IR Laser makes glowing spot on radiometer vane

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SUMMARY

The discussion centers on the behavior of an infrared (IR) laser when directed at a radiometer vane, specifically addressing the phenomenon of the glowing spot produced on the metal surface. Participants analyze the thermal effects, questioning the roles of conduction, convection, and radiation in this process. Key points include the assertion that convection diminishes the effect in air, while the thinness of the vanes may limit conduction cooling. The Stefan-Boltzmann law is referenced to calculate the temperature based on the laser's output power and dot size, emphasizing the need to consider absorbed power rather than incident power.

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Swamp Thing
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Is the explanation at the end of the video correct?

 
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What's the explanation ? maybe give a timestamp. First things that come to mind are either a) gets hot, glows red, or b) frequency upshift.
 
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@Swamp Thing , it's not like you are new here. "Here's a video, 'splain it to me" is a bad thread start. You've been here long enough to know what you get out of PF is proportional to what you put into PF.
 
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Vanadium 50 said:
@Swamp Thing , it's not like you are new here. "Here's a video, 'splain it to me" is a bad thread start. You've been here long enough to know what you get out of PF is proportional to what you put into PF.
Sorry, yes the question should have included more detail...

At 03:15 he points the laser at his palm, apparently without feeling the least bit of discomfort. Yet the laser is able to produce a white-hot incandescent spot on the metal.

Moreover, I would have thought that the metal would have a bit of thermal time constant when cooling down, at least enough to produce a short trail as the spot moves around over it (likely going through white, orange and red while cooling).This does not seem to be the case.

Finally, he is saying that convection can pretty much kill the phenomenon but conduction and radiation don't matter enough to prevent it.

These points led me to wonder if a non-thermal explanation should be considered, such as the one he mentions earlier.
 
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Conduction requires cross-section ; those vanes look pretty thin. I'm not sure why "radiation doesn't matter" : where does he think the glow is coming from ?
 
hmmm27 said:
Conduction requires cross-section ; those vanes look pretty thin. I'm not sure why "radiation doesn't matter" : where does he think the glow is coming from ?
He says the laser is heating the metal to incandescence. He also says that convection is the reason that it doesn't work in air.

The "radiation doesn't matter" is not something that he actually says, but it follows if we accept the hypothesis that it is a heating effect. If radiation mattered, then it wouldn't have glowed even in vacuum.

If the thinness of the vanes is helping to eliminate conduction cooling, then it should also slow down the cooling process when the laser spot moves away from a given position, and we should then see a cooling trail as the spot moves around. But we don't see a trail.
 
Swamp Thing said:
If radiation mattered, then it wouldn't have glowed even in vacuum.
Unless you've perfect conduction, it's going to glow. Whether that glow extends far enough into the visible spectrum to see is another thing.

If we can ignore conduction and make enough other gross simplifications and assumptions to make a few eyes twitch here and there, we can stick the output power and dot size of the laser pretty much directly into the Stefan Boltzmann law to get the temperature.
$$T=\left( \frac P{A\sigma}\right)^{\frac14}$$
dimensions:
##T## Kelvin
##P## Watts (= Joules per second)
##A## square meters
##\sigma## Stefan-Boltzmann Constant :
##5.670374419...\times10^{−8} W⋅m^{−2}K^{−4}##
 
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hmmm27 said:
If we can ignore conduction and make enough other gross simplifications and assumptions to make a few eyes twitch here and there, we can stick the output power and dot size of the laser pretty much directly into the Stefan Boltzmann law to get the temperature.
I think P needs to be the power absorbed and so will be incident power minus reflected power.
 
tech99 said:
I think P needs to be the power absorbed and so will be incident power minus reflected power.
Oh, definitely, but that would require knowing the composition of the leaf (which I imagine is carbon-deposited aluminum, but really have no clue).
 

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