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Sunshine’s Crazy Sloppy Path to You

This here?

Drawing of an arrow pointing to a little yellow speck in a large black blot
Drawing by Robert Krulwich
Drawing by Robert Krulwich

This is a bit of sunshine. It’s made of pure energy. It has no mass—nothing you can hold, touch, or (accurately) draw. It’s called a photon. Some think of it as a little rat-a-tat of energy packets.

Drawing of a black blog with four little yellow spots in a straight line
Drawing by Robert Krulwich
Drawing by Robert Krulwich

But, being pure energy, it goes so fast (at the speed of light) that its true nature is hard to detect—unless it bumps into something. Here it is colliding with an atom, kicking electrons up into higher, more energetic orbits …

Drawing of star bursts in a line
Drawing by Robert Krulwich
Drawing by Robert Krulwich

… and then, an instant later, those electrons settle back, energy is released, and —whoosh!—our photons are off again. This is what photons do: They get passed from atom to atom, absorbed and spit out, absorbed and spit out. For those of you who like spice in your lives, be grateful you’re not a photon, especially when you remember that most photons are gathered in dense clumps of burning plasma called stars.

Stars are crammed so tight that atoms get crunched, their electrons stripped away to form vast, free-floating electron clouds. So if you’re a typical photon, you spend most of your time slamming into electron after electron. Fwwaaack! Fwwaack! Fwwack! Your energy is absorbed, then released. You may be able to fly at the speed of light, but because you’re stuck in the middle of the sun, when you finish with one electron, you get to swoosh less than 1/63rd of an inch before you’re absorbed again.

Drawing of a sunshine surrounded by pink and purple circles with the capital letter ''E'' on each of them
Drawing by Robert Krulwich

If there were ever a shy photon, one that didn’t like crowds, I imagine that it would yearn to escape the madcap crunch of electrons. I see it deep inside the sun, crawling closer and closer to the surface, electron by electron, until—with any luck—it gets scooped up by one of those giant solar flares and then flung …

Gif of a solar flare
Image courtesy of NASA
Image courtesy of NASA

… Whoosh! … across the quiet, empty highway of space, careening along at a crazy 670,616,629 miles per hour, free, free at last, like a happy racehorse.

This happens to real photons. Some do get free of their stars, do escape into space as solar radiation. And if they happen to crash into Earth, we call them “sunshine,” and when we go to beaches, lie down, and rub ourselves with lotions, we wait for them to bang into atoms and warm us up.

But consider this: We don’t appreciate how long it takes for sunshine to escape the sun. Every bit of sunshine warming your skin has a long history—wonderfully, fantastically, ridiculously long. Next time you’re at the beach looking up, think about this story.

Picture of sunlight filtering through a tree in Yellowstone
Photograph by Tom Murphy, National Geographic Creative
Photograph by Tom Murphy

Imagine A Photon

Let’s start at the very center of the sun, where it’s blazingly hot (10 million kelvins). Here, hydrogen atoms are slamming into each other so hard that their protons fuse and form helium, and with every crash, little bits of pure energy are released. Those are our heros, the photons.

In the crowded middle of the sun, a photon can only move a little way before bumping into another atom. We don’t really know how dense it is in there, but scientists figure our photon hero will zing between a tenth of a millimeter (four thousandths of an inch) to a centimeter (four-tenths of an inch) before its next crash. That’s a crazily small step considering that our photon has to travel 700,000 kilometers to get to the edge of the sun.

Drawing showing the sun and measuring it's radius of 700,000 km with an arrow
Drawing by Robert Krulwich

That’s almost twice the distance from the Earth to the moon. So how many steps is it to the surface?

The Nightmare Begins

You don’t want to know. Because here’s the nightmare: When a photon exits an atom, it can go in any direction. It can go up toward the sun’s surface. Or back to the sun’s center. Or sideways. Or any way. Its moves are totally random.

Mathematicians have a name for this: a drunkard’s walk. It describes a guy so drunk that every step he takes is totally arbitrary, and mathematicians have figured out how long it would take this guy, who’s totally blotto, to get from point A (lamppost number one) to point B (lamppost number two).

