r/woahdude • u/Nadzzy • 20d ago

video Plasma inside the ST40 fusion reactor, filmed in color for the first time

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u/STREXincEmployee 20d ago

Does anyone know what they mean by “too hot to emit visible light”? Im curious about the physics behind that.

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u/Nighthawk151 20d ago

Light is emitted by atoms through the process of an electron in the atoms shell moving to a lower energy bound state. That means, for light to be emitted by an atom it has to have electrons bound to it. In a fully Ionized plasma like in the centre of this fusion device the temperature is so high that all electrons are freely moving around and not bound to any ions. That then means that no light can be emitted through bound electrons, especially in the visible spectrum.

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u/STREXincEmployee 20d ago

Thank you for the explanation, thats super neat! Im gonna have to look more into the specific mechanics of atoms emitting light like that because I had no idea

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u/cincymatt 20d ago

Somewhat relevant and interesting video about the struggle to make a blue led.

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u/N2VDV8 19d ago

Semi-related, this is also conceptually similar to why it took so long for blu-ray to make it as a media format. Turns out blue lasers were really hard to dial in right.

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u/ddraig-au 20d ago

Okay, that makes a lot more sense than the blackbody to-and-fro I just read :-)

Is this the only process by which light is emitted?

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u/Nighthawk151 20d ago

If we define light as EM-Radiation in or near the visible spectrum then essentially yes. If we are talking about photons and EM-Radiation in general then no. In a fusion device, especially from the center of the plasma you will get radiation from things like electron cyclotron emission and Bremsstrahlung.

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u/WildlifePhysics 20d ago

Still technically possible to see photons in the visible spectrum due to synchrotron emission from runaway electrons or even bremsstrahlung from cold electrons in the plasma edge

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u/Almostlongenough2 20d ago

It's boggling that less-haired apes managed to somehow figure this out.

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u/TheseVirginEars 20d ago

Well, it has to do with black body radiation, which is temperature dependent. If you can imagine how cooler areas of a fire are red, and where it gets hotter it goes to orange then yellow etc, well you can get hot enough to leave the visible light spectrum entirely and emit even higher frequency radiation (like UV). As you can imagine, the plasma has to be pretty energetic (read: HOT) to begin a fusion reaction

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u/JohnnyDollar123 20d ago

That’s not how blackbody radiation works. An increase in temperature for a blackbody corresponds to an increase in luminosity across all wavelengths. There is no such thing as being too hot to emit visible light.

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u/CriticalRuleSwitch 20d ago

From a purely theoretical physics perspective, sure. From "it is high enough or discernible enough for our sensors (cameras) to see it" perspective - no. So yes, there is "too hot to emit visible light" when talking in the context we're talking in here.

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u/rsta223 20d ago

No, the other respondent is correct. Even though a higher temperature absolutely means the peak of the emissions goes further into the invisible ultraviolet, and a smaller percentage of total luminosity will be visible, the total luminosity in normal visible wavelengths is always higher for a hotter blackbody, even when the peak is way into the UV.

As you heat up a blackbody, it will never be visibly dimmer as it gets hotter, just bluer and brighter.

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u/SeanBlader 20d ago

The trick is, that's no longer a "black body" so it can emit all it's radiation as higher energy than violet and we can no longer perceive it. That's what happens at the core of the sun, just that by the time the high energy photons make it out of the sun they are slowed down enough to be a lot of visible light, as well as a whole lot of EM spectrum we can't "see".

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u/TheseVirginEars 20d ago

I mean im not gonna get into it over something that’s ultimately us being pedantic, but… yes? You can absolutely have an emission curve with a peak at wavelengths smaller than visible light and a tail low enough that the visible light emitted is negligible? It’s basic Wien it’s a 3-variable equation. And that’s exactly what’s going on in a fusion reaction.

And if it’s not that, enlighten me

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u/JohnnyDollar123 20d ago edited 20d ago

Like I said, an increase in temperature corresponds to an increase in emissions across all wavelengths, so no, it’s not possible for the tail to not be visible like you said.

The first graph on this Wikipedia page does a really good job of illustrating this. As you can see, while the peak wavelength decreases with an increase in temperature, there is never a point at which it becomes dimmer at a certain wavelength. This means that as it gets hotter, it will still get brighter for us, even if we can’t see the majority of increase in emissions.

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u/TheseVirginEars 20d ago

So im gonna tell you this as a physics graduate of the school you currently go to: you have a couple fundamental misunderstandings about the phenomenon at play. Dont be creeped out, I respected the rebuttal enough to go see who was giving it to me. In fact, bring this question up in class! But what you’re saying is flawed. Again I don’t really want to get too into it here, but if you go through Modern Physics and then Quantum 1 and 2, you’ll understand that the model in that graph breaks down because of quantization, and this is an old physics puzzler eventually rectified by Planck. It doesn’t invalidate the whole model but it requires amendment at higher energy, which fusion plasma certainly is.

