Week 3: WS5 #2

2. 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 \(F_\nu(T)\) over all frequencies to obtain the bolometric flux emitted from a blackbody, F(T). 

We were given that \(F_\nu(T)=\frac{4\pi\nu^2}{c^2}\frac{h\nu}{e^{h\nu/kT}-1}\), which we then have to integrate.
\(\frac{4\pi}{c^2}\int^\infty_0\frac{h\nu^3}{e^{h\nu/kT}-1}d\nu\)
We define \(u=h\nu/kT\).
\(\frac{4\pi}{c^2}\int^\infty_0\frac{h(ukT/h)^3}{e^u-1}(kT/h)du\)
\(\frac{4\pi k^4T^4}{c^2h^3}\int^\infty_0\frac{u^3}{e^u-1}du\)
The integral is defined as \(\frac{\pi^4}{15}\), so we can pull that out and define \(\frac{4\pi^5k^4}{15c^2h^3}=\sigma\) (a constant).
The final equation is \(F(T)=\sigma T^4\)

b) The Wein Displacement Law: Convert the units of the blackbody intensity from \(B_\nu(T)\) to \(B_\lambda(T)\). 

The relation between the blackbody intensity in terms of frequency and the intensity in terms of wavelength is defined such that the integral of the two equations are equal.
\(\int^\infty_0B_\nu d\nu=\int^0_\infty B_\lambda d\lambda=-\int^\infty_0B_\lambda d\lambda\)
We can get rid of the integral such that \(B_\nu d\nu=-B_\lambda d\lambda\).
Since we know that \(B_\nu(T)=\frac{2\nu^2}{c^2}\frac{h\nu}{e^{h\nu/kT}-1}\) and that \(\nu=\frac{c}{\lambda}\) (and therefore that \(d\nu=\frac{-c}{\lambda^2}d\lambda\)), we can simply plug in the values to do this conversion.
\(B_\lambda(T)d\lambda=\frac{-2(c/\lambda)^2}{c^2}\frac{h(c/\lambda)}{e^{hc/\lambda kT}-1}\frac{-c}{\lambda^2}d\lambda\)
\(B_\lambda(T)d\lambda=\frac{2}{\lambda^5}\frac{hc^2}{e^{hc/\lambda kT}-1}d\lambda\)
We can cancel out \(d\lambda\) to get the final answer:
\(B_\lambda(T)=\frac{2}{\lambda^5}\frac{hc^2}{e^{hc/\lambda kT}-1}\)

c) Derive an expression for the wavelength \(\lambda_{max}\) corresponding to the peak of the intensity distribution at a given temperature T. 

In order to do this, we want to take the partial derivative in terms of wavelength of the equation we derived in part (b) and set it equal to zero to find the maximum wavelength.
\(\frac{\partial}{\partial\lambda}[\frac{2}{\lambda^5}\frac{hc^2}{e^{hc/\lambda kT}-1}]\)
By u-substitution (\(u=\frac{hc}{\lambda kT}\)), we can condense this expression:
\(\frac{\partial}{\partial u}[\frac{2}{(hc/\lambda kT)^5}\frac{hc^2}{e^u-1}]\)
After a bunch of calculus, we have found the derivative and can set it to zero.
\(\frac{2(kT)^5(hc^2)}{(hc)^5}\frac{(e^u-1)(5u^4)-(e^u)(u^5)}{(e^u-1)^2}=0\)
Solving this equation gives us \(5=\frac{ue^u}{e^u-1}\).
If we assume that u is very large, we can simplify to \(5=\frac{ue^u}{e^u}\) yielding the final result that \(u=\frac{hc}{\lambda kT}\approx5\). Solving for \(\lambda_{max}\) and plugging in the constants, we end with the equation \(\lambda_{max}=\frac{0.283}{T}\).

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 \(e^{h\nu/kT}\) to derive a simplified form of \(B_\nu(T)\) in this low-energy regime. 

If we assume that \(\frac{h\nu}{kT}\) is very small, we can assume that \(e^{h\nu/kT}\approx1+\frac{h\nu}{kT}\).
\(B_\nu(T)=\frac{2}{c^2}\frac{h\nu^3}{e{h\nu/kT}-1}\)
\(B_\nu(T)=\frac{2}{c^2}\frac{h\nu^3}{1+\frac{h\nu}{kT}-1}\)
\(B_\nu(T)=\frac{2\nu^2kT}{c^2}\)

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

We've already figured out that \(F(T)=\sigma T^4\). Since the luminosity is simply all of the flux over the entire surface area of the sphere of expanding radiation, we can simply multiply the flux by the surface area of a sphere to yield \(L=\sigma T^44\pi R^2\).