Drawing of a drunk man leaning on a light post with a straight arrow drawn to a nearby light post
Drawing by Robert Krulwich

The answer, writes Richard Gaughan for the blog Synonym, is “that if his starting point and ending point are separated by 10 steps, it will take him, on average, 100 [undirected] steps to get there—that’s 10 squared.” Ten times the steps required.

Drwaing of a drunk man leaning on a light pole with a crazy zig zagged path of arrows leading him to a nearby light post
Drawing by Robert Krulwich
Drawing by Robert Krulwich

The same goes for our photon. By the time it has zigged then un-zigged, zagged then un-zagged its way to the sun’s edge, it will have had billions, maybe trillions of collisions in every direction.

So to our big question: how long will that take?

49 Trillion Trillion Collisions

Well we can figure this out. If we assume the sun is dense with electrons, and each little “step” is a tenth of a millimeter, to go straight from the center of the sun to the edge will take 7 trillion steps.

Drawing showing a zig zagged path out from the center of the sun, creating a tangled mess until the path reaches the edge of the sun and moves in a straight line
Drawing by Robert Krulwich
Drawing by Robert Krulwich

But because our photon is a drunk, its true path will take the square of 7 trillion steps, which works out to 49 trillion trillion collisions before it reaches the surface. Even moving at the speed of light, that will take, more than half a million years.

Half A Million Years!

That’s a long time to wait to become sunshine.

Or …

On the other hand, if we assume a slightly emptier sun, with the traveling distance between electrons a bigger whole centimeter, that works out to fewer steps (only 490 billion trillion) to the sun’s surface—and that in turn works out to journey that lasts roughly 5,000 years.

5,000 years?

But even 5,000 years is a long time. A photon created 5,000 years ago landing on Earth today started its virtual journey around the time the great pyramids were being constructed in Egypt.

If you choose to think about sunshine this way (and I realize photons, being massless, aren’t individuals, and strictly speaking, they can’t be characters in a drama), but if you let the poet in you dance a little, you can go to the beach this weekend, feel the sunshine on your face, and think about how long it took for that to happen.

And whether you’re getting a hit of warmth that’s half a million or 5,000 years old, remember this: Roughly eight minutes ago those photons were still part of the sun, still banging their way through dense throngs of electrons. But once they got flung into space, they raced across the cosmos at the speed of light, wind in their photon hair, and eight or so minutes later, they banged into Earth, and after bouncing through our atmosphere, they settled on you.

Yes, it took a ridiculously long time for photons to get to the edge of the sun, but that last leap to you?

It was short. Crazily, joyously short.

Thanks to Aatish Bhatia for trying to help me with the physics. If I made mistakes, they’re mine, not his, but like a good friend, he tried valiantly to keep me out of trouble. For those of you want a denser look at the math, I recommend “Ask the Space Scientist”, from NASA, where Dr. Sten Odenwald concludes: “…it takes a LONG time for light to leave the sun’s interior.” But here’s a couple of paragraphs…

“The interior of the sun is a seething plasma with a central density of over 100 grams/cc. The atoms, mostly hydrogen, are fully stripped of electrons so that the particle density is 10^26 protons per cubic centimeter. That means that the typical distance between protons or electrons is about (10^26)^1/3 = 2 x 10^-9 centimeters. The actual ‘mean free path’ for radiation is closer to 1 centimeter after electromagnetic effects are included. Light travels this distance in about 3 x 10^-11 seconds. Very approximately, this means that to travel the radius of the Sun, a photon will have to take (696,000 kilometers/1 centimeter)^2 = 5 x 10^21 steps. This will take, 5×10^21 x 3 x10^-11 = 1.5 x 10^11 seconds or since there are 3.1 x 10^7 seconds in a year, you get about 4,000 years.

Some textbooks refer to ‘hundreds of thousands of years’ or even ‘several million years’ depending on what is assumed for the mean free patch. Also, the interior of the sun is not at constant density so that the steps taken in the outer half of the sun are much larger than in the deep interior where the densities are highest. Note that if you estimate a value for the mean free path that is a factor of three smaller than 1 centimeter, the time increases a factor of 10!”

