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My guess would be that light with a higher energy such as visible or UV would feel hotter, but this is not the case!

Is this something to do with human senses or is there a physics explanation?

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    $\begingroup$ We "feel" visible light, in the sense that we have cells that detect it and send the information to our brains for processing. But since the cells in question are different from the ones that detect infrared light, and since our brains process the two sorts of information differently, we use different terminology. To "feel" visible light, in the sense described above, is called "seeing". $\endgroup$ – Andreas Blass Dec 12 '16 at 2:34
  • $\begingroup$ Interesting question. Would I feel the heat from strong visible light? Like a strong white LED directed at the skin? Probably I would get a burn before. $\endgroup$ – Trilarion Dec 13 '16 at 9:43
  • $\begingroup$ Would one not feel microwaves as well, seeing that we've got a lot of water to absorb them? If I was a brave person I'd test it myself, but... I'm not feeling like it. Besides that, any sufficiently intense laser will surely be felt in much the same way as infrared is... $\endgroup$ – Vendetta Dec 15 '16 at 17:31
  • $\begingroup$ Closely related to physics.stackexchange.com/questions/228508/…. The current accepted answer of @Quantumwhisp is wrong. See freecharly answer(s). He is correct and straight to the point. $\endgroup$ – thermomagnetic condensed boson Mar 12 '18 at 17:54
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You are probably guessing that because you have learned that light of higher frequency carries more energy. Stated like that, this is not true.

The statement is that a photon of frequency $\nu$ carries an energy $h\nu $ proportional to the frequency. Yes, the energy per photon increases with the frequency. But this doesn't make a statement about the overall energy of a light ray:

The energy of a light ray is given by its intensity. The whole concept of photons is just taking light and dividing it into little packages. Those little packages are the photons, which carry the energy of the light ray. The size of the photons ($h\nu$) doesn't determine the overall energy of the ray, but instead simply in what packages it comes.

To go further: Why use infrared-radiation to warm up things like your body? Because what depends on the frequency of the light also, is the absorption coefficient of a medium, like for example your body. And if you look at this:

little graphic

You can see, that for water (which makes up the most matter of our body) that infrared radiation is absorbed better than blue or UV light. You could of course also use blue or UV light to heaten up, and it would just take longer, but you would also suffer from UV light, because, unlike infrared light, UV-light destroys your DNA and your body cells.

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    $\begingroup$ This light absorption in water does not correspond to the UV, visible and IR light energy entering and being absorbed in your body. (1) your body with lateral dimensions on the order of $10-50 cm$ would be largely transparent to visible light and the near UV ($\alpha= 6·10^{-5} to 2·10^{-2} /cm$), which is obviously not the case. (2) you have to calculate the radiation transmitted into the body (and not reflected) before you can calculate how much energy is absorbed in the body. (3) you have to consider the energy absorption at a depth ∼1mm of your skins temperature nerve receptors. $\endgroup$ – freecharly Dec 11 '16 at 22:18
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    $\begingroup$ You might want to spell it out that intensity is the combination of photon energy and photon count rate, because the greater number of infrared photons is also, frequently, a factor. $\endgroup$ – Sean E. Lake Dec 11 '16 at 23:16
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    $\begingroup$ @freecharly : You are right, I'm ignoring the skin here, water was just an example, and with more effort I could have found some data providing the absorption behaviour of human skin. I will edit my post to point this fact out. $\endgroup$ – Quantumwhisp Dec 12 '16 at 2:15
  • $\begingroup$ I agree with @freecharly and this is why I downvoted. I also think this answers doesn't answer the original question. The answer given by freecharly is correct and to the point. $\endgroup$ – thermomagnetic condensed boson Mar 12 '18 at 19:15
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Black-body RadiationThis is probably due to Planck's radiation law and Wien's displacement law giving the wavelength of maximum energy emission, which shows that for temperatures of usual very hot bodies on the order of (a couple) $1000K$, the radiation energy emitted in the infrared/visible ($\lambda>380 nm$)region is much larger than in the ultraviolet region ($\lambda<380 nm$).

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  • $\begingroup$ By far the best answer so far. Correct, simple, concise and to the point. The current accepted answer doesn't answers the question and it is wrong, and utterly upvoted. $\endgroup$ – thermomagnetic condensed boson Mar 12 '18 at 17:53
  • $\begingroup$ But hey, stand in direct sunlight and appreciate all the 6000K energy you're absorbing in the visible $\endgroup$ – LLlAMnYP Apr 14 '18 at 16:45
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The short answer is: of course we do.

The long answer has a few parts.

