Tag Archives: light

Of God and Hubble

4 Replies
Edwin Hubble and Andromeda Photograph

Edwin Hubble and Andromeda Photograph

Perhaps my favorite proof of God is that, as best we can tell using the best science we have, everything we see today, popped into existence some 14 billion years ago. The event is called “the big bang,” and before that, it appears, there was nothing. After that, there was everything, and as best we can tell, not an atom has popped into existence since. I see this as the miracle of creation: Ex nihilo, Genesis, Something from nothing.

The fellow who saw this miracle first was an American, Edwin P. Hubble, born 1889. Hubble got a law degree and then a PhD (physics) studying photographs of faint nebula. That is, he studied the small, glowing, fuzzy areas of the night sky, producing a PhD thesis titled: “Photographic Investigations of Faint Nebulae.” Hubble served in the army (WWI) and continued his photographic work at the Mount Wilson Observatory, home to the world’s largest telescope at the time. He concluded that many of these fuzzy nebula were complete galaxies outside of our own. Most of the stars we see unaided are located relatively near us, in our own, local area, or our own, “Milky Way” galaxy, that is within a swirling star blob that appears to be some 250,000 light years across. Through study of photographs of the Andromeda “nebula”, Hubble concluded it was another swirling galaxy quite like ours, but some 900,000 light years away. (A light year is 5,900,000,000 miles, the distance light would travel in a year). Finding another galaxy was a wonderful find; better yet, there were more swirling galaxies besides Andromeda, about 100 billion of them, we now think. Each galaxy contains about 100 billion stars; there is plenty of room for intelligent life. 

Emission from Galaxy NGC 5181. The bright, hydrogen ß line should be at but it's at

Emission spectrum from Galaxy NGC 5181. The bright, hydrogen ß line should be at 4861.3 Å, but it’s at about 4900 Å. This difference tells you the speed of the galaxy.

But the discovery of galaxies beyond our own is not what Hubble is most famous for. Hubble was able to measure the distance to some of these galaxies, mostly by their apparent brightness, and was able to measure the speed of the galaxies relative to us by use of the Doppler shift, the same phenomenon that causes a train whistle to sound differently when the train is coming towards you or going away from you. In this case, he used the frequency spectrum of light for example, at right, for NGC 5181. The color of the spectral lines of light from the galaxy is shifted to the red, long wavelengths. Hubble picked some recognizable spectral line, like the hydrogen emission line, and determined the galactic velocity by the formula,

V= c (λ – λ*)/λ*.

In this equation, V is the velocity of the galaxy relative to us, c is the speed of light, 300,000,000 m/s, λ is the observed wavelength of the particular spectral line, and λ*is the wavelength observed for non-moving sources. Hubble found that all the distant galaxies were moving away from us, and some were moving quite fast. What’s more, the speed of a galaxy away from us was roughly proportional to the distance. How odd. There were only two explanations for this: (1) All other galaxies were propelled away from us by some, earth-based anti-gravity that became more powerful with distance (2) The whole universe was expanding at a constant rate, and thus every galaxy sees itself moving away from every other galaxy at a speed proportional to the distance between them.

This second explanation seems a lot more likely than the first, but it suggests something very interesting. If the speed is proportional to the distance, and you carry the motion backwards in time, it seems there must have been a time, some 14 billion years ago, when all matter was in one small bit of space. It seems there was one origin spot for everything, and one origin time when everything popped into existence. This is evidence for creation, even for God. The term “Big Bang” comes from a rival astronomer, Fred Hoyle, who found the whole creation idea silly. With each new observation of a galaxy moving away from us, the idea became that much less silly. Besides, it’s long been known that the universe can’t be uniform and endless.

Whatever we call the creation event, we can’t say it was an accident: a lot of stuff popped out at one time, and nothing at all similar has happened since. Nor can we call it a random fluctuation since there are just too many stars and too many galaxies in close proximity to us for it to be the result of random atoms moving. If it were all random, we’d expect to see only one star and our one planet. That so much stuff popped out in so little time suggests a God of creation. We’d have to go to other areas of science to suggest it’s a personal God, one nearby who might listen to prayer, but this is a start. 

