Black Bodies and Thermal Radiation

 Worksheet 3.1

Mike Cushing discovered a new type of astrophysical object called a Y dwarf. Y dwarfs are a subclass of brown dwarf. Consider a Y dwarf, with a temperature of about 350 K and a radius of roughly the size of Jupiter's radius, residing near a sun like star.

Brown dwarfs have a large mass but not large enough to be a star. They are mainly composed of hydrogen gas and have no internal energy source that stars have. They emit very little visible light so they are hard to detect, even in the Infared. (Info from NASA - Brown Dwarf Detectives)

a) At what wavelength $\lambda_{max}$ should you observe to have the best chance of detecting the Y

 dwarf?

$\lambda_{max}$= $\frac {hc}{4kT}$

$\lambda_{max}$ = $\frac {0.29 cm K}{350K}$

You should observe a wavelength at $8.3 \times 10^{-4} cm$ for the best chance of detecting the Y Dwarf

b) As measured at $\lambda_{max}$ how many photons per second, per $cm^2$ emitted from a Y dwarf at a distance of 30 lightyears would reach the Spitzer space telescope? Assume you observe over a narrow range of wavelengths such that $\delta \lambda = 1 \mu m$

Our variables are:

$\delta \lambda = 10^-4 cm$

$d= 30 ly= 3 \times 10^{19} cm$

$R_{\lambda}=0.1 R_o= 7 \times 10^9 cm$

$h= 6.6 \times 10^{-27} erg s$

$K=1.4 \times 10^{-16} erg/K$

$c=3\times 10^{10} cm/s$

T=350K

We have the equation for $B_{\lambda}= \frac{2c^2}{\lambda^5} \frac{h}{e^{\frac{hc}{\lambda KT}}-1}$. This equation has units $\frac{energy}{time \times area \times wavelength \times solid angle}$ We want an answer with units of $\frac{number.of.photons}{time \times area}$ To get these units we can multiply $B_{\lambda} \times \delta wavelength \times \delta solid angle \times \frac{1}{energy .of. a. single. photon}$

$\delta \lambda$ which was given to us in the problem, is $10^-4$cm

$\delta solid angle$ can be found by using the ratio of the radius of the dwarf R to the entire sky $4 \pi d^2$. $\delta \omega = \frac {\pi R^2}{\pi d^2}\times 4 \pi = \pi \frac{R^2}{d^2}$

$\frac {1}{energy of a photon} = \frac{\lambda}{hc}$

This gives us the equation $\frac{2c^2}{\lambda^5} \frac{h}{e^{\frac{hc}{\lambda KT}}-1} \times \delta \lambda \times \pi \frac{R^2}{d^2} \times \frac{\lambda}{hc}$

We have all of these variables so if we plug everything in, we get $1.4 \times 10^{-2} \frac{photons}{s cm^2}$

c) Assume the star has a radius equal to that of the sun $R_* =R_o$. How many photons arrive from the sun like star (T=5800K) in a 1 micrometer wide wavelength interval near $\lambda_{max}$ of the Y dwarf

This part of the question is very similar to part b. We could plug in all of the new numbers, but many of the variables remain the same so and the only variables that change are T and R. This means that, since we divided by the term with T, we could multiply by this term to remove it, and since we multiplied by the term involving R, we could divide by it. Then we would have to multiply by the new T term ($T_2$ and divide by the new R term ($R_2$)

$\frac{1.4 \times 10^{-2} \times (e^{\frac{hc}{\lambda KT_2}}-1)R_2^2}{R^2 {\frac{hc}{\lambda KT_2}}-1}$

This gives us $7 \times 10^3 \frac {photons}{cm^2 s}$

d) What is the flux ratio of the Y dwarf to the star near $\lambda_{max}$? This should illustrate why it's so difficult to detect substellar companions around sun-like stars.

The flux ratio can be found by taking the ratio of the answer found in b to the answer found in c. These answers are not flux, but they would both need to be multiplied by the energy of a single photon to get flux and this would cancel out anyway since it is a ratio.

The ratio is $\frac{1.4\times 10^{-2}}{7\times 10^2}$ = $2 \times 10^{-3}$

The flux of the sun like star is so much greater than the flux of the brown dwarf that it is difficult to detect the brown dwarf.