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.

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 \(F_v(T)\) over all frequencies to obtain the bolometric flux emitted from a blackbody, F(T). 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 \(\sigma \approx 5.7 \times 10^{-5} erg*s^{-1}*cm^{-2}*K^{-4}\). Otherwise, commit this number to memory.

The beauty in treating objects in space as blackbodies is the ability to characterize their appearance entirely by temperature. For this problem, we get to examine several ways to quantize their radiation. Here, in part A, we are considering the "bolometric flux" which tells us the amount of energy passing through an area over time, independent of frequency. To arrive at this, we begin with the blackbody flux, \(F_v(T)\), and integrate over all frequencies.

First, let's consider the amount of radiation emitted from a given blackbody. This is given by: \[B_{\nu}(T) = \frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}\]
Now, to arrive at the amount of flux for a black body, we need to integrate over the face of one hemisphere. While a blackbody radiates evenly in all directions, we are only concerned with measuring the radiation toward us, since the radiation from the opposite hemisphere will be equal and opposite.

Similar to the latitude and longitude coordinates of the earth, a blackbody radiates out at solid angles, which are characterized by angles (\(\theta, \phi\)) with respect to our selected origin. As we consider how the radiation that we as an observer receive, we realize that for each point on the surface, only a portion of the radiation actually reaches us. By selecting the north pole of the black body as our origin, the solid angle conveniently behaves as a function of \(cos\phi\). To cover the full hemisphere, we will integrate from the north to south pole, or 0 to \(\frac{\pi}{2}\), and then around the circumference of the body, or 0 to \(2\pi\). Completing this we get:
\[F_{\nu}(T) = \int_{\phi=0}^{\frac{\pi}{2}} \int_{\theta=0}^{2\pi} B_{\nu}(T)cos(\theta)sin(\theta)d\theta d\phi\]
Evaluating this, we discover that: \[F_{\nu}(T) =\pi B_{\nu}(T)\]
Moving on to calculate the bolometric flux, we must now integrate over all frequencies, \(\nu\). \[F(T) = \int_{\nu=0}^{\infty}F_{\nu}(T) d\nu = $\int_{\nu=0}^{\infty}\pi B_{\nu}(T) d\nu\]\[F(T) =\int_{\nu=0}^{\infty} \frac{2\nu^2}{c^2}\frac{h\nu\pi}{e^{\frac{h\nu}{kT}}-1}d\nu\] To tackle this integral, we can utilize u-substitution, where \(x=\frac{h\nu}{kT}\). \[F(T)=\int_{x=0}^{\infty} \frac{2(\frac{xkT}{h})^2}{c^2}\frac{xkT\pi}{e^{x}-1}\frac{kT}{h}dx\] By setting this complicated integral with its constants to a new variable, \(\sigma\), known as the Stephan-Boltzmann constant, we get a bolometric flux of: \[F(t) =\sigma T^4\] In this, our constant \(\sigma \approx 5.7 \times 10^{-5} erg*s^{-1}*cm^{-2}*K^{-4}\).

B) The Wien Displacement Law: Convert the units of the blackbody intensity from \(B_{\nu}(T)\) to \(B_{\lambda}(T)\). 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.

We are now considering the blackbody's radiation in terms of wavelength, rather than frequency. We can begin by knowing that: \[\int_{\lambda = 0}^{\infty} B_{\lambda}(T) d\lambda = $\int_{\nu = 0}^{\infty} B_{\nu}(T) d\nu\] We also know the relationship between \(\lambda\) and \(\nu\) to be \(\nu = \frac{c}{\lambda}\). Taking the derivative of this, we also obtain: \(\frac{d\nu}{d\lambda} = \frac{-c}{\lambda^2}\). By inserting this into our equality, we arrive at: \[\int_{0}^{\infty} B_{\lambda}(T) d\lambda = \int_{0}^{\infty} B_{\nu}(T) \frac{-c}{\lambda^2}d\lambda\]Considering our earlier assertion that the integrands are equal, this reduces to: \[B_{\lambda}(T)=B_{\nu}(T) \frac{c}{\lambda^2}\] Finally, expanding \(B_ {\nu}(T) \frac{c}{\lambda^2}\), we arrive at our final answer of: \[B_{\lambda}(T) = \frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda k T}}-1)}\]
C) Derive an expression for the wavelength \(\lambda_{max}\) corresponding to the peak of the intensity distribution at a given temperature T. (HINT: How do you find the maximum of a function?) Once you do this, again substitute \(u = \frac{h\nu}{kT}\). The expression you end up with will be transentental, but you can solve it easily to first order, which is good enough for this exercise.

