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Why don't metals glow from red to yellow to green to blue etc.? Why only red, then yellow and then white? Shouldn't all wavelengths be emitted one by one as the temperature of the metal increases?

If some metals do glow at with different colours, could you give me examples of such metals and the reason why this happens in specific cases?

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    $\begingroup$ Possible duplicate of What causes a black-body radiation curve to be continuous? $\endgroup$ – Rob Jeffries May 9 '17 at 6:52
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    $\begingroup$ See also here physics.stackexchange.com/q/44664 and physics.stackexchange.com/q/207579 $\endgroup$ – Rob Jeffries May 9 '17 at 6:55
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    $\begingroup$ No answer below say why metals never glow blue. Because the temperature corresponding to a blue blackbody radiation are above vaporization temperatures of any metal. Some stars have blackbody radiation. $\endgroup$ – galinette May 10 '17 at 7:37
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    $\begingroup$ And why not green? The temperatures for which the blackbody radiation has a maximum in the green is around 6500K and then there is still a lot of light emitted in the other wavelength, thus the result is white. $\endgroup$ – galinette May 10 '17 at 7:38
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    $\begingroup$ @Luaan : Sure, that's 100% true. Even without waiting for evolution, the human vision is able to do chroma adaptation, which means we calibrate our perception of white to what we perceive as the ambient light. Behind building green glass, which is clearly green viewed from outside, but not too saturated, people don't feel being in green light. Until you switch on an artificial light source, or half-open the window. $\endgroup$ – galinette May 10 '17 at 11:57
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The physics of why the heated metal glows like a black body has already been thoroughly covered in the previous answers. However, in order to completely bridge the gap with the physiology of color perception (which has been alluded to in some answers), it is worth showing a picture of the Planckian locus:

Planckian locus in the CIE 1931 chromaticity diagram

Plot by PAR, from Wikimedia Commons

This is the set of all the colors a black body can have, plotted in a chromaticity diagram. It is computed by combining the black body emission formula with the color matching functions, which are a mathematical model of our color vision.

This graph clearly shows the path of a black body going hotter: red → orange → yellow → white → blue. Now, one may wonder by which coincidence it hits right into the white, rather than going slightly above (through the greens) or below (through the purples). That question, however, is backwards. The good question would be “Why have we chosen to name ‘white’ a color from the Planckian locus”. This is the question of the definition of white, and it is not straightforward.

In the physicists jargon, the name white is often used to mean a “flat spectrum”, i.e. one in which the power per unit frequency does not depend on the frequency. When talking about actual visible colors, however, it has a completely different meaning:

  • A surface is said to be white if it bounces back almost all the visible light that is shed to it.
  • Light is said to be white if it looks like the light typically coming from a white surface.

This leaves the notion of while light ill defined: the light coming from a white surface has the same spectrum as whatever illuminant (meaning: light source) was shined to it. Then the light from any typical illuminant could be considered, in some sense, to be “white light”.

In practice, in the realm of color science, there are some so called “standard illuminants” which are deemed white. Most notably D65 and D55. These are meant to model natural daylight. The choice of daylight as a reference light source is obvious given that our species has evolved in a world where daylight has always been the standard light source, and thus our natural white reference.

The spectrum of daylight varies with the weather and with the height of the sun above the horizon, but it is never too far from a black body spectrum. Which is probably not very surprising given that the Sun itself is a pretty good black body.

