Does light have a temperature?

[J.C.MacSwell]

Right. Same temperature, but fewer photons. Temperature tells you the average energy of the ensemble. So just like a drop of water and a bucket of water can be at the same temperature, the bucket contains more thermal energy because there's more water. If you added a fixed amount of energy to either, the drop's temperature would go up more.

 

LOL, I am proving my points made in post 11 about my lack of grounding in this subject a little stronger and faster than expected!

 

Thought experiment:

 

We have a one cubic meter black body cube with a "sun" 93 million miles away and perpendicular to each face. (So we have 6 "suns")

 

When equilibrium is reached (constant temperature for the cube) what will be the cube's temperature?

 

And of course the follow up/more important question would be: Why would that temperature not be considered the temperature of the unobstructed sunlight at that point?

[YT2095]

Hmmm...

 

both have the same freq, and the same width of beam, over the same distance, so if both lasers are the same power both will be the same (ignore the semantics and QM stuff) both are effectively the same, now if One is the same as the other in all aspects but has 2x the power (in Joules) there will be more photons hitting per second from it, is this correct?

[x__heavenly__x]

wow,

is it 1250000000000000000 (125e16)??? photons per second!!!

but how manytimes a laser can reflect before it looses its energy?

[swansont]

the picture here' date=' with the laser, is more analogous to the box with a candle burning at one end and an ice-cube melting at the other----the over all temperature of the system is not well-defined. lasers are very non-equilibrium-type creatures. (swansont please correct if wrong)

[/quote']

 

 

The temperature can be well-defined: it's negative.

 

That's what you get when you have a population inversion (at least in two-state or other simple systems). But it's very non-equilibrium, to be sure.

[swansont]

Secondly, the milliwatt rating of the laser beam unit would not be an accurate estimator of the number of photons in the beam either. These numbers (usually on the housing of the laser unit) are indications of power consumption and suppy alone, and give no indication of the efficiency of the laser design.

 

No, the power output is generally the amount of light. That's been the convention for every laser I've ever used - it makes it much easier to classify the eye safety issues, since you don't then have to worry if your "1 W" laser is 50% efficient or 0.5% efficient to make your choice on whether or not safety goggles are necessary. (500 mW definitely yes, 5 mW generally no)

[x__heavenly__x]

So can there be an infinite reflection with good lasers and isolated reflectors??

[YT2095]

wow' date='

is it 1250000000000000000 (125e16)??? photons per second!!!

but how manytimes a laser can reflect before it looses its energy?[/quote']

 

I was thinking more along the lines of something Not pulsed, like a HeNe laser for example, with say a 20% permeability.

 

does doubling the output power increase the photon count, if all other factors like wavelength and beam dia remain the same?

[J.C.MacSwell]

I was thinking more along the lines of something Not pulsed' date=' like a HeNe laser for example, with say a 20% permeability.

 

does doubling the output power increase the photon count, if all other factors like wavelength and beam dia remain the same?[/quote']

 

It would have to double it. (Macswell getting brave again)

[x__heavenly__x]

better if more consistant. but as a general rant, how much energy does a photon(monochrome or white)(or a group of photons) looses after it reflects from:

1. a white surface

2. surface the color of the laser.

3. opposite color of the laser.

[YT2095]

better if more consistant. but as a general rant' date=' how much energy does a photon(monochrome or white)(or a group of photons) looses after it reflects from:

1. a white surface

2. surface the color of the laser.

3. opposite color of the laser.[/quote']

 

although your post is reasonably short, I fear the answer to ALL these permutations will be quite lengthy!

especialy when you take into account the color "White" as it`s a mix of All colors and each will give a differing result!

[x__heavenly__x]

ok in general tearms, if a red laser (which we get in stores) is made to reflect once off a totaly reflecting surface(practically not possible, u told me this long time back thanks)

then is it possible to calc the ammount of energy lost?

[Martin]

...One is the same as the other in all aspects but has 2x the power (in Joules) there will be more photons hitting per second from it, is this correct?

 

essentially correct. (I have to duck in and out very quickly here YT because swansont is knocking heads together and I dont want to be caught in the fray)

 

if two lasers are the same in all respects except one has twice the power (joules per second, watts, whatever)

then the more powerful one delivers twice as many photons per second

 

you said you werent going to worry about semantics but joules is a measure of energy, a single pulse of a laser could be measured in joules or microjoules.

 

power is the rate that energy is delivered, so and so much joules per second.

 

we want YT to be a shining example of semantic rectitude, and so we correct him when he says " in all aspects but has 2x the power (in Joules) "

 

now let us get back to watching swansont

[swansont]

So can there be an infinite reflection with good lasers and isolated reflectors??

 

No mirror is 100% reflective.

[swansont]

I was thinking more along the lines of something Not pulsed' date=' like a HeNe laser for example, with say a 20% permeability.

 

does doubling the output power increase the photon count, if all other factors like wavelength and beam dia remain the same?[/quote']

 

As Martin said, by definition doubling the output power doubles the photon count.

 

The thing is that you don't necessarily get that by doubling the input power. There are a lot of nonlinear processes going on in a laser.

[swansont]

ok in general tearms' date=' if a red laser (which we get in stores) is made to reflect once off a totaly reflecting surface(practically not possible, u told me this long time back thanks)

then is it possible to calc the ammount of energy lost?[/quote']

 

Energy lost to what? If you approximate it as a perfect reflector, there are no absorption or transmission terms.

