If visible light have so much more energy than infrared, why does infrared feels hot?

p3t3r1

Sorry, kind of stupid question, but just wondering. Thanks.

John Creighto

Visible light has more energy per photon but the light that comes from the sun is spreads out so it is at a much lower intensity then it would be from a black body emission. If you think that infrared light is hotter then try putting you hand on a light bulb, a red hot stove burner or a white hot welding rod. Well, actually don't but note that if you see an object turn red blue or white because of thermal emissions then it is visible light you see and not infra red. Of course there will also be infra red light and the proportion of each frequency will depend on temperature. Better idea, if you think visible light is not hot then take a trip to the sun.

NoTime

The sun closely resembles a blackbody in its emission spectrum with the peak emission in the yellow green area.
The difference in what you feel is primarily due to how many photons get adsorbed instead of reflected.
You might have noticed that a black surface (say a car in the parking lot) which doesn't reflect much light is much hotter in sunlight than a white surface which reflects most of the light falling on it.

lightarrow

It depends on which distribution you are considering: if you consider the distribution in wavelenght, the maximum corresponds to ~ 502 nm (green), but if you consider the distribution in frequency, the maximum corresponds to ~ 884 nm which is in the infrared.
I took 5778 K as sun's surface temperature (wikipedia).
The topic was also discussed here:
http://www.physicsforums.com/showthread.php?t=132258

Andy Resnick

Think about the total amount of absorbed energy, which is spread over a range of wavelengths. The sun is nearly a 6000 K blackbody, but walking around on a sunny day doesn't feel like the inside of a 6000 K oven because a lot of the energy is absorbed by the atmosphere (and also that hardly any of the radiated energy actually falls on the Earth).

Same for a lightbulb filament or the arc in a short-arc bulb: most of the thermal energy is dissipated before reaching you. Even though a filament may reach 3000K, and an arc around 6000 K, the total amount of radiated energy in Watts is very little.

Also, infrared absorption is qualitatively different from visible and UV absorption: the first goes to vibrational energy in a material, while the second and third go to electronic transitions.

NoTime

Oh my. It's been forever that I've looked at something like that.
However, since wavelength is just 1/f then at a crude level this is just wrong.
The distinction is that peak in your formulas represent two different things.
Someone will have to correct me if I'm wrong, but in one case I think you end up with max photon count while the other is total momentum.

Edit: the latter being the normal way of looking at this (green).

Jakell

IR feels hot because the nerve cells in our skin reports large quantities of that frequency of photon as heat. You can get hit with higher doses of midIR or microwaves (both just outside the thermal IR range) and not feel anything. Similarly, our skin is fairly transparent to visible light, and it takes a good deal of it to heat up our skin.
The same answer applies to our eyes seeing visible light, and not thermal IR. You can heat something up, or just produce thermal radiation (as in a CO2 laser), and our eyes will not be able to see it, at least until that something is hot enough to give off visible photons.

p3t3r1

Cool, thanks for the great responses.

John Creighto

So what are you saying? It just heats up our insides? Obviously not much gets through us or our shadow would be kind of hard to see. I would think how well our skin reflects light would be more important then how well it transmits light in terms of how hot it makes us feel. Moreover, if the skin was really good at transmitting light then wouldn't we be able to see through it?

russ_watters

Though hopefully you can figure it out from the responses, no one directly addressed the question asked in the OP. The question is wrong: infrared doesn't feel hotter than an equivalent amount (# of photons) of visible radiation.

lightarrow

Integrating in dv (first case) or dl (second case) you have total energy density in both ways; this only is the meaningful physical concept.

uart

Actually it dedends on how close to "black" your body is at different wavelengths. My skin looks fairly pale at visible light so presumably it relects some part of the visible light falling upon it. I don't know what is the "albedo" of my skin at infra-red wavelengths, but if its significantly less than at visible light wavelengths then that could explain the observation. I agree with you however that no compelling case has so far been made that the original "observation" is even true.

uart

Yeah one maximizes the energy per incremental wavelength band and the other maximizes the energy per incremental frequency band, that's why they are two different things.

In other words, say you split all the light from l=200nm through to l=20,000nm into many bands of say 10nm increment, then the band corresponding to approx hc/lkT = 5 will contain the most energy out of all those bands. If on the other hand you split the light into frequency bands of say 1THz per band then the band corresponding to approx hf/kT = 2.9 will contain the most energy out of all those bands.

When you look at it like this it's easy to see why they are two different things. At the lower end of the spectrum under consideration (20,000 nm) each 1THz band corresponds to approx a 1250 nm wavelength band whereas at the upper end (200nm) each 1THz band corresponds to a mere 0.13nm wavelength interval. That is, each 1THz frequency band corresponds to about 125 of the 10nm bands at the lower end of the spectrum under consideration but only about 0.013 of one 10nm band at the upper end of the spectrum.

So yes in each case we are comparing energy per band, however there is no direct correspondence or consistency of the widths of the bands between the two cases which is where the discrepancy arises. Actually I'm pretty sure that you (lightarrow) were already aware of this, I'm just explaining this in case anyone else is confused.

Andy Resnick

No, Jakell did, actually- as he said, the nerves in skin responsible for sensing temperature are more sensitive to IR radiation than visible, so indeed IR does *feel* hotter, even given equivalent amounts of power.

russ_watters

This isn't about any specific way a ray of light happens to come to be. All I said is that if the number of photons is equal, the one with the higher frequency has more energy. These photons can come from a radio tower, lasers, reflection, filtration, or, yes, black body radiation. But this is not just about black body radiation.

uart

I think you may have misread what I said Russ. By "your body" I was literally referring to your body (that is your skin and flesh etc) as the absorber. I was not making any reference at all to the radiation source. If you include the very next sentence after the one you quoted it says,
.
That's the nice thing about context, when you read more than one sentence it often makes more sense.