Blog #7 - Y Dwarfs

Y Dwarfs

Recently, a new type of astrophysical object known as the Y dwarf was discovered. Y dwarfs are a subclass of brown dwarfs that we'll learn about later in the term. Consider a Y dwarf residing near a Sun-like star (where near refers to a small angular separation between the two objects on the sky). Y dwarfs have a temperature of about 350K (for reference, 0 degrees Celcius corresponds to 273K), and as a result they are extremely faint.

a) At what wavelength should observe to have the best change of detecting the Y dwarf? 

Using Wein's displacement law:

and having the information the Y dwarfs have a temperature around 350K, we find a wavelength of \[\lambda_{max} = 8285nm \] which puts the peak wavelength in the long-wavelength infrared range. 

b) Assume the star has a radius equal to that of the Sun, \(R_{\star} = R_{\odot} \), and that the Y dwarf has a radius equal to Jupiter's \(R_Y = R_{Jup} \). As measured at \(\lambda_{max} \), how many photons per second, per \(cm^2\) emitted from a Y dwarf at a distance of 30 light years would reach the Spitzer space telescope in a 1 \(\mu m\)-wide wavelength interval near \(\lambda_{max} \)? 

We begin by solving for the standard blackbody equation at our given Y dwarf temperature (350K) and \(\lambda_{max} \) (8285 nm). 

which, when plugging in constants: \[h = 6.626*10^{-34} \quad c = 3*10^8 \quad k = 1.38*10^{-23} \] gives us \[B_{\lambda}(T) = 7.14*10^6 \frac{W}{Sr * m^3} \] Converting this to new units we get \[B_{\lambda}(T) = .714 \frac{erg}{s*cm^2* A * sr} \] With this information we can now solve for the total number of photons that will strike at a 1 \(\mu m\)-wide wavelength on our space telescope. 

Now we must solve for the steradian size of the planet based on its radius and the distance from the telescope. Doing simple trigonometry with the distance and radius (similar to Jupiter's radius) at a small angle approximation we get \[tan(\theta) = \frac{2*R_{Jup}}{31 \space light \space years} \] where \[R_{Jup} = 69.911*10^6 m \quad and \quad 31 \space light \space years = 2.93*10^17m \]

This gives an angle of \[\theta = 4.77*10^{-10}\] which using the relationship of \(steradian = radian^2 \) we get \[d \Omega = 2.277*10^{-19} \] Now we can multiply our \(B_{\lambda}(T) \) term by both the steradian value and the value of our 1 \(\mu m\)-wide wavelength range to get a value in terms of \(\frac{erg}{s * cm^2}\)

Now our value becomes \[B_{\lambda}(T) * 2.277*10^{-19} \space steradians \space * 10,000 angstroms = 1.625*10^{-15} \frac {erg}{s*cm^2} \] But we want our value in terms of how many photons not how much energy in the form of ergs. So we find the average energy value of a photon at our \(\lambda_{max} = 8285nm \) which comes out to be \[E = \frac{h*c}{\lambda} = 2.4*10^{-20}\frac{joules}{photon} = 2.4*10^{-13}\frac{ergs}{photon}\] We can now finally do our the last calculation to find the number of photons that strike the Spitzer space telescope and find \[\frac{1.625*10^{-15} \frac{erg}{s*cm^2}}{2.4*10^{-13} \frac{ergs}{photon}} = .0067 \frac {photons}{s*cm^2} \]

c) How many photons arrive from the sun-like star (T = 5800K) in a 1 \(\mu m\)-wide wavelength interval near \(\lambda_{max}\) of the Y dwarf? 

Following the same process but for a sun-like star with temperature 5800K. From there we can compute our \(B_{\lambda}\) We find \[B_{\lambda} = .029 \frac{erg}{s*cm^2* A * sr}\] We continue to follow the same procedure as part B and find that the steradian value is \[d \Omega = 2.255*10^{-17}\] where \[R_{star} \approx R_{sun} = 695.7*10^6m\]Again multiplying through our \(B_{\lambda}\) value by the steradian value and the 1 \(\mu m\) wavelength range value we get \[B_{\lambda}(T) * 2.255*10^{-17} \space steradians \space * 10,000 angstroms = 6.57*10^{-15} \frac {erg}{s*cm^2} \]And again finding the energy at our peak wavelength of 8285nm, we get \[E = \frac{h*c}{\lambda} = 2.4*10^{-20}\frac{joules}{photon} = 2.4*10^{-13}\frac{ergs}{photon}\] and get a final result of \[\frac{6.57*10^{-15} \frac{erg}{s*cm^2}}{2.4*10^{-13} \frac{ergs}{photon}} = .027 \frac {photons}{s*cm^2} \]

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 sub-stellar companions around sun-like stars. 

To find the flux ratio between the two bodies we simply divide out their fluxes found in parts b) and c). \[\frac{.027 \frac {photons}{s*cm^2}}{.0067 \frac {photons}{s*cm^2}} \approx 4\] This explains why its so hard to detect sub-stellar companions around stars given that the star typically emits almost 4 times as many photons at the Y dwarf's peak wavelength (even though it is nowhere near the star's peak wavelength).