I could rather boringly sum up the topic of this post as: “what happens when I shine light at stuff”. However, in talking about this we’ll also look at some things that maybe you never thought about: how one can easily see through solid objects; or what are the true colour of nebulae; and how we can harness near-miraculous state-of-the-art wonder materials to marginally improve your TV so you can watch Westworld in peace. We’ll also answer some more mundane questions like why are things the colour they are and how you can have things like shiny black paint.
But it Ain’t All About the Surfaces, Innit?
Okay, in our last post we talked all about what happens when light hits a surface between two mediums. The key take-away was that, depending on how “optically heavy” a material is, you’ll get some amount of reflection and some amount of transmission of the incident electromagnetic wave (i.e. light). In this post, we want to think about what happens after light has moved from one medium to another. In this way, the interface is kind of like a bouncer to the nightclub of the material. If you’re the wrong type of light, there’s a chance you’ll be turned away at the door. (Oh, no, how will it ever spend $300 for a $30 bottle of “bottle service”!?) But what happens when you make it in?
Well, first of all, no material is perfectly transmittive. What is meant by this, is that a material is made of “stuff”, specifically atoms, and those atoms can absorb the energy of EM waves and put it to other uses, like heat or pushing electrons about. What atoms in a medium can or can’t do with energy they steal is the entire topic of our next couple posts, so we won’t dwell on this now. For now we will simply take it as stated fact that they can potentially absorb energy from an EM wave. When this happens, the intensity of the EM wave is diminished. Energy is conserved and if the atoms stole some then there’s less in the wave. To concretely describe this behaviour, there is the very useful concept of “absorption per unit length”, which is often simply called “absorption”.
Absorption is a quantity that is often expressed in the fairly intuitive units of “per meter” (or “per centimeter” or “per nanometer” or what have you, the point is “per unit of length”) and is generally quoted as a value between 100% and 0%, or simply 1 and 0. If a material has an absorption of 0.3 “per meter” it then means: “If light travels through one meter of this material then 30% of it will have been absorbed, and 70% will remain”. This quantity thus has a natural sibling, which I’ll call the internal transmission (per unit length), which means the “chance of surviving per unit length travelled within the material”. Last time we also talked about the chance of there being internal reflections or scattering, so talking only in terms of an internal transmission is probably not super conceptually useful. Instead we might think that for each unit length that light travels in a material, it is either absorbed, reflected in a random direction, or continues to transmit forward. The sum of all these probabilities must be 100%; one of them has to happen.
As a result of absorption, light is (exponentially) diminished the longer it travels through a material. So if a material absorbs light quite well you’ll see a certain characteristic “extinction” curve of the light’s intensity with the distance it travels into the material. In other words, if you could do a cross-section of the strength of the electric field of an incident EM wave as it enters such a medium you would get something like this:
This decay is actually quantitatively described by what is called the “Beer-Lambert law” named after August Beer (and Johann Lambert). I only mention this because I have the maturity of an infant and it makes me smile that there was someone named August Beer (can’t wait for the September pick-of-the-month!) who made an important contribution to physics.
Anyways, this idea also then naturally leads to the idea of “absorption length/depth” or “penetration depth” which basically means: “how far, on average, does light make it into this material before a majority of it can be expected to be absorbed”. It’s worth pointing out that the flip-side to this is that all materials are transparent if they’re sufficiently thin; absorption is a probability whose odds depend on total thickness. If you make things thin enough you can always achieve negligible absorption, like this:
Wonder Materials for Cheap TVs
On that note it’s possible you may have heard about new classes of materials, like graphene, that are two-dimensional materials that are only one atom thick and the marvelous properties they have.
You may also have heard that, despite these amazing properties like outrageously high electrical conductivity and insane tensile strength, that technological application and consumer products have not materialized despite decades of big promises. A big reason for this is because we simply don’t quite know how to grow or make these materials in large sheets in a way that is both cheap and ensures that the final sheet is of a very high quality. However, one of the most straightforward application of graphene and other such materials that is very forgiving to our current engineering and manufacturing issues is related to this Beer-Lambert decay of light.
