Undergrad Can somebody explain this: Planck's Law in action

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The discussion centers on the implications of Planck's Law regarding high-temperature plasmas in fusion reactors. Participants express skepticism about claims of achieving temperatures of 1,000,000 K, arguing that such conditions would produce lethal levels of radiation, contradicting safety measures in place. They highlight that confined plasmas emit various types of radiation, not solely blackbody radiation, and emphasize the importance of understanding the relationship between energy input and thermal energy output. The conversation also touches on the complexities of measuring plasma temperatures and the potential misconceptions surrounding emissivity and absorption in different plasma states. Overall, the consensus is that the reported high temperatures in fusion experiments raise significant concerns regarding energy conservation and radiation safety.
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When you calculate a plasma with 1.000.000 K then you get according to planck's formula a plasma, which emits x-Ray / Gamma-Radiation in a quite deadly intensity. Can somebody explain all this ultra-high temperatures posted in all the fusion experiments?
Plotted is the Irradiance over Wavelength.
Please check for logarithmic scaling.
As you can see, there are 4 curves.
Blue AM 0 as measured
yellow Planck for 5777 K
green Planck for, 5777 K after free space expansion
red Planck for 1.000.000 K

To me the idea of a gamma-Ray-source on earth, below the magnetic field, which protects life on earth from solar radiation, in an intensity, which is way way way outer hand, makes no sense to me. If they really get these high temperatures realized in fusion reactors, we would, according to Planck, already been dead.

nuclear fusion.webp
 
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Matthias_Rost said:
TL;DR: When you calculate a plasma with 1.000.000 K then you get according to planck's formula a plasma, which emits x-Ray / Gamma-Radiation in a quite deadly intensity. Can somebody explain all this ultra-high temperatures posted in all the fusion experiments?
Interesting question- one I had not thought of.

First, confined plasmas have a variety of radiation emission mechanisms, not just blackbody but also synchrotron and "bremsstrahlung". There are also emitted neutrons in addition to photons. I did find a brief summary of the safety measures in "Fundamentals of Magnetic Thermonuclear Reactor Design":

"The MFR technological and biological radiation protection is performed by a combination of the blanket and special structural components.
In addition to fusion energy utilisation and tritium breeding, the blanket provides a considerable attenuation of the plasma radiation flux. Physical protection components that absorb neutrons and gamma radiation from nuclear reactions in structural materials are located behind the blanket. Behind the radiation shielding, the parameters of the irradiation effect (particularly onto the magnet materials) must not exceed the following specified limits:
• total absorbed dose for insulators: not higher than 5 × 107Gy,
• nuclear reaction thermal power absorbed by a superconductor: within 10 kW,
• fast neutron fluence on superconducting coil: within 1019n/cm2.
The shielding material should contain light elements acting as neutron moderators and elements with large atomic numbers absorbing the gamma radiation. The well-reputed heterogeneous iron-and-water medium is generally used for this purpose. Where a thin shielding is necessary, an advanced material based on, for example, zirconium hydride can be used.

In the DEMO and FPP projects, the blanket + shielding thickness is close to 1 m. It is the key component in the gap between the plasma and the TF coil. Approximately 2-m-thick concrete bioshield is used to protect personnel."

Interesting stuff!
 
Yeah sure, quite interesting. I met some guys online, who worked at I guess Korean fusion Laboratory. And they said: "We suppress thermal radiation." And I said: "how?". No answer.

See, the main problem in understanding this is sure there are many types of radiation of an ultra hot plasma, when it's rotating in an accelerator. Formally, synchrotron radiation is one kind of decellaration-radiation or called "bremsstrahlung". And yeah, 1 m lead can absorb many of the radiation. But that's not the point. I think there's a misconception. See the Energy, with which they pump the reactor, is one thing. But the true temperature they reach, so the thermal energy is something different. Since thermal energy is defined by Maxwell-Boltzmann-Distribution. And therefore it accounts only relative movement of particles and not aligned movement of particles together in a circle.

