# Dimensional Analysis. Or: why is the sky blue?

"Why is the sky blue?" is one of those canonical little kid questions that is shorthand for "inquisitive tiny human confounding their adults." While your average grownup may not know why the sky is blue off the top of their head, this is the Future, and the Internet does have lots and lots of useful explanations of the very interesting reasons behind it.

One thing that always bothers me about the explanations though is that they have to punt over one of the most important pieces (namely, why does high frequency light scatter more than short frequency light. If that doesn't make sense to you, don't worry, I'll cover it shortly). Skipping over why this happens means that the answer to "why is the sky blue?" will just cause another Why? question. Of course, I am reliably informed that children in the depths of the Why-ness questions will just keep doing this anyway, but some of us never grew out of it and become scientists.

Anyway, I thought it would be interesting to explore the Why? that the blue sky raises, especially since one way of looking at the answer gets at a of very important tool in the theorist's toolkit: dimensional analysis. Talking about this will certainly raise a lot more Whys?, but that's most of the fun of physics: there is always another why.

OK, first let's go through the answer to the question "Why is the sky blue?" that you'll find if you type that into the all-knowing google. I'll point out the bit that, to me, begs for more explanation.

During the day, the sky (sans clouds) is blue. How can this be? After all, if you are not staring at the Sun and burning out your retinas, when you stare at the sky, you are not staring at anything but empty air. There is nothing up there that reflects or emits light: there is no source of photons to hit your eye when you look into empty sky

The proof of this is simple. Go look at the sky at night, and yep, nothing up there but moons, planets, and stars, none of which are bright enough to light up the sky in any significant way.

OK, but we are definitely seeing something when we stare at empty space: we're seeing blue light. Where is it coming from? The only source bright enough is the Sun, and we know that when the Sun goes down, the illumination from the blue sky goes with it. So the light we're seeing from the blue sky is originating from the Sun, and is somehow being redirected to hit our eyes.

So what must be happening is that light from the Sun passing through the atmosphere, light that is going in the wrong direction to hit your eyes, is being scattered by the air. That scattered light travels out in all directions, some of which the hits your eye, allowing you to see light seemingly originating from "mid-air," a place that has no intrinsic source of photons.

Spectrum of visible light, relating perceived color with wavelength in nanometers. (Image from Wikipedia)

Now, here's where the punt usually comes in. For the sky to seem blue, you must accept the following fact: higher frequency (shorter wavelength) light scatters more in the atmosphere than low frequency (long wavelength) light.

I've included picture of the visible spectrum here. Long wavelength light - light where the distance between "peaks" of the electromagnetic wave is relatively large - corresponds to low frequency and lower energy and is perceived by our eyes as redder colors. Red is about 650 nm (650 billionths of a meter) or about 2 "electron-volts" (eV) in energy. Short wavelengths are higher energy, and are bluer colors. Violet at 400 nm wavelength is about 3 eV. The conversion factor is $$E_\gamma = \frac{hc}{\lambda} = \frac{1240~\mbox{eV nm}}{\lambda},$$ where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the light.

So, blue light scatters in air more than red light. If you accept that, then the blue sky makes sense. The Sun itself is pretty much white light: an approximately even mix of all wavelengths. Given that blue light scatters more than red light, when the Sun is high in the sky, and you look at a random point in the sky (not directly at the Sun), then the light that hits your eye has more blue photons in it than red photons, because more blue light is redirected by the air.

When the Sun is setting or rising, and you look in the direction of the Sun, the sky is red. This is because you're seeing light traveled through more air (it has to not just travel from space down to the surface, but sideways all the way over the horizon too). More of the blue photons have been scattered away due to the longer travel time; the people over the horizon looking up at the sky are seeing your missing blue, leaving only those beautiful reds, oranges, and yellows.

