This question is, in essence, a look at the math underlying my post from a couple weeks ago on the Ultraviolet Catastrophe, examining how the Planck function describes a blackbody. We're going to jump right in with that "frightening mass of algebra" from before, the equation for intensity of radiation emitted from a blackbody. $I_{\nu}(T)$.\[I_{\nu}(T) = \frac{2 \nu^2}{c^2} \frac{h \nu}{e^{\frac{h \nu}{kT}}-1} \equiv B_{\nu}(T)\: (a.1)\] It's good to have a sense of what this looks like graphically, so here's a quick refresher on what the Planck function actually looks like for a few different values of T.
This is a good image to have in the back of your head as we work through this. (Source)
Integrating this over a hemisphere centered on the blackbody turns intensity into the blackbody flux (energy per unit time per unit area) $F_{\nu}(T) = \pi I_{\nu}(T)$. Given \[F_{\nu}(T) = \pi \frac{2 \nu^2}{c^2} \frac{h \nu}{e^{\frac{h \nu}{kT}}-1}\: (a.2)\] we can make the substitution $u = \frac{h \nu}{kT}$, which we can solve for $\nu$ and $d\nu$ such that $\nu = \frac{ukT}{h}$ and $d\nu = \frac{kT}{h}du$. This gives us \[F_{\nu}(T) = 2\pi h c^{-2}(\frac{ukT}{h})^3(e^u-1)^{-1}\:(a.3)\] We now want to integrate this across all frequencies, from $\nu = (0, \infty)$. Upon setting up the integral we will pull out all constants not dependent on $\nu$ (note that because $u = \frac{h \nu}{kT}$, our bounds do not change).\[\frac{2\pi k^4 T^4}{h^3 c^2} \int_0^\infty \frac{u^3}{e^u-1}du\:(a.4)\] While it is true that $u$ is dependent on temperature, and so we have not, in the strictest sense, separated $T$ out into its own term, the integral actually works out in such a way that the dependence on $T$ is irrelevant when integrated from zero to infinity (note that if we were integrating between two specific frequencies this will not be the case). As it happens, the integral combined with all of its coefficients save for $T^4$ comes out to $\sigma$, leaving us with $F_{\nu}(T) = \sigma T^4$.
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 $d\nu$ has to be exactly equal to the amount of energy in the corresponding wavelength interval $d\lambda$.
What we're doing here is deriving a crucially important equation, one which allow us to tell the wavelength of an object's peak emission based solely on its temperature. As the graph of the Planck function above shows, the warmer a blackbody is, the more the peak of its emission is shifted left (to shorter wavelengths). Since we can approximate pretty much any object in space as a blackbody, this paves the way for a pretty powerful mathematical relationship, which can tell you how hot an object is--whether it's a star, a planet, or something else--just by looking at where it gives off the most light.
A comparison of the peak emissions of the Sun and Earth. (Source)
This would at first seem like a straightforward problem, where we would need to apply the old, familiar relation $\nu \lambda = c$. However, the last part hints at an additional layer of complexity. Rather than simply plugging in $c/\lambda$ for every $\nu$ in $B_{\nu}(T)$ and calling it $B_{\lambda}(T)$, we must actually relate $B_{\nu}(T)d\nu$ and $B_{\lambda}(T)d\lambda$ in order to account for the fact that, because $\nu = c/\lambda$, $d\nu$ is actually equal to $-\frac{c}{\lambda^2}d\lambda$. Also worth noting is that, because an increase in frequency corresponds to a decrease in wavelength, $B_{\nu}(T)d\nu$ is actually equal to $-B_{\lambda}(T)d\lambda$.
We'll once again start with $B_{\nu}(T)$, this time tacking a $d\nu$ onto the end of it.\[B_{\nu}(T)d\nu = \frac{2 \nu^2}{c^2} \frac{h \nu}{e^{\frac{h \nu}{kT}}-1}d\nu \: (b.1)\] Now, taking into consideration the complications noted above, we'll substitute for $\nu$ and $d\nu$, giving us \[B_{\lambda}(T)d\lambda = -\frac{2 (\frac{c}{\lambda})^2}{c^2} \frac{h \frac{c}{\lambda}}{e^{\frac{hc}{\lambda kT}}-1}(-\frac{c}{\lambda^2}d\lambda)\:(b.2)\] That looks like a huge mess, but fortunately we can simplify it into something a little more useful. Combining like terms and canceling in the first part gives us \[B_{\lambda}(T)d\lambda = -\frac{2hc}{\lambda^3(e^{\frac{hc}{\lambda kT}}-1)}(-\frac{c}{\lambda^2}d\lambda)\:(b.3)\] Finally, we can distribute the negative signs and toss the $\frac{c}{\lambda^2}$ in with the rest, giving us \[B_{\lambda}(T)d\lambda = \frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)}d\lambda\:(b.4)\] Canceling out the $d\lambda$ on each side gives us an expression for $B_{\lambda}(T)$.
