### This Week's Astro Nutshell: How Many Photons?

(Actually, this is from Astro Nutshell two weeks ago)

Each week I work with first-year grad students Marta and Becky on "order of magnitude" problems at the blackboard. I put that in quotes because we tend to do many more scaling arguments than true OoM. The idea is for them to draw on what they've picked up in class and apply it to common problems that arrise in astronomy.

How many photons per second per cm$^2$ (photon number flux) do you receive from a star as a function of its temperature $T$, radius $R$ and distance away $d$?

Having such an equation would be extremely handy for observation planning. When determining the feasibility of a new project, observers tend to start with a statement of the expected signal-to-noise ratio (SNR) for an observation of an astrophysical object. In the limit of a large expected number of photons, the signal is the number of photons $S = N_\gamma$, and the noise can be approximated as $N{\rm oise} = \sqrt{N_\gamma}$. So SNR$= \sqrt{N_\gamma}$. So this week's question comes down to, "What is $N_\gamma$ for a star of a given temperature, radius and distance away?"

We start with Wein's Law, which states that the wavelength at which a star's (blackbody's) emission peaks is inversely proportional to the star's temperature
$\lambda_{\rm max} \sim \frac{1}{T}$    (1)
This is Astro 101. The flux level at this peak wavelength can be evaluated using the blackbody function (Planck function), which is given by
$F_\lambda(T) = \frac{2hc^2}{\lambda^5} \frac{1}{\exp{\frac{h c}{\lambda k_{\rm B} T} + 1}}$    (2)
This gives the energy per unit time (power), per area, per wavelength per unit solid angle, as a function of temperature and wavelength. If we evaluate this at $\lambda_{\rm max}$, and approximate the total flux, which is an integral over all wavelengths, as a box of height $F_{\lambda_{\rm max}}$ and with a 100 nm width (standard observing bandpass). We also need to multiply by the solid angle subtended by the star of radius $R$ at a distance $d$, which is $R^2/d^2$. This leads to
$F_{\rm tot} \sim \frac{T^5 \Delta\lambda}{e^{\rm const} - 1} R^2 d^{-2}$   (3)
Since $\lambda_{\rm max} \sim 1/T$, then $|\Delta \lambda| \sim 1/T^2$ (Wow, check out that calculus slight-of-hand! However, the same scaling falls out of actually doing the integral over $d\lambda$). Finally, the energy per photon near $\lambda_{\rm max}$ is $E_{\lambda_{\rm max}} = h c / \lambda_{\rm max} \sim T$. Dividing Equation 3 by the energy per photon, and replacing $\Delta \lambda$ we get the flux of photons
$F_\gamma \sim T^5 T^{-2} T^{-1} R^2 d^{-2}$
$F_\gamma \sim T^2 R^2 d^{-2}$
Increasing the temperature or radius of the star results in more flux, which should seem fairly intuitive: hotter, bigger stars emit more photons. Also, there's the familiar inverse-square law with distance. Evaluating for the Sun at 10 pc results in
$F_{\gamma,\odot} = [5\times10^{4} {\rm photons}] T^2 R^2 d^{-2}$
(please check my math on this!)

For an M dwarf with 1/5 the Sun's radius and roughly half the temperature, at 10 pc it would emit 100 times less light, but most of these photons will be down near 1 micron in the near infrared. An A-type star like Vega, with twice the Sun's radius and twice the temperature will emit 16 times as many photons, most of them in the ultraviolet.

I hope you find this scaling relationship as handy as I do!

### On the Height of J.J. Barea

Dallas Mavericks point guard J.J. Barea standing between two very tall people (from: Picassa user photoasisphoto).

Congrats to the Dallas Mavericks, who beat the Miami Heat tonight in game six to win the NBA championship.

Okay, with that out of the way, just how tall is the busy-footed Maverick point guard J.J. Barea? He's listed as 6-foot on NBA.com, but no one, not even the sports casters, believes that he can possibly be that tall. He looks like a super-fast Hobbit out there. But could that just be relative scaling, with him standing next to a bunch of extremely tall people? People on Yahoo! Answers think so---I know because I've been Google searching "J.J. Barea Height" for the past 15 minutes.

So I decided to find a photo and settle the issue once and for all.

I then used the basketball as my metric. Wikipedia states that an NBA basketball is 29.5 inches in circumfe…

### Finding Blissful Clarity by Tuning Out

It's been a minute since I've posted here. My last post was back in April, so it has actually been something like 193,000 minutes, but I like how the kids say "it's been a minute," so I'll stick with that.
As I've said before, I use this space to work out the truths in my life. Writing is a valuable way of taking the non-linear jumble of thoughts in my head and linearizing them by putting them down on the page. In short, writing helps me figure things out. However, logical thinking is not the only way of knowing the world. Another way is to recognize, listen to, and trust one's emotions. Yes, emotions are important for figuring things out.
Back in April, when I last posted here, my emotions were largely characterized by fear, sadness, anger, frustration, confusion and despair. I say largely, because this is what I was feeling on large scales; the world outside of my immediate influence. On smaller scales, where my wife, children and friends reside, I…

### The Force is strong with this one...

Last night we were reviewing multiplication tables with Owen. The family fired off doublets of numbers and Owen confidently multiplied away. In the middle of the review Owen stopped and said, "I noticed something. 2 times 2 is 4. If you subtract 1 it's 3. That's equal to taking 2 and adding 1, and then taking 2 and subtracting 1, and multiplying. So 1 times 3 is 2 times 2 minus 1."

I have to admit, that I didn't quite get it at first. I asked him to repeat with another number and he did with six: "6 times 6 is 36. 36 minus 1 is 35. That's the same as 6-1 times 6+1, which is 35."

Ummmmm....wait. Huh? Lemme see...oh. OH! WOW! Owen figured out

x^2 - 1 = (x - 1) (x +1)

So $6 \times 8 = 7 \times 7 - 1 = (7-1) (7+1) = 48$. That's actually pretty handy!

You can see it in the image above. Look at the elements perpendicular to the diagonal. There's 48 bracketing 49, 35 bracketing 36, etc... After a bit more thought we…