### This Week's Astro Nutshell: It's full of stars!

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.

Suppose you have a magnitude-limited survey such that all stars have magnitudes $m < m_{\rm max}$. What will be the most common type (mass) of star in your survey?

This question is pretty much the same as "What types of stars visible in the night sky are most numerous?" This type of problem was first addressed by Swedish astronomer Gunnar Malmquist back in the 20's, which led to what we now refer to as the Malmquist Bias.

Initially, one might thing: well red dwarfs are the most common stars in the Galaxy, so M dwarfs will be the most common in our survey (or sky). However, M dwarfs are very faint (low luminosities). If the Sun is a 1000 Watt lightbulb then a typical M dwarf would be a Christmas tree light (thanks to Dave Charbonneau for the analogy). Since we're magnitude-limited (brightness-limited), we might not see many M dwarfs.

If we denote the number of stars in our survey as a function of mass $N(M)$, then
$N(M) \sim$(density of stars) $\times$ Volume
Where the "Volume" is characterized by the distance $d_{\rm max}$ out to which you can see a star of a given mass ($V_{\rm max} = d_{\rm max}^3$). Let's denote the density of stars by $\phi$, which is the number of stars per unit volume. This results in
$N(M) \sim \phi \times d_{\rm max}^3$   (1)
The density of stars in a given volume is given by the present-day mass function. Note that this is different from the initial mass function (IMF) because the stars in our survey will not be newly born, but  will instead represent a well-mixed sample of stars of all ages. Since massive stars die young, there will be even fewer massive stars than predicted by the IMF. The PDMF has the form $\phi \sim M^\alpha$, where $\alpha = -1.35$ for stars less massive than the Sun (the standard Salpeter IMF), and $\alpha = -5.2$ for stars more massive than the Sun. Plugging into Eqn 1 gives:
$N(M) \sim M^\alpha \times d_{\rm max}^3$   (1)
As for $d_{\rm max}$, we can use the handy equation that we derived a couple weeks ago (I'll blog about it later), which gives the scaling of the flux received from a star at the peak of its spectral energy distribution. The peak shifts to longer wavelengths for cooler stars, and shorter wavelengths for hotter stars. This all encompassed by the simple scaling relationship
$F \sim T^2 R^2 d^{-2}$ (1)
As the temperature $T$ increases, the flux increases. The same as when the star's radius $R$ increases. Move the star further away, the flux drops. Since our survey is sensitive only up to a limiting magnitude, we can only observe stars with $F < F_{\rm min}$. This means
$d_{\rm max} \sim T R F_{\rm min}^{-1/2}$    (2)
From stellar structure, we recall that $R \sim M$, and $T \sim M^{1/2}$. Subbing into Eqn 2, we get
$d_{\rm max} \sim M^{1/2} M F_{\rm min}^{-1/2}$
And since our flux (magnitude) limit is fixed, there is a maximum distance out to which we can see a star of a given mass, given by
$d_{\rm max} \sim M^{3/2}$
We can now evaluate Equation 1 in terms of stellar mass, $M$:
$N(M) \sim M^\alpha \times M^{9/2}$

For the different mass regimes, the present-day mass function has different values of $\alpha$:
$N(M) \sim M^{3.2}$      for  $M < 1~M_{\rm sun}$
$N(M) \sim M^{-0.7}$  for  $M > 1~M_{\rm sun}$
Given that $N(M)$ has different slopes on either side of 1 $M_{\rm sun}$, then it's clear that stars like our Sun will dominate your stellar sample. Even though M dwarfs are 75% of the stars in the Galaxy, you won't see many of them. This is why so many of the planets found in wide-field transit surveys such as HAT and WASP show up around Sun-like stars. The visible sky is full of G2-F8 stars!

At higher stellar masses, there aren't many stars formed, and those that do form die young because stellar lifetimes scale as $M^{-3}$ or so. But the effect isn't as severe as on the low-mass side. It's a gentle fall-off toward A and B dwarfs.

### 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…