If you want to go through the Hubble calculations yourself, you can find pictures and spectra of galaxies here for the 24 or so original galaxies studied by Hubble: http://astro.wku.edu/astr106/Hubble_intro.html. Based on your analysis, you’ll likely calculate a slightly different time for creation from the standard 14 billion, but you’ll find you calculate something close to what Hubble did. To do better, you’ll need to look deeper into space, and that would take a better telescope, e.g.  the “Hubble space telescope”

Robert E. Buxbaum, October 28, 2018.

Why are glaciers blue

Leave a reply

i recently returned from a cruse trip to Alaska and, as is typical for such, a highlight of the trip was a visit to Alaska’s glaciers, in our case Hubbard Glacier, Glacier bay, and Mendenhall Glacier. All were blue — bright blue, as were the small icebergs that broke off. Glacier blocks only 2 feet across were bright blue like the glaciers themselves.

Hubbard Glacier, Alaska. Note how blue the ice is

Hubbard Glacier, Alaska. My photo. Note how blue the ice is

What made this interesting/ surprising is that I’ve seen ice sculptures that are 5 foot thick or more, and they are not significantly blue. They have a very slight tinge, but are generally more colorless than glass to my ability to tell. I asked the park rangers why the glaciers were blue, but was given no satisfactory answer. The claim was that glacier ice contained small air bubbles that scattered light the same way that air did. Another park ranger claimed that water is blue by nature, so of course the glaciers were too. The “proof” to this was that the sea was blue. Neither of these seem quite true to me, though there seamed some grains of truth. Sea water, I notice, is sort of blue, but isn’t this shade of blue, certainly not in areas that I’ve lived. Instead, sea water is a rather grayish similar to mud and sea-weeds that I’d expect to find on the sea floor. What’s more, if you look through the relatively clear water of a swimming-pool water to the white-tile bottom, you see only a slight shade of blue-green, even at the 9 foot depth where the light you see has passed through 18 feet of water. This is far more water than an iceberg thickness, and the color is nowhere near as pure blue and the intensity nowhere near as strong.

Plymouth, MI Ice sculpture -- the ice is fairly clear, like swimming pool water

Plymouth, MI Ice sculpture — the ice is fairly clear, like swimming pool water

As for the bubble explanation, it doesn’t seem quite right, either. The bubble size would be non-uniform, with many quite large resulting in a mix of scattered colors — an off white– something seen with the sky of mars. Our earth sky is a purer blue, but this is not because of scattering off of ice-crystals, dust or any other small particles, but rather scattering off the air molecules themselves. The clear blue of glaciers, and of overturned icebergs, suggests (to me) a single-size scattering entity, larger than air molecules, but much smaller than the wavelength of visible light. My preferred entity would be a new compound, a clathrate structure compound, that would be formed from air and ice at high pressures.

An overturned ice-burg is remarkably blue: far bluer than an Ice sculpture. I claim clathrates are the reason.

An overturned ice-burg is remarkably blue: far bluer than an Ice sculpture. I claim clathrates are the reason.

Sea-water forms clathrate compounds with natural gas at high pressures found at great depth. My thought is that similar compounds form between ice and one or more components of air (nitrogen, oxygen, or perhaps argon). Though no compounds of this sort have been quite identified, all these gases are reasonably soluble in water so that suggestion isn’t entirely implausible. The clathrates would be spheres, bigger than air molecules and thus should have more scattering power than the original molecules. An uneven distribution would explain the observation that the blue of glaciers is not uniform, but instead has deeper and lighter blue edges and stripes. Perhaps some parts of the glacier were formed at higher pressures one could expect that these would form more clathrate compounds, and thus more blue. One sees the most intense blue in overturned icebergs — the parts that were under the most pressure.

Robert Buxbaum, October 12, 2015. By the way, some of Alaska’s glaciers are growing and others shrinking. The rangers claimed this was the bad effect of global warming: that the shrinking glaciers should be growing and the growing ones shrinking. They also worried that despite Alaska temperatures reaching 40° below reasonably regularly, it was too warm (for whom?). The lowest recorded temperature in Fairbanks was -66°F in 1961.

