Tag Archives: science

Valentine’s Day Special: Romeo and Juliet Exterminate Hedgehogs

Two of the fairest stars in all the heaven,
Having some business, do entreat her eyes
To twinkle in their spheres till they return.
What if her eyes were there, they in her head?
The brightness of her cheek would shame those stars,
As daylight doth a lamp; her eyes in heaven
Would through the airy region stream so bright
That birds would sing and think it were not night.


It’s all very romantic and all, but this is a really bad idea. And not just because removing her eyes would blind Juliet; it’s a metaphor. Besides, that part doesn’t affect the hedgehogs at all.

For some reason, Wikipedia doesn’t say exactly how far away the stars’ spheres were from Verona in the 1590s. Taking Tycho Brahe’s estimate from around that time, the closest of the fixed stars was 14,000 earth-radii from the nearest hedgehog. (55.5 million miles, for those of you who don’t think in multiples of the radius of Earth.) It’s pretty close to where the orbit of Mars is now. Needless to say, this came before the 1698 expansion of the universe.

The reason why this is a bad idea is simple: Global warming. Right now, the hedgehog is classified as endangered in some countries. Having a second sun appear in the heavens, bright enough that it looks like the regular one, could drive the species extinct in short order. Hedgehogs are nocturnal, so the addition of a second sun would mean that they only have half as much night to spend doing whatever it is hedgehogs do.

If there’s a second main source of light (Juliet’s eyes), then it would be doubling Earth’s black-body radiation. I swear by the Stefan-Boltzmann law that this would increase Earth’s effective temperature by a factor of the fourth root of two. Earth would have a base temperature of 134° F, not counting greenhouse effects. (Up from its current—say it with me—42.) Pre-Juliet Earth has reached temperatures as high as 136, and that took place in a location that does contain a native species of hedgehog. They can take some pretty extreme temperatures, though they’d usually sleep through it and they don’t have that option anymore. Add on another dozen or two degrees from the atmosphere keeping heat in, and Romeo’s metaphor might not completely kill all of them. But I wouldn’t count on it.

It Gets Worse. There was at least a chance that some of the hardier desert species might survive Earth’s average temperatures jumping above previous record highs. But the next thing is that, what if Juliet’s eyes, instead of streaming so brightly through the airy region, don’t? What about before they accept the stars’ entreaties, when they’re down on the surface of Earth in her head?

Back before everything Got Worse, Earth was receiving one eight-hundred-millionth of the brightness streaming from the spheres where the fairest stars aren’t. And that was apocalyptic enough. Thanks to Romeo’s hedgehog-killingly well-thought-out metaphor, her eyes are beaming out over a third of the amount of radiation that the Sun does. (One eight millionth of her luminosity equals one part in 2.21 billion of the Sun’s, if the back of this envelope is correct. If.)

And we just put that back on the hedgehogs’ home planet

According to the back of the same envelope (plugging in E= .36 times the luminosity of the Sun and 12.5 mm for the radius of an eye into this thing copied off Wikipedia: E=4πr^2σT^4, sigma is a constant), her eyes are over thirty-one million Kelvin. Because Juliet’s eyes are much smaller than the sun [citation needed], they have to be much hotter than its surface temperature to radiate a third as much energy.

According to my back-of-a-different-envelope numbers, if an eyeball at that temperature were to appear, the explosion from the heat energy alone would be almost the size of the shockwave from air being pushed away from four spontaneously appearing Olympic swimming pools. (We can wish that it had just been drops of Jupiter.) It’s a disaster—though it would probably settle the Capulet-Montague feud in favor of the people who didn’t get their house blown up—but at least it’s not a threat to the hedgehogs.

But Juliet’s eyes aren’t just that hot. Normal superheated eyeballs would cool down much too fast. These are also staying that hot, by the power of metaphor. Like Romeo said, they have to keep glowing at least long enough to fool the birds into thinking it’s daylight.  Which means that before they accept the entreaties to twinkle in the stars’ spheres, her eyes are approximately one third as destructive as having the Sun on the balcony. This will almost certainly exterminate hedgehogs.

Why would you do that? They’re adorable!

It gets worse. All that? That was based on the size of the universe in the 1590s. Today, the “spheres” where the stars twinkle are much further away, so for her eyes to be visible (let alone bright as daylight) from there, they’d have to be implausibly hot. “Hot” in human terms is actually a good thing. Hedgehogs prefer warm temperatures of 72-80° F. This is going to be completely off that chart.

The nearest star (I’m not even going to try for “fairest”) is 4.2 light-years away. Earth’s great circle has an area of 5.11*10^14 m^2, which is a lot, but at the same distance from Juliet’s eyes there’s 1.98*10^34 m^2 of area that isn’t Earth. (We can ignore the possibility of extraterrestrial hedgehogs.) So one two-point-two-billionth of the Sun’s total output equals one thirty-eight-point-eight quintillionth of the energy from Juliet’s eyes. Well, that’s not too bad. Her eyes are only seventeen and a half billion times as luminous as the sun.

