I never understood the relationship between Infrared and Heat. Is IR emitted when heat is generated, is heat generated when IR is made, and how do the two relate?

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    $\begingroup$ Does this answer your question? $\endgroup$
    – pho
    Commented Oct 16, 2013 at 17:21
  • $\begingroup$ @PranavHosangadi no, this isn't even technically a correct definition of heat. Heat can only be defined when there is a movement of thermal energy, and IR light cannot be emitted from "anything above absolute zero" it must be emitted from a system which has enough energy (and the correct transition rules) to generate a photon in the IR region. $\endgroup$ Commented Oct 16, 2013 at 17:36
  • $\begingroup$ IR is NOT "light"; it is EM radiation. Anything above zero kelvins, radiates EM energy, some will be IR, but not necessarily much. The approximately 3k background radiation left over from the big bang, will have a spectrum peaked at about 3,000 microns, or 3mm. So 98% of that radiation energy lies between 1.5 mm and 24 mm in the microwave region. At 600 microns in the far IR, or 1/5th of the peak, the spectral radiant emittance is down five orders of magnitude from the peak, and something less than 10^-7 of the total energy is at shorter wavelengths; undetectable; but it isn't zero. $\endgroup$
    – user26165
    Commented Oct 18, 2013 at 0:30
  • $\begingroup$ Related: physics.stackexchange.com/q/665946/247642 $\endgroup$
    – Roger V.
    Commented Jul 5, 2022 at 11:57
  • $\begingroup$ Even though there are great answers on the physical details here, I missed an answer that clearly states the conversion from heat to IR (at the light source) and conversion from IR to heat (on our bodies). This very detailed answer was what made it click for me: physics.stackexchange.com/a/298284/240490 $\endgroup$ Commented Oct 1, 2022 at 4:49

6 Answers 6


The energy of electromagnetic radiation (particularly a photon) is

\begin{equation} E = h \nu = \dfrac{h c}{\lambda} . \end{equation}

where $h$ is Planck's constant, $\nu$ is the frequency, $c$ is the speed of light, and $\lambda$ is the wavelength. So you can see that as wavelength increases, the energy goes down.

Different wavelengths of electromagnetic radiation correspond with different energies of quantum states. Photons can be absorbed by a molecule to gain energy, but only of specific wavelengths (depending upon the molecule and the two states being transitioned to/from).

Turns out, a photon from the infrared region has an energy on the order of the energy of vibrational transitions in molecules. I mentioned that when a photon is absorbed, the molecule gains energy, well it can also emit a photon and lower its energy to another of its allowed vibrational quantum states. This being said, the reason why IR light is produced and associated with heat is that you are seeing molecules go from one vibrational quantum state to a lower vibrational quantum state by giving off a photon of appropriate energy (in the IR region).

Note that vibrational states are only really accessible at higher (near room temp generally) temperatures.

  • $\begingroup$ So basically infrared radiation vibes with the frequencies in (most) molecules that produce effects we sense as "hot", and vice versa. Right? $\endgroup$
    – user
    Commented Nov 28, 2020 at 6:32
  • $\begingroup$ A few points need correcting. "Note that vibrational states are only really accessible at higher (near room temp generally) temperatures."... This isn't correct. Any object that acquires more energy than its equilibrium state, will attempt to release the excess. $\endgroup$
    – KSK
    Commented Jul 9, 2023 at 6:35
  • $\begingroup$ Without having been promoted there is nothing to release from those DOF. And that promotion only occurs at certain quanta of energy. I don't think this answer (my original) is something anyone who has read any thermal physics, statmech, or QM texts would contest? You have to keep in mind that there are other places to store energy besides the vibrational states....which was sort of the original point. $\endgroup$ Commented Jul 11, 2023 at 13:55

All matter in bulk radiates (approximately) as a black body radiator, approximately because there are coefficients of emissivity depending on the constituents. For gases the functional form is different.

The radiation has a specific spectrum and intensity that depends only on the temperature of the body

enter image description here

As the temperature decreases, the peak of the black-body radiation curve moves to lower intensities and longer wavelengths. The black-body radiation graph is also compared with the classical model of Rayleigh and Jeans.

Note that most of the radiation is in the infrared.

Heat may be defined as energy in transit from a high temperature object to a lower temperature object. An object does not possess "heat"; the appropriate term for the microscopic energy in an object is internal energy. The internal energy may be increased by transferring energy to the object from a higher temperature (hotter) object - this is properly called heating.

The energy loss by a black body is given by the Stefan-Boltzman law


Thus the energy carried away by the infrared radiation reduces the heat content of the radiating body. This is the connection of infrared to heat.