To put it simply, because the amount of light absorbed is a function of material thickness, all 2D materials are transparent. Even if they have very high absorptions, they’re only 1 atom thick so there’s just not much thickness for absorption to occur over. Given this, there is one technological application where it is essential that a material be both transparent and conductive: the front-contact of LCD and “LED” televisions. (Note: in the United States especially, most TVs marketed as “LED” TVs, are in fact just LCD TVs with an LED back-light rather than true LED TVs)
I’m not really going to talk about how such TVs function, though perhaps another time, however, the key thing is that they are made of arrays of semiconductor devices that, like any such devices, must be attached to metallic contacts (just like the leads on a battery). However, because of the design, it is necessary that the light released by the devices must pass through one of these metal contacts.
[Adapted From: https://www.pbslearningmedia.org/resource/508170367-science-technology/cross-section-of-an-lcd-display-science-and-technology]
So to design such a TV you need a material that is “metallic” (or at least has an “okay” electrical conductivity) and is also transparent. And, you know, metals are reflective and not at all transparent. The current technique is to use “transparent conducting oxides” or TCOs that are transparent and… kind of okay conductors. They’re enough to make it work.
So what’s the problem? Well the best TCO that exists is called indium titanium oxide or (ITO) and you want to know something funny about indium? It’s one of the rarest elements on Earth. The indium we have doesn’t even come from mining, it comes out as a manufacturing by-product of the production of zinc sulfide. So the abundance of a material we need to make TVs is, rather bizarrely, held hostage by the amount of zinc sulfide we make globally.
So, this creates a very easy application for 2D materials like graphene. They’re amazing conductors, they’re transparent and even if the quality of the large-size sheets is low, it’s still easily better than TCOs which are thoroughly poor conductors relative to true metals. Graphene is also just literally made of charcoal. It’s extraordinarily abundant.
Just something I thought was interesting.
Surviving the Gauntlet
Okay, let’s move on from that aside about graphene and return to talking about how light moves through a medium. The thing we really need to understand is that a given medium will treat each and every wavelength of an incident EM wave differently. Absorption, transmission and reflection aren’t just single-valued quantities but rather are wavelength dependent properties of a material. (Remember in Part 2 we learned that any “pulse” or “ripple” of the electromagnetic field can be conceptually decomposed into a sum of waves of different wavelength; a point we’ll re-iterate momentarily.) Each wavelength, when it meets the surface of a material, will instantly face a random die (puns!) roll to “survive” (well, maybe not survive, but not be turned away), which represents whether it is reflected or transmitted at the surface. Again, it’s crucial to understand that the odds in this die roll depend on the material and the wavelength. If it passes the “surface reflectance” die roll (saving throw versus death magic!) it then enters the material and keeps going.
Once it’s in a material, light’s troubles are only beginning. Every meter it progresses in the material, it needs to make another survival roll, this time for absorption. I say “a meter” but it is, of course, a continuous on-going chance, that works out to be a certain chance per meter. The odds of every absorption roll are also material and wavelength dependent. The how and why is an upcoming topic, all we need to know now is that it’s some chance per unit length.
On top of this, if the material is not pristine and perfect, there will also be internal surfaces and impurities and that means another die roll, this time to see if it scatters or reflects and has its direction randomized. Of course, should it be scattered or reflected it still exists, it doesn’t lose amplitude/intensity, but it’s still messing with its day. It’s also worth adding that the longer light spends in the material, the greater the chance it will eventually fail its absorption die roll and be absorbed. So for a given material with a certain absorption, you can make it more absorptive just by adding internal impurities to scatter off of so that your average beam of light gets knocked around a bit and spends more time in the scary evil forest that is life in a medium and thus has a greater chance of never making it out. (poor light 😦 )
Should light keep getting lucky and continually make its survival rolls, then it may make it to one of the boundaries of the material and gets its freedom. Of course, due to internal reflections and scattering the boundary it finds may well be the one it came in from. If this is the case the light will be observed as being reflected from the material. This is just sub-surface scattering, which we already talked about in the last post, but it is important to understand moving forward, so I’ll belabor the point: If I shine white light on a material, much of the reflected light I see will have spent time inside the material and thus I will often see that reflected white light is missing the wavelengths that the material absorbs. Reflection often carries information about absorption.