So if some couple today—let’s call them Alexandra and Nick, after a couple of superheroes I’ve read about—were to accurately describe one another’s eyes this way, they’d end up with a set of eyeballs at 14.5 billion Kelvin. Come to think of it, even if it wipes out every living member of subfamily Erinaceinae, “literally hotter than a supernova” is quite a compliment.

Juliet’s eyes would be glowing a bit more brightly than the rest of the galaxy combined. If transported to Verona, this would kill off every array and prickle of hedgehogs on Earth, as a side effect of making there not be an Earth. But it’ll have a Goldilocks zone with a radius of four light-years, so it might make some other planets more habitable. We’d better hurry up and launch some hedgehogs into space just in case. (Not to be confused with hedgehog space, which is not actually a fit habitat for any three-dimensional hedgehogs.)

And finally, for the good of everyone’s favorite spiny species, can we please stick with “my mistress’ eyes are nothing like the sun”?

P.S. Happy Birthday!



Long-distance teleportation is hard, you guys.

I was not expecting this to be this hard.

The question is: Given only teleportation and no other superpowers, can you reliably travel transcontinental distances without instantly dying?

The reason it’s hard is that the point you start from and the point you end at are not stationary relative to each other. The Earth is rotating on its axis. It’s also revolving around the sun and whirling around the galaxy and for all I know it’s disco dancing through the cosmic microwave background, but the rest of those aren’t relevant.

Here’s why the earth’s rotation is a problem, helpfully explained in comic form:

Little did Calvin know that if he ever invented teleportation his life would depend on this very thing.

The point toward the middle of the record is a point toward one of the poles, closer to the axis of rotation. The point on the edge is a point on the equator, away from the axis. They’re moving at different speeds. Say you’re in San Antonio, and you teleport to Mexico City. Well, Mexico City is moving seventy-eight miles per hour faster than San Antonio. The nearest wall rushes you at 80mph from the west, and I don’t like your chances. And that’s two cities that are pretty close together.

But that’s with cities that are basically on the same line north to south. If they’re not, then it’s both more complicated and way more dangerous. Places at different latitudes are moving at different speeds, but places at different longitudes are moving different directions.

Suppose this diagram represents the earth, with us looking down on it from above the North Pole. (This is in fact exactly what it represents.) The earth rotates counterclockwise around its axis, which is helpfully labeled A. There’s nothing at point C; don’t bother going there. It’s spinning counterclockwise: If you’re at point D then your current movement is directly left.

Your speed is based on your latitude: at the equator it’s about 1080 miles per hour, where I am in Los Angeles it’s about 895, at the North or South Pole it’s zero. So if you’re at point D, you’re moving to the left very fast. If you then teleport to point B, you’re still moving to the left very fast. You smash into the earth instantly and die.

The obvious solution is to teleport to the point precisely ninety degrees east of you. For instance, from B to D on that diagram (Just pretend they’re ninety degrees apart.) Then the direction you’re currently traveling in is straight up. You go up, eventually gravity turns you around, and you come back down into the middle of the Atlantic. Bring a parachute and a life raft. Or, if you’re Indiana Jones, just a life raft.

This doesn’t work. Partly because the very fast speeds in question are in fact faster than the speed of sound. You’re not surviving that. More importantly, even if you do use that trick, it just cancels out the momentum you already had. It doesn’t get you moving in the direction Point D already is. Stopping only helps if you’re aiming for one of the poles; you need to match the speed and direction of your destination.

OK, next obvious solution (spoiler: this one doesn’t work either). You can try to get to the point 180 degrees across from you like this: Teleport 90º east, as before. You fly up. When gravity remembers to pull you back, you start falling. Note that, if you started at the bottom of the circle, you are now on the far right, and “falling” means falling left. At some point before you hit the ground, teleport straight up and keep falling. You speed up, gaining 22 miles per hour each second. (I’m starting to regret not using metric for this post.) Eventually you’re moving at the right speed. Since you’re already going in the right direction, you can now safely teleport to your destination. (This can get you to one of two points: The farthest point on Earth from where you started, or the point that corresponds to that one but on the other side of the equator.)

Except that it doesn’t work. It doesn’t work because of one of the most annoying things in the history of physics, air resistance. Human terminal velocity is around 120 mph, and you can’t get much faster than that while falling through air on earth. Depending on where you want to go, you might need to be moving several times that.

So you do it from higher up, outside Earth’s atmosphere. On the bright side, there’s no air. It won’t interfere with your attempts to speed up. On the less bright side, there’s no air. If you know enough to not hold your breath, you’ll survive with no permanent damage for up to ten uncomfortable seconds. But by fifteen you’ll lose consciousness, presumably lose your ability to teleport, and then die. So let’s keep it under ten seconds of vacuum exposure.