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    $\begingroup$ Very beautiful. I'd never thought to plot the Planckian locus before - such a simple and succinct way of thinking about this problem. $\endgroup$ – WetSavannaAnimal May 9 '17 at 11:28
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    $\begingroup$ For those not familiar with chromaticity diagrams it may be pointed out that all the spectral colors are located on the edge with blue text indicating the wavelengths in nanometers. All the mixed colors are located in the area between. If you mix the colors from two points in this diagram you can achieve any of the colors along a line between the points, depending on the ratio. $\endgroup$ – jkej May 9 '17 at 12:10
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    $\begingroup$ @Dieblitzen It is the convolution of quantum mechanic derived black body radiation curves based on tempurature and our eyes interpretation of mixed sets of photons as a particular colour. It has nothing to do with the radiationg material, but rather the tempurature. It has to do with our eyes, and how our brain interprets messages from them. Both of these subjects can have entire academic careers dedicated to them. $\endgroup$ – Yakk May 9 '17 at 13:06
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    $\begingroup$ To follow on, something like an O-Type star would be hot enough to appear blueish to our eyes. Since all knows materials vapourize, in absence of the huge crushing gravity of such a star, we generally cannot get a solid, or even liquid, material to the temperatures required to emit "blue" blackbody radiation. Once in vapour phase, at that, too many other colour-producing reactions take place that, if we pardon the pun, colour the result of the observation (ie: plasma emission lines, etc). $\endgroup$ – J... May 9 '17 at 14:43
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    $\begingroup$ People evolved under the Sun, eyes and brain can adjust and perceive non-white color as white (compare white balance in photography). Therefore a body that reflects all the sunlight is white to us. It's no coincidence the white spot at your picture corresponds to the temperature of about $5770 K$ which is characteristic to the Sun. If the Sun were hotter or cooler, we would call another color "white" in the first place. $\endgroup$ – Kamil Maciorowski May 10 '17 at 8:12
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When metals (or any materials) get very hot, they emit "black body radiation". It's a funny name, because even "non black" bodies emit this thermal radiation. There may be some (small) deviations from the general law due to surface emissivity, but not so much that you would notice.

Black body radiation is not a single wavelength, it is a broad spectrum given by Planck's law. I gave a number of examples of this in this earlier answer. I will just reproduce one of the curves from there: enter image description here

This shows that the general shape of the emission curve is always the same - the only thing that changes is the position of the peak (which follows a 1/T law known as Wien's Displacement Law), and the area under the curve, which increases with the fourth power of the temperature (Stefan-Boltzmann law).

As I mentioned, in a "real" material there may be some "bumps" in the spectrum due to changes in emissivity with wavelength - but to the best of my knowledge this is not enough to stop the appearance of red-yellow-white as things get hotter. However, when things get very, very hot they might appear blue. However, at that temperature the intensity of light is so great (and with so much UV and shorter wavelength light) that I would highly recommend not trying to look at the source. But you can see this in the T=10000 curve above, which has much more blue than red light in it.

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    $\begingroup$ If I understood correctly, when the metal starts getting heated, it emits red because that has the least energy. But as more and more energy is supplied, why does it have to emit the photons in a spectrum? Shouldn't the greater energy supply start causing the metal to emit high energy photons that correspond to the magnitude of energy supplied (like in quantised levels)? If say 10J is for red, then as we increase to 20J red should be phased out and replaced by yellow, and to 30J blue. Why do red and yellow still exist at 30J? $\endgroup$ – Dieblitzen May 9 '17 at 6:19
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    $\begingroup$ Great question, and one of the original ones that lead to quantum mechanics. Without getting into that, the atoms are all moving and bumping into each other. The energy of a bunch of atoms at any given time is a distribution, as shown by the spectra. The "missing" colors are there, but for the eye they are overwhelmed by wavelengths that are produced in much higher quantities. For example, there are no green stars. There is plenty of green light in starlight, but there is no temperature that produces a distribution that is dominant in green. $\endgroup$ – C. Towne Springer May 9 '17 at 7:04
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    $\begingroup$ You can see the blue quite clearly when you're watching a gas flame on the stove, no need for special shielding against intense radiation. $\endgroup$ – Johan May 9 '17 at 7:59
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    $\begingroup$ @Johan But the blue color of a gas flame from a stove is not due to black body radiation, but instead caused by emission lines in the Swan bands of the intermediary products of combustion of hydrocarbons. A gas flame can not be assumed to be in local thermodynamic equilibrium (LTE). Hence its emission spectrum cannot be explained solely by black body radiation. $\endgroup$ – jkej May 9 '17 at 10:10
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    $\begingroup$ @jkej Ha! I never knew about those, and now, aged 53, I do thanks to you! I can almost see a blue gas flame in that spectrum and I was always under the misunderstanding that a gas flame's color came from atmospheric gasses. $\endgroup$ – WetSavannaAnimal May 9 '17 at 11:22
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When I refer to a colour, eg green, I an referring to a range of wavelength which on entering the eye are perceived to be a range of colours depending on the intensity of each of the wavelengths within that range eg yellowish green, green, bluish green.