So when you now equate the magnetic pumping energy with the kinetic energy derived from Maxwell-Boltzmann, you get quite high temperatures. E_magn = E_kin = 3/2 k_b*T. Truth is, that the relative movement in a plasma of particles, which have a velocity of nearly 1/3 of c in a magnetic field, which is nearly homogenous is almost zero.

That's why I don't believe all these published high Temperature values, unless somebody posts a serious temperature measurement. Doesn't have to be a protocol. But as far as I know after 5000 K every material is at least molten. So the only way to measure the surface temperature of an ultra hot plasma would be optical, like you do at the sun. And how such a spectrum should look according to Planck, I plotted as the red curve.

And I still don't get how to be able to suppress thermal radiation. I mean, this would contradict the 2nd law of thermodynamics. Since, Planck is a quite complex derivative of this fundamental law. So I don't get these guys at all.
 
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Matthias_Rost said:
..
Plotted is the Irradiance over Wavelength.
Please check for logarithmic scaling.
..
The absolute values of irradiance on your graph is applicable only to a enough large volume of plasma which absorb as a Black Body. Small volume would be transparent and have much much less emissivity.
 
Gleb1964 said:
The absolute values of irradiance on your graph is applicable only to a enough large volume of plasma which absorb as a Black Body. Small volume would be transparent and have much much less emissivity.

The Volume V doesn't matter at all. If you use a tungsten oxide wire and heat it up to 5777 K it will emit almost the same spectrum as the sun. Difference will only be the absorption bands of WO_3 and H/He. And the different prefactors for ε and A. And it's also not about transparency. Transparency would matter if you used a lamp to shine through a plasma. But this plasma is the lamp. Therefore, everything that matters is the surface between the plasma and the vacuum. I = ε*A*σ*T^4. also accounts for gray bodies, when ε<1.

You apparently didn't get Planck's law. For instance, what is ε?. ε(λ) =1-R(λ). So yeah, if the Reflectivity at the interface is greater than zero for the concerning Wavelength, then it's a gray body. Which for perpendicular emission can be written by Fresnel: R =[(n_1-n_2)/(n_1+n_2)]². Vacuum has an n of 1 and most gases have n around 1. And a plasma can sometimes have a higher n, but these plasmas in the reactors mostly have a low pressure, that's why they are not optically dense and that's why n is also small. Example: water --> air. R =[(1,5-1)/(1,5+1)]² =0,04. So (1-R) would be ε = 0,96.

Let's have a look at the blue and green curve. As you can see, there is almost no deviation between Planck using ε = 1,00 and the AM0 Spectrum. Plus, the sun is an optical dense plasma due to its high gravitation. So I would say ε ~ 1,00. The differences between blue and green curve don't originate from Emissivity/Reflectivity issues. These are the absorption bands of He, H, D and T.

See it is the standard procedure to measure temperatures of plasmas optically, because it's the only way. Have you ever seen a measurement of T from a fusion reactor, or do they only publish their calculated values?
 
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Matthias_Rost said:
..You apparently didn't get Planck's law..

What do you think the emissivity coefficient would be if the mean free path of photons (e.g., gamma photons) is much greater than the size of the volume? The emissivity would be about zero. For a black body, the mean free path of photons needs to be much smaller than the volume size, as in the case of a tungsten wire.
 
Matthias_Rost said:
The Volume V doesn't matter at all. If you use a tungsten oxide wire and heat it up to 5777 K it will emit almost the same spectrum as the sun.

Yes, but the amount of energy radiated by the wire will be much lower than the sun. The amount of energy you need to absorb to protect you from the amount of UV radiation emitted from a tungsten oxide wire is considerably less than required to block UV from the sun.
 
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It’s not the volume itself that matters, but the relationship between the volume size and the photon’s mean free path. That’s what I’m referring to. If radiation passes through a small volume with low absorption, the emissivity of that volume will be proportionally low.
 