Unfortunately, this can’t work. Earth’s gravity can give you 22 miles per hour extra speed for each second you spend in freefall, but you only have ten seconds. Say you’re falling at 120 miles per hour the regular way, then get another 220 from doing that insanely dangerous thing for ten seconds. That’s still not fast enough. If you’re aiming for somewhere equatorial, you need to be able to reach 1080 mph.

The problem here is that Earth’s gravity can’t accelerate you enough. (And no, you don’t get to bring rockets with you.) So, it’s time for another obvious solution: Teleport to Jupiter! This involves some major headaches, because it turns out that your departure and destination cities are not actually in the same orbit as Jupiter. So now we actually do have to think about position relative to the sun as well as position relative to Earth’s axis.

The plan is the same as before: Teleport to a specific point above Jupiter. Spend up to ten seconds letting its gravity pull you toward the planet. This will never take more than ten seconds thanks to Jupiter’s gravity; exactly how long it takes depends on how close to Earth’s equator your destination is. Pick the point above the planet in such a way that the line from where you are toward Jupiter is the vector that will, in a few seconds, make your velocity match the velocity of your destination. Once you’re going at the right speed in the right direction, teleport to where you want to go. You’re travelling at insane speeds, your destination is travelling at insane speeds, and if you did the math right then you’ll match up perfectly.

We can’t always do this. Earth and Jupiter are moving at up to 100,000 mph relative to each other depending on the time of year, and it would really suck if you forgot about that and burned up in its atmosphere. In order to avoid complicated math, let’s say you only do it when Earth and Jupiter are moving along parallel lines. This happens about twenty-two times every twelve years.

With the two planets moving in the same direction, or in directly opposite directions, you just have to pick the time of day when that direction is the vector you want. (Or, once again, the exact opposite direction.) Then you teleport directly in front of or directly behind Jupiter, get the velocity to match your target city, and there you go. Because “teleport interplanetary distances instantaneously and use a gas giant’s gravity to change your velocity at 10g for a few seconds in hard vacuum before teleporting back to the planet you started from” is totally a reasonable thing.

I bet he broke the sound barrier

(Spoilers for the Dark Knight Rises.)

At the end of the movie, Batman and his allies fail to stop the nuclear bomb, so he flies it out over the bay to prevent it from wiping out Gotham. People on the Internet correctly noted that this is definitely underestimating the range at which nuclear weapons can cause Bad Stuff to happen. But there definitely exists some distance such that the bomb going off at that range wouldn’t hurt anyone in Gotham to any instantly noticeable degree.

At the football scene, Dr. Scientist states that it’s a neutron bomb with a blast radius of six miles. Bane said it was a four-megaton bomb, and those numbers do not go well together. I’ll take the nuclear physicist’s word over Bane’s, and we can find out how far Batman must have gone.

I don’t fully understand why the equations I’m using are the right ones (I would have expected inverse squares instead of weird fractional exponents), but Wikipedia cited this as a source so it probably works. Anyway, for a neutron bomb (unlike most nuclear weapons) half the energy goes to radiation and does not affect the size of the blast radius. If the blast radius (in km) is (Yield/2.5kt)^.33, then a six-mile blast radius means 2410 kilotons. Double that, because half the energy is going into radiation instead of blast, and it’d take a 4.82 megaton bomb. So Bane wasn’t that far off after all.

We can determine how far Batman would have to fly it. It was obviously over six miles, or everyone watching it would have been blasted backward and very possibly killed just from the air. But it must have been even farther than that, because there were unprotected humans with a clear line of sight to the blast. When it went off, they cheered. They didn’t appear to have been covered in burns or struck blind or anything.

For a 4.8 Mt neutron bomb, the thermal radiation would hand out third degree burns within about a ten-mile radius. But the other effects of the thermal radiation reach go further than that.

The people on the bridge were all watching Batman fly the bomb away. And they were looking directly at it when it went off. (Note: Never do that.) Fortunately for them, it went off during the day instead of at night. Dilated pupils would be a bad thing. Unfortunately for them, it was a pretty clear day. A one-megaton nuke would temporarily blind people from 13 miles away. That’s for a regular nuclear weapon where 5% of the energy goes to radiation. For a neutron bomb, that number is 50%, leaving proportionately less for thermal radiation. (To make up for that, the neutron radiation is worse, but that has a smaller range anyway.)

Since a two-megaton neutron bomb would blind people from a bit over 13 miles, a 4.82-megaton one would blind people from 13√(4.82/2) miles away.  That’s a bit over 20 miles. Hopefully he flew it farther than that, but any closer and the kids in the school bus would definitely not have been cheering. So this gives us a good lower bound on his speed.

