Blackbody Flux and Intensity

 Worksheet 3 Question 2

A blackbody is a hypothetical object that absorbs radiation and does not reflect anything. The absorbed radiation becomes standing waves which bounce back and forth inside the blackbody. The absorbed radiation causes the blackbody to reach an equilibrium temperature. A true blackbody does not exist, but a spherical cow can be approximated as a blackbody and a better approximation of a blackbody is the entire universe (Figure 1). Stars and other objects can also be approximated as blackbodies and their intensity and wavelength of light emitted can be plotted on the blackbody curve to learn information such as temperature and wavelength using Wein's displacement Law and temperature and energy using Stefens Law. The blackbody curve is represented by Raleigh Jean's Law, but this law does not line up with experimental data for short wavelengths of light which is known as the ultraviolet catastrophe (Figure 2). Max Planck had the idea that inside the blackbody cavity walls, the oscillations of the waves only transferred energy in discrete quanta which works when light is described as particles. (Info from 6.2: Blackbody Radiation - Physics LibreTexts)

a) Bolometric flux is the energy per area per time, independent of frequency. Integrate the blackbody flux $F_v(T)$ over all frequencies to obtain the bolometric flux emitted from a blackbody F(T). You can do this by substituting the variable u=$\frac {hv}{kT}$. 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 known as the Stefan Boltzmann constant. 

$F_v(T)= \pi \times \frac{2 \nu ^2}{c^2}\times \frac{h \nu}{e^u-1}$

Now we integrate over all frequencies

$\int_{0}^{\infty} F_\nu (T) ,d\nu$

We can take out the constants and substitute the variable u=$\frac {hv}{kT}$

$ \frac {2h \pi}{c^2} \int_{0}^{\infty} \frac{\nu^3}{e^{\frac{h \nu}{kT}}-1} ,d \nu$

$u= \frac{h\nu}{kT}$ $\nu= \frac{ukT}{h} d\nu =\frac{kT}{h}du$

$F_v(T)=\frac{2h \pi}{c^2}$ $\int_{0}^{\infty} \frac{u^3 \frac{K^3T^3}{h^3}}{e^{u}-1} ,du \times \frac{KT}{h}$

$F_v(T)= \frac{2k^4T^4 \pi}{c^2h^3} \times \int_{0}^{\infty} \frac{u^3}{e^u-1} ,du$ $= \sigma T^4$

The Stefen Boltzmann constant is 5.7 $\times 10 ^{-5} erg s^{-1} cm ^{-2} K^{-4}$

b) Convert the units of the blackbody intensity from $B_v(T)$ to $B_{\lambda} $ The amount of energy in a frequency interval dv has to be exactly equal to the amount of energy in the corresponding wavelength interval d $\lambda$

$B_{\nu}=\times \frac{2 \nu ^2}{c^2}\times \frac{h \nu}{e^{\frac h \nu}{KT}-1}$

Given that the amount of energy in a frequency interval dv has to be exactly eqyal to the amount of energy in the corresponding wavelength interval d $\lambda$, we can make this equation:

$B_{\nu}(T) d\nu = B_{\lambda} (T) d \lambda$

$B_{\lambda}(T)= \frac {B_{\nu}(T) d\nu}{d_{\lambda}}$

So we should solve for $\frac{d\nu}{d\lambda}$

$c= \lambda \nu$

$0=\lambda \times \frac {d\nu}{d\lambda} + \nu$

$\frac{d\nu}{d\lambda}= \frac {-\nu}{\lambda}$

So to get $B_{\lambda}(T)$ we multiply $B_{\nu}(T) by \frac{d\nu} {d_{\lambda}}$

$B_{\lambda}$= $\frac{2\nu ^3}{c \lambda}$ $\frac{h \nu^2}{e^{\frac{h\nu}{KT}}-1}$

Now we can get rid of $\nu$ by substituting $\nu=\frac{c}{\lambda}$

$B_{\lambda}$= $\frac{2c^2}{\lambda^5}$ $\times$ $\frac{h}{e^{\frac{hc}{2KT}}-1}$

c) Wein displacement law: Derive an expression for the wavelength $\lambda_{max}$ corresponding to the peak of intensity distribution at a given temperature T. 