A simple, classical view of and into black holes

1 Reply

Black holes are regions of the universe where gravity is so strong that light can not emerge. And, since the motion of light is related to the fundamental structure of space and time, they must also be regions where space curves on itself, and where time appears to stop — at least as seen by us, from outside the black hole. But what does space-time look like inside the black hole.

NASA's semi-useless depiction of a black hole -- one they created for educators. I'm not sure what you're supposed to understand from this.

NASA’s semi-useless depiction of a black hole — one they created for educators. Though it’s sort of true, I’m not sure what you’re supposed to understand from this. I hope to present a better version.

From our outside perspective, an object tossed into a black hole will appear to move slower as it approaches the hole, and at the hole horizon it will appear to have stopped. From the inside of the hole, the object appears to just fall right in. Some claim that tidal force will rip it apart, but I think that’s a mistake. Here’s a simple, classical way to calculate the size of a black hole, and to understand why things look like they do and do what they do.

Lets begin with light, and accept, for now, that light travels in particle form. We call these particles photons; they have both an energy and a mass, and mostly move in straight lines. The energy of a photon is related to its frequency by way of Plank’s constant. E = hν, where E is the photon energy, h is Plank’s constant and ν is frequency. The photon mass is related to its energy by way of the formula m=E/c2, a formula that is surprisingly easy to derive, and often shown as E= mc2. The version that’s relevant to photons and black holes is:

m =  hν/c2.

Now consider that gravity affects ν by affecting the energy of the photon. As a photon goes up, the energy and frequency goes down as energy is lost. The gravitational force between a star, mass M, and this photon, mass m, is described as follows:

F = -GMm/r2

where F is force, G is the gravitational constant, and r is the distance of the photon from the center of the star and M is the mass of the star. The amount of photon energy lost to gravity as it rises from the surface is the integral of the force.

∆E = – ∫Fdr = ∫GMm/r2 dr = -GMm/r

Lets consider a photon of original energy E° and original mass m°= E°/c2. If ∆E = m°c2, all the energy of the original photon is lost and the photon disappears. Now, lets figure out the height, r° such that all of the original energy, E° is lost in rising away from the center of a star, mass M. That is let calculate the r for which ∆E = -E°. We’ll assume, for now, that the photon mass remains constant at m°.

E° = GMm°/r° = GME°/c2r°.

We now eliminate E° from the equation and solve for this special radius, r°:

r° =  GM/c2.

This would be the radius of a black hole if space didn’t curve and if the mass of the photon didn’t decrease as it rose. While neither of these assumptions is true, the errors nearly cancel, and the true value for r° is double the size calculated this way.

r° = 2GM/c2

r° = 2.95 km (M/Msun).

schwarzschild

Karl Schwarzschild 1873-1916.

The first person to do this calculation was Karl Schwarzschild and r° is called the Schwarzschild radius. This is the minimal radius for a star of mass M to produce closed space-time; a black hole. Msun is the mass of our sun, sol, 2 × 1030 kg.  To make a black hole one would have to compress the mass of our sun into a ball of 2.95 km radius, about the size of a small asteroid. Space-time would close around it, and light starting from the surface would not be able to escape.

As it happens, our sun is far bigger than an asteroid and is not a black hole: we can see light from the sun’s surface with minimal space-time deformation (there is some seen in the orbit of Mercury). Still, if the mass were a lot bigger, the radius would be a lot bigger and the density would be less. Consider a black hole the same mass as our galaxy, about 1 x1012 solar masses, or 2 x 1042  kg. This number is ten times what you might expect since our galaxy is 90% dark matter. The Schwarzschild radius with the mass of our galaxy would be 3 x 1012 km, or 0.3 light years. That’s far bigger than our solar system, and about 1/20 the distance to the nearest star, Alpha Centauri. This is a very big black hole, though it is far smaller than our galaxy, 5 x 1017 km, or 50,000 light years. The density, though is not all that high.

