magnetic poles, a smaller area and thus a higher temperature. Fortunately, these uncertainties enter T only to the power 1/4.

   It is not obvious that the infalling gas really creates a Boltzmann velocity distribution within the already present gas at the surface of the star. Thus it is not obvious that the spectrum will have a "black body" Planck character. Rather, the gas may radiate at the moment of impact and first collision, before it has mixed with surface gases. The collisions and radiated photons then are more powerful (problem 4.4). In the extreme, if the energy of an infalling proton is converted directly into escaping photons, then the energy emerges as gamma rays.

   Several x-ray sources have been suggested to be isolated neutron stars accreting gas from their surroundings. In 1997 the Hubble Space Telescope identified an extremely faint optical object with one of these x-ray sources, thereby establishing with virtual certainty that the x-ray source is an isolated neutron star. However, it is not yet clear that the x-rays are caused by gas accretion onto an old neutron star. They may merely represent the cooling of a hot young neutron star. A few more sources are awaited eagerly in order to better learn how neutron stars evolve and possibly interact with their surroundings.

Didactics: In a rough approximation, gas flowing past a star can be captured with an impact parameter a such that v2 < GM/a. Thus the largest impact parameter of gas that can be captured is proportional to v-2, and the capture cross section is proportional to v-4. The mass capture rate is proportional to the mass flux of gases past the star, rhov, times the capture cross section, so dM/dt is proportional to rhov-3.


4.4 The solar corona and clusters of galaxies
      Wien's law, Bremsstrahlung

The problem: Assume that "black-body" radiation is emitted by thermal electrons colliding with protons. Evaluate hv in terms of kT at the peak of the Planck spectrum I(v,T) = (2hv3/c2)exp(-hv/kT). What does this value of hv imply about the minimum energy necessary for the colliding electrons that emit this photon? Compare to the mean electron energy 3/2 kT. Given that the electrons have a Boltzmann velocity proportional to exp(-mv2/2kT), estimate a likely maximum energy for the colliding electrons.

   Now assume that you have a gas, with a Boltzmann velocity distribution at some temperature T, that is transparent to its own radiation. Therefore, the intensity of the Planck spectrum definitely does not apply. But you can still make a prediction about the shape of the spectrum : Given what you have just learned about colliding electrons and the resulting radiation, predict the typical energy of the photons emitted by this transparent gas. For T in the range of 106 to 108oK, what is the wavelength band emitted most intensely (radio, infrared, visible, ultraviolet, x-ray, gamma-ray)?

The setting: Many astrophysical gases have temperatures of millions of degrees, for instance the solar corona (about 2x106oK), accretion disks around neutron stars (107oK), and the gases between galaxies in clusters of galaxies (about 108oK). (See problems 5.2, 3.2, and 3.5, respectively.) These gases are clearly not opaque to their own x-rays, so they do not radiate like "black bodies". At what photon energies might one expect these gases to radiate? The radiation is presumably due to collisions of electrons with protons, a process usually called Bremsstrahlung. This problem asks the students to interpret Wien's law in terms of collisions, with the conclusion that the most intense emission from such transparent gases also follows Wien's law. Therefore, these very hot gases radiate mainly x-rays.

The solution: The maximum of the Planck spectrum occurs at hv = 3kT, twice the mean particle energy. The energies of colliding electrons emitting photons with hv = 3kT must have been greater than the photon energies, 1/2 mv2 > 3kT. Yet the energies cannot have been much greater, because much faster electrons are rare. The Boltzmann velocity distribution falls off sharply at about 1/2 mv2 > 6kT. Thus the peak of the Planck spectrum is due to electrons with kinetic energies roughly between 3kT and 6kT.

   The same collisions will occur even if the medium is transparent. Therefore, the x-ray spectrum from a hot transparent gas should have a maximum near hv = 3kT, just like the Planck spectrum does. For T in the range of millions of degrees, the photon energy is in the x-ray range. The x-rays must be observed from above the Earth's atmosphere.

Interpretation: The estimate made here for the peak of the spectrum is quite good for T > 107oK. At these high temperatures, the x-radiation is nearly a continuum, and one determines the gas temperature by the ratio of x-ray intensities in two wavelength bands, much like one determines the temperatures of stars by measuring the relative intensities in red and blue. For the less hot solar corona, the details of the spectrum are quite complicated because atomic transitions cause emissions at many wavelengths with intensities that exceed the collisional continuum.