Monday, February 27, 2017

Snow Line and the Dwarf's Seven

I'm really sorry about the title. Not sorry enough not to use it, of course, but a little sorry.

So you may have heard about the recent discovery of a nearby solar system (a mere 39 light years away!) with seven planets all packed very close to the star (an M-dwarf). The discovery is significant because (a) some of the planets look to be rocky, Earth-sized, and in the habitable zone; (b) the relative nearness of the system makes it a prime target for further investigation; and (c) it's super rad. The occasion gives me the opportunity to explain a bit about how discoveries like this get made while waxing philosophical about the nature of astronomy itself. As a guy with an astronomy degree (I don't feel comfortable calling myself an astronomer) who (kind of) teaches an intro astronomy class, this is basically my job.

Conveniently, last week's discovery does an excellent job of illustrating three aspects of astronomy that I think set it apart from other sciences. (Or possibly my own confirmation bias leads me to see these aspects expressed, but let's leave that for another post.) These features are encapsulated in a kind of motto for astronomy that I've been using recently.

It goes like this: astronomy is the science of what you see when you look up. This sentiment conveys that astronomy is ancient and public, because for thousands of years, anyone could do astronomy just by turning their heads skyward and paying attention. Secondly, astronomy is bound (mostly) by sight, which is a limitation that forces astronomers to be both careful and creative. And finally, “up” is a pretty wide direction, and astronomy encompasses everything from the moon to other stars to the birth of the universe itself and anything else we find along the way.

All of this ties together into something truly remarkable. Astronomy has the power to transform points of light—the ever-present night sky that we rarely stop to consider deeply—into a story about exploding stars and merging galaxies and dark matter halos all under the spell of gravity in a dance that goes back billions of years and will probably continue for many orders of magnitude longer than it's lasted so far. And what's more, we have good reason to be confident in this story. How does astronomy manage to do this? Well, let's take a look at those seven newly discovered exoplanets.

While we've only known about exoplanets for a couple decades now, the study of planets more generally is, like the rest of astronomy, incredibly ancient. There are five planets visible to the naked eye (Mercury, Venus, Mars, Jupiter, Saturn) that have been known into antiquity. The first person to discover a new planet (Uranus) was William Herschel, using a telescope he constructed himself. Neptune followed, after Urbain le Verrier noticed that, after adding up all the known gravitational influences on Uranus, its calculated position on any given night was a little off from its observed position. He predicted that a planet farther out was gravitationally tugging on Uranus, so the astronomer Johann Gottfried Galle looked where le Verrier said to and found another new planet.

I'm giving this brief (and incomplete) history lesson because the fact of the sky always being up there makes astronomical discoveries collaborative and open. There's a parallel in last week's exoplanet discovery both in terms of that public nature and gravitational perturbations. Moreover, discovering new planets used to be a once in a generation kind of thing, but now we've discovered thousands of them and just found seven in one system. Astronomy is a gigantic, ever-expanding field; whenever we look somewhere new or look in a new way, we find new stuff.

So let's talk about TRAPPIST-1. While NASA had a big press conference about the discovery (and they were involved), this was a remarkably international effort, involving astronomers and telescopes from all over the world. Most exoplanets discovered so far have involved space telescopes because the atmosphere makes detecting subtle changes in a star's light curve difficult. A relatively cheap solution being used now is to image the same star many times either with multiple ground-based telescopes or the same scope repeatedly. This lets you produce a single, high quality light curve and means that anyone can get in on the exoplanet discovery game. With a small telescope that spends all its time looking at large patches of the sky, you can detect (and re-detect) the faint signatures of exoplanets. Once TRAPPIST and the other telescopes involved made those initial findings, NASA pointed the Spitzer Space Telescope at TRAPPIST-1 to confirm the discovery.

Okay, but how did these telescopes actually discover the seven exoplanets? This is where the central limitation of astronomy—sight is (just about) our only tool—leads to very creative solutions. The way that we transform TRAPPIST-1 from a point of light into a star with seven worlds is by performing high-precision photometry to construct a light curve of the star. A light curve is the change in a star's light over time. To get an accurate one, you need to get high quality images on short timescales. This runs counter to a very useful tactic in astronomy, which is to collect light from a source over a long period of time to produce a single, bright image. But if you do that, any deviations during that integration time get smeared out and missed.

To detect exoplanets, the deviations you're looking for are dips in the star's brightness at regular intervals. If your telescope, the star, and a planet happen to line up exactly, then every time the planet passes in front of the star from your perspective, the star gets a little bit dimmer. It's just like a solar eclipse here on Earth, except that these planets are much too far away from us to block out all the light of their parent star. Instead we see a tiny drop in brightness.

But these transits reveal a lot of information. First, the duration of the transit and the time between transits tell us how long the planet's year is. Combined with an educated guess about the star's mass (by taking its spectrum), we can figure out how strong gravity's pull on the planet is, and consequently the distance it needs to be from its star to complete an orbit in the observed time. The more massive the star, the faster a planet orbits at a given distance. Finally, the percentage of light blocked by the planet, combined with its distance, tell us how big the planet is compared to the star. Another educated guess about the star's size tells us the actual physical size of the planet.

So by looking very precisely at how a star twinkles, we can deduce the presence of a planet and make a reasonable guess as to how big it is and how close it is to the star. We can do this despite not actually being able to see the planet itself, which is much too small and dim next to its parent star to resolve. But I've been talking about one planet this whole time, and these astronomers discovered seven. You might think sussing out the details of seven different transits while also accounting for anything else that might mess up your photometry would be difficult, and you'd be right. The primary way the team identified seven different planets was through a statistical analysis of the transit times to come up with a chart that looks like this:

Credit: ESO/M. Gillon et al.
As a rule, planets don't share orbits. Doing so isn't stable. And each orbit has a definite period, and each period corresponds to an orbital speed, which tells you how long the transit should last. So if you identify a transit of a particular duration that repeats regularly, then you've found yourself a planet. If you see six or seven different regular transit times, you've found six or seven different planets.

There is a snag in all this, however, called TTVs—transit timing variations. That is, sometimes a transit happens earlier or later than expected. In this case, the variation could be up to half an hour. But it turns out this snag contains even more information, because this sounds an awful lot like the error le Verrier noticed in the orbit of Uranus. The planets weren’t where astronomers thought they would be given just the gravitational influence of the star, which means the planets—all extremely close to each other—are tugging on each other significantly.

Because so much is unknown about the system, the problem is much more complicated than the orbit of Uranus. Le Verrier was able to do a laborious calculation by hand using perturbation theory, but the complexity of TRAPPIST-1 require a slightly faster technique if you want to publish before the stars all die and we’re left in darkness. So instead the team constructed simulations of the system where they plug in the laws of physics and then vary the unknown orbital parameters to see what kind of planetary systems evolve that match the one they observed. In the end, they’re left with a set of possible masses that could produce the tugging required to account for the transit timing variations.

