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 \(F_{\nu}(T)\) over all frequencies to obtain the bolometric flux emitted from a blackbody, \(F(T)\). You can do this by substituting the variable \(u\equiv h\nu /kT\). 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 \(\sigma\), which is also known as the Stefan-Boltzmann constant. If you're really into calculus, go ahead and show that \(\sigma \approx 5.7\times 10^{-5} \text{ ergs s}^{-1}\text{ cm}^{-2}\text{ K}^{-4}\).
A blackbody is an object which absorbs all radiation incident upon it and then re-radiates this energy (isotropically) in a spectrum that is uniquely determined by the blackbody's temperature.
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| Planck Curve for a Blackbody Image Credit: http://quantumfreak.com/wp-content/uploads/2008/09/black-body-radiation-curves.png |
We know that the equation for the intensity of radiation from a blackbody is given as:
\(I_\nu (T)=\dfrac{2\nu ^2}{c^2}\dfrac{h\nu}{e^{\frac{h\nu}{kT}}-1}\)
This equation gives us the energy per time, per area, per frequency, per solid angle. We want the equation for flux, \(F_\nu (T)\), which gives us energy per time, per area, per frequency (without solid angles). So to 'cancel out' the solid angles, we must integrate \(I_\nu (T)\) over all solid angles. Solid angles can be thought of as angles with a \(\theta\) dimension and a \(\phi\) dimension. Also, from Lambert's cosine law, we must scale the intensity by \(\cos{\theta}\) since the blackbody is spherical, causing a surface area on the sphere to appear smaller to an observer who is not seeing the area from a head-on angle. Due to this fact, we multiply the intensity by \(\cos{\theta}\), and since this is a spherical integral, we multiply by \(\sin{\theta}\). So, integrating all solid angles viewed by an observer (a hemisphere), we get:
\(F_\nu (T)=\displaystyle\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi} I_\nu (T)\text{ }\cos{\theta}\sin{\theta}\text{ } d\phi d\theta \)
\( =\displaystyle\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi} \dfrac{1}{2}\text{ }I_\nu (T)\text{ }\sin{2\theta}\text{ } d\phi d\theta\)
\(=\pi\text{ } I_\nu (T)\)
And now to get the bolometric flux, which is the energy per time, per area (independent of frequency), we must integrate over all frequencies.
\(F(T)=\displaystyle\int_{0}^{\infty} F_\nu (T) \text{ }d\nu\)
\(=\dfrac{2\pi h}{c^2} \displaystyle\int_{0}^{\infty} \dfrac{\nu^3}{e^{\frac{h\nu}{kT}}-1} d\nu\)
Now let \(u\equiv \frac{h\nu}{kT}\), where \(\nu=\frac{ukT}{h}\) and \(d\nu=\frac{kT}{h}du\). Substitute to get:
\(F(T)=\dfrac{2\pi h}{c^2}\left( \dfrac{kT}{h}\right)^4 \displaystyle\int_{0}^{\infty} \dfrac{u^3}{e^{u}-1} du\)
To complete this integral, which is called a Bose-Einstein integral, I used Mathematica.
\(F(T)=\dfrac{2\pi^5k^4T^4}{15c^2h^3}\)
Let's define the coefficient of \(T^4\) to be \(\sigma\), which is known as the Stefan-Boltzmann constant:
\(\sigma=\dfrac{2\pi^5k^4}{15c^2h^3}=\dfrac{2\pi^5(1.38\times 10^{-16}\text{ erg K}^{-1})^4}{15(3\times 10^{10}\text{ cm s}^-1)^2(6.626\times 10^{27}\text{ erg s})^3}\)
\(\sigma\approx 5.7\times 10^{-5}\text{ erg s}^{-1}\text{ cm}^{-2}\text{ K}^{-4}\)
b) The Wien Displacement Law: Convert the units of the blackbody intensity from \(B_{\nu}(T)\) to \(B_{\lambda}(T)\). Remember that the amount of energy in a frequency interval \(d\nu\) has to be exactly equal to the amount of energy in the corresponding wavelength interval \(d\lambda\).