The black body emission spectrum at a number of temperature is shown below.

enter image description here

The colours shown on the graph illustrate the colours our eye would perceive if illuminate by the range of wavelengths spanned by a particular colour.

The brain interprets the information from the colour receptors (cones) and so you perceive a colour which sometimes can be "unexpected".
For example a combination of red and green light the eye perceives as yellow.

At a temperature of $3000 \, \rm K$ the vast majority of the visible light entering the eye is red and so the object is seen to be red.

As the temperature increases there is more red light but also orange light so the colour of the object changes and from a dull red it becomes a much brighter red.

At higher temperature there is a greater range of wavelength each stimulating the cones and because of this range the colour of the object is moving towards white not the individual colours which are emitted.
Raising the temperature even more which makes the proportions of the colours roughly equal the object makes the object look white.
An even higher temperature when it is very dangerous to look at the hot object directly because of the uv light which is also emitted, the white has a bluish ting.

A metal rod heated in an induction furnace shows the changing colours extremely well.

enter image description here

If you wanted to see a green hot object what you need to do is filter out the red and blue ends of the light from the object.

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    $\begingroup$ "As the temperature increases there is more red light." Why does this happen? As the temperature increases, shouldn't the number of photons corresponding to an energy greater than that of red (e.g. green) increase while the photons representing red decrease? Shouldn't green "overwhelm" the red light so that we start seeing some green? $\endgroup$ – Dieblitzen May 9 '17 at 12:32
  • $\begingroup$ @Dieblitzen The point is that a whole range of wavelengths is emitted as shown in the graph. It is not individual wavelengths which are seen but an amalgam of a whole range of wavelengths. So there is no overwhelming green but red and green. $\endgroup$ – Farcher May 9 '17 at 13:07
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    $\begingroup$ @Dieblitzen While the number of photons representing green does increase, the number of photons representing red also increases, but less so. This is because a hotter object emits more photons in general (or in other words more energy flows away from it per second, quantized in photons, which is the Stephan-Boltzmann law mentioned in the other answer). $\endgroup$ – Graipher May 9 '17 at 13:18
  • $\begingroup$ Sure, there may be more green light than red, but the "more" is tiny. On the 600K line, there is only about 20% more yellow light than red. We instinctively think of the visible spectrum as being all there is to light, but it is actually a tiny sliver of the entire EM spectrum. Look at @Farcher 's graph, how narrow the colour bands are compared to the vast swathe of infra red at the right hand side of the graph. $\endgroup$ – PhilHibbs May 10 '17 at 12:49
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All wavelengths are included in what you call "white", thus all wavelengths are there.

The evolution you are describing is the evolution of the black body radiation as heat is increased.

You can see in the link that as temperature increases the curve moves to the left, to smaller wavelengths.

When the temperature of a blackbody radiator increases, the overall radiated energy increases and the peak of the radiation curve moves to shorter wavelengths.

Here is the left side of a black body radiator, which a hot metal is.

BBlow temp

Backbody radiation in the visible spectrum at 800, 850, 900, 950, 1000, 1050, 1100, 1150, 1200 and 1250 K.

As the temperature of the sample rises , the spectrum moves to the left, and more and more higher energy photons are released.

As heat increases , to the left, more and more photons at visible energies appear, first the red, then the yellow, then by the time temperature reaches green in the spectrum you already see "white" because that is the way color perception works in our eyes. From then on it is white for our perceptions.

red green blue

color mixing

The curves are almost flat at this level of detection and at temperature s where metal is incandescent , still the number of photons emitted is biased towards the infrared, there will always be more photons at a lower frequency than the next band of the spectrum thus blue cannot dominate over the "white" soup our eyes see.