It’s All About Spectrums (I know it’s “spectra”! Yeesh.)
Since the time of, at least, Newton it’s been known that if you shine white light through a prism, it will be decomposed into its component EM waves revealing all the colours of the rainbow. (I feel like a “double rainbow” reference is already too dated.) When people realized this they went about decomposing the light from all kinds of things: the sun, flames, light that had passed through materials like stained glass, etc. We’ll talk about this a lot more in a later post, but this was the beginning of a field of physics that what we now call “optical spectroscopy” or breaking down incoming light into its composing spectrum of EM waves.
We’ve gotten very good at optical spectroscopy and can do it very accurately with a very high spectral resolution. Though it’s worth noting that modern optical spectrometers don’t really use prisms anymore, but rather what is called a “diffraction grating” (or if they do use prisms they’re like super-duper fancy space-age-technology prisms.) It’s also important to point out that when I’m talking about spectroscopy and maybe you’re thinking in your head about the rainbow that comes out of a prism, remember that I mean decomposition of the entire EM spectrum, not just the ROYGBIV, 0.39 to 0.7 micron wavelengthed light we can see. So radio, microwave, infrared, visible, etc. We’re concerned with it all. If you’re unfamiliar with the EM spectrum we discussed it at length in Part 2.
Up to now I’ve been pretty haphazard when talking about EM waves and not making a distinction between an EM wave with only a single wavelength and “white” light, made of a combination of all wavelengths, without ever really being clear of what I meant and what “white” was. By saying an EM wave “has a specific wavelength” I’ve been implicitly assuming mono-chromatic (one wavelength, or “colour”, though remember, remember, remember that we’re not only talking about the visible spectrum). In reality, a given light source will emit a final wave made from a sum of different coloured EM waves of different intensities, like so:
The thing I’m trying to show in that diagram is that I have, in that example, three different mono-chromatic EM waves of set wavelength/frequency and different amplitudes (the maximum height of each of the three is different) and I’m combining or adding them to make one wave that, in general, has some fluctuating intensity and not-well-defined wavelength. This might look very complicated and in general the final wave shape will be complicated but we can very easily think of situations like this by introducing the concept of a ”spectral distribution” of some incoming light:
In such spectra we have two bits of knowledge: what wavelengths are present in the beam and with what relative intensity. Curves such as these are what are detected by optical spectrometers. However, it is really important to understand that, for the same material, you will get a very different curve depending on the experimental set-up. Specifically, one talks about absorption spectra, reflectance spectra, transmission spectra and emission spectra.
Just by the names you can probably get some idea of what they’re all about. In absorption, reflection and transmission spectra, you, yourself, provide an external light source and shine it on the object. What you want is your light source to be perfect white light, emitting all EM waves with equal intensity (remember, including non-visible EM waves). The reason you want perfect white light is because you’re trying to find out something about the material of the object you’re illuminating, not the specific peculiarities of your external light-source. It shouldn’t complicate the picture and should illuminate all wavelengths without any special character of its own.
Of course, no such experimental set-up that can simultaneously illuminate at all wavelengths with equal strength really exists. Generally you actually have a mono-chromatic source whose wavelength is tunable, (i.e. it only emits one wavelength) but you can change what that tuned wavelength is on-the-fly. With this, you sweep through the wavelengths one-by-one collecting data as you go. However, on top of this, generally one needs an entirely different machine to do spectral measurements for different ranges of the EM spectra. A source of visible light, for example, like a tunable-laser, won’t easily emit radio waves. You need a different set-up for that. However, we won’t concern ourselves with such pragmatic difficulties and simply assume that you can totally sweep the whole EM spectrum with a single perfect source.
