Astro 201 – Radiative Processes – Problem Set 1

Due in class.

Readings: Rybicki & Lightman Chapter 1 except the last section 1.8.

Problem 1. Blackbody Flux

Derive the blackbody flux formula

F = σT^{4} ,  (1)

from the Planck function for specific intensity,

B_{ν} = \frac{2hν^{3}}{c^{2}(e^{hν/kT} − 1)} .  (2)

You may use the integral ∫_{0}^{∞} \frac{dx}{x^{5}(e^{1/x}−1)} = π^{4}/15.

Those who know what they have to do can stop reading here. Those who need a bit of help should read on.

Imagine a perfectly flat, blackbody patch at the center of a sphere of radius R. The patch has tiny surface area dA and is lying flat at r = 0 in the θ = π/2 plane (in spherical coordinates, where r, θ, ϕ are the radius, polar angle, and azimuth). Only one side of the patch is warm; the other side is at absolute zero. Assume that the warm side radiates as a perfect thermal blackbody.

A thermal blackbody emits radiation isotropically—i.e., with no preference for direction. In other words, imagine each point on the patch beaming out a tiny, uniform, hemispherical dome of rays (like an umbrella). It is a hemispherical dome and not a full sphere because only one side of the patch is warm. Note that the Planck blackbody function has no variable inside it that specifies direction of emission; the direction of emission doesn’t matter for purely thermal emission. The only variables that need to be specified for the Planck function are temperature and frequency (wavelength).

Despite the isotropic nature of the emission, because the patch is geometrically flat, has definite area, and emits only on one side of itself, a detector glued to the inside of the sphere at r = R detects different amounts of radiation depending on where it is placed. For example, if the detector is glued on the half of the sphere that can’t see the bright side of the patch, the detector detects nothing. If the detector is glued to the sphere directly above the patch, the detector picks up the most number of photons, because it sees the full face-on area of the patch (namely, dA). If the detector is glued at an angle to the pole, then it sees less than area dA. If it is glued in the plane of the patch, it sees nothing but a infinitesimally thin line segment.

This problem asks you to place detectors everywhere on the inside of the sphere and sum up the radiation collected; this operation is equivalent to “integrating the Planck function over all solid angles into which the radiation is beamed.”

Calculate the total luminosity (in units of energy/time) emitted by the patch of area dA. You must integrate the Planck function over the entire emitting area (which is maximally dA but in general will be less, depending on the viewing geometry), over all solid angles into which the radiation is beamed, and over all frequencies. Think about placing tiny detectors all over the inside of the sphere and asking how much energy/time each of the detectors receives.

Then divide the luminosity by dA to calculate the total flux (in units of energy/time/area) emitted by the patch. You should recover the usual blackbody flux formula, σT^{4}. By definition, σT^{4} is the total amount of energy radiated per time per unit area of a blackbody surface, radiated into all solid angles and over all frequencies.

Problem 2. Flat Disks

This problem forms the foundation for understanding the spectral energy distributions of circumstellar disks, e.g., those surrounding pre-main-sequence stars and AGN.^{1}

Consider a perfectly flat, blackbody disk encircling a blackbody star. The star has radius R_{∗} and effective temperature T_{∗}. The disk begins at a stellocentric radius far from the star, i.e. the inner disk radius r_{i} ≫ R_{∗}. The disk extends to infinity.

Calculate the temperature of the disk, T (r), as a function of radius, r. Make whatever approximations you deem necessary in light of the fact that r_{i} ≫ R_{∗}. Be sure you get the scaling of T with r correctly. It is less important that you get the numerical coefficient correctly.

This disk is PERFECTLY flat. Do not consider individual particles in the disk. It is a blackbody SHEET.

Also, neglect radial transport of energy. Each annulus is independent of neighboring annuli. This is a fine approximation for thin disks since they transport more radiation vertically than radially (the temperature gradient is steeper vertically than radially).

1The problem has the further distinction that for some reason, whenever I teach this course, it is the one problem that drives some students crazy and they never forget it.

OPTIONAL Problem 3. Albedos

Optical astronomers detect rocky bodies in the solar system by virtue of their reflected sunlight. The received flux is proportional to aA, where a is the albedo (reflectivity) of the object in some optical passband, and A is the area of the object. Typically, albedos of minor bodies in planetary systems range from ∼0.01 to ∼0.7.

The proportionality of the flux to aA presumes that all of the optical light received from an object is from reflected sunlight. Of course, that is not quite true; there is some contamination at optical frequencies from the object’s thermal emission (the fact that it is warm).

Give a quantitative explanation as to why this contamination is not worth troubling over. If you wish, you can adopt as your “case study” a Kuiper belt object of optical albedo 0.07 and heliocentric distance 40 AU, observed in the visual (V) passband. Flesh out the problem as you deem appropriate (“make it up as you go along.”) You may find tables of the integrated Planck function, found in the old edition of Allen’s Astrophysical Quantities, useful. Such tables might also be found in the new Allen, but I have not checked.

Assume that the efficiency of emission at the (infrared) wavelengths at which most of the object’s thermal power is radiated is one. Of course, its efficiency of emitting at optical wavelengths is assumed to be less than one (by Kirchoff’s law).

Problem 4. Make like a tree and ...

If all the leaves of a tree fall to the ground, how thick is the layer of leaves on the ground? Give your answer in leaf thicknesses, to order-of-magnitude.

What is the order-of-magnitude of the optical depth presented by the leafy tree to someone lying below the tree? Why does this answer make sense from the perspective of the vegetation? (Be the tree.)

The above two paragraphs are flip sides of the same question. You can try the second paragraph before the first, or vice versa.