Even doing this produced a wide range of possible answers, which led to a great quote in the article: "The system clearly exists, and it is unlikely that we are observing it just before its catastrophic disruption, so it is most probably stable over a significant timescale." The relevance is that the system's existence is itself a piece of data, which means that as more observations are done, the assumed stability of the system can help to rule out orbital parameters that would produce an unstable system.

With those uncertainties understood, the team was able to estimate that most of the planets are in the neighborhood of Earth's mass. If you know the size and the mass, you also know the density. The worlds of TRAPPIST-1 are all rocky (high density) as opposed to gassy (low density). The proximity to the star itself is also important. If planets are too far out from their star—past the snow line—then water and other volatiles condense into ice. Far enough inside that line, however, and water can remain a liquid. Too close, and the liquid evaporates. These planets are all at the right distance to have liquid water.

An entire system of rocky, Earth-sized worlds warm enough to have liquid water—this is why everybody is so excited and why astronomers are going to keep watching these planets. The Kepler Space Telescope is currently looking at the system, and the James Webb Space Telescope will too when it launches. The relative nearness of the system to us means that it is fairly easy to observe. As new observations come in, we could learn about the planets' atmospheres—their density, composition, and variability—and whether they experience tidal heating and geological activity. Are these complex, intriguing worlds like the moons of Jupiter and Saturn or airless rocks scoured dry by the flares of their parent star? We just have to look up to find out.

Thursday, January 12, 2017

When You Think Upon a Star

Among the sciences, astronomy benefits from widespread public appeal. Hilariously large numbers and gorgeous images make it an attractive source for science news. The result is that some difficult notions from astronomy have managed to penetrate successfully into public awareness. For example, this meme, which I've run across several times:

I got this image here (which, incidentally, is a blog post doing exactly what I'm about to do), but I've seen this meme in other forms elsewhere and have no idea what its original source is.
I'd like to say that I feel conflicted about this meme—that I'm happy the joke relies on knowledge of astronomy (the immense size of the universe versus the finite speed of light), despite the specific fact it calls upon being incorrect (visible stars are almost certainly still alive)—but that would be a lie, because I'm an enormous pedant.

However, in this post I'm going to steer my pedantry in what I hope is a slightly more interesting (and less annoying) direction, toward mathematical reasoning. That is, while I think it's great that the public has been able to learn certain specific facts about astronomy (and other sciences), I think it would be far more valuable if the public learned how to apply mathematical reasoning to claims they encounter.

Here's why: as a recent graduate with an official degree in astronomy and all that jazz, I happen to simply know the fact that, in general, the stars we can see with the naked eye are close enough, and live long enough, to still be alive by the time their light reaches us.

But even if I didn't know that fact, I could arrive at it by constructing an argument from some more readily available facts. And this argument, although mathematical in nature, doesn't involve anything more complicated than a bit of algebra, such that anyone who gets out of high school should be able to reach the same conclusion.

Now, the joke's humor relies on some common facts from astronomy: stars are far away, light is slow compared to the size of the universe, stars eventually die. But before we get into the mathematical meat of evaluating this claim, let's think about another common fact: our sun is 4.5 billion years old (give or take), and it's roughly halfway through its life, so it's got another several billion years to go.

In order for the sun to be dead by the time an alien civilization see its light, that civilization would have to be farther away in light years than the sun's remaining age in years. That is, the alien civilization would have to be many billions of light years away. So if we take our sun as typical, then the above meme is only true if we can see, by the naked eye, stars that are billions of light years away. We can't, and as I'll show in a bit, we don't even have to assume our sun is typical for this argument to work (which it isn't, really). But this is the structure of the mathematical argument: compare the lifetimes of stars we can see with the naked eye to their distances from us.

A few more astronomical facts are necessary to work this out, some of which can be gotten by a bit of googling, and one which, I admit, most people probably aren't aware of. This fact, which makes evaluating the claim very easy, is that the more luminous a (main sequence) star is, the shorter it lives. This means the most luminous stars (which are the most likely to be visible by the naked eye at great light travel times) are the best candidates for stars that are dead by the time their light reaches us. If the claim fails for these stars, it fails for all stars.

The most luminous stars live about a million years and are about a million times brighter than the sun. Now, it's always possible that a star we're seeing just happens to be at the end of its life, but all else being equal, if we pick stars at random out of the sky, then on average they will be halfway through their lives, just like (coincidentally) our sun (not strictly true, because there is some selection bias to the stars we can see).

To be visible by the naked eye, a star needs to have an apparent magnitude of 6 or lower.

For the sun to be magnitude 6 (it's currently an obscenely bright -27), it would have to be about 60 light years away. (There's some math involving logarithms here, but there are tools online that could get you this answer.)

How bright a star appears to us is proportional to its intrinsic brightness and inversely proportional to the square of its distance from us. That is, if star A and star B are identical but star B is twice as far away, it looks 1/4 as bright as star A.

And that's everything we need to evaluate the claim. Now here's how we construct the argument. A star is dead by the time its light reaches us if its remaining lifetime in years is less than its distance in light years. It is visible with the naked eye if its intrinsic brightness relative to the Sun is greater than the square root of its distance relative to the Sun's distance at magnitude 6.

Let me unpack that second statement a bit. Say a star is intrinsically four times as bright as the sun. If it's also magnitude 6 (just visible), then it needs to be farther away than the sun. Specifically, a star four times as bright as the Sun will be just visible at twice the distance (square root of 4) of the magnitude 6 sun: 120 light years. If it is farther away, it is too dim for us to wish upon it.

The brightest stars are 1,000,000 times more luminous than the sun, which means they are the same apparent brightness as the Sun when they are 1,000 times farther away. If the sun is just visible at 60 light years, then the brightest stars are just visible at 60,000 light years. Is 60,000 light years greater than the (on average) half a million years the star will have left to live? No. At that distance, the star could only be dead by the time we see it if it were already 95% of the way through its life. For less luminous stars which live longer, that percentage gets even higher, which makes it much less likely that we ever see such a star.

When we learned algebra via word problems, we were supposed to be learning how to solve problems like these. And while most of us probably managed to get through those word problems successfully, it's been my observation that most of us don't apply this kind of analysis outside of school, to things like evaluating claims that have mathematical content. While it's not vital to the health of a democracy that we be pedantic about random Facebook memes, it might be useful for us to be able to think carefully about scientific claims, at least when the facts and math involved don't require a PhD.

Outside of learning a bunch of astronomical facts, one of the most valuable (academic) lessons I acquired while getting my degree was learning how to bring mathematical tools to bear on a problem. I'm sure this blog post doesn't really have what it takes to impart that same lesson on others, but I hope it reveals a bit of the process. If I could wish upon a star (and I were feeling altruistic), I might wish for an educational system that did a better job of that.

Thursday, September 1, 2016

Live From Low-Earth Orbit!

It looks like I disappeared again. Or maybe I was just too faint to detect above the noise of the internet. Sorry about that. To make up for my absence, this post will have a whole bunch of pictures. After all, there is a favorable exchange rate between pictures and words.