Gleb1964 said:
It’s not the volume itself that matters, but the relationship between the volume size and the photon’s mean free path. That’s what I’m referring to. If radiation passes through a small volume with low absorption, the emissivity of that volume will be proportionally low.
Yeah, but that's internal wave propagation. Planck's on the contrary has nothing to do with internal wave propagation in a bulk medium. Planck's law only accounts for the emission of radiation from one object to another or in into free space.
Absorption can be calculated via Lambert-Beer:
Which is I/I_0=exp(-a*c*d).
Whereas I, is the intensity, I_0 is the starting intensity, a is the absorption coefficient, c is the concentration of the absorbance and d is the thickness of the material.

And btw. when light is absorbed, then the Energy conservation holds. Therefore, when a photon is absorbed, E_phot=h*v adds up to E_therm = 3/2*k_b*T. Therefore, the internal light Absorption has no consequence on Planck's law and therefore also not for Stefan-Boltzmann-Law. Since they only depend on Temperature T, Surface Area of the Object A_s, Stefan-Boltzmann-constant σ and the emissivity ε(λ) (= 1-R(λ) with R the reflectivity for surfaces, of course for internal propagation ε(λ) = 1-A(λ), with A the Absorption part, but that's a no-brainer)
I=ε*σ*A_s*T^4. See, the Irradiance does not depend on the Absorption or density of the medium at all, since it's the emission out of the body, which has nothing to do with internal wave propagation.

And we are far far far away from effects like the Casimir-Effect, where the mean free path length as a certain effect. But which holds only true, when the distance light has to overcome is smaller than the size of the wavelength. Then your right, then such effects can happen, so on a nanoscale then yes there some effects that might happen, but not for Objects in the range of thicknesses of some microns and above.
And yeah ε(λ)=α(λ) will still hold true even in that case. But that was never the point. You see.

For instance, if you look at the plot, than you can see a deviation from the blue to the green curve for small wavelength. Which is due to He, and H-absorption bands. But nevertheless, the Rest of the curve still holds true. So yep, I also accounted for absorption effects inside the volume. The point is, as you can see the sun can also have a higher emission than Planck, since when hydrogen absorbs energy it also has to be deposited somewhere, which will then either be at lower energies fulfilling Planck or at higher energies, due to some specific emission energy band, which is also in the plot. Point is, energy conservation holds true. And that is the main point I stress out. Temperatures of 1 Mio °C or above for manmade plasmas are physical nonsense, since they violate energy conservation and Planck's law. Because if they really ever had built any Plasma with 1 Mio °C in Temperature, we would all be dead yet. Plus, such a small Torus would emit more radiation than the whole sun per second. Now get it? It's bull. But as long as you can read it in the newspapers or on your smartphone, nobody ever will question it.
 
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  • #10
Planck’s radiation law applies to an ideal Black Body, where electromagnetic waves are absorbed and re-emitted multiple times within the material. This requires that the mean free path of the radiation is much shorter than the size of the volume, ensuring thermal equilibrium. If this condition is not satisfied, Planck’s law does not accurately describe the system.

Similarly, the Beer–Lambert law is valid only for media with low absorption. In highly absorbing materials, the assumptions behind the law break down, and it no longer provides reliable results.
 
  • #11
Matthias_Rost said:
Temperatures of 1 Mio °C or above for manmade plasmas are physical nonsense, since they violate energy conservation and Planck's law. Because if they really ever had built any Plasma with 1 Mio °C in Temperature, we would all be dead yet.
Seeing as we've achieved 500 million C and higher your statement is obviously nonsense. Unless, of course, everyone involved in plasma research and fusion energy is outright lying. But I doubt it.
 
