# How do photons have temperature?

On the internet, I found

"Temperature is a measure of the average kinetic energy of the particles in an object."

(source)

So, this temperature should be a result of friction.

Light is also known to have some temperature depending on its kind.

So, does it mean that the particles of light (photons) somehow (maybe create friction and) generate the temperature?

This makes me ask: Shouldn't massless things also not have temperature?

Note that I am not talking about the wave nature of light.

• how is the kinetic related to friction? Commented May 19 at 12:26
• The short answer is that temperature is not always a measure of the average kinetic energy of the particles in an object. The long answer is an entire course in statistical mechanics at the upper undergraduate level. Commented May 19 at 12:34
– rob
Commented May 19 at 12:36
• Many seemingly simple questions in physics have deep and complicated answers. Physics is an exciting and important science, however, it does take a long time to learn. One must often start out a long way from the questions that interest them the most. However, the effort is well worth it. Commented May 19 at 13:15
• @Brondahl Light from the sun is warm and I have heard about high energy light vaporising astronomical objects. So I assumed that light is always associated with temperature. I don't really know if it is wrong to say so. If you have any information regarding this I'll be glad to read about it. Commented May 22 at 15:36

Your quoted statement is an oversimplification which is often correct enough to be useful, but which has led you astray in this case.

Fundamentally, temperature is a relationship between energy and entropy. If two systems are in contact and can exchange energy, the energy is more likely to flow in the direction that makes the overall system "more disordered," simply because there are more ways to be "disordered" than there are ways to be "tidy." Again, I'm oversimplifying: all of these terms have careful mathematical definitions. One result of this definition of temperature is that, in a system where the only available degree of freedom is the kinetic energy of its constituent particles, the temperature ends up proportional to the average of these kinetic energies. But that is a result, not an assumption.

An object without any internal degrees of freedom, like a single photon, can't really have a temperature. But an ensemble of photons can have a temperature. If you put an ensemble of photons in contact with a more familiar object with a well-defined temperature, such as considering the "blackbody" photons inside of an oven with only a very small opening, the photons will come to have a particular distribution of kinetic energies as they interact with the walls of the oven.

• 1. No objects can change their momentum in free space. Momentum changes when particles scatter from one another. In the example of photons in the cavity of an oven, they scatter from the walls, in random directions. Particles which are oscillating in a solid are, in a sense, scattering from their neighbors. 2. A parallel beam of photons (especially if they all have same wavelength, like you get from a laser pointer) is a "non-thermal" distribution, and may not have a well-defined temperature.
– rob
Commented May 19 at 13:41
• @Shristeerupa They all move at the same speed. As for ensemble of photons; It took me years to accept stimulated emission so don't start reading about that until you got this well understood... Commented May 19 at 21:39
• @An_Elephant Your favorite textbook on statistical mechanics answers this question. But instead of a single coin, imagine 100 coins in the back of a pickup truck on a bumpy road. Every time the truck hits a bump, all the coins randomly flip. There are $0.99\times10^{29}$ ways to have 49 heads, only slightly fewer than the $1.01\times10^{29}$ ways to have a 50-50 split. The three options 49-51, 50-50, 51-49 will occur about 24% of the time, with approximately equal probability. But a 90-10 split will only occur with probability $10^{-17}$. If you start at 90-10, you move towards 50-50.
– rob
Commented May 20 at 2:56
• @rob Yes I understand it. But it's just probability, which is an artificial and abstract concept , rather than reality. Also , what do you mean by my favourite book of statistical mechanics ? ( I even haven't studied it as I haven't reached college yet. ) Commented May 20 at 5:54
• @An_Elephant The "artificial and abstract concept" of probability theory describes heat flow astonishingly well, because the uncertainties in probability theory become less important as the number of participants gets larger. Consider our pickup truck with 100 coins, where the number of 49-51 configurations is 98% the number of 50-50 configurations. If instead we do a million $=10^6$ coins, the equivalent 490k-510k split has $10^{-87}$ times the probability of a 500k-500k split. A microgram-scale dust mote might have $10^{18}$ atoms. Probabilities on large collections are ironclad.
– rob
Commented May 20 at 15:55

The short answer to your question is "Photons do not have a temperature."

The long answer: unfortunately, almost everything you've quoted from the internet is either manifestly untrue or true only of a specific case. Thermodynamics and statistical mechanics are among the most poorly taught and understood concepts in science, so while unfortunate, I'm not surprised how far wrong the internet goes.

So let's review the statements and fix/amend them as needed.

Temperature is a measure of the average kinetic energy of the particles in an object

This is not true in general, and in fact, gets it backwards! It's like saying "An oven has a gas supply because it is hot."

Temperature is a concept of systems at equilibrium with respect to exchanging energy with another system.

• a system: without getting too philosophical, draw a box around your photon, and other things of interest. That box and its contents and their interactions are your system
• equilibrium: means some property of your system, such as temperature or total energy or pressure, that is not changing in time

The definition of temperature $$T$$ is $$T = \left(\frac{\partial U}{\partial S}\right)_{V,N}$$

In other words, temperature is a measure of how much the total energy $$U$$ of your system needs to increase to get another unit of entropy $$S$$.