The difference between the different spectra that involve you providing your own external illumination (i.e. reflection vs. absorption vs. transmission) is basically where you’re standing relative to the source. If you’re looking at your wall, which is being lit by the sun or your house lights, you’re seeing the visible portion of the reflection spectrum of the material your wall paint is made of. The external light is hitting the wall with a selection of wavelengths (i.e. white light) and only those that reflect make it to your eyes. If you’re looking at curtains, which are between you and the Sun, such that sunlight is passing through them, you’re seeing the transmission spectrum of the material your curtains are made out of. (Like in my case where they’re made out of an old dish rag!)
What about the absorption spectrum? The absorption spectrum can’t be determined directly, but rather indirectly based on the fact that the probability that a given bit of light will be either reflected or transmitted or absorbed is 100%. One of them has to happen! Thus, if you know the reflection and the transmission spectra, you can subtract them from 100% and determine the absorption. All light that didn’t make it out, either through reflection or transmission, must have been absorbed. In fact, it’s actually fairly common practice to ignore reflection when talking about things like gases, which aren’t particularly reflective, and just talk about absorption as the opposite of transmission.
The last spectra we need to talk about are emission spectra. This is the light you receive if you didn’t bring your own external source. We already talked about in Part 1, some of the reasons objects will emit light, like black-body radiation, cyclotron emissions and Bremsstrahlung, though we’ll learn about some more ways in the next couple posts when we talk about things like neon lights and sodium-vapour lamps.
Putting Things to Use
Alright so let’s put this understanding to use. The best way to see the importance of these concepts is to just look at some examples of everyday objects. However, before we do that, an important thing to understand is that reflection, absorption, emission and transmission spectra are, in principle, all different from one another but often they all have a deep connection representing a unifying set of material properties. It’s all the same material being probed in related, but only slightly different, ways. So, it’s worth pointing out that there are a number of, let’s called them, “rules of thumb” in terms of the relationship between the spectra of a material. However, I do want to stress that these are only things that tend to be true. They are not indisputably rooted in fundamental physics, but rather general trends in material properties.
The first of these rules of thumbs is:
For weakly absorbing materials that aren’t very shiny, reflection spectra and transmission spectra tend to be very similar and are approximately just the “compliment” or “opposite” of the absorption spectra. In other words, if the material weakly absorbs a given wavelength then that wavelength will be fairly equally missing from both reflection and transmission spectra.
Now, why might this be true? It seems a bit funny. Well, basically what is being assumed is that most or all reflected light is really from sub-surface scattering and not from direct scattering off the surface a la Fresnel or glass. In other words, both light that reflects off the material and light that passes through the material has spent a lot of time bouncing around in the material. This is what we mean by “weak” absorption. It seems odd to say “weak” when clearly a lot is absorbed, but all we mean is that light has plenty of opportunity to bounce around in the material and work its way back out. Because of this, most of the light that reaches our eyes, either through transmission or reflection, has been robbed of the wavelengths that the material strongly absorbs.
This is honestly the case for a lot of situations, like things like dyes, pigments and matte paints. A particularly strong example is also plants. Think of something like a leaf. Plants get their colour primarily due to two pigments called chlorophyll-a and chlorophyll-b. Let’s look at the absorption spectra of those pigments:
This is the first time I’ve really thrown down a real spectra so really be sure you take the time to understand what you’re seeing as you’ll be seeing many, many more as we continue on with our story of light. The first thing we do is look at the x and y axes. Here the x-axis stretches from 400 nm on the left, which is the colour violet, to 700 nm on the right, which is the colour red. So this graph is only showing how chlorophyll absorbs visible light and shorter wavelengths are on the left. Remember, what that means is that the shown data is incomplete and this spectrum actually does continue outside of the ranges humans can see. For example, it is clear that chlorophyll-a (green line) especially, is also going to be reasonably absorbing in the ultraviolet which would be just off the graph to the left.
Secondly, we look at the y-axis. In this case it is absorption and there’s no real numbers. As we;ve already said, typically units of absorption are either 0 to 1 (or 0% to 100%) or “per units of length”, like say (1/meters). The former is a total probability of light being absorbed while traveling through an entire object – what percent made the whole trip – and will change depending on the thickness of the object. The latter is a “per length travelled through the material” measure and thus is a true “material property”.