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 $u \equiv h\nu/kT$. 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.
Straining back to first-semester calculus, we can recall that in order to find the maximum (or minimum) of a function, we find where its first derivative is equal to zero. We can ignore all the complicating factors that calculus teachers like to throw at you, since (having seen the Planck function before) we know that there is only one point on any curve where the derivative is zero--at its maximum.\[\frac{d}{d\lambda}B_{\lambda}(T) = \frac{d}{d\lambda}\frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)} = \frac{d}{d\lambda}2hc^2\lambda^{-5}(e^{\frac{hc}{\lambda kT}} -1)^{-1}\:(c.1)\] The second step simply being a rearranging in order to make the equation a little more derivative-friendly. We're going to have to pull out all the stops here (power rule! product rule! chain rule!), so let's simplify things visually for the time being by letting $a \equiv hc/kT$. Taking derivatives gives us \[\frac{d}{d\lambda}B_{\lambda}(T) = 2hc^2[-\frac{5}{\lambda^6(e^{\frac{a}{\lambda}})} + \frac{ae^{\frac{a}{\lambda}}}{\lambda^7kT(e^{\frac{a}{\lambda}}-1)^2}]\:(c.2)\] We now set this equal to zero, divide by $dhc^2$, and add across, giving us \[\frac{5}{\lambda^6(e^{\frac{a}{\lambda}}-1)} = \frac{ae^{\frac{a}{\lambda}}}{\lambda^7(e^{\frac{a}{\lambda}}-1)^2}\:(c.3)\] We can now multiply both sides by $\lambda^6(e^{\frac{a}{\lambda}}-1)$, which simplifies things down to \[5 = \frac{ae^{\frac{a}{\lambda}}}{\lambda(e^{\frac{a}{\lambda}}-1)} \: (c.4)\] Finally, let's pull $\frac{hc}{\lambda kT}$ back out of $a$ so that our equation becomes \[5 = \frac{\frac{hc}{\lambda kT} e^{\frac{hc}{\lambda kT}}}{e^{\frac{hc}{\lambda kT}}-1} \: (c.5)\] That looks a whole lot messier, but we can now employ the substitution mentioned in the instructions, $x \equiv \frac{hc}{\lambda kT}$, which is dimensionless. That turns the ugly mess above into something a lot less frightening: \[5 = \frac{xe^x}{e^x-1} \: (c.6)\] That might be a lot prettier, but unfortunately it's not actually solvable, so let's turn to a Taylor approximation to give us something to go on. The Taylor expansion of $e^x$ is $1+x+x^2/2...$ but we're only going to bother with the first order approximation, and plug in $1+x$ for our two $e^x$ terms. That turns $(c.6)$ into \[5 \approx \frac{x(1+x)}{1+x-1} \approx {1+x} \: (c.7)\] Or, more simply, $x \approx 4$. From there, we can plug $\frac{hc}{\lambda kT}$ back in for $x$, then solve for $\lambda$ (which is really $\lambda_{max}$), giving us, at long last\[\lambda_{max} \approx \frac{4hc}{kT} \: (c.8)\] Having come all this way, it's worth making sure that at least our units make sense, so let's do a quick sanity check. Things will need to work out to units of m (length), since we solved for wavelength. On top, we have (erg $\times$ s $\times$ m $\times$ s$^{-1}$), while on the bottom we have (erg $\times$ T $\times$ T$^{-1}$). Ergs cancel, as does temperature (T) and time (s), leaving us with just the m term. Looks good!
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.
Astonishingly, some people feel like the full expression for $B_{\nu}(T)$ is a little bit unwieldy, so it's not a bad idea to have some approximations on hand that we can use for specific regions of the Planck function. The Rayleigh-Jeans law attempted to use classical mechanics to arrive at the Planck function as a whole, but only succeeded in developing an approximation for low-energy photons--classical mechanics actually dictated that as wavelength dropped, energy would go to infinity (and, probably fortunately, we have yet to observe an infinite-energy photon). The explanation for the discrepancy between classical mechanics and observation (known, of course, as the ultraviolet catastrophe), was an early stage in the development of quantum mechanics.