What's brought me out of hiding today is a very cool new account on Twitter. Because the Hubble Space Telescope kind of belongs to the American public, it has started live tweeting where it's looking, what tools it's using to do that looking, and who told it to look there. So you get stuff like this:
The picture is not what Hubble was looking at right then but an image pulled from the Sloan Digital Sky Survey. Hubble can't usefully beam images directly to us, because everything Hubble (and all other telescopes) looks at has to be processed. This notion makes people grumble, because they want to see the raw, unmanipulated data in its purest form rather than rely on whatever artistic license NASA has exercised.

But raw images in astronomy (and raw data more generally in science) simply aren't useful. In fact, they don't even exist, because any contact with an instrument inevitably distorts the data. The purpose of processing images, then, is to remove the imprint of the instrument on the image and hopefully recover what's actually there.

The coolest part about Hubble_Live is that it tweets out this process, too. There are many ways astronomers attempt to extract the true signal from the data collected, but I want to talk about three of the big ones I've learned about and which Hubble employs. These are:

Hubble performs these calibrations in order to figure out how it's interfering with the pictures it's taking. To see what these calibrations do, I want to show you some data my classmates and I took with a much smaller, terrestrial telescope last fall. We were looking at the Ring Nebula, which Hubble has an obnoxiously gorgeous picture of here for reference:

NASA, ESA, and the Hubble Heritage (STScI / AURA)- ESA / Hubble Collaboration
The Ring Nebula is faint, so to image it we tracked it for two minutes, letting the charge-coupled device (CCD) at the bottom of the telescope count up the photons streaming in from space. But a CCD is not really a camera. It's more accurate to think of a CCD as an electron counter.

At each pixel, there's the electric equivalent of a little bucket that collects electrons and converts them into a voltage that can be measured and manipulated digitally by a computer. Ideally, the way the CCD counts electrons is by having them knocked into the pixel bucket by incoming photons. But there are other sources of electrons, too. If you don't take them into account, you end up with an image that doesn't correspond to what you were looking at. So here's the raw data of the Ring Nebula taken by our telescope:
Ignore the numbers.
As you can see, well, err. Now I said this is the raw data, not an image, because what I'm really showing you is a two dimensional matrix where the intensity at each pixel is proportional to the number of electrons that were counted at that pixel. There's no sense in which this is representative of what a human would see if they had eyes as big as a telescope and could store light for two minutes. It's just a graphical representation of the electrons counted. All pictures you see--whether from Hubble or your smartphone--are just that. The difference is sometimes we want to modify that matrix so it looks something like what people see.

I'm being a little disingenuous here, though. The Ring Nebula is in this data, but because it is very faint compared to some of the pixels in the image, it's not apparent. I can turn up the contrast by bounding the brightness levels you're allowed to see, and then the nebula does appear.
Color photography is so overrated.
But I haven't done anything scientific here. I haven't calibrated the data at all, just chosen what data to show you. This isn't a more accurate or useful representation of the data. To get a scientifically meaningful image, we have to account for all the extra electrons our CCD has picked up.

One electron source is the instrument itself, which because it is not at a temperature of absolute zero consists of vibrating molecules that can occasionally knock an electron into the pixel bucket. This is called the "dark current," because it shows up even when the telescope isn't looking at anything. The warmer your telescope, the larger the dark current will be, which means that weak signals can be lost in the noise of the telescope's heat. You can minimize that heat and detect faint signals by keeping your telescope cold (like, say, by putting it in space).

To determine the dark current, Hubble does a dark calibration, which essentially amounts to taking a picture of the same exposure length as your actual picture but with the lens cap on. That way the only electrons detected will be those coming from the heat of the instrument. Once you know what this average amount of heat is, you can subtract it from the electron counts of your image. Here is an image of the dark frame from our observations:
Think TV static.
The intensity of our dark frame is about 60% of the intensity of our image, which means that by subtracting it from the image, we're losing a lot of information on faint sources. But if we don't subtract the dark current, we're overestimating the true brightness of the Ring Nebula by a factor of roughly 2.5, which would lead to some pretty bad science on our part.

Another source of electrons is the electronic components of the CCD. To operate properly, a CCD requires a certain voltage to be coursing through it constantly. For Hubble, this is the BIAS calibration, because you can think of the CCD voltage as being a bias introduced into the electronics in order to produce usable data. Telescopes acquire a bias frame by taking a zero-second exposure that doesn't let in dark current electrons or photoelectrons. Hubble does this separately from taking its dark calibration, but in certain situations you can also simply assume that your dark current includes the bias electrons. In that case, subtracting out the dark frame gets rid of the bias electrons, too. That was the case for the data we collected. If you look at what's left over after this subtraction, this is the image you get:
The Thumbprint Nebula (I've zoomed in a bit here.)
While this looks worse than the artificially contrasted version up above, the Ring Nebula does pop right out when the telescope's heat and bias is removed without manually adjusting the contrast. By fiddling with contrast, you can create spurious images that don't represent anything actually out there. No artificial images happened to be produced in this case, but we can't be so sure that the structure we see in the Ring Nebula is the true structure except by removing the dark current and bias electrons.

Finally (for our scenraio), the individual pixels in the CCD might have varying levels of light sensitivity. Since we want each photon to count equally, we have to adjust for these effects. Balancing the variable sensitivity is known as flat fielding, and you produce a flat field by shining a light of uniform intensity across the CCD. When you do this, the CCD should record the same number of electrons (more or less) at each pixel. If some regions of the CCD are too bright or too dim, you know this corresponds to unequal sensitivity. To remove the effects of this sensitivity, you divide your image by the (normalized) flat field, so that the brightness at each pixel is adjusted by a factor proportional to its sensitivity.

In space, unfortunately, it's difficult to shine a uniformly bright light on Hubble. You might think the Sun would work, but the Sun is way too bright and even a very short exposure would saturate Hubble’s sensors. Saturation causes electrons to bleed over into neighboring pixels and gives you electron counts that are not proportional to the number of photons detected. Instead, Hubble takes flat fields by looking at the Earth, which (with a lot of processing, aided by the fact that the Earth moves beneath Hubble very quickly, blurring any image it takes) can reproduce a flat field. So the DARK-EARTH calibration is Hubble's way of adjusting for the varying sensitivity of its equipment.

On Earth, flat fields are usually produced by shining a light on the dome of your observatory and having the telescope look at that, or looking at a small region of the dark sky before any stars become visible. Here's the flat field we produced:
I think the telescope has floaters.
I suspect we actually did a very poor job of shining light uniformly as I think you can see our light source positioned on the right side there. The smudges, however, probably are true variations in the pixel sensitivity, so producing the flat field removed those. (The ring-like smudge in the middle is an eerie coincidence.) After dividing our image by the flat field, we get this picture:

Possibly the Eye of Sauron (More zooming done.)
The main visual advantage seems to be increased clarity of the inner region of the nebula.

None of these, of course, look like the beautiful pictures we see from Hubble or APOD. There are two reasons for this. One, our telescope simply doesn't have the resolution (or other exquisite features) that Hubble does, so there's a limit to how nice a picture it can take. The second reason, however, is that pretty pictures are created to be pretty, not for doing science. As I said above, this is just a representation of the data, but there are other representations.