I can’t stress how important it is to always check the axes. For one, whether the short wavelength is on the left or the right is frequently a matter of taste and changes from graph to graph. Secondly, it’s always super important to know what range of the EM spectrum you’re looking at. It’s pretty rare to find graphs that attempt to do more than a couple ranges (for example, radio-microwave-infrared or visible-ultraviolet-soft x-ray) in a single graph. So always be clear what range of the spectrum the graph is showing and always remember that there’s more to light than the visible range. Thirdly, the units of the y-axis can sometimes be stated as things that are a bit funny, like “absorption cross-section”, which we won’t really talk about, but they all qualitatively tell you very similar things: what wavelengths of light the material absorbs more than other wavelengths.
So, what does our absorption spectra for our chlorophylls tell us? Well, looking at this what we see is that the chlorophylls mainly absorb blue and red light. Based on our rule of thumb for weak absorbers, which chlorophylls are, that means that it should both reflect and transmit white light minus red and blue. In other words, green. This is why greenery is… well… green. It is an absorber of red and blue and its reflection and transmission are based on a lot of light getting actually inside and sub-surface scattering about before getting out. Which makes most leaves terrible mirrors. Yes, I hear you pointing out that some leaves actually have a waxy sheen and are quite reflective, well you’ll understand that in just a moment. But I’m talking about more regular matte-coloured green leaves.
Plasma and the Colour of Gold
Our second rule of thumb will be a bit hard to understand without getting deeper into the physics of electromagnetism than we’d like to but it is quite important so I’m still going to state it, even if I’m not going to justify it to the extent it deserves:
Materials that, in principle, are very, very strong absorbers of a given wavelength tend to actually reflect that wavelength very, very heavily.
This is very paradoxical. Make sure you understand what I’m saying. In our previous rule we said that a weak absorber tends to have fairly similar reflection and transmission spectra which are the compliment or opposite of the absorption. Here I’m saying that very, very strong absorbers have no transmission and have reflection spectra that are similar to their absorption spectra. Not the compliment or opposite.
Now, as I said, the reasons for this are quite complex, but in a very hand-wavy nutshell, if something is so absorbing of a given wavelength such that that wavelength basically makes it nowhere in getting into the material and is just gobbled up immediately, then it will actually be turned away at the interface. In a way, if a given EM wave can’t survive at all in a material, it is actually reflected at the surface. Again, I’d just take this as something that is true.
The reason I wanted to bring this up, is that it’s actually what is happening with metals. In a certain sense, a signature of the metallic phase is that the electrons of the materials are fairly free to move from atom to atom, or more accurately are fairly shared amongst many, many atoms and thus can move and flow like a fluid in response to an external electric or magnetic field. Specifically, it’s a property of such a charged fluid that it will always flow such to cancel the effect of an externally applied electric field. In a perfect metal, free of defects and of infinite electrical conductivities, there can be no electric fields in its interior. Saying there can be no electric fields means that if I have an incident oscillating electric field (i.e. an EM wave) it is extinguished at the surface of a metal (over a distance called the “skin depth” in a manner similar to our Beer-Lambert law, but on steroids).
So, in a nutshell, a perfect metal is an infinitely good absorber of all light, and because of this paradoxical effect I’m talking about, is a reflector of all light. That’s what makes metals shiny.
However, I do stress that this only the case for a “perfect” metal. In reality, this notion of a free-flowing fluid of electrons is only valid if the electric fields are trying to push the electron around at speeds that they can easily flow at. More to the point, if I’m saying that this fluid of electrons can flow such that it can produce its own electric field that cancels an external one, think what happens when an EM wave is incident on the metal: that fluid has to constantly haul-ass and redistribute itself continually because the electric field is constantly changing in time and thus a new cancelling configuration of the fluid must be adopted to cancel it moment to moment. Now, I realize that this is hand-wave-city and I apologize for that, but bear with me because I just want to make one final point: for a given metal there is a certain frequency, called its “plasma frequency” above which the electric field of an EM wave will be moving too fast for the electron fluid to keep up and continually cancel the field. Thus, metals are only reflective to EM field that don’t oscillate faster than their plasma frequency (or plasma wavelength, again, remember that in a vacuum wavelength and frequency can be talked about interchangeably).
