Does light have a temperature?

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  • #1
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?
 

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  • #2
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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  • #3
samalkhaiat
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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.
 
  • #4
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.
 
  • #5
Nenad
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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,

Nenad
 
  • #6
SpaceTiger
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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.

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).
 
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  • #7
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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:

[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).

Very True.
 
  • #8
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:

[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.

AM
 
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  • #9
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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  • #10
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
 
  • #11
Tide
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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.
 
  • #12
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?
 
  • #13
partha_26786794123
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reply to Andrew Mason

yup , light have temp. , i believe 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.
 
  • #14
Edgardo
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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
 
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  • #15
Antiphon
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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).

http://www.physics.rutgers.edu/ugrad/351/Lecture 25.pdf
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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  • #16
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
 
  • #17
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
 
  • #18
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.
 
  • #19
Tide
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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.
 
  • #20
Andrew Mason
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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
 
  • #21
LeonhardEuler
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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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  • #22
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.
 
  • #23
samalkhaiat
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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?
 
  • #24
LeonhardEuler
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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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  • #25
SpaceTiger
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samalkhaiat said:
Now,hit a thermometer with a laser, the reading will certainly change. So what do you call that reading?

That reading is the temperature that the mercury (or whatever other device you're using) reaches after gaining the energy it absorbed from the laser. Can you then say that the laser has that temperature? No! If you shine that same laser (for the same amount of time) on another object, it will reach a different temperature. This is because what you're really measuring is the amount of energy transferred to the object you're shining it on. Another way of saying this is that your definition of temperature doesn't satisfy the zeroth law of thermodynamics. It's true that the photons in the laser could come into thermodynamic equilibrium with the target, but they would have to be placed in a closed system in which there was no communication with the outside world and they could freely exchange energy until settling into equilibrium.

As you said in your first post in this thread, temperature is a statistical concept. True enough, in astrophysics as well, we often speak of "equivalent temperatures" or "effective temperatures" in which we equate some measure of energy to one of temperature. This is good for back-of-the-envelope arguments, but it does not mean that they are true temperatures in the thermodynamic sense.
 
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  • #26
Chronos
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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]. To paraphrase Einstein - temperature is what thermometers measure.

But physicists often convert the frequency of photons into their energy equivalent in units of electron volts using Planck's Law. And the temperature equivalent of 1 eV is 11,605 K. Most physicists would routinely respond to a question about the temperature of a photon in this fashion, even though they are well aware a photon has an equivalent mass and temperature, but does not intrisically possesses either property in the classical sense. You can't put a photon on a scale, or stick a thermometer in its mouth. A photon, like it's fellow bosons, is only observable through it's interactions with matter.
 
  • #27
LeonhardEuler
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I don't think anyone is talking about the temperature of a photon, but the temperature of a group of photons, just like one does not speak of the temperature of an atom, but of a gas. Please help me understand what you mean by 1eV being equivalent to 11605K. Equivalent in what sense? The units don't match up, so they can't be equal.
 
  • #28
SpaceTiger
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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].

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.
 
  • #29
SpaceTiger
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LeonhardEuler said:
Please help me understand what you mean by 1eV being equivalent to 11605K. Equivalent in what sense?

He's just dividing the energy by Boltzmann's constant. Classically, a gas in equilibrium has an average energy per degree of freedom of

[tex]E=\frac{1}{2}kT[/tex]
 
  • #30
Chronos
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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.
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:
http://www.temperatures.com/wit.html
I really didn't give boson scattering much thought. I think it's fair to say it's only detectable by virtue of its effect on fermions. This is true of all bosons, so far as I know - we can only measure their properties in terms of their effect on fermions. In that sense, there is no need to extend the concept of temperature to bosons since we can only talk about it in terms of fermionic reactions. Anyways, since only fermions make useful thermometers, I thought 'temperature is what thermometers measure' is a simple, but effective definition.
 
  • #31
Andrew Mason
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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.
Two gases with different specific heats will have two different temperatures when placed in the same blackbody radiation field.

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?

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
 
  • #32
SpaceTiger
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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.

The above definition I think applies equally well to radiation as to matter. The only possible difference is that if you took two fields of radiation, say the CMB and the thermal radiation from a neutron star, and put them in a region of space, they might never interact at the level that they can acquire the same "property" (temperature). The reason I say it's only a possible difference is that I don't think it's known whether photons can interact at the level required to reach equilibrium. If so, it would take a finite (though admittedly enormous) amount of time for them achieve the same "property" and they would satisfy the definition entirely. Even if this is not the case, then one could still make the argument that the photons are not in "physical contact", and are therefore not violating the definition.

However, there is no question that when put into thermal contact with matter in a closed system, the photons will reach equilibrium with the matter. To me, it makes the most sense to think of temperature in terms of statistical equilibrium and, regardless of whether or not the photons do interact with one another, I would say they can have a temperature.


Andrew Mason said:
Two gases with different specific heats will have two different temperatures when placed in the same blackbody radiation field.

Not once they've reached equilibrium. The specific heat tells you how much energy is required to change the temperature of an object (per unit mass), it does not tell you about the final equilibrium temperature in such a situation. In the example of shining a laser on an object for a certain amount of time, the final result would depend on the specific heat because you're giving it a set amount of energy. In the case of embedding a body in a blackbody radiation field, if given enough time, it can take any amount of energy it needs to reach equilibrium with the radiation. When the input energy per unit time is equal to the output energy per unit time, the body is in eqilibrium and will no longer change its temperature. The time it takes for this to occur will depend on the specific heat, but the final temperature will not.



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?

Lasers require non-equilibrium conditions to be created. In an atomic gas, the atoms are not in complete equilibrium unless the distribution of their energy levels is given by the ratios of the Boltzmann factors (see http://www.chemsoc.org/exemplarchem/entries/pkirby/exemchem/Boltzmann/Boltzmann.html [Broken], for example). If the energy levels of the source atoms satisfied this criterion, stimulated emission could not occur and there could be no population inversion. Thus, if you put the laser and its target in a closed system (this is crucial for equilbrium to occur), the source of the laser emission would eventually reach an equilibrium population of energy levels and, at this point, would begin to emit a blackbody spectrum rather than a monochromatic beam.



As Chronos points out, the definition of temperature is somewhat elusive.

Yes, I will certainly concede that point. Like I said earlier, I think we'll just have to agree to disagree, since the physics community doesn't seem to be in agreement about the definition either.
 
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  • #33
Chronos
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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
I think temperature is simply the sum of kinetic energy exchanges between fermions in a coordinate system.
 
  • #34
SpaceTiger
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Chronos said:
I think temperature is simply the sum of kinetic energy exchanges between fermions in a coordinate system.

I haven't seen that definition anywhere, including on the site you linked. Care to elaborate? Did you perhaps mean the average value of the translational kinetic energy of the particles? The latter could certainly follow from kinetic theory.
 
  • #35
Chronos
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ST is entirely correct. It is the average, not sum. as I erroneously stated. That illustrates how easy it is to tell the difference between a post doc and a neophyte drowning in his own ignorance.
 
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