What would a blackbody look like? - Page 1
Forum section: electronics / beginners

author: fubar.gr
author_rank: Unknown
timestamp: April 01, 2015, 08:36:02 pm
content: A black body is a theoretical object that absorbs all incoming radiation.What would such an object look like in reality? Is there any man made substance that comes close?

author: suicidaleggroll
author_rank: Super Contributor
timestamp: April 01, 2015, 08:49:20 pm
content: Any flat black finish is going to get relatively closeHere's a table of emissivity values for various materials. A black body would have an emissivity of 1.http://www.infrared-thermography.com/material-1.htm

author: photon
author_rank: Frequent Contributor
timestamp: April 01, 2015, 08:58:55 pm
content: A black body would look black, ie no light reflection. Your eyes see no photons as black.A more interesting question is how did studying black bodies lead to Plank's constant h.

author: AG6QR
author_rank: Frequent Contributor
timestamp: April 01, 2015, 09:06:40 pm
content: A black body would only look black if it were cold. Get it warm enough, and just like the heating element of an electric stove, it starts to glow red. Get it really hot, and it turns white, or even bluish-white.Stars (including the sun) are pretty close to being ideal black bodies. Stars of different temperatures have different colors.But yeah, at ordinary temperatures, an object painted with very flat black paint comes reasonably close to being a black body.

author: Ian.M
author_rank: Super Contributor
timestamp: April 01, 2015, 09:13:47 pm
content: Amorphous carbon deposits (e.g. soot) can come very close to being an ideal black body but if its hot enough will glow red or even shorter wavelength colours. This can frequently be seen in an ordinary fire that has sooted up some object followed by the flames increasing till the soot starts to burn off.

author: owiecc
author_rank: Frequent Contributor
timestamp: April 01, 2015, 09:36:35 pm
content: Check out Vantablack:

author: dannyf
author_rank: Super Contributor
timestamp: April 01, 2015, 09:43:26 pm
content: QuoteBlack?Quotethe interior of a non-reflective ball, with a tiny opening.

author: TimFox
author_rank: Super Contributor
timestamp: April 01, 2015, 09:59:51 pm
content: The black body used in the pre-1979 definition of the "candela" is a small hole leading to a cavity inside a block of tungsten, held at the melting point of platinum.For photographic purposes, the sun, as seen from the earth on a clear day at high noon, is a black body with an effective temperature of 5500 K. Although it doesn't look black, it has the same spectrum (in the visible range) as a black body at that temperature.

