this post was submitted on 11 Aug 2026
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[–] Eheran@lemmy.world 2 points 2 days ago (1 children)

You are mixing different types of IR and on top are wrong for both cases too. We have materials for both NIR and thermal IR that absorb or reflect them.

[–] Cocodapuf@lemmy.world 0 points 2 days ago (1 children)

I've been wrong before, this wouldn't be the first time. But what part of it was wrong? Genuinely, can you clear it up?

As far as absorbing IR light, I'm not even sure how that would work, wouldn't it just be emitted again as heat?

[–] Eheran@lemmy.world 1 points 16 hours ago (1 children)

So if we focus on thermal IR: shiny metal surfaces hardly emit or absorb it. A black surface emits and absorbs as much as possible. We quantify this via the emissivity (and also reflectivity, shiny metal is a mirror). Emissivity of 1 = theoretical maximum of a prefect absorber/emitter (both are directly linked), everyday stuff like skin or wood is about 0.95, shiny metal surfaces roughly 0.1, a rusty or painted surface again about 0.95.

You only emit and absorb the same in thermal equilibrium. If the surface is hotter, it emits more. If it is colder, it absorbs more.

[–] Cocodapuf@lemmy.world 1 points 5 hours ago

So with that last part, I get that in theory, but practically, you can only really absorb IR temporarily, right? Because it all gets released again as heat eventually. Like, say you have a mostly inert object, that isn't creating any meaningful amount of heat. If you hit that object with 10 joules of IR, over time, wouldn't nearly all of that energy be either reflected or emitted off it as IR radiation?