• low temperature: it doesn't take much energy to add a lot of entropy. Example: a very cold solid. Adding even just a bit of energy gives your system access to all kinds of new microstates: new vibrational modes, torsion, maybe even magnons and defect movement
• high temperature: it takes a lot of energy to gain more entropy. Example: plasma. The particles already have access to a lot of positional arrangements, so it will take a LOT more energy to give them access to even more microstates (the meaning of microstates is relevant, but beyond what's needed here).

This is why heat energy flows from hot to cold: to maximize entropy. The cold object gains a lot of entropy with a bit of heat input, while the hot object loses little entropy with a bit of heat loss. So entropy is maximized by heat flowing from hot to cold. (why maximize entropy? It maximizes the number of microstates, and thus makes the maximum entropy state the most likely one at any given time - and the odds increase exponentially with entropy).

For the particles: when you have an ideal gas, the particles don't interact. At thermal equilibrium, i.e. constant temperature, the definition of temperature above means we've fixed $$\frac{\partial U}{\partial S}$$. The math gets a bit complicated but in the end

• fixed temperature means the relative probability of each state (in this case: velocity of the ideal gas particles) is a function of their energy and the temperature
• so we can calculate the average velocity

So the statement is

• a special case, for ideal gasses - temperature is about the whole system and all its interactions and possible configurations. An ideal gas is special because only velocities affect the total energy, by definition. Hence the (average) velocity-temperature relationship. If you were looking at another system, like a rubber band (another famous statistical mechanics demo problem) then velocity is completely beside the point!
• backwards: the particles have a known average velocity because they're at a fixed temperature. If the particles weren't in equilibrium, then their velocities would change on average until they reached thermal equilibrium again

Temperature should be the result of friction

As we reviewed above, we now know that temperature is a concept completely separate from friction. You can have friction without being in thermal equilibrium and vice-versa.

Friction between two objects is the force you need to apply to move one object relative to the other. Since it's force, we know that force times distance is work: whatever moves the objects relative to each other is doing work on the system of two objects. That work can result in heat (admittedly extremely common), but it can also result in

• an electric field: rubbing a balloon on a sweater. See also: triboelectricity, triboelectric series
• light: smashing two quartz crystals releases a flash of light. See also: triboluminescence
• vibrations/sound: moving the bow over a violin string

So, does it mean that the particles of light (photons) somehow (maybe create friction and) generate the temperature?

So now that we understand entropy better: nope. Moving a photon doesn't require energy, so no friction force, and thus no friction. Photons can be absorbed and otherwise interact with things, but it's generally not a function of position. Photons don't even have a well-defined position! Their notion of position is even less defined than particles, which themselves don't have well-defined positions (see also: Heisenberg Uncertainty Principle, Schrödinger Equation). The probability of finding a photon somewhere is a function of all routes a photon can take to reach that point and their relative phases. See also: Fenyman's path integral formulation of quantum mechanics.

Shouldn't massless things not have a temperature?

Correct! 1. because temperature is a property of a system at thermal equilibrium, and 2. because a particle by itself typically is not a system of interest. We're generally interested in particles interacting with other particles.

Final item: So what's the deal with light and heat, anyway? Obviously, hot things glow, hence Edison and lightbulbs!

The reason is that

• unlike particles, which are generally conserved because mass is generally conserved (we're not at nuclear physics energies!), the number of photons in a system is NOT conserved
• in fact, the entropy of a system (how many states it can be in, and how likely those states are) is related to
• the number of photons in the system
• and the color (frequency/wavelength) of those photons
• These states are called "photon modes"
• and since we know that temperature is related to energy and entropy, the probability of each number of photons and their energies is a function of the temperature!
• The derivation is extremely similar to deriving the average velocity of ideal gas particles. Except the degrees of freedom are photon number and wavelength instead of velocity and wavelength
• and this is why hot things glow

What's the deal with "color temperature?" This question in the end is probably about reading about color temperature. For example, why can a lightbulb be marked "5000K" for "bright daylight" and "2700K" for "soft white?"

Just like ideal gas particles have an average speed that is a function of temperature, those photons from a hot object have an average frequency. Photons of all frequencies are present in various numbers, but they also have an average. The average also happens to be the most frequent. So just like an ideal gas has a 1:1 relationship between temperature and average speed, a hot object has a 1:1 relationship between its temperature and the average photon it emits.

For much more on this, search up "blackbody radiation." While you didn't want information about the wave nature of light, its wave nature is part of the derivation of blackbody radiation.

For a famous related paradox to blackbody radiation and a fun connection to Einstein's Nobel Prize for the photoelectric effect, you can also search up "the ultraviolet catastrophe." You might like that because it refers to the particle nature of light, in particular, the fact that photons are quantized: you either have one (or two, three, ..) or 0. Never 0.1 photons.