author: SeanB
author_rank: Super Contributor
timestamp: April 01, 2015, 11:37:32 pm
content: A good approximation that will be at 39C is the inside of your month. If you are not breathing at the time it is a good black body, as your skin is quite good at keeping a constant temperature.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 02, 2015, 12:22:14 am
content: I doubt any material is truly a black body; there's always something -- plastics are usually:Transparent (i.e., capacitive or nonconductive) at DC to RF,Transparent to opaque (absorptive) in THz,Opaque in IR (with holes depending on the chemical base),Transparent in visible (unless pigmented),Opaque again at UV,And various peaks and dips from EUV to x-rays and beyond (for reasons ranging from atomic absorption to x-ray diffraction, nuclear gamma resonances, and weirder phenomena).The same is very true of most insulating solids, including inorganic salts (which encompasses ceramics).Metals are very opaque at all frequencies below the plasma frequency (in the visible to UV range -- in essence, how fast the electrons can move and respond in the material itself), and absorptive above there (until the EUV+ atomic junk starts to kick in). However, they are also very conductive / reflective.The reason a chamber is often specified is, if you have a region where radiation can bounce around forever, eventually it will be absorbed. Therefore, we can factor out the reflectivity of a material pretty easily, without having to use a lot of it. Whereas, transparency cannot be entirely factored out -- it takes an infinite thickness of a transparent material to absorb truly everything.Of course, metals aren't perfectly reflective-absorptive either, and they do transmit some. But the transmissivity is so much smaller that it rarely matters. (It's an exponential thing, so if you start with a large base to your exponent, it grows very quickly indeed.) The metal film on a Compact Disc is a few microns (or maybe it's even less, I forget); blue light can just barely be seen through it.So how do you make such an "echo chamber", as it were? Traditional methods include acute angle corner reflectors. A stack of razor blades is a great example: the polished knife edges are fairly reflective. Most importantly, the edge is too thin to produce a specular reflection, and there are no scratches on the sides to do the same. At the inside corner, where two blades touch, the incident light keeps reflecting between blades, until it gets buried and dissipated in the root. If the contact point is not quite perfect, there will eventually be some residual reflection. A metallic inside corner looks like a solid waveguide wall to incident light, when it gets to the ~1um width near the root. The challenge is to have deep enough grooves (or pits or whatever) that light gets trapped and dissipated before it has a chance to reflect.Scale is always a concern; the same structures, tuned for radio frequencies, are usually constructed from loaded foam pyramids. (The "loading" is some radar-absorbent material, like magnetite, powdered metal or carbon black. Foam is used because its index of refraction is close to 1, so little reflects due to impedance mismatch.)The "blackest black on Earth" materials (like pictured above) are constructed from a similar, but microscopic, structure: usually something like nickel metal, plated or etched into a chaotic nanoscopic structure. It's black for much the same reason activated charcoal is black (it's conductive and full of holes), as well as the above reason (most of the angles between those micro- and nano-scopic structures are acute, trapping light and absorbing it).The last challenge is to make a surface that, although it may be black, it should still act flat. That is, what light it does reflect, tends not to be scattered randomly. Such a material or coating can be applied to an acute corner or cavity structure (or all three in whatever fractal ordering you like!) to achieve an arbitrarily black environment.Related thought:If you ever have the pleasure of using an IR (thermal) camera, try it on bare copper (PCB or otherwise). Whether the surface is (optically) bright, matte, or coarsely sanded, it will still look bright and shiny to far IR, because the waves are just too fat to be scattered by scratches. In other words, it still looks perfectly flat and shiny -- shiny is in the, not the eye so much, as the wavelength -- of the beholder. Black bodies made of the same materials exhibit the same property, so that the Vantablack stuff is probably a fine reflector somewhere below the THz range (depending on just how thick and "lumpy" the surface actually is).So, in summary: what's an ideal black body? A cavity, where light (generally speaking, EM radiation of *any* frequency) can bounce around to anywhere but the aperture, at all frequencies of interest.Quite possibly the only true black bodies are astronomical in origin: cosmic background is the random crackle of the early universe, red-shifted into oblivion; and black holes -- assuming you can find one that doesn't have an accretion disc to spoil your measurements -- should be very black indeed, though at a much lower temperature than usual (solar mass+ black holes are in the uK range, quite cryogenic indeed). Downside being, we're not entirely sure what "particles" should be emitted from black holes -- anything is possible, but it should be that, at that temperature, photons of rather low energy (UHF to microwave?) are by far the most probable.Tim

author: miguelvp
author_rank: Super Contributor
timestamp: April 05, 2015, 09:57:52 am
content: As owiecc linked the Vantablack video, the people that've seen it say that it messes with your perception because no light seems to come out from it and is very unfamiliar at first glance and you can't tell the shape.But as I've heard, if this was made into clothing material all our wifes/girlfriends will purchase them, although I don't know how we will like dresses that reveal no shape at all, just the silhouette contour, myself I think I won't like it.

author: Rerouter
author_rank: Super Contributor
timestamp: April 05, 2015, 11:31:18 am
content: at my tafe they had a simulation of one, one of the black chambers with special ridges to trap any incoming light, even when you shone a torch down it, you could not focus or make sense of what you where looking at, it was just a black void,

author: jpb
author_rank: Super Contributor
timestamp: April 05, 2015, 11:43:33 am
content: As people have pointed out, black bodies have a well defined radiation spectrum depending on temperature.The original question is probably more appropriate to a black hole from which light can't escape.

author: wiss
author_rank: Frequent Contributor
timestamp: April 05, 2015, 10:01:10 pm
content: The sun! Do not stare at it, will damage your eyes!