Now let’s consider a black hole comprising 15 billion galaxies, the mass of the known universe. The folks at Cornell estimate the sum of dark and luminous matter in the universe to be 3 x 1052 kg, about 15 billion times the mass of our galaxy. This does not include the mass hidden in the form of dark energy, but no one’s sure what dark energy is, or even if it really exists. A black hole encompassing this, known mass would have a Schwarzschild radius about 4.5 billion light years, or about 1/3 the actual size of the universe when size is calculated based on its Hubble-constant age, 14 billion years. The universe may be 2-3 times bigger than this on the inside because space is curved and, rather like Dr. Who’s Tardis it’s bigger on the inside, but in astronomical terms a factor of 3 or 10 is nothing: the actual size of the known universe is remarkably similar to its Schwarzschild radius, and this is without considering the mass its dark energy must have if it exists.

Standard picture of the big bang theory. Dark energy causes the latter-stage expansion.

Standard picture of the big bang theory. Dark energy causes the latter-stage expansion.

The evidence for dark energy is that the universe is expanding faster and faster instead of slowing. See figure. There is no visible reason for the acceleration, but it’s there. The source of the energy might be some zero-point effect, but wherever it comes from, the significant amount of energy must have significant mass, E = mc2. If the mass of this energy is 3 to 10 times the physical mass, as seems possible, we are living inside a large black hole, something many physicists, including Einstein considered extremely likely and aesthetically pleasing. Einstein originally didn’t consider the possibility that the hole could be expanding, but a reviewer of one of his articles convinced him it was possible.

Based on the above, we now know how to calculate the size of a black hole of any mass, and we now know what a black hole the size of the universe would look like from the inside. It looks just like home. Wait for further posts on curved space-time. For some reason, no religion seems to embrace science’s 14 billion year old, black-hole universe (expanding or not). As for the tidal forces around black holes, they are horrific only for the small black holes that most people write about. If the black hole is big, the tidal forces are small.

 Dr. µß Buxbaum Nov 17, 2014. The idea for this post came from an essay by Isaac Asimov that I read in a collection called “Buy Jupiter.” You can drink to the Schwarzchild radius with my new R° cocktail.

Do antimatter apples fall up?

13 Replies

by Dr. Robert E. Buxbaum,

The normal view of antimatter is that it’s just regular matter moving backwards in time. This view helps explain why antimatter has the same mass as regular matter, but has the opposite charge, spin, etc. An antiproton has the same mass as a proton because it is a proton. In our (forward) time-frame the anti-proton appears to be attracted by a positive plate and repelled by a negative one because, when you are going backward in time, attraction looks like repulsion.

In this view, the reason that antimatter particles annihilate when they come into contact with matter –sometimes– is that the annihilation is nothing more than the matter particle (or antimatter) switching direction in time. In our (forward) time-frame it looks like two particles met and both disappeared leaving nothing but photons (light). But in the time reversal view, shown in the figure below, there is only one normal matter particle. In the figure, this particle (solid line) comes from the left, and meets a photon, a wiggly line who’s arrow shows it traveling backwards in time. The normal proton then reverses in time, giving off a photon, another wiggly line. I’d alluded to this in my recent joke about an antimatter person at a bar, but there is also a famous poem.

proton-antiproton

This time reverse approach is best tested using entropy, the classical “arrow of time.” The best way to tell you can tell you are going forward in time is to drop an ice-cube into a hot cup of coffee and produce a warm cup of diluted coffee. This really only shows that you and the cup are moving in the same direction — both forward or both backward, something we’ll call forward. If you were moving in the opposite direction in time, e.g. you had a cup of anti-coffee that was moving backward in time relative to you, you could pull an anti -ice cube out of it, and produce a steaming cup of stronger anti-coffee.

We can not do the entropy test of time direction yet because it requires too much anti matter, but we can use another approach to test the time-reverse idea: gravity. You can make a very small drop of antimatter using only a few hundred atoms. If the antimatter drop is really going backwards in time, it should not fall on the floor and splatter, but should fly upward off the floor and coalesce. The Laboratory at CERN has just recently started producing enough atoms of anti-hydrogen to allow this test. So far the atoms are too hot but sometime in 2014 they expect to cool the atoms, some 300 atoms of anti hydrogen, into a drop or two. They will then see if the drop falls down or up in gravity. The temperature necessary for this study is about 1/100,000 of a degree K.

