Does light have a temperature?

[Juan R.]

Juan, I cannot prove you wrong. At the same time I am inclined to agree with Mr. Mason, and support the notion that "temperature is fundamentally a macroscopic quality".

I'm sorry to say this but i think that you are rather confounded in those topics.

Let me simply to say

"Frontiers of Quantum and Mesoscopic Thermodynamics"

Quantum and mesoscopic =/= macroscopic.

Of course, you can claim that the only known thermodynamics is that of 19th century that you appears to know very superfitially. Temperature is not defined in terms of heat transport or similar. I already said which was the definition. Of course you can ignore it.

No problem

I wonder like people measure heat capacities for single nucleus!

 

[SpaceTiger]

Juan, I cannot prove you wrong. At the same time I am inclined to agree with Mr. Mason, and support the notion that "temperature is fundamentally a macroscopic quality".

What bothers me most is that you are arguing with me, yet you don't understand my objection. It has nothing to do with macroscopic vs. microscopic (AM's quote was taken out of context), it has to do with one type of particle to another. Photons and electrons (for example) are both elementary particles and they both have equilibrium distributions corresponding to a certain temperature. The primary difference, as I said before, is their relative self-interaction rates. There are many particles that can interact (such as the neutrino), but only do so very seldom. As a result of this, their behavior (in this respect) is very similar to that of light, yet they are matter and they would qualify for most of the definitions of temperature that light doesn't fit.

Maybe you guys will find some experimental evidence someday that conclusively proves that light (for example) can gain and lose heat/temperature as it travels through space independent of any kind of interaction with molecules (hence, conclusively proving that it behaves identically to temperature and the transfer of heat as they are defined in thermodynamics). If/when you can do that, I'll owe you both a drink.

Despite your misunderstandings and misuse of the word "molecules", I would say that this is a good way to conclude the argument. As far as we know right now, light cannot interact with other light without some influence from matter. I don't think this is qualitatively important for the definition of temperature, but you're free to use whatever definition you choose.

 

[Gir]

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

the light is stimulating atomic movement, so technicly, it is generating heat.

 

[Jonny_trigonometry]

I've always thought that a photon is at absolute zero, since it doesn't perceive time, and therefore has to movement in it's rest frame. Being that temperature is a measure of average kinetic energy or simply movement of the things contatined within the sample, if the sample is a single photon, then it's at absolute zero because all kinetic energy, velocity, momentum is time dependant. hmmm, come to think of it, it's division by zero, so it could very well be at infinate temperature.

 

[leright]

the light is stimulating atomic movement, so technicly, it is generating heat.

You don't understand this thread. There's a difference between light GENERATING heat by stimulating atomic movement (therefore producing a temperature in the surrounding atoms or molecules), and the light itself having a temperature. Think of light in a vacuum...does this light have a temperature in the absence of molecules and atoms? I don't think the idea of a thermometer heating up in this light-containing vacuum necessarily means the light itself has a temperature. I think to answer this you need to know if the photons themselves have mass, because to have kinetic energy (and therefore temperature) you need to have mass. IF photons have mass, then I suppose light has temperature. The question of whether or not photons have mass is beyond me.

 

[timeformation]

one word, superconductivity, study it, you will understand... cosmic foam.

 

[ZapperZ]

one word, superconductivity, study it, you will understand... cosmic foam.

Can you please explain how what you wrote here have any relevance to the thread?

Zz.

 

[tdunc]

If I had to define temperature I would Include somewhere in my definition the magnitude of potential energy, such that whenever it happens to collide or however it happens to transfer its energy it does so according to potential energy. So Gamma rays have a potential energy of x, radio waves a potential energy of y being considerbly less than x. A photons temperature can be said to be its energy level with reference to the EM Spectrum, correction its wavefunction, and is inherently a value determined by what electron emitted it. So in other words we can go further to say that an electron itself has a temperature and it can gain or lose temperature by emitting a photon such that the photons energy is the "temperature lost equivilent". This photon will inevitibly be absorbed later on by another electron, governing the transfer of heat per say or it is at least one definitive aspect of the quantum mechanical process of thermodynamics and statistical temperature.

Temperature can be measured by humans. Temperature in that context is purely an invention by our mind to know when a change in normal thermodynamic modes in our bodies molecules occurs-- this information handled by our nervous system. That is the top of the "what is temperature" pyramid along with anything that can show the difference in Temp for a Nbody system, thermometor of course. Breaking that down one level I think we all know how that works, interaction between molecules/ atoms blah blah, Thermodynamics. Breaking it down one more level to atomic particles themselves... "Do photons have an temperature?" Why Not? Of course you could say that, absolutely. But an GOOD point was made...

"I think to answer this you need to know if the photons themselves have mass, because to have kinetic energy (and therefore temperature) you need to have mass. IF photons have mass, then I suppose light has temperature."

Not to mention it would need some kind of mass to EXSIST and to SCATTER.