In fact, one purpose of this lab was to determine the three dimensional structure of the nebula. That is, is it a donut or a shell? A picture alone can be deceiving. But other methods of interpreting the data might be more useful. So here's another representation, plotting the brightness of the nebula along a particular axis in different wavelengths of light:
Graphs are the best, you guys.
Doing some math on graphs like these, we were able to show that the Ring Nebula is probably more like a thin shell of material than a donut, despite its visual appearance. The ring is a bit of an illusion. But a graph like this is only accurate because of the processing done to the remove observational artifacts, even though that processing does not produce an aesthetically pleasing picture.

Nevertheless, what's astronomy without cool pictures? In addition to looking at the nebula with a clear filter, we also used filters that passed only red light from glowing hydrogen and blue/green light from doubly-ionized oxygen. When you clean up that data, assign a color to each filter, and plot them on top of each other, you get this:
Insert riff on Beyoncé lyrics here.
That's not really what the Ring Nebula looks like, but it is one way of seeing it.

Thursday, March 17, 2016

On Guessing

This is a follow-up to my Lagrange point post. At the end, I briefly mentioned the L4/L5 Lagrange points, which are stable and form equilateral triangles with the masses of a three-body system. I'd like to delve into the physics of these points a bit to illustrate something about how physicists solve problems.

That is, physicists (in general) do not like doing calculations. They don't want to sit around all day crunching numbers to arrive at an answer. When you solve a physics problem, the goal is to build as simple a model as possible that captures the essential features of what you're studying. (This is where the spherical cow jokes come in.) That way, if you're lucky, you can avoid having to do a lot of math. Instead you can arrive at the answer you want by symmetry, or dimensional analysis, or guessing.

Guessing is an important part of the physicist's toolkit and some of what makes doing these problems fun (for me, at least). It's easy to stare at a problem for hours and feel overwhelmed by the complexity of it. I liken this to how it feels when you've just begun to write something. You have a blank screen and a blinking cursor in front of you and there's nothing more terrifying or paralyzing.

In writing, sometimes the solution is to just start writing and see where the story takes you. And so it follows with physics. If you have a complex problem, at times the best strategy is to just guess at the answer and see where the physics takes you. In this way, doing physics can be a lot like playing a game or solving a puzzle. It's fun, and I seriously wouldn't still be in school if I thought otherwise.

So let's return to the L4/L5 Lagrange points. In class, when discussing the three-body problem, our professor performed enough derivation to get us to believe that stable orbits can exist. He went through the same argument I used about rotating frames and centrifugal force. So a test mass is in a stable orbit when gravity and centrifugal force cancel out. He then gave us the punch line, telling us where the Lagrange points are, but didn't go through the math of actually finding them. Why not? Because if you do the derivation, the equations of motion you end up having to solve are:

I should probably credit Massimo Ricotti for this.
I'm not going to attempt to explain what all that means. It's ugly, and you wouldn't want to solve that unless you had no other choice. But there is another way. Our professor mentioned that when thinking about the 5 Lagrange points, you can guess where 2 of them (L4/L5) must be.

This intrigued me, which is why we're here today. What makes it possible to guess these locations? As we saw with the L2 point, its exact location is related to the square root of the ratio between the two big masses. This is (probably) not something you could just pull out of thin air. But that's not the case for L4 and L5. The location of one of these points is at the vertex of an equilateral triangle that has the two large masses at the other vertices. Flip this triangle over and you get the other one. How massive the objects are isn't relevant at all; distance is the only important variable (and two masses can basically orbit each other at any distance they like). So you could conceivably guess the answer just by looking at the problem.

There are a lot more MS Paint illustrations coming. You've been warned.
But what makes equilateral triangles, as aesthetically pleasing as they are, physically appealing? Let's consider a special case and then move on to a more general scenario.

Forget the Earth-Moon system and consider two stars of equal mass in circular orbits about each other. In that case, the stars are actually orbiting their center of mass, which is halfway between the two for equal mass stars. A third body that's motionless in the rotating frame also orbits the center of mass, which means centrifugal force pushes away from that center. To make the problem even simpler, let's put the third body equidistant from the two stars.

I'm a big fan of purple.
Then the forces of gravity to the left and right cancel out, leaving only gravity pulling down and centrifugal force pushing up. To get our Lagrange point, we just need those forces to balance. This means we have to guess how far up from the center the Lagrange point is.

First, let's consider gravity. The total strength of gravity depends on the inverse square of the distance to the stars, d. But we don't want the total force, only the vertical component. That part is a fraction of the total, and that fraction is equal to h/d. This means gravity now depends on the distance to the center of mass and the inverse cube of the distance to the stars.

On the other hand, centrifugal force depends on the distance to the center of mass, h, and the inverse cube of the distance between the stars, a. Our gravity and centrifugal terms are nearly the same, except one uses a and the other d. But we're trying to find d, so let's just guess that d=a. Then all the lengths of our triangle are equal and we've found a point where all the forces cancel out--a Lagrange point. (This guess works because the constants in each equation are the same. Otherwise, d might just be proportional to a.)

So there we have it. Using a few reasonable assumptions, a simple model, and nothing more than geometry, we've found the Lagrange points. Where do we go from here? How about back to the Sun-Earth system, where one of the two masses is much, much bigger than the other. If that's the case, then the center of mass moves to the sun, and centrifugal force points directly away from it.

It's a trap!
If we maintain our equilateral triangle guess, where does that leave us? With a problem. The problem is that if you rotate the above picture so that the sun's gravity vector and the centrifugal vector are horizontal, you're left with the Earth's gravity vector at an angle of 60° away from horizontal. This is bad because the "vertical" component of the Earth's gravity isn't balanced by anything else, which means that no matter what values you insert into your equation, there is no equilibrium point. Uh, oh.

But our graph has fooled us here. You see, by moving the center of mass directly on top of the sun, we are implicitly saying that the Earth has no mass whatsoever. And if that's the case, then it has no gravitational force, which means it doesn't need to be counteracted at all. In the limit where the Earth has no mass, the three-body problem reduces to the one-body problem. So there is a point of stability at the equilateral triangle, but also at any point along the same circular orbit.

This wasn't a totally useless exercise, however. It shows us that it's reasonable to expect L4/L5 to be stable from one extreme of equal masses to the other extreme of just one big mass. But we haven't yet proven that the L4/L5 points exist where they do for any arbitrary masses. How do we do that? First, let's make a generic diagram describing the situation.

You made it.
Let's say that Star A has a mass of m and Star B has a mass of km, where k is some fraction between 0 and 1. This means we can vary between the two extremes of equal mass (k=1) and one dominant mass (k=0). The smaller k is, the farther to the left the center of mass moves, the smaller Star B's gravity vector is, and the more horizontal the centrifugal vector gets. This should mean that the forces pointing to the right stay balanced. Additionally, as k gets smaller, there is less overall gravity pointing down, but because the centrifugal force is getting more horizontal, that gravity has less it needs to counteract. So our equilateral triangle still looks good.