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Matthias_Rost said:
The Volume V doesn't matter at all. If you use a tungsten oxide wire and heat it up to 5777 K it will emit almost the same spectrum as the sun. Difference will only be the absorption bands of WO_3 and H/He. And the different prefactors for ε and A. And it's also not about transparency. Transparency would matter if you used a lamp to shine through a plasma. But this plasma is the lamp. Therefore, everything that matters is the surface between the plasma and the vacuum. I = ε*A*σ*T^4. also accounts for gray bodies, when ε<1.
If you're suggesting that the plasma inside a reactor is emitting the same amount of radiation as a solid blackbody at the same temperature and size, then you are very wrong. A very, very thin and sparse medium at temp T will obviously emit less radiation than, say, a blackbody at that same temp T. You can easily test this at home. Heat up something until it's red hot (a stovetop burner will also work in a pinch) and then place your hand near it and feel how warm it is. Then move your hand up and place it just out of the hot air plume. The hot air emits so much less thermal radiation that you almost can't feel it.

Matthias_Rost said:
See it is the standard procedure to measure temperatures of plasmas optically, because it's the only way. Have you ever seen a measurement of T from a fusion reactor, or do they only publish their calculated values?
I don't know of every method used, but one is by using a Langmuir probe.

Matthias_Rost said:
That's why I don't believe all these published high Temperature values, unless somebody posts a serious temperature measurement. Doesn't have to be a protocol. But as far as I know after 5000 K every material is at least molten. So the only way to measure the surface temperature of an ultra hot plasma would be optical, like you do at the sun.
Nonsense. You can physically stick a probe into the plasma without it quickly melting. The heat transfer to the probe is much lower than you might expect thanks to the very low density of the plasma. This is also why you can stick your hand into an oven without burning it but as soon as you touch any metal inside you almost immediately get burned. The air is so much less dense than the metal that it simply cannot conduct heat into your skin fast enough to give you a burn. The density of plasma in a magnetic confinement reactor (about 10^-9 g/cm3 or less) is vastly less than that of the air in your oven, so even the extremely high temperatures will not immediately melt solid materials.
 
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Gleb1964 said:
Planck’s radiation law applies to an ideal Black Body, where electromagnetic waves are absorbed and re-emitted multiple times within the material. This requires that the mean free path of the radiation is much shorter than the size of the volume, ensuring thermal equilibrium. If this condition is not satisfied, Planck’s law does not accurately describe the system.

Wrong Planck only applies to black bodies, when epsilon = 1, then its a black body, otherwise it|s a grey body. So you then have to check for epsilon(lambda). This doesnt mean that physics is not applicable anymore. Since Planck can be directly derived from the 2nd law of thermodynamics. So as soon as you state, that Planck is not applicable for bodies with a) a high temperature, b) a high& low pressure, c) different color, d) thermodynamic gradients inside the material, you are wrong.

You dont need any multiple reabsorbtion to be inside thermodynamic equilibrium, so that Planck holds true. For instance. Take thin Glassrod, or even a thinner glassfiber. Standard Soda-Lime-Glass has a quite remarkable low thermal conductivity and a quite good transparency over a broad wavelength range. And you can colour it from the outside as you like and if you put some colouring elements like Fe, Cr, Ni, Ti inside the glassmelt, when recycling soda-lime-glass shards, you can also do this on the inside volume. And now you can do a trick. You take an IR-Camera and view the Glassrod from the Top, so zou only see the temperature variation in the cross section of the glassrod, at the edge where it was cut. And since the thermal conductivity is quite low you can also put the glassrod inside an oven, preferrably an inductively coupled oven with a platinum crucible and heat the crucible up. Then you will see, that the glassrod is definitvely not in thermal equilibrium, since it is cold inside and hot on the outside. And as long as you dont heat up over 400 degrees Celcius the glassrod wont melt. What do we learn about that? Well we can now use a spectrometer and measure the emitted spectrum. And guess what if paint the surface black, for instance with vanta black, works quite well or you also just a black Edding, which has almost constant reflectivity of 2%. Then you get a measurement curve and what comes out? Ah Plancks law. And yep it is then not consistent to the average Volume Temperature of the Object, since it|s not in thermal equilibrium. But guess what, its consistent to the surface Temperature.