The timer on the bomb showed 1:57 when Batman attached it to his flying car. Then he kissed Catwoman and told Commissioner Gordon his not-remotely-secret-anymore identity and started the car, and by the time he took off it had been over 40 seconds. That leaves less than 77 seconds for him to fly it more than 20 miles. Apparently the Bat can fly at 935 miles an hour, which is well over the speed of sound. So I win my bet, muahaha.

Maybe there are other effects with wider reaches. Like, Gotham is probably going to have some severe fallout problems later. If there are effects that would show up on screen as soon as the bomb goes off and have a wider range than the flash blindness, Batman would have had to take the bomb even farther. But 20 miles in 77 seconds gives us a lower bound: He must have gone at least that fast.

(And it just occurred to me that I should be timing from when he passed the bridge instead of when he took off, since the bridge is where the people were. But the movie didn’t show the timer position for that, so I’ll just say it’s definitely significantly higher than 935 and use that number anyway.)

Incidentally, if he covered that much distance in that little time, then his average acceleration was 20/77*3600/77 miles per hour per second. Or 5.4 m/s^2, about half a gee. Which is much more survivable than I was expecting.
But the air resistance against the giant spherical bomb would be .5*1.225 kg/m^3*(418 m/s)^2*.47*π*.75^2 = 88.885 kiloNewtons. That’s about ten tons of force just from the air pushing back against the bomb. And the hovercraft can apparently fly at over 935 miles an hour while dragging that behind it. Too bad it got nuked, because that must be a seriously awesome machine.

Drops of Jupiter

I heard a song the other day. Half the lyrics sounded like the sort of meaningless but positive-sounding things you expect out of song lyrics, and the other half were astronomy. Bad astronomy. I utterly failed to take the song seriously, because whoever it’s being sung to is very extremely dead.

“She’s back in the atmosphere with drops of Jupiter in her hair.”
Believe it or not, saying “drops of Jupiter” actually makes sense. There exists a substance that, while not unique to Jupiter, is characteristic of it. It makes up a large fraction of the planet and is in fact liquid. Perfect. That material is metallic hydrogen. So naturally the question is, what happens if a drop of metallic hydrogen from Jupiter were to appear in someone’s hair?

Not pictured: Safety.

At the shallowest relevant depth in Jupiter’s atmosphere (there are much bigger numbers available, but we want to give the lyric-writers the benefit of the doubt), the temperature is 10,000 Kelvin, and the pressure is hundreds of thousands of gigaPascals, millions of times Earth’s atmospheric pressure. We’re talking center-of-the-earth type pressure and temperature hotter than the surface of the Sun. It should go without saying that you do not want that in your hair.

Wolfram|Alpha tells me that a “metric drop” is a unit of volume. I needed a number, so I may as well use the actual one. So this person has a couple of drops of metallic hydrogen in her hair, each drop being .05 cm^3. To find the actual amount of hydrogen in each drop, PV=nRT. Yes, that’s an equation for gases not liquids, but this temperature and pressure is a transition point where it’s going between those phases, so I’m hoping it’s still valid to use it here. (Despite the term “metallic,” this is kind of like a gas and kind of like a liquid and not remotely solid.)

If this equation doesn’t work here because the material is too liquid, well, the liquid phase is obviously denser than the gas anyway, so this will be an underestimate rather than an overestimate. So just pretend I added another clause on to each paragraph saying that it’s at least this bad.

PV=nRT. (2*10^8 kPa)*(.05 cm^3)=n*(8.31×103 cm3*kPa*K-1*mol−1)*(10000 K). Solve for n, and it’s 0.12 moles of hydrogen. Thanks to hydrogen’s excessively convenient molar mass, that’s also 0.12 grams. That…doesn’t sound so bad. Until you remember the temperature and pressure it’s at.

The first thing that happens is that this small drop immediately and explosively expands until it’s at the same pressure as the air around it. Since the pressure is dropping by a factor of 2*10^8 kPa/101.3 kPa =1974000, the volume increases by the same factor. So that’s how many metric drops of volume it occupies, and that means it takes up about three and a half cubic feet. Not a huge amount, but enough that when the explosion happens right next to someone’s head, well, it does not go well for them. I might run those numbers later.

The next thing that happens is that the hydrogen sublimates and cools down. It goes from being in its metallic liquid phase to being diatomic gaseous, and then since it’s surrounded by air that’s cooler than it is, its temperature decreases. Hydrogen has a specific heat of 14.3 J/g/K, meaning it takes 14.3 Joules of energy to raise the temperature of one gram of hydrogen by one degree Kelvin. Conversely, for each degree Kelvin that the temperature decreases, there are 14.3 Joules per gram of hydrogen released. That energy has to go somewhere.

