(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 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.
(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.
(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 = 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.
(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 (T) in this low-energy regime. (HINT: The Taylor expansion of ex « 1 ` x.)
(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.
(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.
(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 = 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.
(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 (T) in this low-energy regime. (HINT: The Taylor expansion of ex « 1 ` x.)
(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.
Let's start with a brief discussion of blackbodies. These are theoretical (yet highly useful) objects that absorb all light that is incident upon them with out reflection it. They also give off radiation based on their temperature. Most things can be be approximated as black bodies, even humans and stars.
(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 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.
We know that frequency specific flux has the formula:
\[B_{\nu}(T)=\pi\frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}\]
We want to integrate this over all frequencies to find \(F_{\nu}\) so:
\[F_{\nu}(T)=\int^\infty_0B_{\nu}(T)d\nu=\int^\infty_0\pi\frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}d\nu\]
Let's make a u-substitution of \(u=\frac{h\nu}{kT}\) and \(du=\frac{h}{kT}d\nu\)
\[\int^\infty_0\pi\frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}d\nu=\frac{2\pi k^4T^4}{c^2h^3}\int^\infty_0\frac{u^3}{e^u-1}du\]
Now comes the fun part: We can set everything except temperature to equal a constant \(\sigma\), so:
\[\boxed{F(T)=\sigma T^4}\]
(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.
So we are back to specific intensity.
Now comes the fun part: We can set everything except temperature to equal a constant \(\sigma\), so:
\[\boxed{F(T)=\sigma T^4}\]
(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.
So we are back to specific intensity.
\[B_{\nu}(T)=\frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}\]
We know that \(\nu=\frac{c}{\lambda}\), but there is a catch, we need to make sure that the energy in a \(d\lambda\) equals the energy in a \(d\nu\). Thus \(B_{\lambda}(T)d\lambda=B_{\nu}(T)d\nu\) or \(B_{\lambda}(T)=B_{\nu}(T)\frac{d\nu}{d\lambda}\)
This gives us \(B_{\lambda}(T)=B_{\nu}(T)\frac{c}{\lambda^2}\)
Finishing the substitution of \(\frac{c}{\lambda}=\nu\) we get:
\[\boxed{B_{\lambda}(T) = \frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)}}\]
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 = 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.
To find a maximum value of a function, we look for where it's first derivative is zero. So:
\[\frac{d}{d\lambda}B_{\lambda}(T) =\frac{d}{d\lambda}\frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)}\]
\[B'_{\lambda}(T)=\frac{-10hc^2}{\lambda^6(e^{\frac{hc}{\lambda kT}}-1)}+\frac{2h^2 c^3 e^{\frac{hc}{\lambda kT}}}{\lambda^7 kT(e^{\frac{hc}{\lambda kT}}-1)^2}=0\]
To simplify, we can remove anything that we can factor from the two expressions since it equals zero simplifying the equation to:
\[B'_{\lambda}(T)=-5+\frac{h c e^{\frac{hc}{\lambda kT}}}{\lambda kT(e^{\frac{hc}{\lambda kT}}-1)}=0\]
That looks slightly less scary. Now lets do another u-substitution where \(u=\frac{hc}{\lambda kT}\) giving us:
\[5=\frac{u e^u}{e^u-1}\]
Taylor approximations tell us that \(e^u \approx 1+u\) so:
\[5=\frac{u+u^2}{u}=1+u\]
\[u=4=\frac{hc}{\lambda kT}\]
So at the end of it all, the maximum wavelength is is given by:
\[\boxed{\lambda_{max}(T)=\frac{hc}{4kT}}\]
(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 ekT to derive a simplified form of Bν (T) in this low-energy regime. (HINT: The Taylor expansion of ex « 1 ` x.)
So, we start with:
\[B_{\nu}(T)=\frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}\]
If \(e^x \approx 1+x, e^{\frac{h/nu}{kT}} \approx 1+\frac{h/nu}{kT}\)
Plugging this in, we get:
\[B_{\nu}(T) \approx \frac{2\nu^2}{c^2}\frac{h\nu}{1+\frac{h\nu}{kT}-1}\]
(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.
Earlier in this problem we already established that if one integrates \(F_{\nu}(T)d\nu\) then the rustle is \(F(T)=\sigma T^4\).
With this as a starting point, this is the flux through every area of the black body. The surface area of a sphere is \(4\pi r^2\), so the total luminosity is:
\[\boxed{L=4\pi r^2 \sigma T^4}\]
This problem was done with collaboration with G. Grell.
This gives us \(B_{\lambda}(T)=B_{\nu}(T)\frac{c}{\lambda^2}\)
Finishing the substitution of \(\frac{c}{\lambda}=\nu\) we get:
\[\boxed{B_{\lambda}(T) = \frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)}}\]
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 = 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.
To find a maximum value of a function, we look for where it's first derivative is zero. So:
\[\frac{d}{d\lambda}B_{\lambda}(T) =\frac{d}{d\lambda}\frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)}\]
\[B'_{\lambda}(T)=\frac{-10hc^2}{\lambda^6(e^{\frac{hc}{\lambda kT}}-1)}+\frac{2h^2 c^3 e^{\frac{hc}{\lambda kT}}}{\lambda^7 kT(e^{\frac{hc}{\lambda kT}}-1)^2}=0\]
To simplify, we can remove anything that we can factor from the two expressions since it equals zero simplifying the equation to:
\[B'_{\lambda}(T)=-5+\frac{h c e^{\frac{hc}{\lambda kT}}}{\lambda kT(e^{\frac{hc}{\lambda kT}}-1)}=0\]
That looks slightly less scary. Now lets do another u-substitution where \(u=\frac{hc}{\lambda kT}\) giving us:
\[5=\frac{u e^u}{e^u-1}\]
Taylor approximations tell us that \(e^u \approx 1+u\) so:
\[5=\frac{u+u^2}{u}=1+u\]
\[u=4=\frac{hc}{\lambda kT}\]
So at the end of it all, the maximum wavelength is is given by:
\[\boxed{\lambda_{max}(T)=\frac{hc}{4kT}}\]
(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 ekT to derive a simplified form of Bν (T) in this low-energy regime. (HINT: The Taylor expansion of ex « 1 ` x.)
So, we start with:
\[B_{\nu}(T)=\frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}\]
If \(e^x \approx 1+x, e^{\frac{h/nu}{kT}} \approx 1+\frac{h/nu}{kT}\)
Plugging this in, we get:
\[B_{\nu}(T) \approx \frac{2\nu^2}{c^2}\frac{h\nu}{1+\frac{h\nu}{kT}-1}\]
\[B_{\nu}(T) \approx \frac{2\nu^2}{c^2}\frac{h\nu}{\frac{h\nu}{kT}}\]
\[\boxed{B_{\nu}(T) \approx \frac{2\nu^2kT}{c^2}}\]
(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.
Earlier in this problem we already established that if one integrates \(F_{\nu}(T)d\nu\) then the rustle is \(F(T)=\sigma T^4\).
With this as a starting point, this is the flux through every area of the black body. The surface area of a sphere is \(4\pi r^2\), so the total luminosity is:
\[\boxed{L=4\pi r^2 \sigma T^4}\]
This problem was done with collaboration with G. Grell.
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