As an aside, sunlight itself is nearly white, due to the temperature of the surface of the Sun. Why we intrinsically see it as yellow is unclear to me. Maybe its because we can't look directly at the Sun when it's high in the sky (when the direct sunlight is scattered less), and only when its low in the sky (when more blue is scattered out, leaving yellows, oranges, and reds). Maybe its an mental processing issue where white on blue is seen as yellowish. Maybe enough blue is scattered away that we see the remainder as yellow/red. This raises an interesting (and worthy of a totally separate blog post) question of how we perceive color. Briefly, there is the intrinsic color of light: the wavelength. Then there's the way those different packets of energy interact with the complex chemical system in our eyes to send signals to our brain. Then there's how our brain processes those signals to see "color." In this post, I'm talking only about the first kind of color, which is really just wavelength or energy; red is just light at around 600 nm; which is only the first part of a long process that tells us whether we see that light as "red."

Green Flash at sunset. Image by Brocken Inaglory, via wikipedia.

As another aside, what about the Sun's green light? The day-sky is blue, the dawn and dusk yellow to orange. Where's our green? Well, apparently due to those complicated chemical and neurological processing issues, it is not possible to see objects that glow due to their temperature as "green." So even if we were around a cooler star than our Sun, a star which put out light mostly in the green wavelengths, our eyes would not see it as green; the chemical processes in our rods and cones would pick up too much of the other wavelengths and it wouldn't seem green to us. This is sort of strange, since as animals evolved on a planet full of green foliage, we're actually incredibly good at picking out different shades of green, but there it is. No green stars, due to the way our eye works and how things glowing from their own heat work.

However, if the atmospheric conditions are just right, at sundown you can occasionally see the "green flash," a brief moment where all the other colors from the Sun can get scattered or refracted away, and you see only the green. It requires the temperature and quality of the air to be just right, and its very rare. I've never seen it.

So that story makes total sense, right? The sky is blue because of one simple fact: blue light scatters more than red. The only thing you can see when looking up during daytime is scattered light, so you see more blue than red. Thus, the sky is blue.

But wait, Why? Why is blue scattered more than red? Why not the other way around? Or why does one wavelength scatter more than the other at all, can't all scatter evenly? All I've done is replaced one problem with the another.

To which I say, welcome to science.

So now I want to try to explain a bit about why the molecules of air will scatter more blue light than red. Some of the concepts are tricky, and it turns out to range over a lot of different areas of particle physics, and the explanation is probably not one that will make a lot of sense to the inquisitive toddler asking "Why?" However, it makes a nice example of how to think about particle physics, so maybe some tiny corner of the internet will find it interesting and useful.

First, let me go back to something I tossed out earlier without much comment: the fact that high energies correspond to small distances and vice versa. The fact that I can convert freely between length and (inverse) energy is made possible by constants of Nature, Planck's constant and the speed of light, in particular. This may seem like a really arbitrary thing, after all, in our macroscopic world things with a lot of energy tend to be big. But in at the level of particles, the more energy packed into a particle, the smaller we can think of that particle as being. This is why we sometimes call the Large Hadron Collider "the world's best microscope." It packs a lot of energy into particles, so it "sees" very small distances.

This connection between energy and distance is incredibly important in physics. It's probably the most important thing to comprehend in this whole post, so I'll even bold it for you: In particle physics high energies correspond to small distances and short times and vice versa. Once you've been doing theoretical physics for a while, in fact, you'll start thinking of all dimensionful quantities in terms of units of energy, and stop thinking in terms of length or time. For me, energy has dimension 1 (that dimension: energy, usually in units of "giga-electronvolts" or GeV since I work on LHC physics a lot). A distance is a length, and so is dimension -1 (inverse energy). A temperature is energy, dimension 1. An interval of time is just like a length (I convert between the two by asking "what distance would light travel in this time," which everyone agrees on because of relativity. Thanks Einstein), so time has dimension -1. And so on.

OK, so higher energy light corresponds to smaller distances. What of it?

Well, now ask how light interacts with objects in the world around us. Light is a wave of electromagnetic radiation. If you wiggle an electic charge, it causes the electric and magnetic fields nearby the charge to be perturbed. This perturbation is self-sustaining and propagates along in space, just like if you perturb the surface of water. That it is, wiggling an electric charge gives you an electromagnetic wave, which we call light.

When that light hits another electric charge, it causes that charge to wiggle; which can allow the light to be absorbed and reradiated. This is what causes light to be scattered or reflected by materials: the wiggling of the charged electrons in the atoms.