• How did you explain: "temperature is a measure of how much the total energy $U$ of your system needs to increase to get another unit of entropy $S$", when $T=\frac{\partial U}{\partial S}$ and not $T = \frac{\partial S}{\partial U}$? Or am I misinterpreting something? Commented May 22 at 17:25
• I think it's simpler than you're thinking. $T = \frac{\partial U}{\partial S}$, so another way of looking at it is as Taylor series: $U(S) = U_0 + T \, (S-S_0) + O(S^2)$. If you increase entropy by a small amount $\Delta S$, then energy increases by about $T \Delta S$. Therefore the energy increases more with temperature. Commented May 23 at 7:54
• @User198 It's a little more correct to write $\frac1T=\frac{\partial S}{\partial U}$ and $S=S(U,N,V,\cdots)$. Here the internal energy $U$ and its friends are the things you vary independently, and the entropy $S$ varies as a consequence. The simpler expression in this answer is correct given some reasonable assumptions which aren't relevant to this question.
– rob
Commented May 24 at 15:37
• @rob Yes, this now makes sense. Because if $T$ is high, than $\frac{\partial S}{\partial U}$ will be relatively small. Because in an already "really random system", a little more "randomness" will not make much big of a difference. But in a system that is relatively well ordered, even small perturbations produce a "visible" change and thus a steeper increase of entropy. Thanks. Commented May 24 at 18:04

There have been a lot of questions about momentum, energy and gravitation of massless objects lately, so I'll make a blanket statement:

Newtonian equations ($$p=mv$$, $$KE=\frac 1 2 m v^2$$, $$\Phi=GM/r$$) do not work in non-newtonian situations.

See: the relativistic triangle:

So, Total energy is the sum of rest mass energy and kinetic energy:

$$E_{tot} = KE + mc^2$$

which is rest mass energy and energy from momentum added in quadrature:

$$E_{tot} = \sqrt{(pc)^2 + (mc^2)^2}$$

Bringing velocity into the situation:

$$p = \gamma m v$$ $$E_{tot} = \gamma m c^2$$

Note kinetic energy:

$$KE = E_{tot} - mc^2$$

for $$v\ll c$$ becomes:

$$KE \approx (\gamma-1)mc^2 = \frac 1 2 mv^2$$

which is ofc the familiar Newtonian result.

That was just a note, as the temperature part has already been answer well.

• At this point you may note that if m = 0, then E = pc, which is relevant for the photons considered by the OP. Commented May 20 at 12:34

The photon energy is

E=h f

Where h is the Planck constant, and f the frequency.

Particles at a certain temperature T have a certain energy. The energy varies, and the probability(rho) of certain energies E is given by the Boltzmann distribution:

rho~exp(-E/(kT))

k is the Boltzmann constant.

When atoms or molecules collide or interact, their thermal energy can be transferred to an electron, bringing the electron to a higher energy level. Because of this process the electrons also have an energy following the Boltzmann distribution. Of course this is also influenced by what energy states are allowed.

The excited electrons can fall back to the ground state, whereby they emit a photon. This is the thermal radiation.

The energy of the photons also follows the Boltzmann distribution, corrected by the allowed energy levels. And the photons can be absorbed again, bringing electrons to higher energy levels.

This is how photons are connected to temperature. Atoms and molecules are in a thermal equilibrium because they exchange energy. And one way how they exchange energy is through photons, thermal radiation. And everything follows the same energy distribution.

• @Peter Mortensen Yes, that makes sense. Well I explained some details of that process, which were only understood after the discovery of black body radiation, and after the formula was found for Planck radiation. Now I start wondering about the differences, or what's missing in my explanation. Of course there is the transition probability for electrons to lower energy levels. The larger the energy difference, the higher the probability of transition and therefore emission. This affects the energy dependence of thermal radiation, which the Planck radiation formula describes Commented May 24 at 14:07
• What I wrote is certainly important when it's not the ideal case of a black body Commented May 24 at 14:14

Temperature is the average kinetic energy of a mass, and since light-photons are massless then there isn't any mass, and no mass equates to no average. The average of nothing is nothing.

Light-photons are pure energy derived from a mass somewhere in time $$E=mc^2$$ and momentum $$p=mv$$, and since the velocity of a photon is at the speed of light, time stops for it and kinetic energy is nonexistent, but only to an outside observer due to the special theory of relativity.

The interpretation of light photons having a temperature is just an illusion of interpretation similar to heat radiation via thermal radiation is nonexistent until it strikes a mass to increase its entropy and thus interpreted as a rise in temperature, but only to the average mass.

For example, gasoline is not hot until it burns in the temperature scales; outside of absolute temperature.

In conclusion, photons are pure energy that don't carry any temperature, but create a temperature rise on the mass that absorbs its momentum or kinetic energy. There isn't any magic, only chance.

Light is invisible until it strikes a mass and the mass radiates a frequency in the human light spectrum, outside the use of infrared, or ultraviolet equipment. Shining a strong powerful flashlight into a clear night void of dust or moisture will be invisible until it strikes a mass.

• What do you mean by "...burns in the temperature scales" (seems incomprehensible)? Commented May 21 at 13:38