12 thoughts on “Sunshine’s Crazy Sloppy Path to You

  1. Really shy photons save up their energy, find a friendly electron, and turn into neutrinos through Neutrino Photoproduction.

  2. Very good article. Puts solar radiation into an easily understood concept. It opens the door to understanding how many photons are released, have been released, and can be released before there finally is a last, beautiful day on earth. Eventually, the gravity of the sun will diminish, our star will expand because of it and it will become a cooler red giant. It may expand to the orbit of the Earth, give or take, and cooler or not, it will still be hot enough to turn our pretty planet into a cinder. Depressing thought. But not the only one this topic conjures up! You mentioned that hydrogen fuses into helium, but you said nothing about making other elements. Since there are 92 known natural elements, and all arose — we believe — by collisions of atoms, then this implies that (a) other elements form in the sun at the same time the hydrogen is fusing into helium or (b) the other elements form AFTER hydrogen fuses into helium. I like the latter implication because the sun might be thought of as a giant still. We know that distillation separates substances based on boiling point and that the temperature remains constant until all of the lower boiling substance has been evaporated. If — and that is a big “IF” — that idea is correct, it means that the other elements have formed sequentially. It means that our sun and the planets of our solar system are not primordial, but rather came from scraps and pieces of prior stars and planets. (“Yes, yes” some of you may be saying impatiently, “we know that!”) OK, so if the age of the universe is pegged at about 14 billion years, and it is “just” middle aged, then perhaps the universe will live on for maybe 30 billion years. But that would just be to fuse all the hydrogen into helium! What if after that, the helium had to fuse to make beryllium? And then into oxygen, and so forth and so on? What if each “cycle” took 30 billion years? (Yes, I’m taking liberties since (a) we still have what we think is primordial hydrogen and (b) I haven’t accounted for odd-numbered elements. But play along with me.) So if we’ve had, say, 6-and-a-half cycles of fusing so far, then maybe the true age of the universe is actually 183 billion years. In which case Robert’s photon is an old, old bit of sunshine at that!

  3. Unless particles are not there if we don’t look at it. It’s quantum mechanics. But also information theory. If something can be described by less. Then the simplest math description would be enough to let it behave as it. Since nature seams very efficient. We cannot rule out that nature might be less tall then we observe it. Or better just as tall as we observe it. It will show us ever smaller detail if we look for that. But it wouldn’t be if we don’t look at it. Think of it as an efficient program which spares memory and execution time.

    Might sound strange but quantum mechanics are just as valid as Newton physics..

  4. Strangely, while that may seem like a long time to us, it wasn’t for the photon. In fact, due to relativity, it wasn’t any time at all. For any object moving at the speed of light, all points are simultaneous. The moment it started off in the center of the sun and the moment it hits the earth are the very same for the photon. Time simply doesn’t exist if you move at c. Vsauce has a good take on it the video “Would Headlights Work at Light Speed?”, available here: https://youtu.be/ACUuFg9Y9dY

    There’s also the matter of identity. Is a photon absorbed then re-emitted the same photon? This is a more philosophical question than anything else.

  5. In reply to John, on July 8: Agree that Einstein’s famous equation E=mc^2 does not have a term for time, so your point that the photon doesn’t age, per se, is rational. But at the same time, aren’t photons massless? Rearrange the equation to solve for “m” and there will be a mass! Consequently, is your hypothesis based on a valid conjecture? Also, let me ask this: Since photons are regarded as bearing the property of “wave/particle” duality, that means there is an electric field propagating a magnetic field which propagates an electric field, and so forth, all at right angles to each other with resultant flow of the particle/wave at right angles to the first two. Since the net vector motion is 186,000 mps, does that mean each field is propagated at twice that speed? (I’ve always wondered about that and no one I’ve ever asked could answer this question!)