Absorption

Different wavelengths have different absorption ratios in the same materials. The typical example is a plastic bag, which is transparent to visible light, but opaque to infrared light. This means that it mostly lets visible light through (no absorption, no heating), while capturing infrared light (absorption, heating).

The human body is mostly transparent to both very high and very low frequency light. Radio passes straight through, and so do e.g. X-rays for the most part (don't try hiding from a nuclear blast behind another human - not a lot of protection). There could be kilowatts of radio waves passing right through your body without you noticing any heating, because your body only absorbs very little of those frequencies. Infrared is very important because it's readily absorbed in water - and there's a lot of water in a human body. Still, visible light is readily absorbed in the human body as well - you do in fact feel the heat of visible light (if you've ever tried focusing a lens on a piece of paper - you're mainly doing this with visible light; infrared light of course has a different focus). However, under normal conditions, this tends to be dwarfed by infrared light, because...

Emission

... most of the light sources around us are pretty close to black body emitters. You may be familiar with the rather distinctive curve derived from Planck's Law for the photon emission of a black body. Now compare the area under the curve in the IR region with the one in the visible or UV region - for low-temperature sources (simple incandescent lightbulbs) IR utterly dominates, and even for sunlight, you can see that even before accounting for all the trickiness of the atmosphere etc., we get a whole lot more IR light than visible light. While the per-photon energy of UV light is much higher than for visible light, the total amount of energy carried by all the photons is much smaller - and most of UV light is absorbed in the atmosphere anyway.

In fact, even modern high-efficiency light bulbs still tend to produce more IR light than visible light; light sources with efficiency higher than 50% are quite rare. A decent LED light bulb might have an efficiency around 20%, which means that for each watt of light, it emits four watts of heat (either direct IR radiation or cascading through its surroundings).

IR is everywhere

The feeling of heat on your skin is a relatively simple matter of comparing two temperatures - the temperature of upper skin, with the temperature of lower skin. If the upper skin is hotter, we feel warm, if it's colder, we feel cold.

All objects emit IR light. All of them - and in proportion to their temperature. That's why IR is commonly associated with heat - the room around you is hot with IR radiation, the computer under your desk is hot with IR radiation, you are hot with IR radiation. That's what makes passive thermal vision work - different objects have different temperatures and different emissivity, which makes them stand out against each other on an IR sensor.

How much heat are we talking about? Let's compare to the Sun, just for fun. Sunlight gives about 1100 W per square meter on ground level (there's plenty of different averages - this is basically the value at noon on the equator with average cloud cover). Out of this, about 55% is infrared light and about 42% is visible (see? Even after all so much IR is absorbed in the atmosphere, it still dominates on ground-level :)). So let's say you get about 500 W of IR heat on surface level per square meter. Not something to sneeze at, certainly. Let's put it in human terms, though.

Take a naked human and angle him to the Sun. The human surface area on average is about two square meters, and one half of that is facing away from the Sun, so on a great day, you might absorb as much as that 500 W of IR light. Close enough for our purposes :) But you have to consider something else - the human body is also an IR emitter, and quite a good one at that. How much energy does a typical human emit when idle? About 1000 W. Yes - almost the entire incoming sunlight in the most sunlit place on Earth at noon. So why do we feel warm anyway?

Because sunlight is not the only source of radiation on Earth. Humans radiate a huge amount of energy, true - but so do our surroundings. If you close yourself in a dark room at room temperature, you'll get about 900 W back. So your net radiative loss is only 100 W, rather than 1000 W. And it so happens that the average idle heat loss of the human body is around 100 W, which is why a 25° C room with no direct sunlight feels comfortable - it's more or less a perfect balance between the inefficiencies of human metabolism and the difference of temperature between the human body and the room. Of course, this changes a lot depending on clothing and other factors. Add a 100 W light bulb, and you're outright warm :)

IR photons have a very low energy

Now, this might sound counter-intuitive, and that's because this mostly targets human thinking, rather than reality. But for completeness: infra-red light has negligible effects beyond heating. It's not energetic enough to affect atoms or chemical bonds. The only thing it maps well to is the random motions of atoms and molecules - which add up to what we call heat.

On the other hand, if you take something like visible light, beyond the vibrations you also get chemical changes - electrons being bumped into excited states, (relatively weak) chemical bonds changing; in fact, that's why we see visible light in particular - it's more or less in the sweet spot of "strong enough to excite electrons, but weak enough not to destroy the photoreceptors and their proteins" (the animals that are sensitive to IR use a different mechanism than electro-chemistry). UV light can be easily absorbed, but it's strong enough to break even rather strong chemical bonds, which leads to substantial damage - that's how UV light destroys the DNA in your cells, for example (though again, there are animals that have UV senses - many insects do).

