Let’s get into the math of Stellar Temperatures

We are bounded in a nutshell of Infinite Space: Worksheet # 5, Problem #2: Let’s get into the math of Stellar Temperatures

2. 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 this term, and throughout tour astronomy career.

a. In astronomy, it is often useful to deal with something called the “bolometric flux”, ot the energy per area per time, independent of frequency. Integrate the blackbody flux $F_\nu (T)$ over all the frequencies to obtain the bolometric flux emitted from a blackbody, $F (T)$. You can do this using u substitution for the variable $u \equiv hv/kT$ . This will allow you to split thing s into a temperature dependent term, and a term comprising an integral over all frequencies. However, rather than solving for the integral, just set all the constants to a new, single constant called $\sigma$ , which is also known as the Stefan-Boltzmann constant, and retain the dependence on the temperature T. If you’re really into calculus, go ahead and show that $\sigma \approx 5.7 \times 10^{-5} erg~s^{-1}~cm^{-2}~K^{-4}$.  Otherwise, commit this number to memory.

b. The Wein 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 $d\lambda$.

c. Derive the 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 transcendental, but you can solve it easily to the 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 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 $e^x \approx 1 + x$ .)

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.

a) As we have seen in less detail at other points, blackbodies are a good approximation for most objects in the universe. The most “perfect” blackbody we have found is actually the CMB (Cosmic Microwave Background), the afterglow of the Big Bang. Based on a relationship between the frequency/wavelength of the observed light and the temperature of the light, we find an equation which can relate these two, as well as be derived and anti-derived to give broader descriptions.  

From a previous problem, we know that a description of the Flux of a blackbody, any object which absorbs and releases thermal radiation, is represented by: $$F_\nu (T) = \frac{2\pi}{c^2} \frac{\nu^3 h}{e^{h\nu/kT} -1} ,$$ so if we integrate over all possible frequencies, going to infinity, we have: $$\int _0 ^\infty F_\nu (T) = \frac{2\pi h }{c^2} \int _0 ^\infty \frac{\nu^3 }{e^{h\nu/kT} -1} d\nu ,$$ where we can substitute $$u = \frac{h\nu}{kT}$$ and $$u\frac{kT}{h} = \nu$$ as well as $$du = \frac{h}{kT},$$ into $$F(T) = \int _0 ^\infty \frac{\left(\frac{kT}{h}\right) u^3 }{e^{u} -1} \left(\frac{kT}{h}\right)du$$ which can be more easily integrated once the constants are completely separated: $$F(T) = \frac{2\pi h }{c^2} \left(\frac{kT}{h}\right)^4 \int _0 ^\infty \frac{ u^3 }{e^{u} -1} du,$$ and knowing a particular integration of a strange equation, this all becomes much easier $$\int _0 ^\infty \frac{ u^3 }{e^{u} -1} du = \frac{\pi^4}{15}$$ $$F(T) = \frac{2\pi k^4 }{c^2 h^3} T^4 \cdot \frac{\pi^4}{15}.$$ As is stated in the problem, all the constants can be grouped together and you would find that we come remarkably close to the actual value of the  Stefan-Boltzmann constant, although I suspect we would get a closer value were we to complete the integration by ourselves, as difficult as that may be. Nevertheless, the constants all become a single term, which simplifies the Bolometric Flux equation into:   $$F(T)= \sigma T^4$$

b) From the problem, we know we are simply converting the specific intensity according to frequency, to specific intensity according to wavelength. Therefore, if they are both the same value, but in different metrics, they can simply be equaled to the other and then attempt to integrate both, with parameters which leave us with their derivatives with respect to the parameter: $$-B_\lambda d\lambda = B_\nu d\nu ,$$ (we add a negative sign to establish how they are inversely proportional) and by placing all on one side:   $$-B_\lambda = -B_\nu \frac{d\nu}{d\lambda} ,$$  and after simplifying it down:  $$B_\lambda = B_\nu \frac{d\nu}{d\lambda} ,$$ we just find the derivative of $\nu = \frac{c}{\lambda}$ in terms of $\lambda$: $$B_\lambda = B_\nu~ \frac{c}{\lambda^2},$$ and so applying it to the original intensity equation (remember to change all $\nu$’s into $c/\lambda$) : $$B_\lambda = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda kT} – 1} .$$