The microscopic interactions that give rise to the photons are explained in the other answers. This answer concerns the thermodynamic framework

  • $\begingroup$ Awesome answer, a pity the image it contains has really low resolution and I can't read it's text, can you clarify what it says, or have another picture? $\endgroup$ Commented Apr 7, 2020 at 3:39
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    $\begingroup$ @Santropedro the image is in the first link, it is in wikipedia and can be expanded there. $\endgroup$
    – anna v
    Commented Apr 7, 2020 at 4:15

I think I understand your confusion. The answer would be: You've been misguided. There is no special link between heat and infrared radiation, except for the fact that most bodies radiate most of their heat in the infrared spectrum because they don't have enough energy (heat) to radiate at a higher frequency. See the graphs in this thread.

So one could claim the same connection between X-rays and heat. In fact, it would be even more so, since interactions with X-rays are even higher energy, except that there aren't that many things radiating x-rays around. Probably.

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    $\begingroup$ So how are infrared thermometers accurate over such a wide range of temperatures and for such a wide range of materials? Would it be just as easy to measure temperature using the color green, or low frequency radio waves, for instance? $\endgroup$
    – vercellop
    Commented Jul 14, 2016 at 4:49
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    $\begingroup$ @vercellop, because objects starts to emmit visible light above 500°C, below that there is only infrared radiations emmited. for example when in photography you say a "5800k white", it's the white that would be emmited by an object at 5800°Kelvin = 5526°C $\endgroup$ Commented Jan 6, 2017 at 16:37

The easiest answer is that below 3,000 Kelvin in temperature, heat radiates EM (often referred to as light by physicists although it is all EM not just visible light) in the infrared. Because most heat generates light in the infrared, scientists often refer to infrared light as heat. This is a generalized term or a convention.

Technically heat energy and light are different things, but heat energy can be measured by its EM radiation. Given enough temperature, heat can produce visible light as in our sun as well as ultraviolet as in our sun. Also, although heat energy and light are different things, heat energy produces radiation in the EM spectrum and most heat produced on earth produces light in the infrared.

  • $\begingroup$ An example to illustrate "most heat" (last sentence): some metalworkers judged the temperature of heated steel and iron by its color: a yellow metal is hotter than red (I guess it's because its atoms/molecules vibrate more quickly). $\endgroup$
    – JinSnow
    Commented Jul 27, 2020 at 11:37
  • $\begingroup$ But ... "Even objects that are "red-hot" or “white-hot” emit a little bit of radiation in the visible spectrum, but a lot more in the infrared." quora.com/Is-infrared-heat-or-does-it-cause-heat/answer/… $\endgroup$
    – JinSnow
    Commented Jul 27, 2020 at 11:52

""""".....How does Infrared Relate to heat?....."""""

I take it, that is your question.

"Infra-red" is electro-magnetic radiant energy generally encompassing the range of wavelengths from about 800 nm (near IR) to about 100 micro-meters (far IR). Infra-red radiation includes emissions from single molecules or atoms, as a consequence of quantum mechanical transitions between energy states of those atoms or molecules. It also includes emission of a continuum spectrum of EM radiant energy from large assemblages of atoms or molecules that have a Temperature higher than zero kelvins; that radiation being entirely due to, and characterized by the Temperature of the material, and unrelated to any quantized energy levels that are characteristic of the emitting material. The origin of the radiation is the acceleration of electric charge in the atoms or molecules of the material, while they undergo distortion as a result of being in collisions with each other; those collisions being characteristic of the Temperature of the material.

"Heat energy" on the other hand, is purely mechanical energy of translation or vibration, rotation etc, of the atoms or molecules themselves, and is unrelated to electric charge or properties of electros or atomic states. Heat energy can be transported through physical materials, as a consequence of the collisions between atoms or molecules and their neighbors. (conduction) It can also be transported, by physical bulk transport of the (heated) medium itself (convection).

"Heat energy" is NOT electromagnetic radiation.


IR radiation is a consequence of the thermal energy of a system, and it is in this form that the energy propagates. Ofcourse, there are various regions in the electromagnetic spectrum and just about any one of them can carry energy (in simple terms). Doubtlessly, you must have heard of the microwave, which uses the radiation in the microwave spectrum to heat food. So, the relation between infrared and heat is that they can be said to be generated from each other.

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    $\begingroup$ To add to this, the microwave region corresponds with the energy gaps between rotational quantum states most generally. This is the reason microwave spectroscopy can resolve rotational quantum states. $\endgroup$ Commented Oct 16, 2013 at 17:31

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