Question about infrared radiation

[Ghost Repeater]

In Serway and Jewett's 'Physics for Scientists and Engineers', the authors state that "People all over the world have skin that is dark in the infrared, with emissivity about 0.900.'

Similarly, in Louis A Bloomfield's 'How Things Work: Physics of Everyday Life', he states that 'Your skin is almost perfectly black at the infrared wavelengths of skin-temperature thermal radiation. Your skin's low-temperature emissivity is approximately 0.97. In fact, most nonmetallic objects have low-temperature emissivities greater than 0.95. If you could see infrared light rather than visible light, all the people and most of the objects around you would look black - you would see that they absorb almost all the infrared light that strikes them and that they glow brightly with their own infrared thermal radiation.'

I am massively confused by this language. What can it possibly mean to be 'black in infrared' or 'dark in infrared'? How can an object 'glow brightly with its own infrared thermal radiation' and yet also 'look black' to a creature that can SEE infrared thermal radiation?!

My understanding of blackness is that an object or region of space looks black when no visible light is coming to your eye from the object/region.

So if we could see infrared radiation, for an object to be 'black in the infrared' would mean that no infrared radiation reached our eyes from the object. But if the emissivity is almost 1, how can it be that no infrared radiation reaches the eyes? Isn't that what high emissivity means, that the object IS giving off lots of radiation?

I get that these objects have emissivity of about 1 in IR, and it's emissivity of 1 that essentially defines an ideal absorber/emitter. But my confusion is how this interacts with appearance. Black objects are better absorbers and better emitters than other objects (at a given temperature), since their emissivity is so high. But the reason they APPEAR black to us (i.e. it appears they emit no radiation) is because our appearances are dictated by the visible spectrum, and low-temperature bodies just don't emit much in that range. (And also of course because they don't reflect any light incident on them.)

But this wouldn't be true if we could see infrared radiation. If we could see IR, then we WOULD see the radiation emitted by low-temperature objects. So then they WOULDN'T be black. Right?

Are these authors using confusing terminology, or am I misunderstanding what it means to be black? Or am I misunderstanding what they are trying to say? Like, maybe they are just trying to say that if we could see IR, then most objects would look the same as objects that currently look black to us would look in IR?

 

[essenmein]

Its two separate things.
What it means is that they absorb all incident IR, so no light is reflected, therefore in the context of colour (ie what wavelengths of light is reflected vs absorbed), they are "black".

Then, if this object then also has some temperature above 0K, it will emit IR of a wavelength based on that temperature, but this is not reflected light, its light generated by the body.

Basically if you take a piece of carbon black, at room temp it looks black, heat it up to ~700C (in absence of oxygen to prevent ignition, a different source of light again) it will start to glow red-orange since its now emitting light. Its still a "black" object that's glowing red...

 

[Cutter Ketch]

Yes, I think the term “black” can be confusing. You want to think of it as “no light comes out”, but what it really means is “acts the way a black surface would. This is confusing because black surfaces do emit light.

Absorption and emission go hand in hand. Something that is strongly absorbing also has a high probability for emission. The large absorption means there is a strong coupling between an external electric field and the allowed states of electrons in the material. Well that coupling goes both ways. Yes, if light is incident it will be absorbed pushing some electron into a higher state, but also if you happen to excite the electronic states by some means they will also easily emit light. You can’t have one without the other. High absorption is also high emissivity.

Surfaces that appear black at some wavelength of interest, that is to say absorb the wavelength but do not emit it, do not emit at that wavelength only because the electronic states have not been excited.

Where would energy come from to excite the states? Well one possibility is heat. Electronic states will be thermally populated in a Boltzmann distribution depending on the temperature. The excited states will emit light in a blackbody spectrum. The higher the temperature the higher energy states are populated in the Boltzmann distribution and the higher energy light that is emitted. That’s why things glow when you get them hot enough. They always glow, but when they get hot enough they start glowing at wavelengths you eyes can see.

Now an object at room temperature has a blackbody spectrum that is peaked around 9 um in the mid IR and has significant emission as short as 4 or even 3 um. This surface may be highly absorptive at 1 or 2 um in the NIR but not emit there seeming to be very black. However at longer wavelengths it’s glowing. In fact it also absorbs even at the wavelengths where it is glowing. However it also glows at those wavelengths. Picture a red hot poker. Is it reflective in the red? No. Does it scatter incident red light? No. It absorbs incident red light. However it also emits red light from its own thermal energy. That’s what black bodies do.

 

[Ghost Repeater]

Its two separate things.
What it means is that they absorb all incident IR, so no light is reflected, therefore in the context of colour (ie what wavelengths of light is reflected vs absorbed), they are "black".

Then, if this object then also has some temperature above 0K, it will emit IR of a wavelength based on that temperature, but this is not reflected light, its light generated by the body.

Basically if you take a piece of carbon black, at room temp it looks black, heat it up to ~700C (in absence of oxygen to prevent ignition, a different source of light again) it will start to glow red-orange since its now emitting light. Its still a "black" object that's glowing red...

Thank you for your reply. But I'm still confused. Bloomfield makes two claims:

1) An object at room temperature would look black to someone perceiving infrared.

2) An object at room temperature glows brightly in the infrared spectrum.

