# Does light have a temperature?

Does light have a temperature? Is it measureable? Often in photography they talk about color temperature...

DOES LIGHT ITSELF HAVE A TEMPERATURE?

SpaceTiger
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natski said:
Does light have a temperature?

Sure does. In pure thermodynamic equilibrium, a body will emit a blackbody spectrum, a distribution of light that has a temperature associated with it. For example, we talk about the cosmic microwave background radiation having a temperature of 2.7 K because it has a blackbody spectrum consistent with a body emitting at that temperature.

The color temperature that you're referring to is the temperature one would have to heat a theoretical blackbody to in order produce that color that you're seeing. Google seems to have several matches on the subject, so I suggest you take a look.

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samalkhaiat
Temperature is a statistical concept. it does not apply to elementary particles like electrons or photons(particles of light) or even single atom. temperature of an extended object is directly related to the kinetic energies of its molecules.
However if "God" presented you with a tea cup filled with light, you can insert a thermometer into that cup and measure the temperature of the light soup. because the thermometer will absorp some light (i.e some energy) its pointer will rise to indicate that gain of energy.

SpaceTiger
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samalkhaiat said:
However if "God" presented you with a tea cup filled with light, you can insert a thermometer into that cup and measure the temperature of the light soup.

This brings to mind another point I forgot to mention. Since photons don't exchange energy with each other in most everyday circumstances, they are seldom in thermodynamic equilibrium with their immediate surroundings, so most light you see in your everyday life is not at a single, well-defined temperature. For example, the light in your room might have contributions from the sun, a light bulb, your body, and the other contents of the room. Each of these things may be an approximate blackbody with a single temperature, but the resulting photon field is nothing close.

So it's probably better to say that light can have a temperature.

I think that elecromagnetic radiation and thermal radiation are being mixed up here. A photon can induce a temperature change as you stated in the first post SpaceTiger, but I don't think that it can have an intristic temperature.

Regards,

SpaceTiger
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A photon can induce a temperature change as you stated in the first post SpaceTiger, but I don't think that it can have an intristic temperature.

As samalkhaiat already said, temperature is a statistical quantity, so a single photon, electron, or atom cannot be said to have a temperature. However, a blackbody radiation field (consisting of many photons) does have a well-defined temperature. The distribution of light with frequency follows the usual blackbody formula:

$$B_{\nu}=\frac{2h\nu^3/c^2}{e^{\frac{h\nu}{kT}}-1}$$

where T is the temperature of the photon field (or emitting blackbody).

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SpaceTiger said:
As samalkhaiat already said, temperature is a statistical quantity, so a single photon, electron, or atom cannot be said to have a temperature. However, a blackbody radiation field (consisting of many photons) does have a well-defined temperature. The distribution of light with frequency follows the usual blackbody formula:

$$B_{\nu}=\frac{2h\nu^3/c^2}{e^{\frac{h\nu}{kT}}-1}$$

where T is the temperature of the photon field (or emitting blackbody).

Very True.

Andrew Mason
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SpaceTiger said:
As samalkhaiat already said, temperature is a statistical quantity, so a single photon, electron, or atom cannot be said to have a temperature. However, a blackbody radiation field (consisting of many photons) does have a well-defined temperature. The distribution of light with frequency follows the usual blackbody formula:

$$B_{\nu}=\frac{2h\nu^3/c^2}{e^{\frac{h\nu}{kT}}-1}$$

where T is the temperature of the photon field (or emitting blackbody).
But the question asked whether LIGHT ITSELF had a temperature. We often say that light has a certain temperature, but what we really mean is that a black body would have to have that temperature in order to emit that wavelength of light. But this does not mean that the LIGHT ITSELF has that temperature.

AM

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SpaceTiger
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Andrew Mason said:
But the question asked whether LIGHT ITSELF had a temperature. We often say that light has a certain temperature, but what we really mean is that a black body would have to have that temperature in order to emit that wavelength of light. But this does not mean that the LIGHT ITSELF has that temperature.

I feel silly discussing semantics, but in case it wasn't clear what was meant by my above posts:

- Does a single photon have a temperature?

No.

- Can a collection of photons have a temperature?

Yes.

- Do all collections of photons have a well-defined temperature?

No.

- Does the light we usually see have a well-defined temperature?

No.

From the above, you can interpret the question however you like and get the associated answer.

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Andrew Mason
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SpaceTiger said:
I feel silly discussing semantics, but in case it wasn't clear what was meant by my above posts:

- Does a single photon have a temperature?