The anti-time view of antimatter is still somewhat controversial. For it to work, light must reside outside of time, or must move forward and backward in time with some ease. This makes some sense since light travels “at the speed of light,” and is thus outside of time. In the figure, the backwards moving photon would look like a forward on moving in the other direction (left). In a future post I hope to give instructions for building a simple, quantum time machine that uses the fact that light can move backwards in time to produce an event eraser — a device that erases light events in the present. It’s a somewhat useful device, if only for a science fair demonstration. Making one to work on matter would be much harder, and may be impossible if the CERN experiments don’t work out.

It becomes a little confusing how to deal with entropy in a completely anti-time world, and it’s somewhat hard to see why, in this view of time, there should be so little antimatter in the universe and so much matter: you’d expect equal amounts of both. As I have strong feelings for entropy, I’d posted a thought explanation for this some months ago imagining anti matter as normal forward-time matter, and posits the existence of an undiscovered particle that interacts with its magnetism to make matter more stable than antimatter. To see how it works, recall the brainteaser about a tribe that always speaks lies and another that always speaks truth. (I’m not the first to think of this explanation).

If the anti hydrogen drop at CERN is seen to fall upwards, but entropy still works in the positive direction as in my post (i.e. drops still splatter, and anti coffee cools like normal coffee), it will support a simple explanation for dark energy — the force that prevents the universe from collapsing. Dark energy could be seen to result from the antigravity of antimatter. There would have to be large collections of antimatter somewhere, perhaps anti-galaxies isolated from normal galaxies, that would push away the positive matter galaxies while moving forward in time and entropy. If the antigalaxies were close to normal galaxies they would annihilate at the edges, and we’d see lots of photons, like in the poem. Whatever they find at CERN, the future will be interesting. And if time travel turns out to be the norm, the past will be more interesting than it was.

Most Heat Loss Is Black-Body Radiation

7 Replies

In a previous post I used statistical mechanics to show how you’d calculate the thermal conductivity of any gas and showed why the insulating power of the best normal insulating materials was usually identical to ambient air. That analysis only considered the motion of molecules and not of photons (black-body radiation) and thus under-predicted heat transfer in most circumstances. Though black body radiation is often ignored in chemical engineering calculations, it is often the major heat transfer mechanism, even at modest temperatures.

One can show from quantum mechanics that the radiative heat transfer between two surfaces of temperature T and To is proportional to the difference of the fourth power of the two temperatures in absolute (Kelvin) scale.

Heat transfer rate = P = A ε σ( T^4 – To^4).

Here, A is the area of the surfaces, σ is the Stefan–Boltzmann constant, ε is the surface emissivity, a number that is 1 for most non-metals and .3 for stainless steel.  For A measured in m2, σ = 5.67×10−8 W m−2 K−4.

Infrared picture of a fellow wearing a black plastic bag on his arm. The bag is nearly transparent to heat radiation, while his eyeglasses are opaque. His hair provides some insulation.

Unlike with conduction, heat transfer does not depend on the distances between the surfaces but only on the temperature and the infra-red (IR) reflectivity. This is different from normal reflectivity as seen in the below infra-red photo of a lightly dressed person standing in a normal room. The fellow has a black plastic bag on his arm, but you can hardly see it here, as it hardly affects heat loss. His clothes, don’t do much either, but his hair and eyeglasses are reasonably effective blocks to radiative heat loss.

As an illustrative example, lets calculate the radiative and conductive heat transfer heat transfer rates of the person in the picture, assuming he has  2 m2 of surface area, an emissivity of 1, and a body and clothes temperature of about 86°F; that is, his skin/clothes temperature is 30°C or 303K in absolute. If this person stands in a room at 71.6°F, 295K, the radiative heat loss is calculated from the equation above: 2 *1* 5.67×10−8 * (8.43×109 -7.57×109) = 97.5 W. This is 23.36 cal/second or 84.1 Cal/hr or 2020 Cal/day; this is nearly the expected basal calorie use of a person this size.