To prove the general validity of our guess, let's see what happens if the interior angles are some arbitrary angle, rather than the 60° they must be. We have to compare the combined vertical force of gravity to the vertical centrifugal force. Using trig, we can find the distance from the test mass to a star in terms of a and θ. Because of the inverse square law of gravity, a is going to be squared. Trig also gets us the vertical component of that force in terms of θ.

On the other hand, centrifugal force depends on the distance to the center of mass, l. But because we only want the vertical component, the actual location of the center of mass is irrelevant and all we need is h, which again can be found in terms of a and θ. As before, centrifugal force also depends on the inverse cube of a, so some canceling of exponents means it's the inverse square of a that shows up.

Because both expressions depend on the square of a, we can get rid of it. Both forces are also equally dependent on the sum of the masses of the two stars, so we can cancel the mass terms, too. This means our equation is now defined entirely in terms of θ. After a little algebra, we can arrive at the following equality:

sin(θ) = 1/2

Everything else in our equation is gone. All that matters is the angle between h and d. Now, I just happen to know that the sine of 30° is 1/2. This means the full interior angle is 60°. With our guess that the test mass is halfway between the two stars, the only possibility is an equilateral triangle with interior angles of 60° and lengths of a. (A similar argument can be made for the horizontal components of the forces.)

I should note that this doesn't prove that there aren't other Lagrange points forming different triangles when the test mass is not half way between. To see that there can't be other points of stability (except on the line joining the two stars), you need to solve for the effective potential of the force fields at work in this system. That can't be done by guessing, but it can be done by drawing! Unfortunately, drawing equipotential surfaces would strain my artistic talents past their breaking point. Here's some computer art instead.

Credit: NASA / WMAP Science Team

Wednesday, March 9, 2016

Lagrange Point 2: Newton's Redemption

This past November, I had the opportunity to tour Goddard Space Flight Center. Although we saw many cool operations (including a gigantic cryogenic chamber!), the most interesting was the under construction James Webb Space Telescope. I had intended to write about the visit at the time, but I spent much of my fall semester trying not to hyperventilate instead. However, we just covered some relevant material in my theoretical astrophysics course, so let's take a look now.

A full-scale model. Credit: NASA
JWST gets called the successor to Hubble, but calling it the sequel would probably be more appropriate. It promises to explore material untouched by the first one, it's going to have even more spectacular visuals, and it's way over budget and behind schedule. The two features that most distinguish it from Hubble are its size (bigger) and its wavelengths of interest (longer).

Longer means infrared. Being an infrared telescope, JWST will see through dust, directly image planets, and peer further back in time at objects redshifted out of the visible range. But infrared telescopes come with some complications. On Earth, we don't do a lot of infrared astronomy, partly because the atmosphere absorbs too much of it, but also because stuff too cold to emit visible light (basically everything on Earth) is usually spilling out lots of infrared instead. We can't do IR astronomy on Earth for the same reason we can't do visible astronomy during the day: it's too bright.

That's why JWST will be in space. But even in space, the Earth and sun loom large. Keep the telescope too near the Earth, and the Earth warms it up, generating noise in the cameras. JWST must be kept cold, much colder than the objects it wants to look at. The only way to accomplish that is to put it far away from the Earth and hold up a shield to block the Earth and sun. The trick is that you want to be able to block both bodies at the same time, which wouldn't work if you just flung the satellite into any old orbit. The farther you get from the sun, the longer your year (Kepler's third law says the cube of your semi-major axis is proportional to the square of your year), so the sun and Earth will change relative positions in the sky.

You need to find an orbit that's far away, stable, and lets you block two objects at once--tricky. Arranging three objects in space is known as the three-body problem in celestial mechanics , and it has a long history. When Newton first formulated his laws of motion and gravity, he was able to solve the one- and two-body problems. That is, he could tell you how a tiny, insignificant planet would orbit a gigantic star (the one-body problem) or how two comparable objects would orbit each other (the two-body problem), but he was not able to count any higher than 2. Newton reasoned that miniscule interactions from nearby planets would build up over time and slowly destabilize orbits, and he assumed the only solution was divine intervention.

Astronomers, physicists, and mathematicians spent a long time looking for more precise answers. It turns out there is no generic solution to the three-body problem, no simple orbit that works for any configuration of three or more masses. Using perturbation theory, you can account for the infinitesimal, cumulative influences of many bodies over time, but in the long run (millions of years), orbits become chaotic. Chaotic doesn't necessarily mean that a planet will be flung from the solar system, but that we eventually can't say with any precision where in an orbit a planet will be at any given time.

A couple mathematicians were able to work out very specific periodic solutions to what gets called the restricted three-body problem, or the 2+1 body problem: two large gravitating masses, one tiny mass that is virtually insignificant. In just the right location relative to the big ones, the small one can be stable. Nowadays these are known as the Lagrange points, in honor of one of the mathematicians who worked them out (Euler already had enough named after him).

This seems perfect for JWST. If there's a line between the sun and the Earth, we want JWST to be on that line out past Earth. Can we find a Lagrange point there?

In space, lines are purple.
Well first let's backtrack just a second. There isn't really a line connecting the sun and the Earth, because the Earth is constantly in motion about the sun at ~30 km/s. The only way to draw such a line is if we imagine ourselves moving along at the same angular speed as the Earth so that it appears stationary.

Notice I said angular speed, which is how long it takes to move a given angle rather than a given distance. If you think about a spinning tire, the outer bits are moving faster than the inner bits, because the bigger the radius, the larger the circumference covered in the same amount of time. But they are both covering the same fraction of a circle in the same time, and thus both have the same angular speed. If different bits moved at different angular speeds, they wouldn’t keep the same relative positions and the tire would spin apart.

We want our frame and JWST to be moving at the same angular speed as the Earth. But in establishing this frame of reference, we have invalidated Newton's laws of motion. We are no longer in an inertial frame, which is one moving at a constant velocity. Circular motion is not constant, because velocity includes direction.

What does it mean for Newton's laws to be invalidated? It means that an object not experiencing any net force will seem to accelerate away. For our rotating frame, maintaining circular motion requires constant force toward the center of the circle. Tie a ball to the end of a string and spin the ball in a circle. The tension along the string is the radial force that maintains circular motion. If the ball comes loose, it will fly off in a straight line. But from the frame of the spinning string, which can continue spinning as long as you supply a force, the ball will appear to curve away. This tendency to accelerate away from a spinning frame can be accounted for if we invent a fictitious force--centrifugal force--that acts in opposition to whatever force maintains circular motion--centripetal force.

So if we look at the Earth from a rotating frame, JWST will seem to experience a centrifugal force pushing it away from the Earth. In order to have the telescope remain stationary in our rotating frame, the force from gravity must balance the centrifugal force.