And please I dont get it, why you dont get Plancks law. Its easy peasy to do. And again it doesnt matter, what the Volume does. Its a solely for surfaces of objects valid law. What do you think how every accredited calibration institute calibrates IR-Cameras or stuff like solar cells. They use a standard black body to calibrate for emission of objects. What does that look like? Its nothing more than a shoe carton painted on the inside with black color. And then the shoe carton is heated up. So what you measures is the temperature of the shoe carton and the emission of radiation through a circular hole inside the box. And what you get then, ahh Plancks law. I reallz dont get, whz zou do theoreticalliriye Plancks law. It ist fundamental and directly derived by 2nd law of thermodynamics. And please do me a favour and try to measure (in praxis) the spectrum of any body with known emissivity (= 1-Reflectivity) under conditions which violate thermal equilibrium. And then if you plot the data over the actual surface temperature, that Planck still holds. And this is not remarkable, this is just the consequence, that any surface is in thermal equilibrium at any time you like, just the volume is not.
Why has the devil killed his grandma? Ahh, because she had no excuses left to talk her self out of the affair.

Similarly, the Beer–Lambert law is valid only for media with low absorption. In highly absorbing materials, the assumptions behind the law break down, and it no longer provides reliable results.

What are the assumptions behind Lambert-Beer?
The assumption is, that you can divide any material into smaller parts.
Lets say we take a sheet of glass and divide (intellectually) in ten parts.
Then we assume, that absorption takes place linearly So I_0-I= a*c*d
Problem is, that for instance after the first sheet of glass 2% of light is absorbed.
So for the next sheet of glass I_1= 98%. So you would have 0.02*0.98 of light absorption.
And this continues for every layer of the objects, assuming that its homogenous.
So you get I=I_0*(1-a*c*d)^(number of layers). And this can then easily
be transformed from a potential into an exponential function (just by applying logarithm laws to switch to another base)

So please tell me, which assumption of all of these from Lambert-Beer will be not valid, when the absorption coefficient gets higher?
Answer non. But it is quite more complicated to measure proper results, when a*c*d is high as a kite.
And circumvent this problem you can easily dilute the Absorbance, so use a lower concentration. That helps for sure.

Come on these are basics in every physical basics practicum at every university.
We even had the Maths and assumptions of behind Lambert-Beer in Math Tutorium before our university education started seriously.
And Lambert-Beer cant ever break down, simply for logical reasons i already pointed out.

But I guess you simply have a different opinion.
 
  • #14
Drakkith said:
Seeing as we've achieved 500 million C and higher your statement is obviously nonsense. Unless, of course, everyone involved in plasma research and fusion energy is outright lying. But I doubt it.

Ahh, so you achieved it. How? And please tell how did you measure the temperature? I mean serious measurement. Not calculation. Because I know all your flawed calculations, since most of you didnt even get Maxwell-Boltzmann-distribution right. And therefore your calculations are most of the times pure nonsens (classical case of math a 1, calculus a 5). So please go on and show me a serious measurement protocoll of achieving temperatures of 500.000 degrees celcius. Thats all I wanted for now over 20 years. One measurment protocoll, that i can check and believe you. But as long as you (in german it is called schwurbeln) just dont come straight to the (G-)spot, I dont believe you anything. Believing is a thing for which you can go to church. Reality takes places on the outside.

So you said, that you achieved it, how can you be so sure. See if I want to heat up water in a bowl to 100 degrees celcius, i have to use thermometer for it, to check if its really 100 degrees celcius, boiling point. At least at 1 bar pressure. If you talk with an alpinist, then he will state in order to cook potatoes on a mountain you will need a pressurized cauldron, since you cant cook potatoes at 80 degrees celcius, which is approx. the boiling point of water according to the barometric formula. So again in order to be able to state that you achieved 500.000 degrees celcius as a temperature you will need a proper thermometer. And as I pointed out multiply and for decades, that the only way to measure temperatures above 5000 K is optically and I plotted how these measurements should look like and you simply dont show any proof of your high temperatures. Then if you just claim it, but in over 20 years never ever have shown any proof, then I guess your a fraud. And yeah there are many frauds (so called academics) calling themselves scientists, not even knowing what science means. Cheeers.
 