c) Now, to maximize the equation, we have to derive the function and set it equal to ):   $$\frac{d ~B_\lambda}{d\lambda} = \frac{d \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda kT} – 1}}{d\lambda} ,$$ which can now be interpreted with partial fractions (only deriving with respect to a specific variable: $$\frac{\partial B_\lambda}{\partial d\lambda} = \frac{\partial B\lambda}{\partial u} \frac{\partial u}{\partial \lambda}$$ and thus we can define u as $$u = \frac{hc}{kT\lambda},$$ and we place thisn into the equation and get:   $$B_u = 2hc^2 \frac{\left(\frac{ukT}{hc}\right)^5}{e^u – 1} ,$$ and now we take the derivative with respect to u on both  sides and get (with some separation of constants and application of the product rule): $$\frac{d B_u}{du} = 2hc^2 \left(\frac{kt}{hc}\right)^5  (5u^4(e^u -1 ) – u^5(e^u)),$$ so we now take it and optimize it by setting it to 0: $$0 = 2hc^2 \left(\frac{kt}{hc}\right)^5  (5u^4(e^u -1 ) – u^5(e^u)),$$ we eliminate all the constants and begin solving the equation: $$5u^4 (e^u -1 ) = u^5 ~e^u$$ $$5 = \frac{ue^u}{e^u -1 },$$ and now we can make a first order expansion and simplify the equation: $$5 = \frac{u e^u}{e^u}$$ $$5 = u$$

Knowing what we defined to be u: $$u = \frac{hc}{kT\lambda}$$ and solved to find $\lambda$, we see: $$\lambda = \frac{hc}{ukT},$$ and now we define our constants, the Plank constant: $$h =6.63 \times 10^{-27} erg \cdot s,$$ Boltzmann’s constant: $$k = 1.38 \times 10^{-16} erg/K,$$ and the speed of light:   $$c = 3 \times 10^{10} cm/s,$$ we solve and find that:   $$\lambda_{max} = \frac{6.63 \times 10^{-27} erg \cdot s \cdot 3 \times 10^{10} cm/s }{ 5 \cdot 1.38 \times 10^{-16} erg/K  \cdot T}$$ $$\lambda_{max} = \frac{0.29 cm \cdot K}{T }$$

d) This part of the problem is simply using a basic equation and expanding it to see how it acts in small frequencies, $$B_\nu(T) = \frac{2\nu^3 h }{c^2 (e^{h\nu / kT} -1)}$$ and the expansion is given to us as: $$e^x \approx 1 + x$$ which, plugged in, would just be:   $$B_\nu(T) = \frac{2\nu^3 h }{c^2} \frac{1}{1 + \frac{h\nu}{ kT} -1},$$ and thus we are left with: $$B_\nu(T) = \frac{2\nu^3 h }{c^2} \frac{1}{\frac{h\nu}{ kT}}$$

e) Now, just applying what we learned in part a), we know how $$F_\nu (T) = \frac{2\pi}{c^2} \frac{\nu^3 h}{e^{h\nu/kT} -1} ,$$ can be integrated over all frequencies and we find: $$F(T) = \sigma T^4,$$ then, we have to understand how Flux is distributed isotopically. This means we have to understand the entire area, this being: $$A = \int^{2\pi}_0 \int^\pi_0 R^2 \sin\theta d\theta d\phi$$ $$A= \int^{2\pi}_0  2R^2 d\phi$$ $$A = 4\pi R^2$$ So by multiplying both sides of the $F(T)$ equation by the complete area of the light expansion, we have: $$4\pi R^2 F = 4\pi R^2 \sigma T^4,$$ and Flux in all directions at all times is simply the Luminosity, as such: $$L = 4\pi R^2 \sigma T^4,$$ the relationship between Luminosity and Temperature.