The specific heat is actually higher, because these are some pretty extreme temperatures and it’s not really a constant. But I’ll round down so the numbers are simpler, and add a third “it’s actually worse than this” disclaimer. We’ve got 0.12 grams here, so each degree of temperature change sends 1.716 Joules of energy into the air or whatever else is unfortunate enough to be close.

Since the temperature starts at a staggeringly high 10000 K, and is going down to regular Earth atmosphere temperatures like 300 K (80º F), that’s a change of 9700 K. So it’s releasing more than 16.6 kJ of energy in the form of heat to whatever happens to be nearby. That’s surprisingly small; Wikipedia tells me it’s about the energy contained in nine M16 rifle cartridges or four grams of TNT, but to be honest I was expecting something more apocalyptic. But it does happen to be centered on a human head, which is bad.

A human brain masses about 1400 g, and has a specific heat of 3.64, so if all the heat were going in to the brain and were evenly distributed, it would only raise the brain temperature a few degrees. Which, sure, could kill the person. And yes, it won’t be evenly distributed. The heat will be concentrated at the skin near the drop. All the skin on that side of the head will be instantly destroyed and I don’t know what will happen to the skull. But it’s a lot less bad than it could have been; I was expecting the heat to leave her head completely vaporized.

Well, that was less bad than expected. Let’s stop worrying about the heat and look back at the force of the explosion.
The center of the blast, of course, was close enough to the person’s head to be described as “in her hair,” so a significant fraction of the incredibly hot exploding gas is going to come into contact with her head. It’ll be less than half, but in about that ballpark.

A sphere with a volume of one drop has surface area 6.56*10^-5 square meters. The pressure in that drop was 2*10^8 kPa= two hundred billion Newtons per square meter. So the force being directed headward is about half of the product of those. (Can I use these numbers this way? It seems like I should be able to.) If that much force hits the head of the person in question, that could be millions of Newtons. Like, the kind of force like when you’re minding your own business and suddenly there’s a blue whale on top of your head. That kind of force. It’s even worse than I was expecting.

To make things even worse, the singer said drops. Plural. It is very safe to say that the subject of this song no longer has a head, and the rest of her body is probably obliterated as well. So, as a general rule of thumb, you should probably keep drops of Jupiter away from your hair.

Sunburned by height

“Maybe you got more sunburned because you were closer to the sun.”
—Seen on Facebook last week.

Random act of math: At what point would height determine which of two people gets more sunburned?
It’s not going to be literally closeness to the sun that determines it, because then you’d have to be tall enough to reach some significant percentage of the way to the Sun. Instead, you just have to be tall enough to be some significant percentage of the way through Earth’s atmosphere. Then it’ll block less of the sun’s rays and interfere less with the sunburning. The “significant fraction of the distance into the atmosphere” bit says right away that it won’t matter much under normal circumstances.

Assume the two people are at the same latitude, same time of day and year, and so on. Also assume they have precisely the same tolerance to UV. And despite this being one of the times when “how’s the weather up there” is actually a reasonable question, we’ll just say that’s the same, too. A perfectly clear sunny day, both at ground level and head level.

There’s a way of measuring sunburn danger, of course, and of course nobody uses it. The UV index measurement, unusually for this kind of thing, is linear instead of logarithmic. So a 9 on that scale actually is about three times as much sunburn risk as a 3 for the same amount of time, and a being out in the sun when it’s a seven actually is equivalent to the amount of UV as a six plus a one. Which is really shockingly convenient and it’d be nice if more scales did this.

A 0-1 on the UV index means there’s no UV radiation to speak of that day. Maybe wear sunglasses if you’re walking on a reflective surface like snow or a giant mirror, but you don’t really have to worry about sunburn. In the spirit of erring on the side of conservative assumptions, I’ll use 1 to mean the difference between slight risk of a small sunburn versus definitely no sunburn. Thanks to the really convenient linearity property of the UV index, the difference between a 0 and a 1 is about the same as the difference between a 3 and a 4 or a 16 and a 17. Sure, that sounds obvious, but don’t ever make that assumption if you’re working with the Richter or Kardashev scales. Not if you want your city to continue to exist.

The day in question, there was a UV index of about 8. That’s for a clear, sunny day with sunburning amounts of UV B radiation. So for the tall person to get more burned than the vertically challenged one, she’d have to be high enough that there’s an index of maybe around 9. Since the difference between slight and none should be the same as the difference between some and slightly more.

That means there’d be an eighth more radiation, an increase of 12.5% While there is more to the UV index than just measuring amount of radiation, most of that “more” (according to the compendium of human knowledge) is just involved in weighting the appropriate types of radiation. We care more about some than others because only some are dangerous. So we multiply them all by different amounts to reflect that. But as we go upward, they all increase by the same percentages.

