Worksheet 5, Problem 2: Setting up important equations

The problem: 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.

Part 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 Fν(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 \(σ \approx 5.7 x 10^5 erg s^{-1} cm^{-2} K^{-4}\) . Otherwise, commit this number to memory.

In question 1 on this worksheet we found that \[F_ν(T) = \frac{2 \pi ν^3h}{c^2 (e^{ \frac{hv}{kT}} - 1 )}\] for this problem we need \[F(T) = \int_{0}^{\infty} \frac{2 \pi ν^3h}{c^2 (e^{ \frac{hv}{kT}} - 1 )}\] where \[ u = \frac{hν}{kT} \: and\: dν = \frac{h}{kT} dν\] so substituting u in for every instance of ν gives \[F(T) = \frac{2 \pi K^4}{c^2h^3} \cdot \int_{0}^{\infty} \frac{T^4 u^3}{e^u - 1} du\] as per the problem's instructions, we can set sigma equal to the integral and all constants:  \[σ  = \frac{2 \pi K^4}{c^2h^3} \cdot \int_{0}^{\infty} \frac{u^3}{e^u - 1} du\] therefore \[F(T) = σ T^4\]

Part B: The Wien Displacement Law: Convert the units of the blackbody intensity from Bν(T) to Bλ(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.

Mathematically, the problem is telling us that \[B_ν \cdot dν = B_λ \cdot dλ\] or \[B_λ = B_ν \frac{dν}{dλ}\] We also know from previous problems that \[ ν = \frac{c}{λ}\] which gives: \[\frac{dν}{dλ} = -  \frac{c}{λ^2}\] Plugging these values into the above equation gives: \[B_λ = -\frac{2 \left( \frac{c}{λ} \right)^3}{c^2} \cdot  \frac{h \left( \frac{c}{λ} \right)}{e^{\left( \frac{hc}{kTλ} \right)} -1} \cdot \frac{c}{λ^2} \] which simplifies to \[B_λ(T) = \frac{-2hc^2}{λ^5 \left( e^{\left( \frac{hc}{kTλ} \right)} -1 \right)} \]

Part C: Derive an expression for the wavelength λmax 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 \(u = \frac{hν}{kT}\). 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.

Beware: There are a lot of equations ahead.

So basically we can take the derivative of the below equation to find the maximum wavelength. \[B_λ(T) = \frac{-2hc^2}{λ^5 \left( e^{\left( \frac{hc}{kTλ} \right)} -1 \right)} \] using \[ u = \frac{hc}{kTλ} and λ = \frac{kT}{hcu} \] which gives us: \[B_λ(T) = \frac{-2K^5T^5u^5}{h^4c^3(e^u - 1)} \] taking the derivative gives: \[B_λ(T) = \frac{h^4c^3(e^u-1)(-10K^5u^4) + (2K^5T^5u^4) + 2k^5T^5u^5(h^4e^3e^u)}{(h^4c^3e^u - h^4c^3)^2} \] which when set equal to zero amazingly enough simplifies to \[e^u(u-5) = -5\]. Because it is difficult to solve, we can approximate u with a taylor expansion \( e^x \approx 1 + x \). \[(1-u)(u-5) = -5\]\[u=4\] Plugging back into the original equation for λ gives the maximum wavelength: \[λ_{max}= \frac {hc}{4kT} \]

Part 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 \(e^{\frac{hν}{kT}}\) to derive a simplified form of Bν(T) in this low-energy regime. (HINT: The Taylor expansion of \( e^x \approx 1 + x \) ).
If \[e^x \approx 1 + x \] then \[e^{\left( \frac{hν}{kT} \right)} \approx 1 +\frac{hν}{kT} \] plugging this into the given equation gives: \[B_ν(T) = \frac{2ν^2}{c^2} \frac{hv}{1 + \frac{hν}{kT} -1} \] which simplifies to:  \[B_ν(T) = \frac{2ν^2kT}{c^2}\].

Part 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.

We know the units of F is \(\frac{erg}{s\:cm^2}\) and L is \(\frac{erg}{s} \) and A is \(cm^2\)

We can therefore conclude that we can use the relationship \[F= \frac{L}{A} \] where A is the area where the blackbody is isotropically emitting energy with the radius, R, being the distance away from the blackbody so \[A_r = 4 \pi R^2\] which gives \[L= \frac{F}{ 4 \pi R^2}\] and \[ L = σ T^4 4 \pi R^2\]

Acknowledgements: A huge thank you to Barra for doing this problem with me and to John for guiding us in TALC.