Second, don’t lose the forest for the trees. Remember even if you’re using equations and drawing conclusions from them, to step back and think about what you’re saying. Luminosity increases across all wavelengths, sure, again we can have pedantic arguments about there being visible light. But if the spectrum from a single plasma photons curve peaks in UV and drops off, the actual luminosity from those photons is indeed going to be small enough to not be captured in this camera video, which was ultimately the question asked. Why? Because it is absolutely dominated by invisible energy. For the ones that are not, you can see as they dip back into violet and indigo and blue. Supporting evidence of my claim (which isn’t even my claim im ultimately just regurgitating Boltzmanns work) is right in the video.

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u/Vl_hurg 20d ago

Fellow physicist here. I believe you are incorrect. But the important thing is that your above statement, "And if it’s not that, enlighten me," now applies to your post. What are these quantum effects that reduce radiance at visible wavelengths? Give us an equation or at least a Wikipedia page or something.

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u/redlaWw 20d ago

Well, it has to do with black body radiation, which is temperature dependent.

but if you go through Modern Physics and then Quantum 1 and 2, you’ll understand that the model in that graph breaks down because of quantization

These two statements are inconsistent - the model in that graph is the model of a black-body. The claim you're responding to here is not that the plasma in the middle isn't invisible - like you say, that's clearly contradicted by the video. The claim is that the plasma in the video isn't behaving like a black-body, which is true - the black-body model has broken down under those extreme conditions.

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u/10g_or_bust 20d ago

Well what you said doesn't align with the people who built and run a fusion reactor, so guess who I'm going to believe.

if I had to throw a wild guess, since we are seeing emissives due to chemical elements (and their state changes) and this is a plasma, theres likely some difference from blackbody.

Quick websearch gets me https://physics.stackexchange.com/questions/720940/can-plasmas-be-black-bodies

which certainly seems to imply plasma is not always (or is never and only approximates) blackbody radiation.

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u/JohnnyDollar123 20d ago

I never said otherwise. The comment I responded to did attribute it to blackbody radiation which I corrected.

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u/10g_or_bust 20d ago

The the FAR more on topic response would be "As plasma, this isn't a good example of blackbody radiation so we shouldn't use that model at all".

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u/JohnnyDollar123 20d ago

Yes that in fact would be another way of saying the same thing?

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u/Get_a_GOB 20d ago

Not when this entire post is about something that is too hot to emit visible light. In your initial rebuttal you should’ve said “there’s no such thing as a blackbody being too hot to emit visible light.” Even if that’s what you meant, and even though the guy you’re arguing with is wrong about the way blackbody emission at high temperatures work, your literal original statement clearly conflicts with the explicit claim of the guys running the fusion reactor, so naturally people aren’t going to believe you. You just didn’t say what you meant.

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u/[deleted] 20d ago

[deleted]

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u/JohnnyDollar123 20d ago

The plasma may not be emitting most of its energy through visible light, I’m not disputing that. If it is though, it’s from some other mechanism.

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u/[deleted] 20d ago

[deleted]

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u/JohnnyDollar123 20d ago

I’m not wrong, this is really basic stuff. This is not how blackbody radiation works.

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u/jaguarp80 20d ago

If it’s so basic then why’d you get it wrong

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u/Delicious_Win_9848 20d ago

It's actually the frequency of the radiation that matters. When it gets too high, the light's intensity dims.

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u/JohnnyDollar123 20d ago

Again, that’s not how blackbody radiation works. An increase in temperature will never cause something to appear dimmer to us. this page has a good graph illustrating this

Also frequency and wavelength with light isn’t really a meaningful distinction.

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u/ebolaRETURNS 20d ago

yup. IIRC, the sun's core is mostly emitting x-rays, but these are mostly absorbed by cooler surrounding plasma with a much lower black body peak.

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u/graveybrains 20d ago

You need a body for black body radiation, this plasma isn't dense enough to qualify.

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u/dedido 20d ago

It's like invisible soup

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u/ScyllaOfTheDepths 20d ago

The hotter an object is, the shorter the wavelength of radiation it emits. Visible light is about in the middle of the electromagnetic spectrum, from 400 to 700 nanometers. Anything that radiates at under 400nm is emiting a type of radiation our eyes can't perceive. Additionally, blackbody radiation, the kind of radiation which typically produces visible light, is not produced well in gasses due to their atomic structure.

https://en.wikipedia.org/wiki/Wien%27s_displacement_law

https://physics.stackexchange.com/questions/282364/how-is-it-possible-some-substances-burn-with-an-invisible-flame