Let me show you the reflectivity of three metals: gold (Au), silver (Ag), copper (Cu) and aluminum (Al):
[Adapted from: https://commons.wikimedia.org/wiki/File:Image-Metal-reflectance.png]
This is our first reflectivity spectra. It isn’t showing absorption, it’s showing the fraction that reflects. However, other than that the x and y axes are very similar.
I bet you could just read off what their plasma frequencies/wavelengths are, based on that graph (about 600 nm for copper (Cu), 500 nm for gold (Au), and 300 nm for silver (Ag)). For light with a higher frequency or shorter wavelength than that cut-off, metals are absorptive or even transparent. Now, if you look at the cut-offs for our metals, we see that silver (with 300 nm, which is in the ultraviolet) and aluminum (plasma wavelength is off the chart) have a cut-off outside of the visible range. But gold and copper don’t. Gold is reflective to all the visible spectrum *except* violet and blue, and for copper also green. Specifically, they actually absorb those colours and thus when you shine white light on them you get white specularly reflected back except for those colours, and thus what you get back is red/orange tinted “silveriness”. This is why copper and gold have a colour, because the frequency where their electron fluids can’t “keep up” falls in the visible range and they are actually absorbers of the bluer part of the spectrum.
I’d also just like to add that, on this issue, there are a lot of wrong, or at least, let’s say, “ill-posed” explanations on the internet. We’ll come to understand this much better later, but I found many sources claiming that the colour of gold was due to, what is called, an atomic absorption resonance. It is not. If it were then gold would only absorb a very specific wavelength (again, you’ll understand this better later), but this isn’t the case. I’d also point out that although gold does have such an “atomic absorption resonance” in the blues, but so do silver, aluminum and copper. Gold atoms also have such resonances in the reds and yellows, so if that were the explanation, why wouldn’t those absorptions be noticed?
The reality is that these explanations are simply not correct because: a) gold/copper will absorb a continuum of wavelengths above its plasma wavelength, and b) as we’ll come to understand, the absorption spectra of materials, like a solid metal, are often entirely different than the spectra of the atoms that make them up. A dilute gas of gold atoms has these resonances; solid gold in a metallic state does not. Only SOLID gold, and not a gold gas, is a reflective system with a gold-ish tint. So anyways, lot of junk on the internet, just wanted to give a warning.
Okay, let’s just do one more material, one that is both black and shiny. An example of this would be something like obsidian (i.e. volcanic glass) or even just black gloss paint like the kind used on pianos and cars.
How can this work? At first glance it might seem contradictory. For something to be black, it means that none of the white light I shine on it reflects. This means it absorbs all visible wavelengths. But then it’s also shiny, which suggests it reflects all wavelengths and it does so specularly (i.e. you can see your reflection). If you don’t understand what I mean by this, see Part 3. How is that possible?
Well to get a first idea, I’d point out that when obsidian is cut very thin, it is in fact transparent. Unfortunately I couldn’t find a good image of thinly sliced obsidian that wasn’t copyright, however I’d encourage you to simply Google image search “thin obsidian” to see some examples.
The fact that obsidian is transparent if thin enough, I think, gives a pretty clear idea of what’s going on. Let me just draw a bit of a cartoon of what something like obsidian or the black polyurethane that they make piano finish out of looks like at a microscopic level:
What I have is little black particles or grains, that absorbs all light, embedded in another medium that is quite transparent. Furthermore, the surface is smooth and can be polished. So I have a lot of specular reflection at the surface and if light never hits a black inclusion/particle it will have a very glass-like trip through the material. However, the thicker the material becomes the slimmer the odds get that that will happen and the more it becomes guaranteed that it will hit one of the black grains. If it does it is absorbed. The more that is absorbed, the blacker it is. So we see why obsidian surface reflects specularly and is glassy when thin but extremely black when thick.