We find the maximum of the function by  finding the derivative and setting it equal to 0 because at the maximum, the slope is 0.

$B_{\lambda}= \frac{2c^2}{\lambda^5} \frac{h}{e^{\frac{hc}{2KT}}-1} d\lambda=0$

$5\lambda= (5\lambda- \frac{hc}{KT})\times e^{\frac{hc}{KT}}$

$e^{\frac{hc}{KT}}$ is approximately $1+\frac{hc}{KT}$

$5\lambda= (5\lambda- \frac{hc}{KT})(1+\frac{hc}{\lambda KT})=5\lambda - \frac{hc}{KT}+\frac{5hc}{KT}-\frac{h^2c^2}{K^2T^2\lambda}$

$4=\frac{hc}{KT \lambda}$

$\lambda=\frac{hc}{4KT}$

$\lambda_{max} T=\frac{hc}{4K}$

d) The Rayleigh-Jeans Tail: Consider photon energies that are much smaller than the thermal energy. Use a first order Taylor expression on the term $e^{\frac{hc}{\lambda kT}}$ to derive a simplified form of $B_{\lambda}(T)$ in this low energy regime.

$e^{\frac{hc}{\lambda kT}}=e^u$ is approximately $1+u$

$\frac{2c^2}{\lambda^5}\frac{h}{(1+u)-1}$

$\frac{2c^2}{\lambda^5}\frac{h\lambda kT}{\lambda ^4}$

$B_{\lambda}(T)= \frac{2kTc}{\lambda ^4}$

e) Write an expression for the total power output of a blackbody with radius R starting with the expression for $F_v$. This total energy output per unit time is also known as the bolometric luminosity L.

$Power$= $\frac{Energy}{Time}$

$F_\nu$= $\frac{Energy}{time \times Area}$ $\times Area = \frac{Energy}{time}$

So we need to multiple $F_v$ by Area to find power.

The energy output per unit time is also bolometric so we will need to integrate over all wavelengths.

As we saw in part a) $F_v(T)= \pi \times \frac{2 \nu ^2}{c^2}\times \frac{h \nu}{e^u-1}$

Now we integrate over all frequencies

$\int_{0}^{\infty} F_\nu (T) ,d\nu$

$F_v(T)= \sigma T^4$

Power= $\sigma T^4 \times Area$

f) You observe two gravitationally bound stars (a binary pair). One is blue and one is yellow. The yellow star is six times 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).

$L_B$= luminosity of blue star

$L_Y$= luminosity of yellow star

$L_B=6L_Y$

$\lambda_B= wavelength .of. blue. star= about. 450.nm$

$\lambda_Y= wavelength. of .yellow .star= about .580.nm$

$L= \frac{surface .area}{Flux}= \frac{surface. area}{\sigma T^4}$

F(T)= $\frac{Energy}{Area \times time \times \nu}$

$\nu=\frac{c}{\lambda}$ so F(T)= $\frac{Energy \times \lambda}{Area \times time \times c}$

$\lambda_{max}= \frac{b}{T}$

The blue star is hotter than the yellow star because it has a smaller wavelength. The blue star has a greater luminosity so the blue star must have greater surface area which means greater radius.

The temperature of the blue star is about 6,400 K

The temperature of the yellow star is 5,000 K

$\frac{6L_Y}{L_Y}$=$\frac{R_B^2}{\sigma T_B^4} \times \frac{\sigma T_Y^4}{R_L^4}$

$\frac{R_B}{R_Y}$=$(\frac{\sigma T_B^4}{6 \sigma T_Y^4})^{1/2}$

The ratio of radii of the stars is

$\frac{R_B}{R_Y}$= 0.7