According to the World Health Organization, UV levels increase by about ten percent for each additional kilometer of elevation. This actually sounds pretty questionable, because I can’t think why it would be a constant ratio at each kilometer instead of something like not much change until you reach the ozone layer, and then more UV quickly as you start going through that. But it’s the WHO, so they’re probably right.
So to increase the UV index from eight to nine by being tall, your height would have to be 42.6 metric PB (Paul Bunyans). And that’s the Disney Paul Bunyan. Moral of the story: No. Just no.

Astrology That Works

Inspired by this excellent post on all the things wrong with astrology, I decided to try to design a universe in which astrology actually does work. My criteria, in order, were:
1) At the end, something similar to modern astrology would be able to make accurate predictions.
2) Explanations should be good enough that it wouldn’t drive me screaming out of a theater if a sci-fi movie used technobabble of approximately this quality. Ideally I wouldn’t facepalm, but that’s negotiable.
3) Within those constraints, it’d be nice to have it be as similar to the real universe as possible. By similar I mean that a casual observer looking at Earth wouldn’t notice the…alterations…I perpetrated in the background.
Disclaimer: I know virtually no astrology, so if any of my blunders were too bad, feel free to correct me and…who are we kidding, if you’re reading this you’re probably not the kind of person to know astrology either.

To start with, I should clarify what astrology can do and what it can’t. The most important thing to recognize is that it’s a statistical effect only. Being a Sagittarius can’t guarantee you’ll be an extrovert, but it can make it more likely. If you’re genetically predisposed to something opposite your sun sign, or raised in a way opposed to it, that can probably overcome it. But being born under a certain sign definitely makes it more probable that you’ll have those characteristics. Current research suggests that the position of the heavens probably has more of an effect than what kind of books or music you grew up with but less of an effect than genetics. (See Malcolm Gladwell’s book Outliers, in which he shows that what time of year someone is born can have huge effects on their life.)

There’s also a lot of information necessary for a good prediction. Those one-line predictions they print in newspapers are barely better than worthless. They even give the same prediction to everyone born within a month of you, so clearly they don’t  take themselves too seriously. “You are dedicated to your goals” might be disproportionately true of Leos, but it’s not reliably true of anyone.

But enough of what it does and doesn’t do, and on to how. As far as mainstream science goes (if you listen to those sheep), there are precisely no forces that could account for astrology. Only gravity even really reaches far enough, and gravity has a pretty well-understood list of effects. (Noticeably absent: mind-controlling humans.) So there must be something else going on.

See, what all those mainstream scientists don’t know about is that there’s a particular type of subatomic particle called the fakeion. The fakeion travels at lightspeed and, like a neutrino, can be hard to detect. But there are more of them than neutrinos. In fact, Earth is perpetually being bombarded by vast numbers of fakeions of all types. (There are twelve types, one coming from each constellation in the zodiac. More on that later.)

The particles themselves have no measurable mass and are hard to observe, but they have detectable effects when zipping through the Earth. They even affect humans. Sometimes it’s just a subconscious sense that something’s different. Sometimes it seems like there’s something in the air when the concentration of different types of fakeions changes. (Perhaps literally true, caused by collisions of fakeions with the nuclei of atmospheric gases?) Changes like this, indirect evidence of fakeion distribution, can be and have been detected by scientists. But they tend to be considered normal seasonal things, since they do vary regularly around the year. So even if they notice a correlation between, say, enthusiasm and the sun being in Leo, they’ll put it down to the fact that people are just more enthusiastic about things in the summer. And they’re not wrong, but there’s more to it than that.

Psychologists know that what you’re surrounded by can have significant effects on your mood. Being in a room painted blue can calm someone, for instance, even though calmness isn’t exactly an essential property of the color blue. It’s just that the human subconscious interprets things weirdly sometimes. In just the same way, when surrounded by Libra fakeions and their effects on the environment, it can subconsciously influence people to be slightly more social. Other types have comparable effects. It’s not like the stars can beam down empathy rays that make humans more empathetic; that wouldn’t make sense. It’s just that just like how being in sunlight can make people happier, being flooded with fakeions can have different effects.

I said earlier that there’s one type of fakeion coming from each constellation. That’s oversimplified, because a constellation isn’t exactly a place that anything can come from. It’s a group of stars, many of which are farther from each other than they are from Earth. Saying something is coming from there is really just talking about the direction, not the location.

Now, there’s one emitter of fakeions in each of the twelve sections of the night sky. They’re all at the same (humungous) distance and intensity, so there are the same amounts of each type. With one exception. At different times during the year, the Sun is in the same direction as one of the signs.

In this picture (from Wikipedia), the Sun is in Cancer, meaning it’s between that constellation’s fakeion emitter and the Earth. That emitter is shooting out its particles in every direction, and only a tiny number of them head toward the Earth. That goes for all twelve emitters, actually. But with the Sun in Cancer, it slightly increases the how many of that type of fakeion reach the Earth. It does this through a phenomenon called gravitational lensing. So you end up with slightly more of the ones from Cancer hitting Earth than the other types. Through the effect of the environment on the human subconscious, this is enough to cause more empathy and sensitivity and that kind of thing. Effects could be different if it were a different type of particle; this is just what happens with the ones from Cancer.

