Worksheet 5, Problem 2

Blackbodies are nice because they’re such simple objects. Their outward appearance is entirely determined by their temperature. If there were cows in space, astronomers would imagine them to be spherical blackbodies (and seriously, it wouldn’t be a bad approximation). In this exercise we’ll take advantage of the relative simplicity of blackbodies to derive some useful expressions that you’ll use during this term, and throughout your astronomy career.

Blackbodies are objects that absorb all electromagnetic radiation. However, they cannot keep absorbing this energy forever because that would break the \(2^{nd}\) law of thermodynamics which implies that heat can't pass from one body to a hotter body. Thus, once the blackbody reaches thermal equilibrium with its environment, it begins to emit any radiation that it receives. This radiation output is measured by Intensity (I) which is measured in units of power. Intensity of radiation varies with the frequency of the radiation being emitted. Since all electromagnetic waves have a speed of c, frequency will change when wavelength is altered because \(v\:=\:\lambda\:\times\:f\). The intensity with respect to frequency (\(I_{\nu}\)) is given by Planck's law for spectral radiance (B): \[I_{\nu}(T)\:=\:\frac{2\nu^{2}}{c^{2}}\:\times\:\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}\:\equiv\:B_{\nu}(T)\]

1. (a)  In astronomy, it is often useful to deal with something called the “bolometric flux,” or the energy per area per time, independent of frequency. Integrate the blackbody flux FνpTq over all frequencies to obtain the bolometric flux emitted from a blackbody, FpTq. You can do this using by substituting the variable u ” hν{kT. This will allow you to split things into a temperature-dependent term, and a term comprising an integral over all frequencies. However, rather than solving for the integral, just set the integral and all constants equal to a new, single constant called σ, which is also known as the Stefan-Boltzmann constant. If you’re really into calculus, go ahead and show that σ « 5.7 ˆ 10 ́5 erg s ́1 cm ́2 K ́4. Otherwise, commit this number to memory. 
   
   To obtain the flux expression F (T), one needs to integrate \(I_{\nu}\) over the surface area of the hemisphere emitting receivable radiation. This multiplies Planck's law (\(I_{\nu}\:\equiv\:B_{\nu}(T)\)) by \(\pi\). This gives \(F_{\nu}(T)\). Then to find the flux of all frequencies, F(T), integrate
   (F_{\nu}(T)\) over all frequencies from 0 to infinity. This gives \[\int_0^{\infty}\:\pi\:\times\:\frac{2\nu^{2}}{c^{2}}\:\times\:\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}\:dv\] 
   
   Then, to simplify this integral, set \(\mu\:=\:\frac{h\nu}{kT}\). With this substitution, \(\nu\:=\:\frac{kT\mu}{h}\) and \(d\mu\:=\:\frac{h}{kT}\:dv\:\rightarrow\:kT\:d\mu\:=\:h\:dv\). Substitute these \(\mu\) based values for all \(\nu\) based values and you end up with \[\frac{2\pi k^{4}T^{4}}{c^{2}h^{3}}\:\int_0^{\infty}\:\frac{\mu^{3}}{e^{\mu}-1}\]
   
   Separating out the \(T^{4}\) and setting the rest equal to a constant \(\sigma\) shows us that the bolometric flux, F (T) is equal to \(\sigma\:\times\:T^{4}\).
   
   
2. (b)  The Wien Displacement Law: Convert the units of the blackbody intensity from BνpTq to BλpTq IMPORTANT: Remember that the amount of energy in a frequency interval dν has to be exactly equal to the amount of energy in the corresponding wavelength interval dλ. 
   
   Since the amount of energy in a frequency interval is equal to the amount of energy in the corresponding wavelength interval and since spectral radiance is measured in terms of energy, it can be said that: \(B_{\nu}(T)\:d\nu\:=\:B_{\lambda}(T)\:d\lambda\). Rearranging this shows us that \(B_{\lambda}(T)\:=\:B_{\nu}(T)\:\frac{d\nu}{d\lambda}\). Since the spectral radiance with respect to frequency is already known from Planck's Law, we just need to find \(\frac{d\nu}{d\lambda}\). Remember that \[c\:=\:\nu\:\lambda\:\rightarrow\:\nu\:=\:\frac{c}{\lambda}\:and\:thus\:\rightarrow\:\frac{d\nu}{d\lambda}\:=\:-\frac{c}{\lambda^{2}}\]
   
   Now plug in \(\frac{c}{\lambda}\) for all \(\nu\). \[\frac{2h\nu^{3}}{c^{2}\:\times\:(e^{\frac{h\nu}{kT}}-1)}\:\rightarrow\:\frac{2h(\frac{c}{\lambda})^{3}}{c^{2}\:\times\:(e^{\frac{h(\frac{c}{\lambda})}{kT}}-1)}\]
   
   Then multiply by \(d\nu\) to reveal Wein's Displacement Law:
   
   \[B_\lambda(T)\:=\:\frac{-2hc^{2}}{\lambda^{5}\:(e^{\frac{hc}{kT\lambda}}-1)}\]
   
   
   
   
3. (c)  Derive an expression for the wavelength λmax corresponding to the peak of the intensity dis- tribution at a given temperature T. (HINT: How do you find the maximum of a function?) Once you do this, again substitute u ” hν{kT). The expression you end up with will be transcendental, but you can solve it easily to first order, which is good enough for this exercise. 
   
   From the previous question, we know that \(B_\lambda(T)\:=\:\frac{-2hc^{2}}{\lambda^{5}\:(e^{\frac{hc}{kT\lambda}}-1)}\). In order to find the wavelength that results in the highest intensity, we need to take the derivative of this function, set it equal to zero, and then solve for \(\lambda_{max}\). Using both the quotient and the product rules, you come up with \(e^{\mu}\:\times\:(\mu-5)\:=\:-5\). Using the Taylor expansion, \(e^{\mu}\:\approx\:1+\mu\), you then have \((1+\mu)(\mu-5)\:=\:-5\). Solving this give \(\mu\:=\:4\). Then plugging this into a modified version of the definition equation for \(\mu\) shows us that \(\lambda_{max}\:=\:\frac{hc}{kT\mu}\:=\:\frac{hc}{4kT}\).
   
   

(d) The Rayleigh-Jeans Tail: Next, let’s consider photon energies that are much smaller than hν

the thermal energy. Use a first-order Taylor expansion on the term ekT to derive a simplified form of Bν pT q in this low-energy regime. (HINT: The Taylor expansion of ex « 1 ` x.) 

 The Taylor expansion of  \(e^{\mu}\:\approx\:1+\mu\) can be applied generally to give us \[e^{\frac{h\nu}{kT}}\:\approx\:1+\frac{h\nu}{kT}\]. 

Plugging this into Planck's law gives the Rayleigh-Jeans Tail:
\[B_{\nu}(T)\:=\:\frac{2kT\nu^{2}}{c^{2}}\]

(e) Write an expression for the total power output of a blackbody with a radius R, starting with the expression for Fν. This total energy output per unit time is also known as the bolometric luminosity, L. 

This part of the problem is relatively simple.  We know that: \[Flux(F_{\nu})\:\frac{L}{Surface\:Area}\] 
We also know that black bodies are assumed to be spherical and thus have a surface area of \(4\pi R^{2}\).
Thus, \(F_{\nu}\:=\:\frac{L}{4\pi R^{2}}\). Thus, 
\[L\:=\:4\pi R^2\:F_{\nu}\:=\:4\pi R^{2} \sigma T^{4}\]