  • #15
Drakkith said:
If you're suggesting that the plasma inside a reactor is emitting the same amount of radiation as a solid blackbody at the same temperature and size, then you are very wrong. A very, very thin and sparse medium at temp T will obviously emit less radiation than, say, a blackbody at that same temp T. You can easily test this at home. Heat up something until it's red hot (a stovetop burner will also work in a pinch) and then place your hand near it and feel how warm it is. Then move your hand up and place it just out of the hot air plume. The hot air emits so much less thermal radiation that you almost can't feel it.

What is this for nonsense your talking about. You have no idea of the three mechanisms of heat transfer, do you? These are conductivity, radiation and convection. Air, can be as hot as it likes, since air almost always has a low heat capacity plus a very very low heat conductivity. A metal on the contrary has a little bit mor heat capacity but a way higher heat conductivity. So in order to get a stove to glow red, is roughly 800 degrees celcius. And you state i should touch 800 degrees celcius stove by hand, or you serious? This is dangerous so I give you a red card for improper behaviour.

Plus if you lets work at a superkanthal oven. It can easily achieve 1500 degress celcius. The cladding insinde the oven is mostly made of Al2O3, since it then is nearly a white body almost emitting no radiation. But as soon as you open the doors of this oven youll have to wear special clothing because of all the radiation which comes out. As you might know Stefan-Boltzman-law is I = epsilon*sigma*A*T^4. thermal conductivity on the other hand is just dQ/dt = lambda*A*(T_2-T_1)/d. So the flow of energy is proportional to lambda - heat conductivity, A - the surface, d - the distance to overcome. And The Temperaturedifference T_2-T_1. So conductivity goes linearly with rising temperature, radiation on the other hand goes with T^4. And you can easily calcculate the radiation is the dominant heat transfer process above 1000 degrees celcius, regardless of how small or large your epsilon is. Simply because hte heat transfer by radiation rises wwith the power of T^4.

I don't know of every method used, but one is by using a Langmuir probe.


Nonsense. You can physically stick a probe into the plasma without it quickly melting. The heat transfer to the probe is much lower than you might expect thanks to the very low density of the plasma. This is also why you can stick your hand into an oven without burning it but as soon as you touch any metal inside you almost immediately get burned. The air is so much less dense than the metal that it simply cannot conduct heat into your skin fast enough to give you a burn. The density of plasma in a magnetic confinement reactor (about 10^-9 g/cm3 or less) is vastly less than that of the air in your oven, so even the extremely high temperatures will not immediately melt solid materials.

Ah now I get it. you use a langmuir-probe with a small exposure to stick inside the plasma. Yep a langmuir-probe is sort of like a capacitor. And yep, if electrons fly around this object, then you can get a charge transfer from the one side of the capacitance to the other side, which gives a voltage. And yep Langmuir, was a bright scientist, who even calculated, the dependence of the voltage of the langmuir-probe to the electron temperature T_e. But T_e is not the Plasma-Temperature. Which can easily be explained by the fact, that electrons simply have a small mass in comparison to the nucleaus of an atom, even if it was just hydrogen. But since you claim your using Deuterium and Tritium the ratio is even worse. So what do you want to prove with your statement. A short time exposure of a probe is possible, and I never said something different.

I said to measure the temperature of a plasma with a physical object is impossible. Since temperature measurment with a physical object only is possible in thermodynamic equilibrium (Electron temperature on the contrary can also be measured transient but has no meaning to the plasma temperature). And most materials are either gaseous or already a plasma, when they reach 5000 K. Plus if your plasma is so diluted, that if you go into it with a langmuir-probe and it can stay there at these high temperatures, because a) of its high reclectivity and b) because the plasma is so diluted, that almost no heat transfer via thermal conduction takes place. The introduction of the langmuir-probe will surely act as a heat sink, which distorts zour measurement results. What sahll I say, dang happens.