The relation between wavelength and frequency is \(\lambda=\frac{c}{\nu}\), and we know that:
\(F(T)=\displaystyle\int_{0}^{\infty} \dfrac{2\pi h}{c^2}\left( \dfrac{kT}{h}\right)^4 \dfrac{u^3}{e^{u}-1} du\)
Changing \(u\) to be in terms of \(\lambda\), we get \(u= \frac{hc}{\lambda kT}\) and \(du=-\frac{hc}{\lambda^2kT}d\lambda\). Plugging these back into the expression above, we get:
\(F(T)=\displaystyle\int_{\infty}^{0}\dfrac{2\pi h}{c^2}\left( \dfrac{kT}{h}\right)^4 \dfrac{\left(\frac{hc}{\lambda kT}\right)}{e^{\frac{hc}{\lambda kT}}-1} \left( -\dfrac{hc}{\lambda^2 kT}\right)d\lambda\)
\(F(T)=\displaystyle\int_{0}^{\infty}\dfrac{2hc^2}{\lambda^5 (e^{\frac{hc}{\lambda kT}}-1)}d\lambda\)
So from this, we can conclude that \(B_\lambda (T)\) is given by:
\(B_\lambda (T)=\dfrac{2hc^2}{\lambda^5 (e^{\frac{hc}{\lambda kT}}-1)}\)
c) Derive and expression for the wavelength \(\lambda_{max}\) corresponding to the peak of the intensity distribution at a given temperature \(T\). Once you do this, again substitute \(u\equiv h\nu /kT\). 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.
First, we must find the maximum of the function \(B_\lambda (T)\) for a given temperature \(T\). To do this, we can take the derivative of the function and set it equal to zero.
\(0=\dfrac{\partial B_\lambda (T)}{\partial \lambda}\)
\(0=\dfrac{\partial}{\partial \lambda}\left(\dfrac{2hc^2}{\lambda^5 (e^{\frac{hc}{\lambda kT}}-1)}\right)\)
\(0=2hc^2\left(-5\lambda^{-6}(e^{\frac{hc}{\lambda kT}}-1)^{-1}+\frac{hc}{kT}\lambda^{-7}e^{\frac{hc}{\lambda kT}}(e^{\frac{hc}{\lambda kT}}-1)^{-2}\right)\)
Cancel out like terms and let \(u\equiv \frac{hc}{\lambda kT}\) so that we get:
\(0=-5+\dfrac{u e^{u}}{e^{u}-1}\)
Letting Wolfram Alpha find an approximation for this value graphically, we find that \(u\approx 4.965\). But \(u= \frac{hc}{\lambda kT}\), so we can solve this for the peak wavelength.
\(u= \dfrac{hc}{\lambda kT}\)
\(\lambda =\dfrac{hc}{ukT}\)
\(\lambda=\dfrac{(6.626\times 10^{-27}\text{ erg s})(3\times 10^{10}\text{ cm/s})}{4.965(1.38\times 10^{-16}\text{ erg/K}) \text{ }T}\)
\(\lambda_{peak}=\dfrac{0.3 \text{ cm K}}{T}\)
And we have now derived Wien's Displacement law.
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 \(e^x \approx 1+x\).
We know that \(B_{\nu}(T)\) is given by:
\(B_\nu (T)=\dfrac{2\nu ^2}{c^2}\dfrac{h\nu}{e^{\frac{h\nu}{kT}}-1}\)
Using the Taylor expansion for \(e^x\), we can simplify this expression to:
\(B_\nu (T)=\dfrac{2\nu ^2}{c^2}\dfrac{h\nu}{1+\frac{h\nu}{kT}-1}\)
\(=\dfrac{2\nu ^2}{c^2}\dfrac{h\nu}{\frac{h\nu}{kT}}\)
Giving us a final answer of:
\(B_\nu (T)=\dfrac{2\nu ^2}{c^2}kT\)
And now we have derived the Rayleigh-Jeans Tail. The Rayleigh-Jeans Tail is only accurate for larger wavelengths, or lower energy, and is quite inaccurate for smaller wavelengths and higher energies, as shown in the graph below. This discrepancy with observations in the smaller wavelengths is called the "ultraviolet catastrophe" since it predicts that blackbodies will emit radiation with infinite power in the shorter, ultraviolet wavelengths. This obviously is not the case, however.
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| Rayleigh-Jeans Tail and the Ultraviolet Catastrophe in comparison to the Planck Curve Image Credit: http://hyperphysics.phy-astr.gsu.edu/hbase/imgmod/uvcatas.gif |
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, \(L\).
Power is given in units of energy per time. We know the bolometric flux, which has units of energy per area per time, so we can just integrate over the area to get energy per time. However, since we already know the surface area of a sphere is \(4\pi R^2\), we can just multiply the bolometric flux by the area to find the bolometric luminosity. So we get:
I would like to thank Charles Law, Daniel Chen, and Shai Szulanski for their help in solving this problem.
Citations:
Power is given in units of energy per time. We know the bolometric flux, which has units of energy per area per time, so we can just integrate over the area to get energy per time. However, since we already know the surface area of a sphere is \(4\pi R^2\), we can just multiply the bolometric flux by the area to find the bolometric luminosity. So we get:
\(L=4\pi R^2 F(T)\)
\(L=4\pi R^2 \sigma T^4\)
Citations:
Hi Deanna, this is a good post, but the reason there is a cosine in the flux integral is NOT because of Lambert's cosine law. The cosine will still be there even if the blackbody is a point source!
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