Beyond the Visible
I bet you’re getting pretty sick of me harping on about not everything being about the visible spectrum. But, like I’ve said, humans are super prejudiced in how we talk about light. Let me throw up an image of a guy holding a garbage bag using a “regular” camera, which takes information about light in the visible spectrum, and again with an infrared camera, which takes in information in the infrared:
Take a look at the garbage bag, but also at the guy’s glasses. We humans say “the garbage bag is opaque” and “glass is transparent”. (And we’re all humans here, right? -nervous laughter-). That’s simply not correct though. The garbage bag is opaque to wavelengths in the visible range, is the correct statement. As we can see, it is actually transparent to wavelengths in the infrared range. Conversely, glass is transparent to visible light but opaque to infrared. So when we talk about absorption or transmission or reflectance being dependent on wavelength or “colour” we have to remember the whole EM spectrum exists and watch ourselves when we making broad-sweeping statements, like “that’s opaque” based on the very limited data we collect about the visible range. An object that is opaque in the visible, may well be see-through in another part of the spectrum. Something we’ll talk about more when we finally come to our discussion of predator-vision.
Your Colours are False Sir! Good day. I Say, Good Day to You!
Of course, an infrared camera records an image using only the light within the infrared range of wavelengths, but if it displayed that information using infrared light humans wouldn’t see it. It’d be like “The Emperor’s New Clothes””, “Nah, boss, trust me. I totally remembered to take pictures at the event last night. They’re just not in the visible spectrum!” So, in order to bestow this information to humans we need to present an image using only visible spectrum light. The way we do this, is by arbitrarily choosing a mapping rule, that maps the range of wavelengths the data was collected from to the range of wavelengths humans can see.
So with that in mind, take a look at this picture from the Spitzer Infrared Nearby Galaxy Survey, or SINGS (love those astro acronyms, mentioned that):
[Source: https://en.wikipedia.org/wiki/File:Sig05-016.jpg and NASA]
So beautiful. So colourful! Exxccceeeeppppttttt…. That image was taken using an infrared telescope. This leads to the issue of false colour. When NASA makes such an image they are free to map their infrared data to the visible however they like. The debate then becomes, should this be done in a way that is, in essence, an honest attempt at being as faithful as possible? What would it even mean to be “faithful” in that case? Or should it be done so that, when the mapping is complete and you have an image in the visible spectrum, that it looks as “pretty” as possible? Throughout these blog posts we’ll come to understand why NASA would be very interested in looking at wavelengths outside the visible. We already talked about the X-ray emission of matter falling towards neutron stars in Part 2. But I just thought I’d make a comment here. Not all of these wonderful images you see were taken with cameras in the visible range. So if you looked at them with your eyes, you wouldn’t see this. But also, the colouration or “mapping” chosen may not necessarily be done in “good faith” in terms of fidelity, but rather to maximize “ooohhhh…. preeeetttyyyy….” factor. I’m not suggesting that’s wrong, but consider yourself informed on what you’re actually seeing.
The Energy of Light
That’s probably enough for this post, but in the next post we’re going to revisit the question of the energy contained in light. In our previous post, with Fresnel, we said that the energy of an EM wave is simply proportional to is amplitude/intensity.
Optional Note for Advanced Readers
As I said in my advanced note in that post, this isn’t strictly true and one can recover some amount of frequency dependence of an EM wave if one shifts from a discussion of E and B fields to a discussion of vector potentials.
One of the first and most important cracks in the facade of classical physics at the end of the 19th century, which helped spawn the age of quantum mechanics, was the realization that this expectation simply isn’t true. This isn’t how it works and the equations of (classical) electromagnetism couldn’t explain why… but Einstein (and others) could… And thus the idea of the “photon” was born. This will be the topic of our next post.
Beyond that we’ll be looking at neon signs and medical tri-corders and predator-vision and exploring the stars from the safety of home and the ozone layer and UV damage and much, much more.
See ya there.