Doing physics really involves making diagrams like this.
So here's our three-body problem. JWST is pulled inward by the gravity of the sun at a distance of a+d and by the gravity of the Earth at a distance of just d. That sum is:

Fg = Gmsunmjwst/(a+d)2 + Gmearthmjwst/d2

And it's pulled outward by the centrifugal force which results from the angular motion of the system. How do we characterize the centrifugal force? It's the square of the angular speed times the distance from the center of mass (the sun, in this case) times the mass of the accelerating object. The angular speed is inversely proportional to the period, the Earth’s year. So centrifugal force involves the square of the period. Using Kepler’s relation between period and semi-major axis, we can substitute in that quantity (a in our diagram). Doing some algebra, that gives us a centrifugal force of:

Fc = G(msun+mearth)mjwst(a+d)/a3

And we want Fg to equal Fc. If we cancel some stuff out, we arrive at the following expression, which is defined purely in terms of the masses and the distances between them:

msun/(a+d)2 + mearth/d2 = (msun+mearth)(a+d)/a3

We're trying to solve for d, the point at which all these forces cancel out. But there's a problem. If we were to multiply all these terms out (FOIL!), we'd find this was a quintic function, which means there'd be a d5. And there is no equivalent of the quadratic formula for quintic equations. So we have to make some approximations. We have to assume that the sun is so much bigger than the Earth (true, in this case) that the Earth can be ignored whenever the two terms are added together. And we also assume that d is much smaller than a, which lets us do some mathematical tricks. If you make those approximations, and then do some more algebra, you eventually find that:

d = a(mearth/3msun)1/3

That is the location of the second Lagrange point (and the first one, but on the other side). Plugging in the relevant numbers, d = 1.5 million km, which is curiously 1/100 Earth’s distance from the sun. The sun is a little more than a hundred times wider than the Earth, which means that from L2, the Earth and sun appear just about the same size--more or less the moon. And that means JWST can easily block both of them with the same shield. (The similarity in angular size really is a happy coincidence that has to do with an accidental congruence of densities, radii, and that factor of 3 up there. Try it with any other planet and it doesn't work.)

So there you have it. When the combined gravitational pull of the Earth and sun cancel out the centrifugal force pushing JWST away, the telescope remains stationary with respect to Earth’s motion about the sun. It sits 1.5 million km behind the Earth and completes an orbit in a year despite being farther away from the sun.

But that's not quite the end of the story. It turns out that L1, L2, and L3 (on the other side of the sun from the Earth) are only metastable, which means a slight push sends an object flying off into a new orbit. So we can put satellites there, but they require station keeping to prevent them from falling away. L4 and L5, which form equilateral triangles with the two big masses of the 2+1 problem, are stable. Consequently, we actually find families of asteroids called the Trojans at the Sun-Jupiter L4 and L5 points. Also, I’ve totally neglected the Coriolis effect here, which is another fictitious force that pops up when… oh dear, look at that word count.

Wednesday, February 24, 2016

When the Moon Hits Your Eye, There's Some Math Using Pi

It was warm and clear this past weekend, so I did some late night observation with my new binoculars. Weather is the bane of astronomers, unless you only work with space telescopes or you do neutrino observation or now even gravitational wave detection. Actually, I'm not so sure about that last one. I imagine a light drizzle could easily be mistaken for colliding black holes.

You're welcome, Celestron.
Anyway, it struck me while I was observing that the moon is very bright. Whenever I found it in my binoculars, I flinched momentarily before I adjusted to the stark change between black sky and white moon. And indeed, at night, the moon is by far the brightest thing in the sky (except for inconveniently placed streetlamps).

But it turns out the moon is pretty dim, too, when considered from another perspective (no, not the dark side). So why does the moon shine in the first place? While it does have a temperature, the vast majority of its thermal radiation is not in the visible range. Instead, of course, the moon borrows its light from the sun, reflecting it back toward us.

Naively, then, you might expect the moon would be roughly the same brightness as the sun. And when you look at a full moon hovering imperiously in the night, washing out all the stars in the sky, it does seem darn bright. However, our eyes (and the rest of our senses) are pretty terrible at discerning objective levels of radiant power. The moon is bright only relative to the sky and the stars. In astronomical terms, the sun is much, much more luminous than the moon.

Measured with fancy equipment, the apparent magnitude of the sun is about -27, while the apparent magnitude of the moon is roughly -13. If you remember from my nerdrage over Star Wars, larger magnitudes are dimmer, the visible stars are around magnitudes 1-6, and the scale is not linear. From this we can tell that the sun is way brighter than the moon, the moon is way brighter than the stars, and astronomers use a needlessly cumbersome system for quantifying brightness.

If you do the math, 1014/2.5, a magnitude difference of 14 is about a factor of 400,000 in brightness. Yes, objectively, the sun is 400,000 times brighter than the moon (as seen from Earth). So when the moon shines its paltry reflected sunlight back at you, what happens to the other 99.99975% of the light? How do we go from a sun’s worth of light to one moon unit (a Zappa)?

It's at this point you may recall that different objects reflect and absorb different amounts of light. That's why color exists, after all. You can also measure an overall amount of reflectivity, which gets called albedo. The bond albedo of an object is just the percentage of light that is reflected rather than absorbed. Freshly fallen snow has an albedo as high as 0.9, whereas asphalt can be as low as 0.04. The moon's average albedo is 0.12, which means 88% of the sun's light is absorbed. But 88% is not 99.99975%. From albedo considerations alone, the moon is still too bright by a factor of 48,000. How does the moon get rid of the rest of its sunlight?

The problem is that we're thinking of the moon as a giant, flat mirror directly reflecting the sun's light toward us. But the moon is not a mirror. You can tell this because it doesn't look like the sun. A mirror exhibits specular reflection, which means incoming light bounces off cleanly at a particular angle. If it comes in 30° one way, it bounces off 30° the other way. And since all the light bounces in the same way, mirrors reproduce an image of what’s reflecting off of them.

Ignore everything about this picture that is ridiculous.
Non-mirrors (everything else) reflect light diffusely, which from the name alone suggests the process is not so orderly. On the moon, the properties of the rough, irregular regolith on the surface determine how light is reflected, but the gist is that it’s very strongly dependent on the phase angle. In fact, the moon has an opposition effect, which does tend to bounce light directly back when light is coming from behind us, i.e. when the moon is full. Even still, the picture above doesn't hold.

I admit I struggled with this problem for a bit before finding a suitable answer. Here's what I did to solve it. How do you account for a factor like 48,000? Well, let's compare some relevant numbers. The moon is 384,400 km away from us on average. Its radius is 1,737 km. The Earth's radius is 6,400 km. The distance from the sun to us is 150,000,000 km. Hmm, I can’t think of anything else that might be important.

The distance from the sun can't matter, because we're dealing with the apparent brightness of the sun, which is how bright it looks to us from here on Earth. Distance already factors into the 400,000 figure. The Earth's radius can't matter, because we're talking about how bright the moon is to our eyes. If the Earth were the size of a pin (and we were still the same distance from the moon), it wouldn't affect the light that hits our eyes. So the only two numbers that can matter are the moon's radius and its distance from us.