The conductive heat loss is typically much smaller. As discussed previously in my analysis of curtains, the rate is inversely proportional to the heat transfer distance and proportional to the temperature difference. For the fellow in the picture, assuming he’s standing in relatively stagnant air, the heat boundary layer thickness will be about 2 cm (0.02m). Multiplying the thermal conductivity of air, 0.024 W/mK, by the surface area and the temperature difference and dividing by the boundary layer thickness, we find a Wattage of heat loss of 2*.024*(30-22)/.02 = 19.2 W. This is 16.56 Cal/hr, or 397 Cal/day: about 20% of the radiative heat loss, suggesting that some 5/6 of a sedentary person’s heat transfer may be from black body radiation.

We can expect that black-body radiation dominates conduction when looking at heat-shedding losses from hot chemical equipment because this equipment is typically much warmer than a human body. We’ve found, with our hydrogen purifiers for example, that it is critically important to choose a thermal insulation that is opaque or reflective to black body radiation. We use an infra-red opaque ceramic wrapped with aluminum foil to provide more insulation to a hot pipe than many inches of ceramic could. Aluminum has a far lower emissivity than the nonreflective surfaces of ceramic, and gold has an even lower emissivity at most temperatures.

Many popular insulation materials are not black-body opaque, and most hot surfaces are not reflectively coated. Because of this, you can find that the heat loss rate goes up as you add too much insulation. After a point, the extra insulation increases the surface area for radiation while barely reducing the surface temperature; it starts to act like a heat fin. While the space-shuttle tiles are fairly mediocre in terms of conduction, they are excellent in terms of black-body radiation.

There are applications where you want to increase heat transfer without having to resort to direct contact with corrosive chemicals or heat-transfer fluids. Often black body radiation can be used. As an example, heat transfers quite well from a cartridge heater or band heater to a piece of equipment even if they do not fit particularly tightly, especially if the outer surfaces are coated with black oxide. Black body radiation works well with stainless steel and most liquids, but most gases are nearly transparent to black body radiation. For heat transfer to most gases, it’s usually necessary to make use of turbulence or better yet, chaos.

Robert Buxbaum

Why isn’t the sky green?

11 Replies

Yesterday I blogged with a simple version of why the sky was blue and not green. Now I’d like to add mathematics to the treatment. The simple version said that the sky was blue because the sun color was a spectrum centered on yellow. I said that molecules of air scattered mostly the short wavelength, high frequency light colors, indigo and blue. This made the sky blue. I said that, the rest of the sunlight was not scattered, so that the sun looked yellow. I then said that the only way for the sky to be green would be if the sun were cooler, orange say, then the sky would be green. The answer is sort-of true, but only in a hand-waving way; so here’s the better treatment.

Light scatters off of dispersed small particles in proportion to wavelength to the inverse 4th power of the wavelength. That is to say, we expect air molecules will scatter more short wavelength, cool colors (purple and indigo) than warm colors (red and orange) but a real analysis must use the actual spectrum of sunlight, the light power (mW/m2.nm) at each wavelength.

intensity of sunlight as a function of wavelength (frequency)

intensity of sunlight as a function of wavelength

The first thing you’ll notice is that the light from our sun isn’t quite yellow, but is mostly green. Clearly plants understand this, otherwise chlorophyl would be yellow. There are fairly large components of blue and red too, but my first correction to the previous treatment is that the yellow color we see as the sun is a trick of the eye called additive color. Our eyes combine the green and red of the sun’s light, and sees it as yellow. There are some nice classroom experiment you can do to show this, the simplest being to make a Maxwell top with green and red sections, spin the top, and notice that you see the color as yellow.