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  • $\begingroup$ I did not understand the graph shown in the answer. I get that as temperature increases the colour of the line changes from red to white, but how do we read the relationship between the x and y axes of the graph? As in, do we say "As wavelength increases, spectral energy density increases"? and if so, what does that mean? Thank you. $\endgroup$ – Dieblitzen May 9 '17 at 6:10
  • $\begingroup$ spectral energy density divided by h*nu , for a interval, gives the number of photons for that interval. The color of the line does not change like a spectrum, it is cumulative. If you look at the full spectrum in the answer by Faecher, you get the full curve. It is for very high temperatures ( the tail on the left at low metal melting temperatures is in the graph I show) and increasing with temperatures. s shown, that energy in that part of the spectrum increases as temperature increases but the total is there in the radiation. When the metal temperatures,as seen in the graph I show $\endgroup$ – anna v May 9 '17 at 8:20
  • $\begingroup$ the number of optical frequency with respect to infrared and lower , increases and become visible as their numbers increase. The hotter the body the more the cumulative photons so there is a possibility of having enough photons at higher and higher frequencies to be seen by our eyes as light. $\endgroup$ – anna v May 9 '17 at 8:22
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    $\begingroup$ So the relevant word is "cumulative", all frequencies are there, but at room temperatures the probaility of a visible photon radiated by black body radiation is very small. In heating metals towards melting, as the heat rises it follows the curves in the number of probable photons at the visible frequencies $\endgroup$ – anna v May 9 '17 at 8:26
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    $\begingroup$ If you continue your graph to 10000K, it will be blue. But any metal is vaporized before reaching that temperature. $\endgroup$ – galinette May 10 '17 at 12:07
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Shouldn't all wavelengths be emitted one by one as the temperature of the metal increases?

Blackbody emission can be grossly oversimplified as "we start with the lowest and then we keep adding more energetic wavelengths as the temperature increases". That's why when we get to the point that everything visible was added to the mix, humans perceive "white". White is the full spectrum.

So it goes like:

  • only red (we perceive red)
  • red and green (we perceive yellow)
  • red green and blue (we perceive white)

If you wanted to see blues and greens, it would have to stop emitting reds first. And it's not how it works, quantum mechanic explains why.

Of course, this is gross oversimplification. Other answers show that as the temperature increases, the emission curve not only gets wider (what I've described here) but also gets slightly shifted up (what OP expected). But the first effect is large while the other is small in comparison, so the widening into white dominates. Only with the hottest of the hot (like arc lamps) we get to the point where we begin to perceive the light as "slightly bluish".

Also, there is no need to limit to metals. The only thing about metals is that you can't get them really hot. Which, accidentally, makes the experiment stop about right where the white is prettiest.

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I made a figure to underscore what I think are the most crucial points to understand why black body radiation is never perceived as green:

  • The widths of and the separations between the spectral sensitivity curves of the human cone cells are much smaller than the width of the Planck curve.
  • The green cone (M cone) is the "middle" one, while the red and blue cones (L and S cones) are furthest displaced to one side each of the visible spectrum.

enter image description here

The result of this is that it is impossible to stimulate the M (green) cone with black body radiation without also stimulating either the L (red) or the S (blue) cone to a similar or larger degree.

The black body radiations that we perceive as most red and blue do not actually peak where the L and S cones peak in sensitivity. The L cone sensitivity peak around $570$ nm while the S cone sensitivity peak around $442$ nm. Using Wien's displacement law we find that the black body radiations for $\sim 5100$ K and $\sim 6550$ K would peak at these wavelengths. But in this figure (from Wikipedia), showing the perceived color of black body radiations for different temperatures, we can see that both of those temperatures would be perceived as more or less white.

enter image description here

The most red and blue black body radiations instead peak at much longer and shorter wavelengths. These are instead the Planck curves for which the slope is at its steepest between the peaks of the spectral sensitivity curves, producing the maximal contrast in stimulation between the different cone types.

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One important thing to note, as galinette did in a comment, is that metals do vaporize soon after they get warm enough to emit a significant amount of blackbody radiation in the visible portion of the electromagnetic spectrum. As graphs posted by others indicate, a metal would emit light perceived as white around 5000 K. It would have to be warmer than this for the graph to be situated such that the emitted radiation in the visible portion of the spectrum appears blue. Iron, for example, has a melting point of 1811 K and a boiling point of 3134 K. If iron got warm enough to have a blue tint, you wouldn't be able to see it for very long.

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protected by Qmechanic May 9 '17 at 8:26

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