    1. In reply to Jay: The experience of time dilation has little to do with E=mc^2 and everything to do with the fact that c is constant, forcing both time and space to be relative. If you’ve ever heard of the Twin Paradox, then you have heard of time dilation.
      The video I linked before has a good explanation with illustrations and whatnot, but I’ll do my best to explain how it emerges here. The classic illustration is two spaceships, A and B: A stationary, and B moving at 99% the speed of light. Suppose they both carry clocks, and have headlights. As B passes A, they both turn on their headlights. Because the speed of light is constant, after one second the light from both ships headlights has traveled precisely one light-second. This seems to cause a problem though: the non-stationary ship has moved in the one second as well, and is closer to the light than the stationary ship. This would seem to imply that the light from ship B traveled less distance. This is where space and time dilation come in. Ship B would not agree with a stationary observer that one second has passed. If ship A looked at ship B’s clock, it would appear to be running slow, and vice-versa. Because ship B hasn’t experienced one second yet, it would expect the light from its headlights to not have reached a light-second yet — after all, light can’t go faster than c! They would also experience a compression of space, and would not agree on the distance that the light had traveled either. These two effects make sure that light always travels at c.
      To something traveling at c, these effects reach their limit. To an observer traveling at c… nothing could be observed (there wouldn’t be any photons to transmit information). But if it could “see” things, all clocks would appear stationary. And to someone not moving at c watching the observer, the observer would appear stationary. At c, time dilates to 0. The experience of time requires moving slower than c. Space, too, would dilate to 0, making the experience of space impossible as well. So while, to us, the photons may take millions, billions, or trillions of years, and travel just as many miles, a photon has “experienced” no time or distance.
      And photons are massless, which is why they can travel at c. If you try using E=mc^2 to solve for photons, you run into the problem that if they have energy they must have mass. This is because that is an abbreviated version of the equation. Einstein’s full equation is E^2 = P^2 * c^2 + m^2 * c^4, where P is momentum and everything else is the usual. Photons have momentum (as they have velocity), but not mass, so the equation for photons would be E^2 = P^2 * c^2, simplified to E=Pc. A stationary object (which everything that isn’t light might as well be) has no momentum, so E^2 = m^2 * c^4, or E=mc^2.

      As for your second question, you seem a bit mislead by exactly what wave-particle duality entails. The photon itself is not emitting any electric field — it is the electric field being emitted. The photon can be thought of as a single, quantized wave in a single direction, and/or as a particle in a single direction. As far as we know, the photon itself emits no field, only serving to propagate the fields of others.
      However, there are interpretations of quantum mechanics where photons emit a “pilot wave”, which does indeed “travel faster than c” (see the de Broglie-Bohm interpretation, my personal favorite). Strangely, this wouldn’t be the first quantum mechanical effect that appears to violate relativity: Quantum Entanglement, which Einstein famously called “spooky action at a distance”, would appear to as well. Technically, both are in line with relativity though, as it does not allow for the transmission of useful information at speeds above c; see the No-Communication Theorem. These pilot-wave theories gain back determinism at the cost of explicit non-locality, in my opinion a small price to pay, especially when Couder has created such a beautiful analogue.

      I hope I was able to answer your questions, or at the very least, give you some concepts to google to further your understanding.

  6. Since the photon is travelling at the speed of light, though, from the photon’s point of view doesn’t it reach us in no time at all?

    Moreover, since our original photon is a gamma ray, and what comes out is much less energetic and massy, namely visible light, I wonder if for every one original gamma ray photon one actually gets a thousand or so photons radiated from the surface. The extra photons would have been generated by processes similar to those which cause a hot body to emit thermal radiation.

  7. Interesting article. This is the first time I’ve seen a discussion about the variance in propagation estimates out of the sun. Ditto for the Drunkard’s walk calculation. One small quibble: photons can be said, as a theoretical matter, to have zero rest mass. However, they are never at rest and consequently have mass, especially at the center of the sun.