author: Zero999
author_rank: Super Contributor
timestamp: April 06, 2015, 08:17:53 pm
content: Just learning a bit about the reflectivity of metals.Quote from: T3sl4co1l on April 02, 2015, 12:22:14 amSo that's why copper is reddish-orange and gold is yellow.http://www.superstrate.net/pv/optical/metal.htmlQuoteThat's odd, it allows some blue light through yet appears reflective to most visible wavelengths.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 07, 2015, 04:44:29 am
content: Quote from: Hero999 on April 06, 2015, 08:17:53 pmAnd cesium is yellow...ish, by most accounts.Though as it's usually sealed in glass, then photographed with a CCD camera under unknown lighting conditions, one could justly question the precision of such color; but then, few have seen it directly (through transparent gas and eyeballs alone), or at least... without having the metal erupt into colors much more luminous!Further brain tease: radio waves reflect off the ionosphere, up to about 30MHz, allowing worldwide propagation on the shortwave band (and the MW band and below, but lightning supplies a lot of noise down there). The ionosphere is made up of charged particles trapped in the Earth's magnetic field. A gas of charge carriers, just like in a metal, but much less dense...QuoteI don't know if the blue color is due to interference (being 1/4 wave thick at reddish wavelengths?), or a material property (wouldn't seem right, as aluminum's "tint" is presumably well into the UV), but as reflectance is in the 99% range, it shouldn't be surprising that, what little does get through (i.e., < 1%!), can be colored strongly. It could very well be the "tail" of the high-pass characteristic -- only a few dB, but still noticeable.This is characteristic of dielectric reflectors too: mundane things like glass, and soap bubbles, reflect some fraction of light incident from air. Bubbles are notably thin enough to exhibit interference effects as well, hence the rainbow of colors reflected/transmitted. You don't normally notice the transmitted tint, because -10dB reflected is like -0.7dB (or whatever it actually is) transmitted."Ideal mirrors" are made from a great many layers of alternating dielectrics; a series of 1/4 wave plates, even if not very reflective by their own, will eventually reflect everything if enough are used (since reflectance/transmittance is exponential with N).This is directly analogous to a series of alternating transmission lines, say a bunch of 1/4 wave stubs of 50 and 75 ohm cable, connected together -- even though the reflection coefficient of each junction is small (0.2), enough of them in series will eventually reflect >99%, limited only by the loss factor of the medium (which, for coax at most frequencies, isn't really all that great an example, but pretty clear optical media are easy to find).Such a structure also makes a lowpass filter (in much the same way as an LC lumped equivalent transmission line has a cutoff frequency, because that's a further analogy), which after some manipulation for layer thickness (and refractive index, if possible), makes possible very precise optical filters of arbitrary wavelength and nearly-arbitrary sharpness, both much better than Doppler-shifted atomic absorption lines characteristic of conventional colored or dyed filters / gels. An added benefit is tunability, because the light can be transmitted at an angle, making the optical path csc(theta) times longer and therefore blocking longer wavelengths.Note that an ideal electronic filter is perfectly reflective in the stop band, perfectly transmissive in the pass band, and never dissipative. Just like shiny optical things, whether dielectric or metallic.Tim

author: Zero999
author_rank: Super Contributor
timestamp: April 07, 2015, 09:58:07 pm
content: After some Googling, I found this paper. I admit I don't understand it all but you'll probably find it interesting.https://ore.exeter.ac.uk/repository/bitstream/handle/10036/48093/Some%20considerations.pdf?sequence=5

author: photon
author_rank: Frequent Contributor
timestamp: April 19, 2015, 11:55:07 pm
content: I was hoping that someone on the forum would add the connection between black bodies and plank's constant h, but so far not. I'm going to add what I know hoping someone adds more.As mentioned in the above posts, a black body is one which absorbs all the incident electromagnetic (EM) waves (like light) and generate only the EM waves that are due to its temperature. The theory being that when matter gets hot, the electrons in it start moving fast, causing EM waves due to these electrons' acceleration only.Why so important? I believe that they were hoping to use the conservation of energy principle to equate the energy in an electromagnetic wave with the energy due to heat (temperature). The tools used would be Maxwell's formula for the energy in an EM field and Maxwell's formula for the distribution of speeds of the molecules in a gas at a given temperature. And this gave Plank's formula for the energy of the EM generated by the temperature of a black body:I think it's the only formula in physics which contains the 3 physical constants c, k and h.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 20, 2015, 11:42:06 am
content: I recall something about requiring a cutoff: strictly speaking, Maxwell-Boltzmann has no cutoff, so the energy at infinity can be infinite (the UV catastrophe). Which makes no sense (black body radiation, at basically any temperature below fusion plasma temperatures, simply doesn't produce detectable gamma radiation).The essence of that cutoff was a suitably chosen, but nonetheless unjustified, point, with the implication that radiation is not carried by continuous waves, but in small packets, photons. Which means, as much as Planck wished to deny quantum theory, he kind of invented it.I used to know much more about this, but it's been so many years since going through that in school!Tim