[MetaFrizzics]

No, the power output is generally the amount of light. That's been the convention for every laser I've ever used - it makes it much easier to classify the eye safety issues, since you don't then have to worry if your "1 W" laser is 50% efficient or 0.5% efficient to make your choice on whether or not safety goggles are necessary. (500 mW definitely yes, 5 mW generally no)

This is quite unusual. Perhaps the descrepancy can be accounted for by the fact that a multitude of lasers in university and research environments are hand-made or manufactured and supplied directly to institutions. The normal CSA regulations and Safety Standards ratings will require power consumption, not laser efficiency to be posted on any device designed to plug into commercial A.C. power, and it must pass other tests and requirements as well.

 

These requirements could not in any way be waived or altered in the case of commercial lasers. So either there are *two* sets of labels or a more detailed description of the laser model in the accompanying operation manual / sales literature, than simply the CSA label required for any Electronic or Radio Frequency device that uses electric power.

[swansont]

This is quite unusual. Perhaps the descrepancy can be accounted for by the fact that a multitude of lasers in university and research environments are hand-made or manufactured and supplied directly to institutions. The normal CSA regulations and Safety Standards ratings will require power consumption' date=' not laser efficiency to be posted on any device designed to plug into commercial A.C. power, and it must pass other tests and requirements as well.

 

These requirements could not in any way be waived or altered in the case of commercial lasers. So either there are *two* sets of labels or a more detailed description of the laser model in the accompanying operation manual / sales literature, than simply the CSA label required for any Electronic or Radio Frequency device that uses electric power.[/quote']

 

I've actually never made a laser - they have all been commerical devices. The thing is many systems are modular and very few have a standard power connection (A HeNe is an exception here). The power supplies have the expected labels on them, but all of the lasers themselves most definitely are labeled with the optical power output.

 

All I can conclude here is that you don't work with lasers much.

[Klaynos]

I'm wondering why when classifying lasers you would care how much energy they use? Surely output is far more important? Every laser I've ever come accross whether little pointer lasers or ones for experimental use have all been classified by their output power, as you'd expect. I don't even see how power consumption could ever be considered important except for general electical safter (emergancy off switches and the like)...

[Martin]

...

 

All I can conclude here is that you don't work with lasers much.

 

I conclude the same about Friz. Having also a modest experience of lasers.

[J.C.MacSwell]

LOL' date=' I am proving my points made in post 11 about my lack of grounding in this subject a little stronger and faster than expected!

 

Thought experiment:

 

We have a one cubic meter black body cube with a "sun" 93 million miles away and perpendicular to each face. (So we have 6 "suns")

 

When equilibrium is reached (constant temperature for the cube) what will be the cube's temperature?

 

And of course the follow up/more important question would be: Why would that temperature not be considered the temperature of the unobstructed sunlight at that point?[/quote']

 

Observations:I am assuming the cube is not at the same temperature as the sun

 

1. I have the same amount of energy crossing each square meter "face" in each direction. (not that the "exiting" photons all exit perpendicular to the face). So at least in some respect an equilibrium condition has been reached.

 

2. I can see my "sun's". In fact I would burn my eye's out if I viewed one unobstructed from any "face" yet if I turned around and stared at the cube I would certainly be fine.

 

3. More photons are "exiting" than entering the cube, though at a lower energy/frequency. In fact the "inputs are at the same frequency signature as they left the sun.

 

4. The "inputs" to the cube are at a lower entropy than the outputs. The inputs are the "cause" and the outputs the "effect", and not the other way around.

 

 

I'm slowly getting it, but I haven't yet jumped out of the bath and run naked down the street shouting "Eureka"!

[J.C.MacSwell]

OK, I think what I have set up across each "face" is an example of a system in "steady state" and not an example of thermal equilibrium at all.

[GeminiinimeG]

isnt temperature the measure of the ammount of energy in something? Lazers with higher frequencies must have more energy coming out of em and must also be different color because diferrent freqencies of light are recognized by us as the different colors of the so called spectrum

 

 

 

(Infra Red ) ( Red ) ( Violet ) (Ultra Violet)

<The frequency changes through the spectrum>

[Klaynos]

Firstly it is LASER, it is an acronym for: Light Amplification by Stimulated Emission of Radiation

 

Temperature is a measure of the average kinetic energy of the particles in a sample of matter, expressed in terms of units or degrees designated on a standard scale. As a photon has no mass it has no KE and therefore no temperature.

 

That is a poor definition of Tempreature, and you really need to read up on the zeroth law of thermodynamics to get a good definition... :|

[swansont]

Firstly it is LASER' date=' it is an acronym for: Light Amplification by Stimulated Emission of Radiation

 

Temperature is a measure of the average kinetic energy of the particles in a sample of matter, expressed in terms of units or degrees designated on a standard scale. As a photon has no mass it has no KE and therefore no temperature.

 

That is a poor definition of Tempreature, and you really need to read up on the zeroth law of thermodynamics to get a good definition... :|[/quote']

 

Photons most certainly have energy - it's all they have. It's just not the classical 1/2 mv². As we've gone over, light from a laser is not in thermal equilibrium, but light from a blackbody can be said to have a temperature.

[Klaynos]

Photons most certainly have energy - it's all they have. It's just not the classical 1/2 mv^{2[/sup']. As we've gone over, light from a laser is not in thermal equilibrium, but light from a blackbody can be said to have a temperature.}

 

^{My appologies I'm not thinking this morning}