No.

- Can a collection of photons have a temperature?

Yes.
Hmmm. Although one can associate a temperature with a certain color of light, I think it would require a fundamental change in the definition of temperature for photons to actually have a temperature. Temperature is fundamentally a measure of kinetic energy. Light does not have kinetic energy.

AM

Tide
Homework Helper
Light does have energy and one can most certainly talk about and define a spread in photon energy about a mean value. In the case of particles, the root mean square of the particle velocity is a measure of the spread in kinetic energy and is related to the temperature.

SpaceTiger
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Andrew Mason said:
Hmmm. Although one can associate a temperature with a certain color of light, I think it would require a fundamental change in the definition of temperature for photons to actually have a temperature. Temperature is fundamentally a measure of kinetic energy. Light does not have kinetic energy.

Are you saying that the photon isn't a boson or that the "temperature" in the bose-einstein distribution function isn't a real temperature?

yup , light have temp. , i belive this quote .
it can be proved two ways.
1.from the principal of law of conservation of energy , we can say every energy is transformed lastly as a heat energy . we know heat is a reason & temp. is its
effect. from this theory we can say this.
2. if we focus through a lense in a paper we wil see the paper will burn. it is proved light have temp.

It should be pointed out that temperature in general is not defined as some sort of kinetic energy of particles. There's even negative temperature if you use a certain definition.

http://www.maxwellian.demon.co.uk/art/esa/negkelvin/negkelvin.html [Broken]

The temperature of a photon gas is treated in "statistical physics".
(Bose gas).

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Edgardo said:
It should be pointed out that temperature in general is not defined as some sort of kinetic energy of particles. There's even negative temperature if you use a certain definition.

http://www.maxwellian.demon.co.uk/art/esa/negkelvin/negkelvin.html [Broken]

The temperature of a photon gas is treated in "statistical physics".
(Bose gas).

The second link there states that presence of a small a mount of matter is
essential for a photon gas to reach equilibrium because photons don't
interact with one another.

If you only have light, the gas can't have a temperature any more
than a single photon can..

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Andrew Mason
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Tide said:
Light does have energy and one can most certainly talk about and define a spread in photon energy about a mean value. In the case of particles, the root mean square of the particle velocity is a measure of the spread in kinetic energy and is related to the temperature.
In the case of freely moving particles, the spread in kinetic energy of the particles is given by the Boltzmann distribution. In the case of thermal radiation, the energies or wavelengths of photons will follow a different distribution curve (Bose-Einstein). One can relate the light wavelength distribution to the Boltzman distribution of the thermal source. But does that give light 'temperature'?

AM

Andrew Mason
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SpaceTiger said:
Are you saying that the photon isn't a boson or that the "temperature" in the bose-einstein distribution function isn't a real temperature?
Neither. The 'temperature' in the Bose-Einstein distribution of blackbody thermal radiation is the temperature of the radiation source. It is not the temperature of the radiation itself. The radiation energy distribution is fundamentally different than the kinetic energy distribution of the molecules in the thermal source.

AM

SpaceTiger
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Andrew Mason said:
Neither. The 'temperature' in the Bose-Einstein distribution of blackbody thermal radiation is the temperature of the radiation source. It is not the temperature of the radiation itself. The radiation energy distribution is fundamentally different than the kinetic energy distribution of the molecules in the thermal source.

Sorry bro, but I think you're really reaching here. It is just an issue of terminology, but I find your definition unsettling. You're claiming that the fact that photons don't usually interact with one another makes them fundamentally different from other particles to the extent that they can't be given a temperature. To be sure, this prevents them from reaching equilibrium while in isolation, but when put in a system with things with which they can interact, they follow the same statistical distribution as any other boson. Furthermore, a region of space (say, within a star or the interstellar medium), is not said to be in complete thermodynamic equilibrium unless the radiation follows a blackbody distribution with the same temperature as the matter. In fact, if the matter is embedded in a blackbody radiation field much larger than itself (like the CMB), then it will eventually reach the temperature of that field, regardless of whether the matter that "created" the field is still around.

There are other reasons I find your definition unsatisfying:

1) It "binds" the radiation field to the matter that created it. Does it really seem reasonable to paramaterize the CMB in terms of a hot gas that no longer exists?
2) As far as I can tell, radiation fields obey the zeroth law of thermodynamics, the basis for the definition of temperature.
3) Virtually every expert and textbook on the subject that I know of has described radiation fields as having a temperature.