Well, what's 384,400/1,737? 221. 221 doesn't look very good, but if we square it, we get about 49,000. That's very close, within a few percent, of our factor of 48,000 (which is a heavily rounded figure). Okay, but why does squaring matter?*

In the illustration above, we're thinking that the moon intercepts the sun's light and shines this perfect sun laser back at us. If that's the case, then we are hit by a circle of light with the area of the moon's disc. The area of a circle is πr2. (I told you π was involved.) If the above relation is valid, then we are really being hit by a circle of light with the radius of the distance between the Earth and the moon. How could that be? Imagine that instead of the moonlight bouncing straight back at us, it spreads out in a cone, with the angle between the edge of the cone and the line connecting the Earth and the moon being 45°.

Jobs I won't get upon completion of my degree include: NASA artist
In that case, the cone forms a right triangle, with half that base being equal to its height, the Earth-Moon distance. And if you turn the cone toward us, you see that the base is a circle. So instead of the light being concentrated into a disc the size of the moon, it's spread out into a disc with a radius of the lunar distance, which dilutes the light by a factor of 48,000 or so, because the Moon is much farther away from us than it is big.

Why would the light reflect back that way? It probably doesn't, exactly. The process by which the moon reflects light is complicated and is modeled with something called a bidirectional reflectance distribution function. But the opposition effect means a full moon tends to reflect light directly back, so everything coming back at an angle of 45° or less seems reasonable. But we're ignoring for a moment that the moon is not a point source, so that right circular cone probably looks different at other latitudes. On average, though, it works out to produce the above picture.

Anyway, that's probably enough MS Paint illustration from me for one blog post. Also, this is a reasonable length, so I better stop now before things get out of hand.

*Update: My solution to the problem posed in this post is almost certainly wrong. I believe I was right about the square relation between the moon's radius and distance from us, but wrong about why that relationship is important. That's the tricky thing about proportionality arguments: without constants, you can fool yourself about what you're talking about. Anyway, I think I've figured out the real answer.

So one of the issues that bothered me about my solution is that it relies on the moon being this weird, hard to study surface, but gives you an answer with a simple and neat geometric interpretation. That seemed unlikely, but the math worked so I accepted the answer anyway. But it turns out that the moon's surface is both harder and easier to analyze than I realized. Before I get to that, however, there's another important issue.

When I first considered this problem, I assumed the answer was that the inverse square law causes the light reflected from the moon to diminish so that it is less luminous than the sun. But after some thought, that didn't seem plausible. You see, when the sun's light travels to us, it loses some intensity because of the inverse square law, just like gravity gets weaker with distance.

For the moon's light, however, that light goes the extra distance from the Earth to the moon and back again (for a full moon). But the distance to the sun is 150,000,000 km, and the distance to the moon is 384,400, which means the additional distance traveled is only .5% more, which is only going to lose you .25% of your intensity from the inverse square law, and not the factor of 48,000 we needed. So I figured that couldn't be the answer.

What I was failing to consider, however, was that light reflecting off of the moon changes the applicability of the inverse square law. The inverse square law isn't mysterious. Rather, it's a consequence of geometry in a 3-dimensional world. If an object emits light radially from a point source, then at any given distance from the source, the light will be spread out on a spherical shell around the source. As the distance grows, the light falls off with the square of the distance, because the surface area of a sphere is 4πr2.

But any real emitter is not actually a point source. The sun radiates the light we see from its surface, which is (almost perfectly) spherical. All this means, however, is that there is some defined power at the surface, and we can imagine that power increasing to infinity as we dip below that surface to a point. But here's the key: the power radiated per unit area has some value at 1 radii out (the surface), and that power drops to 1/4 its original value at 2 radii, 1/9 its original value at 3 radii, and so on. Note that this exactly mirrors (ha) my original answer. At 221 moon radii (384,400/1737), the power has been reduced by a factor of 2212=49,000.

This answer being applicable, however, requires that light reflected from the moon is emitted radially (from the half that is facing the sun, anyway), which seemed implausible to me in the beginning given how complicated the moon's regolith is supposed to be. But it turns out that if you assume the moon is an ideal diffuse reflecting surface, then radial emission is what happens.

For a specular reflecting surface, the incident angle of the light exactly determines the angle of reflection. But for an ideal diffuse surface, the incident angle is not important at all, and the light reflects in a random direction. If the light reflects entirely randomly, then on average the angle of reflection will be exactly perpendicular to the surface, because any angle away from perpendicular will be balanced out. So on average, a diffuse reflector looks like a radial emitter and follows the inverse square law.

The complicated surface of the moon, with its opposition effect, means that the "on average" part up there is not strictly speaking true, but it apparently doesn't have enough of an effect to eliminate the approximately true inverse square relation that shows up. Why radially emitting from the moon seems to drop off more quickly than radially emitting from the Sun is because a radial emitter that has the Sun's apparent brightness at 1 lunar radii is actually a weaker source than the same apparent brightness at 1 solar radii. If you expand the lunar emitter to the size of the solar emitter, then your power/area is reduced accordingly and you have a dimmer surface, so of course its power will fall off more quickly than the solar emitter.

Well, anyway, so much for this post being a reasonable length.

Sunday, February 14, 2016

The Equivalence Post

About twenty years ago--maybe right around the time LIGO was finally getting funding, when the gravitational waves it just detected were still a couple dozen star systems away--my elementary school class did a living wax museum. We researched a historical figure, dressed up as our subject, and, when a "visitor" to the museum pressed a red dot on our hand, recited a first-person speech based on our research. Unrepentant early nerd that I was, I chose Albert Einstein.

I don't really remember anything about the contents of my monologue. I probably gave a brief biographical sketch, but likely left out the part where Einstein bribed his first wife into divorce with Nobel money he'd yet to receive. I probably talked about the theory of relativity and how it merged space and time, but likely didn't include anything about Riemannian geometry and metric tensors.

My knowledge of the scientist and his science was patchy, to be sure, but that didn't stop me from admiring him. Einstein is the model of the lone genius working tirelessly, using nothing more than the power of his mind to change the world. For a long time, I imagined he and I were equivalent. I imagined that I alone knew the secrets of the universe and that my solitude represented nothing more than the gap in intellect between myself and others.

Before the inevitable deconstruction of that paragraph, let's talk a bit about Einstein the genius. While E=mc2 is his most famous equation, it's not the equation that made him famous. Physicists will tell you that general relativity was his crowning achievement.

GR grew out of Einstein's attempt to extend his special theory of relativity to gravity. SR and electromagnetism fit together perfectly, but gravity did not behave. According to Newton, gravity acts instantaneously, and that didn't sit well with light speed being the ultimate limit. To reconcile gravity with relativity, Einstein looked at a subtle difference between the electrostatic force and the force of gravity.

When two charged particles are sitting next to each other, the electrostatic force that one feels is proportional to the product of their charges divided by the square of the distance between them--simple enough. When two masses are sitting next to each other, the gravitational force on one is proportional to the product of their masses divided by the square of the distance between them. The forces are nearly identical, just swapping charge for mass.

But when a particle feels a force, it follows Newton's second law and accelerates by an amount inversely proportional to its mass, which is what inertia is all about. This means the mass term from gravity and the mass term from inertia cancel out and bodies under the force of gravity experience the same acceleration regardless of their masses. We know this; it's just the idea that a hammer and a feather (ignoring air resistance) fall at the same rate.