  8. @ Jay
    You’re right that hydrogen (the nuclei, or protons, really) fuses into helium nuclei and helium nuclei fuse into more massive nuclei such as beryllium and carbon, etc.Your inference that the Sun and other objects in the solar system not being primordial is correct, too. As Carl Sagan used to say, “We are star stuff.” We are the product of the fusion that occurred in previous generations of stars. The differences between what astronomers have discovered and your speculation is the actual mechanism of fusion into the other elements on the periodic table. Stars similar to our sun in mass fuse hydrogen into helium in their cores and then fuse the helium into carbon and oxygen only after the cores have depleted their hydrogen and collapse under gravity. That collapse increases the pressure and temperature to the point that helium nuclei overcome their electrical repulsion to each other and fuse into carbon and some oxygen. That occurs in giant stars, which our sun will become in about 5 billion years when it “leaves” the main sequence of hydrogen fusing stars. Our sun’s core will never reach the extreme temperatures necessary to fuse the core’s carbon and oxygen into more massive elements, however. That can only occur in supergiant stars much more massive than our sun where temperatures are in the hundreds of million kelvin. Those stars as they evolve can fuse less massive nuclei into more massive nuclei as they age, but only up to the creation of iron nuclei (where temperatures climb into the billion kelvin range in the core as the successive fusion steps occur.) Iron nuclei are extremely stable, having the highest binding energy of any nuclei, and fusing them into more massive nuclei is an endothermic process that robs the core of energy. Splitting of iron nuclei (such as when a gamma ray photo causes fission) also is endothermic. Iron in the core of a star is a dead end resulting in core implosion as the star cannot sustain itself against gravitational collapse. That implosion of the core triggers a violent rebound as the core collapses to the density of nuclear matter and a supernova results. It is the intense energy released during the supernova that allows fusion processes to create the massive nuclei up to uranium. The rarity of these events and the likelihood that the steps occur is what makes these heavy elements rare. I’ve left out a number of details cover in any introductory astronomy text, but this is the gist of it and is backed by decades of computer modeling and comparing results to the known properties of different stars.

    These reactions that produce the elements on the periodic table occur sequentially, but they occur rapidly as massive stars evolve and eventually destroy themselves as supernovae (supergiant stars such as those in the Belt of Orion may last only a few tens of thousands of years as opposed to an 11 billion year lifespan for a star like our sun.) The elements up to oxygen that we find in abundance on Earth and in the rest of the solar system are the product of evolution of sun-like stars and are left-overs from the earlier generation of stars. The universe can be 13.8 billion years old and still filled with 92 naturally occurring elements because the stars that forge most of these elements have very short life spans.

    That light see see is very old and it’s amazing that we are made of material created by earlier stars and able to observe and investigate stars that exist today.

  9. Thank you, Brent, for your clear explanation. I do understand and — even though I’m just a dumb chemical engineer — had the general gist of the process. It is, after all, chemistry. But my REAL question had to do with the age of the universe. We say it is about 14 billion years old — and when giant stars collapse, more dense elements do form. I do get that. But my (playful) question really is couldn’t the more dense elements also form by prior cycles of “big bangs” with expansion and collapse? If that could occur — and if we could detect it — then that could suggest that the known universe is more like a yo-yo than a big ball. And, in such a case, it would mean that whatever is playing with the yo-yo has produced many “cycles” since he/she/it started to play! And we could wonder how many more cycles are left before he/she/it is called in to supper by mommy! So, in other words, maybe the “true” age of the universe is actually 183 billion years at this point and has trillions more to go. Please also understand that the silliness I’m writing brings new meaning to the words of the song, “I’ve got the world by a string, sittin’ on a rainbow. . . “

  10. I am confused.

    If the photons are absorbed and then expelled in random directions so that they actually travel 5000 to a half million light years to reach the surface of the sun how many times are they redirected as they travel the 8 light years from the surface of the sun to our earth. I understand that the suns mass is compressed but would the protons not be absorbed and rejected many times encountering the few molecules in the vacuum of space before they reach earth. A direct route would be about 8 minutes of travel but what would the mean distance be? A direct transit to the earth would be a shorter time than say hitting the moon and being redirected towards earth.

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