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natski
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Does light have a temperature? Is it measureable? Often in photography they talk about color temperature...
DOES LIGHT ITSELF HAVE A TEMPERATURE?
DOES LIGHT ITSELF HAVE A TEMPERATURE?
natski said:Does light have a temperature?
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.
Nenad said: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.
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:
[tex]B_{\nu}=\frac{2h\nu^3/c^2}{e^{\frac{h\nu}{kT}}-1}[/tex]
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.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:
[tex]B_{\nu}=\frac{2h\nu^3/c^2}{e^{\frac{h\nu}{kT}}-1}[/tex]
where T is the temperature of the photon field (or emitting blackbody).
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.
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.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.
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.
The second link there states that presence of a small a mount of matter isEdgardo 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).
http://www.physics.rutgers.edu/ugrad/351/Lecture 25.pdf
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'?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.
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.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?
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.
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
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.SpaceTiger said:3) Virtually every expert and textbook on the subject that I know of has described radiation fields as having a temperature.
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.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
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.
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?
samalkhaiat said:Now,hit a thermometer with a laser, the reading will certainly change. So what do you call that reading?
Chronos said:Temperature, in the classical sense, is a property of matter. Hence it is not meaningful to talk about properties like the temperature or mass of a photon [since it is not composed of matter].
LeonhardEuler said:Please help me understand what you mean by 1eV being equivalent to 11605K. Equivalent in what sense?
I was hoping to clarify the difference between temperature in the thermodynamic vs quantum sense. It appears I was not entirely successful. This pretty much sums up the definition I had in mind:SpaceTiger said:Nor is it meaningful to talk about the "temperature" of an electron, so I'm not really seeing the where you're going with this line of reasoning. Where is your definition coming from? I've searched the web and my textbooks and found a variety of definitions. Clearly, the concept of temperature would only have applied to matter when it was first devised because physicists were not even aware of the particle aspects of light, but we live in a world of modern physics and it's not clear to me why you think the concept of temperature does not extend to bosons. Boson-boson scattering does occur, so they can reach equilibrium just as fermions do.
Two gases with different specific heats will have two different temperatures when placed in the same blackbody radiation field.SpaceTiger said: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.
What is your definition of thermodynamic equilibrium as it applies to pure radiation?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.
What is your definition of thermodynamic equilibrium as it applies to pure radiation?
[PLAIN said:http://www.grc.nasa.gov/WWW/K-12/airplane/thermo0.html]The[/PLAIN] [Broken] zeroth law of thermodynamics begins with a simple definition of thermodynamic equilibrium . It is observed that some property of an object, like the pressure in a volume of gas, the length of a metal rod, or the electrical conductivity of a wire, can change when the object is heated or cooled. If two of these objects are brought into physical contact there is initially a change in the property of both objects. But, eventually, the change in property stops and the objects are said to be in thermal, or thermodynamic, equilibrium. Thermodynamic equilibrium leads to the large scale definition of temperature.
Andrew Mason said:Two gases with different specific heats will have two different temperatures when placed in the same blackbody radiation field.
The laser source will reach a stable temperature and emit a blackbody spectrum typical of that temperature. It will also emit a swack of identical photons whose frequency depends on the energy level of the excited electrons in the laser material. If blackbody radiation has a temperature, why don't the laser photons?
As Chronos points out, the definition of temperature is somewhat elusive.
I think temperature is simply the sum of kinetic energy exchanges between fermions in a coordinate system.Andrew Mason said:Two gases with different specific heats will have two different temperatures when placed in the same blackbody radiation field.
What is your definition of thermodynamic equilibrium as it applies to pure radiation?
The laser source will reach a stable temperature and emit a blackbody spectrum typical of that temperature. It will also emit a swack of identical photons whose frequency depends on the energy level of the excited electrons in the laser material. If blackbody radiation has a temperature, why don't the laser photons?
As Chronos points out, the definition of temperature is somewhat elusive.
AM
Chronos said:I think temperature is simply the sum of kinetic energy exchanges between fermions in a coordinate system.
Yes, light does have a temperature, although it may not be in the traditional sense that we think of temperature. Light is made up of photons, which are particles that carry energy. This energy can be measured in terms of temperature, but it is not a physical temperature like we experience with objects.
The temperature of light is measured using a unit called Kelvin (K). This unit is based on the absolute scale of temperature, where 0 K represents the complete absence of thermal energy. The temperature of light can also be measured using the Planck's law, which relates the temperature of an object to the intensity of light it emits.
Yes, the temperature of light does affect its color. This is because the color of light is determined by its wavelength, and the temperature of light affects the wavelength of the light. For example, a higher temperature light source will emit more blue light, while a lower temperature light source will emit more red light.
Yes, light can have a negative temperature. This is known as negative temperature, which occurs when the majority of particles in a system have more energy than the minority of particles. This is a rare occurrence and is usually only seen in highly specialized laboratory conditions.
The temperature of light does not affect its speed. The speed of light is a constant, which means it does not change regardless of the temperature. However, the temperature of the medium through which light is traveling can affect its speed, as light travels slower through denser materials.