Space Tiger, I just want to clarify something. You say

"Boson-boson scattering does occur"

Which I was going to ask you about without a doubt, cause I got to know if its true, but then I keep reading and you said

"As far as we know right now, light cannot interact with other light without some influence from matter."

So now I am confused, seems like a contradiction of statements? Or this some influence with matter part the part that makes it not contradict?, care to elaborate? I hope you don't mean it being absorbed kinda interaction...then again, what else is there? No mass, no charge...

FYI I do not think photons can scatter NOR interfere with other photons DIRECTLY if we assume no mass nor charge exsists. Which goes to say that I agree with your statements regarding various light sources being able to be present without interaction with each other to become in equilibrim of energies such that light could never obey a law defined by thermodynamics with interaction in itself. You could mathmatically find an average temperature ie. average kinetic or otherwise energy level, But light will never find an average temperature with itself, because light does not and cannot interact with other light. The very definition of a boson is such that it can occupy the same space Without Interference -- not enough localized mass. Only in context of a very large almost finite degree of decoherence could the argument be made that a potential exsists for light - light interaction. This is how I attempt to deal with how light is able to scatter with Any particle, because it and whatever little mass it has becomes localized ie. collapsed wavefunction. - Adding to that the Decrease in probability amplitude of P defines the classical particle nature at that time such that a classical elastic collision can and will occur given a mutual collision course with another particle whose wavefunction happens to be or at the moment of wavefunction interference, collapse ie. Free ionized electrons in air, Compton scattering.

Back to topic, You'll find that the temperature as it relates to photons is ALSO related to the density of said photons per unit area, such that for example focusing light by means of a magnifying glass, you increase the energy density of an arbitrary unit of space. This energy is absorb as is and translates into a greater change in kinetic energy for a greater energy density. Meaning if the light was more spread out whatever absorbs it will absorb fewer photons per unit area. Photons that are absorbed transfer their energy perhaps not even in the form of kinetic energy, but it is then translated or converted into kinetic vibrational energy ie. EXCESS vibrational energy that the electron is given and maintains so long as it keeps interacting* with photons, absorbing photons, then after it stops absorbing photons it normalizes and settles into its normal new wave function after having perhaps jumped a level or several. Since atoms are connected together in a material or substance By electrons in some fashion... whatever may not have absorbed any photon energy directly will still feel the residual effects following some sort of inverse square law according to material density and composition ie. #electrons # exchanges, I would assume...Thermodynamics. ;)

*if the photon does not match the electron resounance and does not fullfill the requirement for absorption, thus a new electron level, it will simply pass by perhaps giving a jolt to the electron which is translated into excess vibrational kinetic energy that the electron will "work off" and settle into its NMWR. This can be related to temperature perhaps, such that we always have the option to say well maybe temperature is several things one of them being not necessarily anything to do with the electron radiating energy - a purely semi-classical mechanical explanation.

 

[SpaceTiger]

If I had to define temperature I would Include somewhere in my definition the magnitude of potential energy, such that whenever it happens to collide or however it happens to transfer its energy it does so according to potential energy. So Gamma rays have a potential energy of x, radio waves a potential energy of y being considerbly less than x.

Are you simply saying

$$x=h\nu$$?

That would seem to be equating temperature with energy, as mentioned earlier in the thread. I think there's more to it than that...

A photons temperature can be said to be its energy level with reference to the EM Spectrum, correction its wavefunction, and is inherently a value determined by what electron emitted it. So in other words we can go further to say that an electron itself has a temperature and it can gain or lose temperature by emitting a photon such that the photons energy is the "temperature lost equivilent".

People will often quote "effective temperatures" in which they basically equate it to energy, but I don't think this is a very practical way to define temperature in the more general sense. For example, if I'm going to say that an elementary particle has a temperature, from which property will I determine it? A free monatomic particle has three degrees of freedom, each of which will correspond to a different temperature (assuming E=0.5kT). I can assign a temperature based on the total energy, but it still cannot be distributed evenly amongst its degrees of freedom, so an ensemble of systems, each containing a free elementary particle, cannot be used to define an "average" energy per degree of freedom. We can take your definition to the macroscopic level and say that the individual (identical) particles of a gas in equilibrium each have their own temperature, T, and when they're combined, they yield a gas of temperature, T₀. Then

$$T_0=<T>$$

meaning that the temperature of the combined system is the average of the individual components. However, since all of these systems (i.e. the individual particles) are in thermal contact, the zeroth law implies that the equilibrium state should be one of equal temperatures. That is

$$T_0=T$$

which will only occur in the macroscopic limit.

In other words, if we use the zeroth law to define temperature, it must be a statistical property, not one of individual particles.

Temperature can be measured by humans. Temperature in that context is purely an invention by our mind to know when a change in normal thermodynamic modes in our bodies molecules occurs-- this information handled by our nervous system. That is the top of the "what is temperature" pyramid...