So there's this weird bias in the human mind - you see all those different kinds of light, and they all have interesting properties... except for IR. It just heats stuff, and not much more. Go to even deeper IR (like microwave or radio waves), and you get other interesting behaviours - and a lot less direct heating, since they are less easily absorbed.

Conclusion

We mostly care about infra-red radiation in terms of heat, simply because there's so much of it everywhere, and most of the sources of visible light also involve a higher amount of infrared light. However, take a pure visible light source of enough wattage (say, a cold, high power LED bulb) and point it at yourself, and you'll feel the heat. We use a lot of high-powered visible light lasers, and they're quite obviously pretty good at heating things.

A typical photo-voltaic solar panel captures most of its power production from visible light, as do photosynthetic plants (while some plants also need UV light, that's really more of an catalyst, rather than the primary source of energy; consider how well your house plant is doing despite getting no UV light at all). You need an energy gradient to do useful work, and that makes visible light a lot more interesting than IR for most plants - take a look at an IR photograph of trees or plants; there's quite a decent chance their leaves are actually reflecting incident IR light rather than absorbing it, simply because it's basically waste heat you do not want. That said, there are photosynthetic organisms that are different - working with IR, red or blue light, depending on their niche.

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We do. Here are two ways to demonstrate it.

First the recommended way: Get a really bright white LED (e.g. a 1200 lumen bike light) and look at the spectrum either on a data sheet or witha spectrometer. If you don't trust that, put some IR-blocking glass in front (e.g. KG1). Put your hand in the beam. You'll feel some warnth especially outside on a cold night. A variation is to get an extremely bright single-colour visible LED.

Now the not-recommended way: Put your hand in the beam of a visible laser of at least 50mW (more if the beam is wide). 120mW of 532nm (green) into a sub-millimetre spot on the back of your hand gives quite a sting. With this sort of power your should be wearing goggles, but then you can't see the beam and can get your hand in it accidentally when aligning. But don't try this at home.

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  • $\begingroup$ Some of the cheap DPSS green lasers have a substantial near IR output too (they're IR-pumped and the IR blocking filter is sometimes omitted). $\endgroup$ – Spehro Pefhany Dec 12 '16 at 19:01
  • $\begingroup$ @SpehroPefhany that's certainly true, but the high power CW ones I've used have been good enough for Raman spectroscopy - very clean output spectra indeed. Both the 1064 nm fundamental and the 830 (IIRC) nm diode pump were no more than picowatts after the filter, or we'd have seen them in the spectra. $\endgroup$ – Chris H Dec 12 '16 at 19:07
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I post this in response to the answer posted by Quantumwhisp, which explains a higher body heating by infrared light(IR) compared to visible/ultraviolet(UV)light by an increase of light absorption coefficient in water with increasing wavelength $\lambda$ from UV to visible and IR light (see graph in cited answer). In my opinion, this is not correct (See my comment to the cited answer.) To compare the heating effect of incident light in different wave length regions, you have to know how much energy of the incident light is absorbed in the human skin. I found a scientific article reporting such measurements of the relative energy absorption of light in skins of different males and females in the wave length range from UV ($\lambda=200nm$) to near IR ($\lambda=1000nm$)(Penjweini et al.2013) which shows a decrease of light absorption with wave length in this region. An example of this decrease of relative light absorption with $\lambda$ (skin of a male) is seen in the graphRelative light absorption in skin of male. This shows that the absorption of incident light energy in the skin decreases with wave length from UV to visible and near IR light and doesn't increase as suggested by the absorption coefficient of water. This supports my earlier explanation that the stronger heating sensed by IR radiation is probably due to the energy emission spectrum of hot bodies which is similar to Planck's blackbody radiation spectrum.

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If I understand your question correctly, you might be confusing two different but related concepts namely that of temperature and that of heat. When I say I feel "heat" I am mixing together the notion of temperature and heat. There is molecular motion in a substance, the more molecular motion then the higher the temperature. But this is not the same thing as heat (in the physics sense). Heat is the spontaneous transfer of energy one system to another that can't be attributed to work done on or by the system. $\textit{The way}$ in which this transfer is made, is in our common experience seen through radiation. So in practice we could have a really hot material (high kinetic energy molecules) some of this energy can then be released in the optical or infrared spectrum.

The sun releases some ultraviolet rays and while they can be dangerous to the skin, a red hot iron is the preferred method of a sadistic torturer. The pain I feel from the red hot iron can be portended by the red hue but the pain I feel is as a result of the molecular motion of molecules and atoms in the iron that transfer their mechanical energy to the atoms on my skin.

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protected by Qmechanic Dec 12 '16 at 0:00

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