To determine the wavelength of maximum intensity, we can use our expression for \(B_{\lambda}(T)\) and find a maximum. To find this maximum, we will take the first derivative and find an inflection point (where the first derivative equals zero). Then, we will need to observe from the second derivative if the point is a maximum or a minimum.

First, we need to take the derivative of \(B_{\lambda}(T)\). \[B_{\lambda}(T)\frac{d}{d\lambda} =  \frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda k T}}-1)}\frac{d}{d\lambda}\] \[B_{\lambda}(T)\frac{d}{d\lambda} =  \frac{-10hc^2}{\lambda^6(e^{\frac{hc}{\lambda k T}}-1)}+\frac{2h^2c^3e^{\frac{hc}{\lambda k T}}}{\lambda^7kT(e^{\frac{hc}{\lambda k T}}-1)^2}\] Before setting this to zero, we can simplify the expression by factoring out common terms: \[B_{\lambda}(T)\frac{d}{d\lambda} = (\frac{2hc^2}{\lambda^6(e^{\frac{hc}{\lambda k T}}-1)})(\frac{-5}{1}+\frac{hce^{\frac{hc}{\lambda k T}}}{\lambda k T(e^{\frac{hc}{\lambda k T}}-1)})\]
Again, we can substitute \(x = \frac{hc}{\lambda k T}\). We now have a much friendlier term of: \[B_{\lambda}(T)\frac{d}{d\lambda}= -5 +\frac{xe^x}{e^x-1}\]
Setting this to zero, we now get:\[B_{\lambda}(T)\frac{d}{d\lambda}= -5 +\frac{xe^x}{e^x-1}=0\]\[x=4\] Now substituting x back in for it's actual values, we get \[x= \frac{hc}{\lambda k T} = 4\]\[\lambda = \frac{hc}{4kt}\]
D) The Rayleigh-Jeans Tail: Next, let’s consider photon energies that are much smaller than the thermal energy. Use a first-order Taylor expansion on the term \(e^{\frac{h\nu}{kT}}\) to derive a simplified form of \(B_{\nu}(T)\) in this low-energy regime. (HINT: The Taylor expansion of \(e^x \approx 1+x\))

Recalling that: \[B_{\nu}(T) = \frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}\] We can use this Taylor expansion approximation to rewrite this as: \[B_{photon}(T) = \frac{2\nu^2}{c^2}\frac{h\nu}{1+x-1}\] This then simplifies to: \[B_{photon}(T) = \frac{2\nu^2}{c^2}kT\]
E) Write an expression for the total power output of a blackbody with a radius R, starting with the expression for \(F_{\nu}\). This total energy output per unit time is also known as the bolometric luminosity, L.

The total power output of a blackbody can easily be found by using the bolometric flux that we found in part A. Since the bolometric flux is given as power per area, we just need to multiply this quantity by the surface area of the blackbody to arrive at the power output. Recall: \[F(t) =\sigma T^4\] Multiplying this by the surface area of a sphere, we get:\[L = F(T) \times surface\,area = \sigma T^4 4\pi R^2\] \[L = \sigma T^4 4\pi R^2\]

Acknowledgements: Johnathan Budd & Willie Pirc