That’s why it’s so important what sign the Sun is in. Gravitational lensing can “brighten” the image of whatever is directly behind the Sun at appropriate distances. I don’t know enough relativity to say precisely what kind of distances we’re talking about here, but it’ll be at the interstellar range if not more.

More information on those emitters. At any given time, a straight line from the Earth through the Sun intersects precisely one of them. There’s always one directly behind the Sun at any point. That means they’re really really huge. Picture a gigantic circle around the Sun, with a radius of plenty of light years. And the gigantic circle is in twelve segments (the signs), each of which sends a different flavor of particle, and whichever segment is currently behind the Sun is overrepresented in the composition of Earth’s current fakeion environment.

(Incidentally, a giant ring like that centered on the Sun would not only prove the existence of some extraterrestrial intelligence, but also that they had some interest in this solar system in particular. And they’re well over a two on the Kardashev scale.)

So, the position of the Sun in the zodiac can have some effect on people’s general mood. But people’s personalities also have to be influenced by where it is when they’re born. Well, as we know, the earliest times in neurological development can be crucial. And if there are huge numbers of these fakeions around (there are), it makes sense that they would affect the youngest the most. These are weakly interacting particles, so it’s not like they’ll do anything to people directly very often, but with this many of them, there will be secondary effects from the rare occasions when they do interact with something.

Most of the subatomic collisions happen inside the Earth of course, since that’s where most of the mass is. But those have little to no effect on the surface. Looking only at the collisions that do happen in the surface range, it produces various types of low-level radiation or temperature change or things like that. Things like that can have a strong effect on an undeveloped brain, so it’s no surprise that the effect the environment has on someone that only just started interacting with it would end up lasting longer. Tangentially, this implies that if a child were raised in a carefully controlled environment, probably with its own air and light supplies, astrology might have no effect on them at all. Nobody has ever tried this experiment, for good reason.

So, this takes care of the Sun and the signs. But the planets have some effect, too. As that Bad Astronomy article points out, each planet has an approximately equally strong astrological effect despite the fact that they’re different in basically every way that planets can be. The obvious answer is that on each planet (or in orbit around it, for the ones where “on” wouldn’t work very well) there is a bit of machinery. A fakeion detector wouldn’t be too hard to build. (I’ve been lobbying the astronomers to build one for years, but they don’t seem to take me seriously after I tell them that a major component is beer.)

There could easily be a high-quality, long-lasting, beerless detector at each planet (plus the moon). Once the machinery knows what section of the zodiac the planet is in, it can fire up its own fakeion emitter and have the standard effects. (Consult any reputable horoscope for what those would be.) So we can say that, for instance, when Mercury is in Aquarius it’s a good time for logical objectivity. Even more so than usual, that is.

The planetary emitters would of course be far weaker and FAR smaller than the interstellar ones, but they’re also closer. The giant ring around the Sun needed to be so big not in order to be powerful enough but so that the Sun would always be between it and the Earth. These don’t have that requirement. Smaller ones are all it would take for closer things like planetary ones.

The biggest thing to note is that this entire array would have to be put there intentionally. It is even more obviously focussed on Earth specifically than the giant ring around the solar system. And if creating astrology is its purpose instead of just an unintended effect, then it would have to be set up by someone who understood human brains better than humans currently do, enough to predict what reactions they’d have right down to the second-order side effects of subatomic particles. The astrology is not even the most important thing any more.

But since I still can’t convince the astronomers to finance my expedition to the moon, we’ll have to settle for predicting the future. The next time I’ll have a good chance at convincing someone would be…sun in Sagittarius…Mercury in Aquarius…pretty soon, actually. With the stars on my side, it’s just a matter of time.

How Many Subatomic Particles Are In That Dandelion?

A lot of septillions.

Phineas and Ferb is a show about a bunch of supergeniuses. It is also awesome. In one episode, the characters build a helmet that makes people smarter, and use it on their smartest supergenius. To make sure it worked, they ask him to find the number of subatomic particles in a dandelion. It takes him a few seconds, and he covers the fence in math-like squiggles.

Here’s how to estimate it simpler and with less effort than the intelligence-enhanced fictional supergenius, and do it in your head.

We’re just going for an order-of-magnitude estimation. That basically means figuring out the number of zeroes on the end of the number. This kind of estimate is not very precise, but is still useful and much easier. In high school I was taught that mathematicians are lazy and always do things the easy way. Never bothered confirming that.

First thing: What’s the mass of the dandelion? Phineas didn’t specify if he meant just the flower or the whole plant, but the calculation is similar either way.