But air molecules, like most materials around us, are electrically neutral. So for light to be scattered by air, the electromagnetic wave must "see" that there isn't a neutral atom of oxygen or nitrogen, but that there really is a bunch of charged electrons separated from the charged nucleus. That is, the light has to be able to resolve small distances. And that takes, you guessed it, high energies (where "high" means here enough energy that the wavelength of light is at least vaguely comparable to the distances inside molecules and atoms).

So that gives us a clue: higher energy photons (bluer light) are smaller wavelength, so they can better "see" that the atoms of air they are passing through are really a bunch of tiny separated charges as compared to the atoms as "seen" by low energy, long-wavelength, red light. And this further makes sense when you think about really long wavelength light like radio waves, which have wavelengths of a meter or more. You can get radio waves inside your house, passing straight through a wall that visible light can't, because the radio wave is too big and too low energy to see the tiny separation between charges in the atoms of the wall that visible light can.

Now lets go a bit deeper. I mentioned in a previous post that physicists like to think about the sizes of objects in terms of "cross sections." So we can try thinking about the "size" of air molecules seen by visible light in terms of the "area" (the cross section, or $\sigma$ that each air molecule has with respect to light. Larger $\sigma$ means that the air is "easier" for the light to interact with, so light would scatter more.

Now, a cross section is an area. An area is length-squared. So by my rules about length and energy, this means $\sigma$ has dimension -2: it needs to depend on $1/\mbox{energy}^2$ for some energy. This sort of thought process, by the way, is called "dimensional analysis." It's really useful to get an idea of what could be important in a physics problem.

So, I need some energy in order to get a cross section. What energy? Well, my first guess might be the energy of the photon. After all, all I have is a molecule of air and a photon hitting it. Not many things to pick from. So my first guess for the size of an molecule to interact with light could be:

$$\sigma(1^{\rm st}~\mbox{guess}) \propto \frac{1}{E_{\rm photon}^2}.$$

But let's think about this a bit more. This guess would say that as my photon energy got smaller, the cross section should get bigger and bigger. But that doesn't track with what our physics-intuition tells us, which is that really low energy photons can't see "inside" atoms to wiggle the charges around. So $\sigma$ should depend on positive powers of $E_{\rm photon}$, not negative ones.

So let's try again. There is another scale in the problem besides the energy; the size of the atom or molecule. I could characterize this as a length, but let me call it an energy instead: the characteristic energy of electrons inside the air molecule. I'm going to call this $\Lambda$. I have no idea what $\Lambda$ is right now, dimensional analysis won't tell us that piece of information, but it lets us move forward.

With this, a second guess for the size of air as seen by light could be

$$\sigma(2^{\rm nd}~\mbox{guess}) \propto \frac{1}{\Lambda^2},$$

but we know that's not right, since this predicts every photon sees the same size air, and we know that really low energy photons shouldn't interact with air at all. So, combining our dimensional analysis with our new understanding of the relation between energy and size, we can say that

$$\sigma(3^{\rm rd}~\mbox{guess}) \propto \frac{E_{\rm photon}^n}{\Lambda^{n+2}}$$

Our dimensional analysis won't tell us from only this level of consideration what that value of $n$ should be.

It turns out that the right answer is

$$\sigma \propto \frac{E_{\rm photon}^4}{\Lambda^6}.$$

Why it is the fourth power of energy is another rabbit hole that I will hopefully go down later (there's a nice explaination in terms of effective operators I've seen by Iain Stewart). However, the key ideas are all here: we have a physics question. It can only depend on a few key parameters (size of the atom or molecule, energy of the photon). The thing I want to calculate has a particular dimensionality, in this case it is an area, so that is -2 units of some energy. Combining that with some intuition about what the answer must be in particular limits of the photon energy gives us a general idea of how the dependence of the answer must depend on the quantities at hand. All the rest of the detailed calculation will "only" give us the constants of proportionality to convert between something that "depends on" quantities (represented by the $\propto$ symbol) to something that equals ($=$) some collection of quantities and constants of Nature. That's a lot of work being done by that "only" there, and it may look like I'm continuing the grand tradiation of punting the hard problem down the road, and in a way I am.

But this idea of dimensional analysis is really useful, and hopefully I've given an interesting demonstration as to why.