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Do photons have temperature? If not, does it mean that photon lose energy while travelling through space? As the planets farther away from the sun are comparatively cooler than the one that are closer to it, does it imply that photon also lose energy?

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I could have sworn we have a existing question on the smallest system to which thermodynamics could be reasonably applied, which I think is very much related, but I can't find it. – dmckee Mar 30 '13 at 4:38

The photons themselves do not have temperature as such. However, photons do contribute to the temperature of objects since they carry energy. A very good example is the microwave background radiation which is known to contribute a temperature to the universe at about 3K. One can work out the frequency of these photons using the basic relation $k_BT_{mwb}=hf$ where $k_B=1.381\times 10^{-23}$JK$^{-1}$ and $h=6.63\times 10^{-34}$Js, so that the requency turns out to be in the microwave part of the electromagnetic spectrum. Photons contribute to the temperature of your body when you sit in the sunshine and absorb the sunlight.

The farther you go from the sun the cooler, correct, but this is because the intensity of solar light is inversely proportional to the square of the distance from the sun. On Earth we receive about 1350 W/m$^2$ of solar power. But on Mars, which is about 1.52 the Earth-Sun distance, it is only about 584 W/m$^2$.

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This is really just an extension to JKL's answer since I wanted to pick up on his point about the microwave background, but first it's worth mentioning that although individual photons do not have a temperature EM radiation can be assigned a temperature. The EM radiation emitted by an object has a spectrum that depends on its temperature through Planck's law. So if you measure the spectrum of radiation it is sometimes possible to assign it a temperature through Planck's law, and indeed this is how the cosmic microwave background is assigned the temperature of 2.7 degrees.

But back to the CMB: I would guess your question is asking if an individual photon can lose energy by radiating away like a cooling object, and the answer is no. However light can cool if the spacetime through which it is travelling is expanding. The light cools because it's energy is spread out over a larger volume of space. This is how the cosmic microwave background has cooled from it's original very high temperature of about 3,000K to its current value of 2.7K.

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Indeed, the maximum in the spectrum connects the radiation wavelength, $\lambda_{max}$ at the maximum, and temperature of the hot object as $\lambda_{max}T=2.9\times 10^{-3}$mK (Wien's Displacement Law). – JKL Mar 30 '13 at 20:33
* The light cools because its energy is spread out over a larger volume of space* : is it a shortcut or a mainstream explaination ? TY – igael Jan 2 at 8:02

Assume the sun emits a certain number of photons, such that, at 1m from the sun's surface 1 million photons go through each square meter. As the photons spread radially out from the sun, their number stays the same, but they have to cover larger and larger areas. At 10m from the sun, those $10^6$ photons will cover an area of $10$ x $10 = 100m^2$. So the density of the photons will be 100 times smaller than at 1 m. This shows how, as you travel away from the sun, the density of photons decreases inversely proportional to the square of the distance from the sun.

That is what causes the temperatures of the planets to reduce with their distance from the sun.

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At thermal equilibrium temperature $T$ is defined by $$\frac{1}{T}=\frac{\partial S}{\partial U} $$ Where $S$ is the entropy and $U$ is the total energy. Since a system composed of photons has a well-defined energy and entropy, photons can be said to have a particular temperature if the whole ensemble could be in thermal equilibrium with some hypothetical environment.

Planets further from the sun are cooler than close ones because the same photon flux is spread out over a greater area compared to the size of the planet: Earth's cross sectional area is a greater fraction of the surface area of a sphere the size of it's orbit than Neptune's for it's orbit. This means fewer photons per unit area hit Neptune than Earth. Photons do lose some energy going out that far (redshift from the sun's gravity, this is from general relativity), but this effect is tiny.

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Can't a cavity in thermal equilibrium be looked upon as a gas of photons at some particular temperature? – Charuhas Mar 30 '13 at 12:59
@Charuhas: Exactly! – Dan Mar 30 '13 at 16:53

Well it is not the total energy alone which determines temperature but the "quality" of energy as well. Temperature is a measure of average kinetic energy or also related to the wavelength of the photon. If you have enormous quantity of microwave range of photons the temperature anywhere may not rise above 3K. Classically and close to reality it is Weins displacement law that determines temperature in a spectrum.

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According to the laws of physics the electromagnetic particles called photons individually have a zero thermodynamic temperature. This astonishing discovery means that they are the coldest quantum particles in existence

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