We may have to agree to disagree.

Tide
Homework Helper
Andrew

Andrew Mason said:
In the case of freely moving particles, the spread in kinetic energy of the particles is given by the Boltzmann distribution. In the case of thermal radiation, the energies or wavelengths of photons will follow a different distribution curve (Bose-Einstein). One can relate the light wavelength distribution to the Boltzman distribution of the thermal source. But does that give light 'temperature'?

AM

I said nothing about the details of the distribution except that the temperature is related to the "width" of the distribution.

Andrew Mason
Homework Helper
SpaceTiger said:
3) Virtually every expert and textbook on the subject that I know of has described radiation fields as having a temperature.
Well, they may ascribe a temperature to a certain spectrum or a certain discrete wavelength of radiation - but that is just the temperature of the blackbody source that would produce a spectrum that peaked at that wavelength. It is a matter of semantics whether you want to say that the radiation has that temperature. I would say that unless you can define temperature in terms of the radiation itself that is independent of the temperature of the matter to which it is associated, then you can't really say that the radiation itself has temperature.

To illustrate the problem, would you say that monochromatic radiation from a laser has the same temperature as a blackbody spectrum of radiation that is peaked at the same wavelength?

AM

LeonhardEuler
Gold Member
Andrew Mason said:
Well, they may ascribe a temperature to a certain spectrum or a certain discrete wavelength of radiation - but that is just the temperature of the blackbody source that would produce a spectrum that peaked at that wavelength. It is a matter of semantics whether you want to say that the radiation has that temperature. I would say that unless you can define temperature in terms of the radiation itself that is independent of the temperature of the matter to which it is associated, then you can't really say that the radiation itself has temperature.

To illustrate the problem, would you say that monochromatic radiation from a laser has the same temperature as a blackbody spectrum of radiation that is peaked at the same wavelength?

AM
In accordance with the zeroth law of thermodynamics the temperature of light should be defined as the reading that a thermometer would give if it was placed in a vacuum with the light. Inside the cavity of a blackbody this would obviously give the temperature of the body, since if it were less or greater you would have a violation of the second law of thermodynamics and a potential perpetual motion machine by exploiting the work you could get by bringing the material in the thermometer in contact with the blackbody and then removing it over and over again. As for a laser, if you shined it on a thermometer I am not aware of any argument which would indicate that the temperature it would read must be the same as the temperature of a blackbody with a spectrum peaking there. The temperature would just be whatever the thermometer says. You ascribe the temperature to a collection of photons, not to a particular part of the spectrum.

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SpaceTiger
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Well, they may ascribe a temperature to a certain spectrum or a certain discrete wavelength of radiation - but that is just the temperature of the blackbody source that would produce a spectrum that peaked at that wavelength.

One could just as easily say that the temperature of a gas is the same as that it would obtain in a certain blackbody radiation field. The argument is completely symmetric.

Andrew Mason said:
To illustrate the problem, would you say that monochromatic radiation from a laser has the same temperature as a blackbody spectrum of radiation that is peaked at the same wavelength?

For any system, temperature is only well-defined in equilibrium. A laser would not be said to be in thermodynamic equilibrium, so it wouldn't have a temperature. The same would be said of a bunch of non-relativistic particles that have a distribution much different from Maxwell-Boltzmann. However, since the relaxation times of gases on earth are generally very short, we almost never see them deviate from equilibrium.

The problem is not qualitative, as I think one would require in order to exclude light from the definition of temperature. Rather, it is purely a quantitative issue concerning how often light interacts.

samalkhaiat
If we think of temperature as a measure of energy, then any physical system has temperature. If it is suitable for my calculations, I can even talk about the temperature of a single electron as T=mc^2/K, and that of "green" photon as T=wh/K.
If you are a theoretical physicist, you can get away with it! otherwise you have to look at your thermometer.
Now,hit a thermometer with a laser, the reading will certainly change. So what do you call that reading?

LeonhardEuler
Gold Member
Temperature is certainly not an energy. If you have a 1kg metal block and a 100 kg metal block both at 20C, and then you raise the T of both of them to 100C, clearly you must add more energy to the heavier block, but both are at the same T. Either they have different energies at 20C or 100C or both because it takes different amounts of energy to change their T. Also think about a block of ice at 0C and 1atm and liquid waer at 0C and 1atm. Energy must be added to the ice to melt it, but the temperature does not change. You can not have a T of one particle, it just isn't defined.

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SpaceTiger
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