Using xkcd’s phrasing: Picture a bundle of ten dandelion plants. That’s something you can pick up in your hand and throw, something you can pick up and throw is one pound, one pound is one kilogram. (That emphasizes that we’re really only estimating, but 100 grams per plant is probably in the right ballpark.) The point is that it’s probably a better estimate than either 10g or 1000g.

What is all that mass made of? Anything smaller than an electron has negligible mass. Those 100 grams are all protons and neutrons. Those have the same mass (well, they don’t, but the difference is too small to matter). What’s the mass? I don’t know. I can never remember that. But we can find it. There is a number called Avogadro’s number, since in science everything needs to have a name. This is so that outsiders know when to be scared. It says that there are 6.022×10^23 atoms in one mole. (Mole as in unit, not mole as in small furry cat food.) And hydrogen has a mass of one gram per mole and a hydrogen atom is just a proton and an electron. So our 100-gram dandelion contains the mass of 100*6*10^23 protons.

But the question wasn’t about mass; it was about the number of subatomic particles. There isn’t enough information so far to find that. Maybe it’s only made of one giant dandelion-shaped particle. (It’s a dandelion, right? So it’s made of dandelion. Obviously.)

Well, we know what dandelions are made of. Carbon, hydrogen, nitrogen, oxygen, stuff like that. All of those atoms except for hydrogen have the same numbers of protons and neutrons. Carbon is six of each, oxygen is eight, etc. So that tells us that it is something like 3*10^25 protons and 3*10^25 neutrons. It’ll be slightly toward the proton side because of the hydrogen, which is a proton without a neutron, but that won’t throw it off much.

The obvious next thing we’re missing is the electron. Since the dandelion is grounded, it most likely has net zero electrical charge. So there are as many electrons as there are protons, meaning 3*10^25 of each. This is where it starts to matter that there are more protons than neutrons. If the dandelion were made entirely of hydrogen, there’d be 12*10^25 particles (half protons, half electrons). If it were entirely made of those other elements with no hydrogen at all, it’d be 9*10^25 (one third protons, one third neutrons, one third electrons). The actual value will be somewhere in between, but it’ll be much closer to that last number. In terms of composition by mass, hydrogen is heavily outweighed (heh) by neutron-containing elements.

The tally so far: 3*10^25 protons, 3*10^25 neutrons, 3*10^25 electrons.
But each proton and neutron is made of three quarks. A proton is two up quarks and a down quark, a neutron is two down one up, but the point is that it’s three each. (This also makes the hydrogen thing from earlier even less important.) So it’s 9*10^25 quarks in protons, 9*10^25 quarks in neutrons, and 3*10^25 electrons. Luckily, none of those particles are made up of smaller ones. We’re probably supposed to count both the protons and the quarks making them up, because they are both subatomic particles. Anyway, the total comes to 2.7*10^26 particles, or two hundred seventy septillion. I promise “quark” is a real word.

I’m not counting gluons. That’s because I have no idea how many gluons are in a nucleus at any given time. (After Googling it, I’m not even sure if it’s a meaningful question.) We’re talking about light elements, so it’ll probably be relatively low, but I have no frame of reference for what that means. Well, we just wanted an order of magnitude, and unless the gluons significantly outnumber everything else (unlikely) the true number is somewhere on the hundreds-of-septillions scale. Plus or minus a couple of orders of magnitude. I never said it was a very precise estimate.

Start with the mass in grams, multiply by 6*10^23 to find the number of nucleons, multiply by three halves to include the number of electrons. Of those nine hundred sextillions, six are made of three quarks each. So multiply those by four (so we count the quarks as well as the nucleons. If you’re asked for the number of elementary particles, multiply by three instead) and add the electrons. You get 6*4+3=27*10^23. Since questions like this come up a lot, here’s how to do it in your head, as promised. For ease of remembering (and mental calculation), round it off to two and a half septillion subatomic particles per gram. Ta-da!

How well this works depends on what it’s made of. Light elements (except hydrogen) work best. This approximation would overestimate stuff with a lot of hydrogen in it; for heavier atoms it’ll underestimate it unless you want to factor in the proton-neutron ratios. Nobody wants to do that, but it won’t matter too much. It’ll always stay within a factor of 1.5 or so and we’re just aiming for an order of magnitude.

Turns out, and I did not know this in advance, this is the same answer, right down to two significant figures, as the Phineas and Ferb writers had Baljeet say. (Page search this transcript for “Heisenberg.”) This is probably not coincidence. In other words, HOLY FLYING FISH, THEY ACTUALLY DID THE MATH.

UPDATE: I emailed Swampy Marsh, and he confirmed that they did indeed do the math. He didn’t say who specifically, but Doing The Research as standard practice is even better. I told you they were awesome.