Today's Problem: Dim Stars at Great Distances

What is the power output of a star that is 100 light years away that you can barely see from a dark site at night?

This problem is, in part, informed by another one that isn't on my blog, so some further background info is called for. Part of the setup of that problem was the following: "The eye must receive ~ 10 photons in order to send a signal to the brain that says, 'Yep, I see that.'" Pretty straightforward, right? Maybe, except you could interpret that a few different ways. 10 photons every yoctosecond is going to be plenty more light than the brain would need (or, I imagine, want). 10 photons every year probably isn't going to cut it. We figured that if the brain can register frames of a movie running at around 30 fps, it needs somewhere on the order of 300 photons per second. Obviously if those photons aren't being emitted in visible light, that won't be too helpful either, so we figured a wavelength of 500 nanometers. The tools we have to work with, then, are as follows:

-The rate $R$ of photons received, 300 photons/s

-The wavelength of our photons, 500nm, or $5\times10^{-7}$ meters.

-The approximate area $A$ of a pupil (appropriately dilated for night observing), 1 cm$^2$, or $10^{-4}$ meters$^2$.

-The distance the the star, $D = 100ly$

-The speed of light $c$, $3\times10^8$ meters/s.

-Equation $(1)$: the energy of a photon, known to be $E = h\nu$, where $\nu$ is the photon's frequency and $h$ is Planck's constant $h = 6.6\times10^{-28} ergs/s$. 

-Equation $(2)$: the relationship between frequency and wavelength, $\lambda = \frac{c}{\nu}$.

-Equation $(3)$: flux, the amount of energy received in a given area over a given time, is given by $F = \frac{L}{4\pi D^2} ergs/s$ for a source of luminosity $L$.

That last equation comes from the fact that light propagates based on an inverse-square law. In other (more useful) words, if you imagine a star emitting light, you can imagine a spherical shell exactly 100 light years in radius consisting of all of the light that star emitted exactly 100 years ago. The amount of light that you observe is equal to the area of your detector (in this case, your pupil) divided by the total surface area of the spherical shell, $4\pi D^2$.

What we're ultimately looking for is the luminosity, $L$, of the star emitting the light we're observing. We can find this by using that last equation--we know $D$, so what we need first is $F$. For that, we'll need a way to translate photons received per second every second into energy per second--or, more precisely, photons into energy. We know that for a single photon, $E = h\nu$, and so our first step will be to deal with $\nu$, the frequency of our photons. Since we have the wavelength of the photons, we can plug it into $(2)$, which allows us to substitute $\frac{c}{\lambda}$ for $\nu$, giving us the following:\[E = \frac{hc}{\lambda}\]For 300 photons per second, we can simply multiply this by $R$ and and units of inverse seconds, giving us power. Dividing that by A gives us our flux as\[F = \frac{Rhc}{\lambda A}\]Now we can take this back to equation $(3)$ and do some algebra to solve for $L$. \[ \frac{Rhc}{\lambda A} = \frac{L}{4\pi D^2}\]\[L = {Rhc\times4\pi D^2}{\lambda A}\]This is worth unit checking before taking it as gospel. The units on the left side are power, or energy$*$time$^{-1}$. On the right we have $\frac{time^{-1}\times energy \times time^{-1}\times length \times time^{-1} \times time^2}{length^2 \times length}$, which also comes out to energy$\times $time$^{-1}$. So we're all set! At this point, all we need to do is plug in numerical values for all of the constants on the right side of the equation. That gives us the following:\[\frac{3\times 10^2 \times 7 \times 10^{-28} \times 4\times \pi \times 10^{36} \times 3\times 10^8}{10^{-4}5\times10^{-7}}\] Which evaluates to just about $2\times10^{31} ergs/s$. That's a big number, but it's actually pretty dim--this is a star with only about one percent of the luminosity of the sun. However, we're assuming otherwise uniformly dark skies, and operating at the bare minimum of your eye's sensitivity. It's a stretch, but in conditions as ideal as the ones we're describing, it might be possible.

We came into this problem with a fair amount of outside knowledge, particularly in the form of our equations (Planck's constant isn't something we'd be likely to estimate well in a casual experiment). However, with just a few relatively simple tools on hand, we were able to determine a pretty significant property of a hypothetical faraway star--pretty cool. As usual, I had help on this one, and once again credit goes to Anne Madoff, Jennifer Shi, and Louise Decoppet for working through this problem with me.