Thank you, NASA.
This quirk of gravity gets called the equivalence principle, because it seems to show that "gravitating" mass and "inertial" mass are equivalent, even though there's no particular reason why they need to be.

As Einstein thought about this peculiarity of gravity, he was struck with what he called "the happiest thought" of his life. He postulated a modification to the equivalence principle, which is that being in a gravitational field is equivalent to be in an accelerated reference frame. What he meant was that gravity is not a real force but an effect we observe, so there's no difference between your car seat pushing up against you when you hit the gas and the Earth holding you down.

The link to the other equivalence principle is that, in free fall, any object falling with you moves at the same rate, and the same thing is true in an accelerated reference frame, because the acceleration you feel is a result of the frame (your car, a rocket) and not your mass.

This happiest thought led Einstein to the conclusion that being in free fall in a gravitational field is just as "natural" as being at rest. When you do feel a force (your car seat, the ground), that's just an object getting in the way of your natural path through spacetime. As usual for Einstein, his next step was to imagine what this meant for light.

Assuming his principle is true, weird things happen in gravity. Say you're in a rocket ship at rest in space. If a beam of light comes in one window, it will trace a straight line through the rocket ship and out another window. If you're moving at a constant speed, you observe the exact same thing, because special relativity says you can't tell the difference between different inertial frames.

If you're accelerating, the light will trace out a parabolic curve, because you're moving faster when the light leaves the rocket than when the light enters it. The equivalence principle says you can't tell the difference between gravity and acceleration, so the same thing should happen if you're in a gravitational field. Light passing near the Sun, for example, will curve.

Now it's all well and good to say this happens because of the equivalence principle, but that's not a mechanism. If there isn't a force causing the light to curve, what's doing it? Einstein says this is the wrong question to ask and that what looks like a force is just light taking the only path available.

Here's an imperfect analogy: imagine you're driving up a mountain, maneuvering through twisting switchbacks. If you veer one way, you fall off the mountain. If you veer the other way, you crash into the side of it. So you stick to one narrow path. To the GPS satellites monitoring the position of your phone (but not the mountain or the road), it looks as if your phone, you, and the car are being pushed around by some mysterious force, but in reality you are simply following the only path available.

Except you might think, well that works for light zooming around at 300,000 km/s, but what if there's nothing propelling me? Why am I following any path at all? And the answer is that we are all following a path constantly through spacetime. We're moving forward through time. But in the presence of a gravitational field, spacetime gets warped, and your straight path through it moves a little bit out of time and into space. The "speed" you had going through time gets converted into speed in space, which is why clocks slow down close to a black hole.

Figuring out the specifics of how mass could warp spacetime took Einstein about a decade, but he finally succeeded in 1915, giving the world general relativity. With it came a number of predictions, including the bending of starlight, the correct shape of Mercury's orbit, and the fact that accelerating masses will send out gravitational waves that stretch and shrink spacetime as they pass by. Finally detecting those waves reaffirmed Einstein's genius one more time a century after he first proposed them. And all of that came from Einstein tinkering around with the fact that all objects fall at the same speed.

I said earlier that I equated myself to Einstein, but the truth is I'm no Einstein. I'm a pretty smart guy, but not a genius, and certainly not one of the greatest scientific minds in history, capable of deducing fundamental and quantitative physical truths about the universe from simple thought experiments. What can I possibly hope to achieve compared to that?

But there is an equivalence between me and Einstein, because in reality he was no Einstein, either. It took him a decade to complete general relativity because, talented though he was at math, he was not a mathematician and had to learn an entirely foreign branch of it to make his theory work. He got help from a mathematician friend of his, Marcel Grossmann, who was familiar with Riemannian geometry. That branch of math was invented in the 19th century by a couple of guys, including Bernhard Riemann.

The idea of looking at space and time as a unified thing was partly inspired by Hermann Minkowski, who applied geometrical concepts to Einstein's special relativity. Before Einstein even got to special relativity, which was critical for getting to GR, he frequently discussed difficult subjects with a group of likeminded friends that maybe ironically called themselves the Olymipa Academy. And most of the pieces for SR were put in place by earlier physicists, such as Hendrik Lorentz and George FitzGerald.

Black holes were first theorized about by Karl Schwarzschild, who found one of the simplest solutions to Einstein's field equations while fighting in the trenches during WWI. Roy Kerr figured out how rotating black holes behave. And many others over the ensuing decades contributed to the theory.

As far as gravitational waves are concerned, Einstein himself waffled as far as whether they even existed. But even so, he originally showed only that they could exist and radiate away energy. Solving general relativity for the shape of gravitational waves emitted by two inspiraling, merging black holes took until the 90s. In fact, it was only accomplished with the help of supercomputers using numerical techniques.

And even ignoring the many contributions from theorists not named Einstein, his prediction about gravitational waves would have meant nothing if we did not have the means to detect them. The feat accomplished by LIGO this past week involved scientists who are experts in interferometry, optics, vacuum chambers, thermodynamics, seismology, statistics, etc. The effort required theorists, as well as experimentalists, engineers, and technicians.

I don't mean to imply that Einstein's work would be for naught without the janitors who cleaned his office, that he couldn't have done it without all the little people supporting him. I mean that Einstein's contribution to the discovery was only one part of a vast web of contributions by a host of extremely talented people, alive and dead, who did things Einstein couldn't have done.

On Thursday, we all learned the magnitude of what they had accomplished. Rumors of the discovery had been swirling around for awhile before it was announced. By the time I arrived at school on Thursday to watch the LIGO press conference, I had a pretty good idea of what they were going to say.

Yet that didn't detract from the occasion. Packed into a lounge in the physics department, students, TAs, professors, and I--maybe a hundred altogether--watched the press conference webcast on a giant screen. We all cheered when the discovery was confirmed and cheered again when we heard the primary paper had already been peer reviewed. Half an hour in, I had to leave to go to my theoretical astrophysics course. There, the professor and TA set up a projector and we all continued to watch the press conference. When the webcast ended, the professor took questions about gravitational waves.

Being a part of that, in the minutest and most indirect way, was thrilling. It was a day when Einstein's greatest theory was confirmed yet again, when a new field of astronomy began, and when a thousand scientists got to tell the whole world about the amazing thing they had discovered.

There's a certain--possibly strained--equivalence to my wax museum Einstein moment from 20 years earlier. School was involved, as well as a story about Einstein. But this time I was listening to that story. My passion for science and learning has remained constant, but the attitude has changed. Back then, and for a very long time after that, I took joy in knowing more than others, in being the smartest guy in the room.

Now I know that's not the case. But I also know it doesn't matter. We just don't learn about the universe by sitting alone and thinking brilliant thoughts. That is, at most, one part of the process. So I don’t have to be a mythical genius to contribute. I can be a part of something amazing, of humanity's quest to understand the world around us, just by collaborating with others who are as passionate as I am. I haven't done it yet, obviously, but just as Einstein's magnificent theory has been reaffirmed, so too has my drive to be a scientist.