The Very Large Array. Approximations based on the Rayleigh-Jeans law are still used for formulas used in radio astronomy. (Source)
We're thus working somewhat backwards in this case, but let's start with the full equation for $B_{\nu}(T)$ and try to break it down into a simpler expression for very low $\nu$. Remembering that \[B_{\nu}(T) = \frac{2\nu^2}{c^2}{h\nu}{e^{\frac{h\nu}{kT}}-1} \: (d.1)\] we use a Taylor expansion to simplify the troublesome $e^{\frac{h\nu}{kT}}$ term. This time we'll think of $\frac{h\nu}{kT}$ as $x$ in a Taylor expansion of $e^x$. This gives us the first-order Taylor expansion of $e^{\frac{h\nu}{kT}}$ as $1+\frac{h\nu}{kT}$. Plugging this in for $e^{\frac{h\nu}{kT}}$ gives us \[B_{\nu}(T) \approx \frac{2\nu^2}{c^2}\frac{h\nu}{1+\frac{h\nu}{kT}-1}\: (d.2)\] Doing a bit of canceling then gives us \[B_{\nu}(T) \approx \frac{2kT\nu^2}{c^2} \: (d.3)\] Which, fortunately, is the Rayleigh-Jeans law in terms of frequency!
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, $L$.
Let's pick up at that nasty integral in a), equation $(a.4)$, recalling that that equation is the blackbody flux of an entire object--its emissions integrated across all frequencies, measured in units of energy per time per unit area. What we need to do now is turn it into luminosity--the total output of the blackbody, irrespective of area. This is simpler than it sounds. We can assume our blackbody to be a sphere, which we know to be of radius $R$. Eliminating the unit area element of blackbody flux is a simple matter of multiplying it by the surface area of our sphere, so all we need to do is put $4\pi R^2$ out in front. Thus, \[L = 4\pi R^2 \times \frac{2\pi k^4 T^4}{h^3 c^2} \int_0^\infty \frac{u^3}{e^u-1}du = 4\pi R^2T^4\sigma\:(e.1)\] Once again employing our favorite method of double-checking (sorry, dimensional analysis), we can see that this is indeed the formula for the luminosity of a spherical blackbody.
f) You observe two gravitationally bound stars (a binary pair). One is blue and one is yellow. The yellow star is brighter than the blue star. Qualitatively compare their temperature and radii, i.e. which is hotter, which is smaller? Next, quantitatively compare their radii (to 1 significant figure).
As we can see visually on the graph of the Planck function, and mathematically from Wien's Displacement Law, the hotter an object is, the shorter the wavelength of the peak of its emission curve. Blue light, at roughly $450$ nm $= 4.5 \times 10^{-7}$ m, has a shorter wavelength than yellow light, coming in at around $550$ nm $= 5.5 \times 10^{-7}$ m. Thus, we would expect the blue star to be hotter and brighter than the yellow one, as is typically the case when looking at main sequence stars.
The Hertzsprung-Russell Diagram: Temperature increases (and wavelength decreases) to the left, while brightness increases moving upward. The main sequence is the line going diagonally across. (Source)
However, we know that in this case, the yellow star is actually brighter than the blue one. If we're assuming that they're the same temperature, that would be impossible (unless Planck, and most of modern astronomy, were way off the mark). However, we can account for this by assuming that the yellow star is significantly larger than the blue one--thus, its temperature might be lower, and thus its blackbody curve shifted to the right of the blue star's, but it manages to be brighter just by pumping out a huge number of lower-energy photons. While this wouldn't really work for main sequence stars, you can see looking at the H-R diagram that this situation, while unusual, is not totally out of the question if we were to be dealing with a dimmer blue star and a yellow giant (or supergiant) star.
A yellow supergiant: Polaris A, an unusually interesting star--as well as being the north star it's a triple star system and the closest Cepheid Variable star to Earth (Source)
The blue star is still hotter--there's no getting around the Planck function--but if we beef up its yellow companion so that it's significantly bigger, it's still possible for it to be brighter.
Excellent post, Tom! Good explanations, flowing writing, clear math and helpful illustrations. It's pretty much perfect but for one very minor point: For the last part of the problem, I know we asked you to do a quantitative comparison even though we didn't provide the ratio of brightnesses - our mistake. Still, it would be interesting to see the dependence of the radius on this quantity (call it A) - is it a log, an exponential, a square root?
ReplyDelete