I still don't see how these two statements can be made consistent with each other. Is it that 'looking black' is purely a matter of reflecting incident light, versus emitting light as a result of the object's own temperature? So that a room temperature object would look black (to someone perceiving IR) even though it is glowing (in IR), simply because it doesn't reflect any of the IR light on it? In other words, is it that perceived color has nothing to do with emitted radiation and is only determined by reflected radiation?

But that can't be true, either, otherwise we wouldn't see stars!

 

[Ghost Repeater]

Yes, I think the term “black” can be confusing. You want to think of it as “no light comes out”, but what it really means is “acts the way a black surface would. This is confusing because black surfaces do emit light.

So in what sense, precisely, does a room temperature object 'look black' to someone who can perceive IR? If it's emitting IR, then how can someone who perceives IR not perceive it to have a color corresponding to whatever IR wavelength its emitting at?

And if black surfaces emit light, what is it that distinguishes a black surface from other color surfaces? Is it just that a black surface emits most of its radiation in a frequency below the visible range?

 

[Cutter Ketch]

Right, like I said, confusing.

I think the most important thing is to distinguish reflection and absorption from emission. A wrought iron poker is pretty much black. It strongly absorbs all wavelengths of visible light. However if you get it hot enough it will glow red or orange or even yellow. Well guess what: it’s still black. If you shine red light at it it will still absorb it even though it is also emitting it. (At least unless the occupation of the ground state drops hugely)

The physical phenomena are clear, and the real problem here is the use of our color words like “black” which are usually about perception and get used a variety of ways. For example what do we mean when we say a car is red? The car does not emit red light. When we say it’s red we mean that if you shine white light on it it absorbs blue and green but reflects and scatters the red. On the other hand when we say a piece of glass is red it too absorbs blue and green but TRANSMITS red. When we say a film of oil is red it isn’t emitting red, but it also isn’t absorbing blue and green. It’s reflecting red toward the observer and blue and green in a different direction. And when we say a red hot poker is red we don’t mean it reflects or scatters red light or that it absorbs blue and green when you illuminate it with white light. It’s actually producing red light. We look at all of these different phenomena and call them “red”. If I tell you something is “red” which of these things do I mean?

Well there is even more variety in the use of the word “black” because we’ve used it even beyond the visible region. However the common theme in the use of the word “black” is that if shine light on it in some specified band it will absorb it. Using it by region is a weird linguistic usage because we could say that a red car is “black” in the blue and the green. Nobody does that when referring to the visible, but that is the logical extension of how it is being used when they say something is black in the infrared.

So, weird, but you know what they mean. That is the flexibility of language

 

[Ghost Repeater]

Well there is even more variety in the use of the word “black” because we’ve used it even beyond the visible region. However the common theme in the use of the word “black” is that if shine light on it in some specified band it will absorb it. Using it by region is a weird linguistic usage because we could say that a red car is “black” in the blue and the green. Nobody does that when referring to the visible, but that is the logical extension of how it is being used when they say something is black in the infrared.

I'm sorry to be obtuse about this, but that seems to be precisely NOT how they are using it in these cases. Bloomfield specifically says BOTH that the object 'looks black in the infrared' AND that it 'glows brightly in the infrared'. So he is not using black in the same sense as you are when you say that a red car is 'black' in the blue and green, i.e. because it doesn't reflect the blue and green light incident on it.

I do think you're right that the matter of perception is really screwing me up here. I don't know what these authors mean by 'looking black'. What does it really mean to 'look black'? This depends on the perceptual apparatus of the observer and the wavelength we're talking about. I get that objects can appear black to us while emitting (or even reflecting) light, because they may be emitting light that is out of our range of perception. But these authors are specifically saying that the radiation being emitted IS IN the range that we can perceive, and yet it still 'looks black'.

It's as if someone said (in your terms) 'If you look at a red car, you'll see that it is 'black' in the blue and green, while also glowing brightly in the blue and green.'

So if I take a picture of someone wearing eyeglasses with an IR camera, their skin will show up, but the glasses will be black, because the glasses are opaque to IR radiation. It seems like what these authors are saying is that the person's skin would be black. But it clearly isn't, as any picture of a person in IR shows.

I think I might just be the victim of careless language use here and muddy conceptualization, similar to when people say that hot air can 'hold more water' than cold air. But I thought I might be missing something important, because these are authors that I've come to trust in other matters. It seems as though blackness has nothing to do with perception but is only a matter of emissivity (in whatever wavelength range we're interested in). But if that's the case, authors should not talk about things 'looking black' or 'appearing dark' in these contexts.

 

[Cutter Ketch]

. Bloomfield specifically says BOTH that the object 'looks black in the infrared' AND that it 'glows brightly in the infrared'.

I didn’t say that is how he used it. I pointed out several varieties of confusing even contradictory uses of the word “black”. My first paragraph covers the usage you were noting. The poker is black, but if you get it hot enough it glows. Most people seeing a red hot poker would not say it is black. They would say it is red. However in a physics sense it’s still black in that it has a high absorption coefficient (and a high emissivity) It acts as a black body.

 

[essenmein]

If its just about the wording of that Bloomfield quote then I would agree it cannot both "look black" and emit light, to me the description "looking black" means an area in my field of view where no light is being received from, reflected or other wise, that's what "looks black".