In order to add some math to the analysis of sky color, I show a table below where I divided the solar spectrum into the 7 representative colors with their effective power. There is some subjectivity to this, but I took red as the wavelengths from 620 to 750nm so I claim on the table was 680 nm. The average power of the red was 500 mW/m2nm, so I calculate the power as .5 W/m2nm x 130 nm = 65W/m2. Similarly, I took orange to be the 30W/m2 centered on 640nm, etc. This division is presented in the first 3 columns of the following table. The first line of the table is an approximate of the Rayleigh-scatter factor for our atmosphere, with scatter presented as the percent of the incident light. That is % scattered = 9E11/wavelength^4.skyblue scatter

To use the Rayleigh factor, I calculate the 1/wavelength of each color to the 4th power; this is shown in the 4th column. The scatter % is now calculated and I apply this percent to the light intensities to calculate the amount of each color that I’d expect in the scattered and un-scattered light (the last two columns). Based on this, I find that the predominant wavelength in the color of the sky should be blue-cyan with significant components of green, indigo, and violet. When viewed through a spectroscope, I find that these are the colors I see (I have a pocket spectroscope and used it an hour ago to check). Viewed through the same spectroscope (with eye protection), I expect the sun should look like a combination of green and red, something our eyes see as yellow (I have not done this personally). At any rate, it appears that the sky looks blue because our eyes see the green+ cyan+ indigo + purple in the scattered light as sky blue.220px-RGB_illumination

At sunrise and sunset when the sun is on the horizon the scatter percents will be higher, so that all of the sun’s colors will be scattered except red and orange. The sun looks orange then, as expected, but the sky should look blue-green, as that’s the combination of all the other colors of sunlight when orange and red are removed. I’ve not checked this last yet. I’ll have to take my spectroscope to a fine sunset and see what I see when I look at the sky.

Why isn’t the sky green and the sun orange?

4 Replies

Part of the reason the sky isn’t green has to do with the color of the sun. The sun’s color, and to a lesser extent, the sky color both are determined by the sun’s surface color, yellow. This surface color results from black body radiation: if you heat up a black object it will first glow red, then orange, yellow, green etc. Red is a relatively cool color because it’s a low frequency (long wavelength) and low frequencies are the lowest energy photons, and thus are the easiest for a black body to produce. As one increases the temperature of a black object, the total number of photons increases for all wavelengths, but the short wavelength (high frequency) colors increase faster than the of long wavelength colors. As a result, the object is seen to change color to orange, then yellow, or to any other color representative of objects at that particular temperature.

Our star is called a yellow sun because the center color of its radiation is yellow. The sun provides radiation in all colors and wavelengths, even colors invisible to the eye, infra red and ultra violet, but because of its temperature, most of the radiated energy appears as yellow. This being said, you may wonder why the sky isn’t yellow (the sky of Mars mostly is).

The reason the sky is blue, is that some small fraction of the light of the sun (about 10%) scatters off of the molecules of the air. This is called Rayleigh scatter — the scatter of large wavelegth waves off of small objects.  The math for this will be discussed in another post, but the most relevant aspect here is that the fraction that is scattered is proportional to the 4th power of the frequency. This is to say, that the high frequencies (blue, indigo, and violet) scatter a lot, about 20%, while the red hardly scatters at all. As a result the sky has a higher frequency color than the sun does. In our case, the sky looks blue, while the sun looks slightly redder from earth than it does from space — at least that’s the case for most of the day.

The sun looks orange-red at sundown because the sunlight has to go through more air. Because of this, a lot more of the yellow, green, and blue scatter away before we see it. Much more of the scatter goes off into space, with the result that the sky to looks dark, and somewhat more greenish at sundown. If the molecules were somewhat bigger, we’d still see a blue sky, maybe somewhat greener, with a lot more intensity. That’s the effect that carbon dioxide has — it causes more sunlight to scatter, making the sky brighter. If the sun were cooler (orange say), the sky would appear green. That’s because there would be less violet and blue in the sunlight, and the sky color would be shifted to the longer wavelengths. On planets where the sun is cooler than ours, the sky is likely green, but could be yellow or red.

Rayleigh scatter requires objects that are much smaller than the light wavelength. A typical molecule of air is about 1 nm in size (1E-9 of a meter), while the wavelength of yellow light is 580 nm. That’s much larger than the size of air molecules. Snow appears white because the size of the crystals are the size of the sun wavelengths, tor bigger, 500-2000 nm. Thus, the snow looks like all the colors of the sun together, and that’s white. White = the sum of all the colors: red + orange + blue + green + yellow + violet + indigo.

Robert Buxbaum  Jan. 27, 2013 (revised)