Yes, I suspect that many of the people in this thread were arguing with an intuition founded on the above notion of temperature. This is fine for everyday purposes, but for doing real physics, I would say that it's insufficient.

Space Tiger, I just want to clarify something. You say

"Boson-boson scattering does occur"

Which I was going to ask you about without a doubt, cause I got to know if its true, but then I keep reading and you said

"As far as we know right now, light cannot interact with other light without some influence from matter."

So now I am confused, seems like a contradiction of statements?

Well, no, the top statement was not meant to be general; that is, I was not saying that boson-boson scattering occurs for all bosons, just that it can occur for some bosons. This was not relevant for most of the discussion in the thread -- I was just responding to Chronos. You're right, however, that I was waffling on whether or not the scattering of two photons could occur. The basic reason was that QED does allow photons to exchange energy, but only with matter as the mediator.

FYI I do not think photons can scatter NOR interfere with other photons DIRECTLY if we assume no mass nor charge exsists.

As far as I know, you're right. I've been careful not to contradict this anywhere in the thread.

 

[Juan R.]

The zero law for small systems is not so simple like

$$T_{A} = T_{B}$$

which is the second law for large systems.

See my work CPS: physchem/0309002 and references therein (it will be freely available online in www.canonicalscience.com in the section of research, when copyright issues solved I wait in 15-50 days).

for example,

4. ČÁPEK, V. Zeroth and second laws of thermodynamics simultaneously questioned in the quantum microworld. Eur. Phys. J. B 2002, 25(1), 101–113.
5.

However Cápek asumes that macroscopic zeroth law would be applicable to microsystems, which is obviously false. See my physchem for the details. The general zeroth law read

$$1/T_{A} = 1/T_{B} + f(1/N)$$

for macroscopic bodies, N is of order of Avogadro and one recovers the usual equality of temperatures.

See also

KOPER, GER J. M; REISS, HOWARD. Length Scale for the Constant Pressure Ensemble: Application to Small Systems and Relation to Einstein Fluctuation Theory. J. Phys. Chem. 1996, 100, 422–432.

for a similar rule for presures.

 

[tdunc]

Well, no, the top statement was not meant to be general; that is, I was not saying that boson-boson scattering occurs for all bosons, just that it can occur for some bosons. This was not relevant for most of the discussion in the thread -- I was just responding to Chronos. You're right, however, that I was waffling on whether or not the scattering of two photons could occur. The basic reason was that QED does allow photons to exchange energy, but only with matter as the mediator.

Example. Or should I know this? ;)

 

[SpaceTiger]

Example. Or should I know this? ;)

There's some nice discussion on it in this thread. I've seen some speculation that it might not require a medium to occur in some situations, but I don't think it has been observed.

 

[tdunc]

SpaceTiger

Would you object to the idea that Gamma rays have a greater magnitude of potential energy per degree of freedom than radio? Degrees of freedom being either an electric or magnetic potential ie. "electric & magnetic field" of the photon; That in turn being related to polarization.

The Displacement or I call Amplitude of the wavefunction is according to mass/energy and its important to note that while radio waves may displace more space per degree of freedom, that the magnitude of potential energy per degree of freedom is less than Gamma rays. So it is the Displacement Potential rather than the actual Displacement of a particle in the context of then relating thermodynamics and temperature.

And your right there is a lot more to it...I wish someone would come in and lay it out for us, if not give me a couple of days ;) I admittedly never concieved or attempted a quantum explanation of temperature, even if in the end its not even appropiate to do such. I suspect there's something to be said though, or not said just considered.

Something else I might add. While the electron energy lost is the energy of the photon, the photons energy is not equal to the remaining energy of the electron. So the statement that regards the temperature of light is equal to the temperature of what emmited it is not necessarily true. In fact, the temperature of the blackbody should Decrease the more energy it loses, so the temperature of the light can theoretically be greater than the temperature of the blackbody, vice versa. Of course that is only in regards to one aspect of how light is radiated, and of course radiation is in turn only one aspect of temperature... you its complicated.

If we can say that wavefunction (includes amplitude and degrees of freedom) of a particle is according to mass/energy, then potential energy is _insert simple equation_ per degree of freedom. Not forgetting that the forward velocity of light, which is one of its degree of freedom, is faster than a free electrons "FV" AND more importantly that forward velocity is not equal to the rate or speed at which a particle oscillates; That is, in simpler terms, it can vibrate up and down FASTER than it can go forward through space at the same time which happens to be C. C being a fundamental speed limit of some aspect inherent to space (dont want to say medium) not the particle, to explain the same speed of light regardless of spectrum.

"For example, if I'm going to say that an elementary particle has a temperature, from which property will I determine it? A free monatomic particle has three degrees of freedom, each of which will correspond to a different temperature (assuming E=0.5kT)."

Not to mention the wavefunction velocity is not constant, meaning that a collision could occur at any point along its wavefunction at any instance of times, so I see little hope to come up with a simple solution.