So again the question, how do you measure a temperature inside a plasma? Answer. Not possible, you can only measure the surface temperature of a plasma. as it is done daily by satellites measuring the AM0 Spectrum of the sun using IR cameras. This is the easiest and most reliable way to measure the temperature of a hot plasma without interacting disturbing with it. Because the volume temperature, how do you wanna measure it? As soon as you bring a cold object into a hot plasma youll induce a heat gradient, distroting the plasma. That is bullshit.

And please dont write anything about measurement tools, if you have no idea how to use them and what they actually measure.
Because if you dont know what is Phase, then you should also avoid to become an electrician, seriously.

And thanks alot, that you try to mob me, because of my easy straight to the point jargon, proclaiming that i was stupid or something.
Guess what I got a PhD in applied Quantum Physics, where I used a Quantum Physical Measurment Routine and advanced it
to the point, where you could measure chemical elements and their corresponding reactions in a silicon substrate.
Most of them were equilibrium reactions, so you could trigger these reactions quite easily in the one or other direction.
So yeah I could prove, that the laws of thermodynamics and kinetics also applied for these ractions.
Why should you care? Because the concentration of these Elements was as low as 10^5 cm^-3.
And if you say now, wooh that is much, than you dont know anything about matter.
Since the density of silicon is roughly, (depends on temperature for instance) 2*10^22 atoms cm^-3.
So it was a chemicla analysis of reactions in the conentraions of ppq to ppp (parts per qudrillion to parts per pentillion)
To compare it an SEM with WDX detector can in the best case give you a resolution limit of 10 to 100 ppm
With Qudrupol-SIMS or TOF-SIMS you get into the ppt range. But even Sims is restricted.
So to get into the ppq Range of concentrations youll already have to use something like DLTS
And in the ppp Range, then you need T-IDLS, preferably with a self calibrating PL-Tool, sop you dont get so much Trapping-Measurement-Artifacts.

Or you could also try out my favourite measurement technique, wwhich is called TMBA (Too man bloody Acronymes)
Say it straight to the point an not sophisticating around, thats not scientific.

So Yep I kn ow what Im talking about. either in physics, chemistry or even biology and programming.;
I know what Im doing and I know what Im talking about. So please take my advice serious!

Give me a measurement protocol, which is capable of measuring plasma temperature (not electron temperature, know the f%$#@ng difference).
Or I dont believe you anything. And yeah this picture is quite old, I would say 5 years or so since I first published it on multiple platforms. And many serious scientists agree with me. So please now its your turn. Deliver results, that are reliable, or in a few decades your fraud will simply be discovered by other scientists who now start to question if you guys even have results at all, or just publish some values for entertainment reasons. Cheeers.
 
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  • #16
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  • #17
Matthias_Rost said:
Deliver results, that are reliable, or in a few decades your fraud will simply be discovered by other scientists who now start to question if you guys even have results at all, or just publish some values for entertainment reasons. Cheeers.
This is an internet forum for learning about science. The vast majority of members here, including myself, are not scientists, so you are addressing the wrong crowd. And doing it poorly.

Matthias_Rost said:
That's why I don't believe all these published high Temperature values, unless somebody posts a serious temperature measurement.
Please see the links below, which show multiple methods that do not depend on measuring the thermal emission of the plasma as if it were a blackbody (which you've botched, hence your wildly inaccurate emission curves in the original post).

Matthias_Rost said:
Or I dont believe you anything. And yeah this picture is quite old, I would say 5 years or so since I first published it on multiple platforms. And many serious scientists agree with me.
I am quite certain this is an outright fabrication.

Matthias_Rost said:
And as I pointed out multiply and for decades, that the only way to measure temperatures above 5000 K is optically and I plotted how these measurements should look like and you simply dont show any proof of your high temperatures.
You are simply wrong, and you've been wrong over and over and over again for decades. But you need to come to terms with that somewhere else, not here on Physics Forums. We do not exist to argue with people who think large swathes of the scientific community are both morons and liars.

Thread will remain locked.

For those interested, here are some sources related to measuring plasma temp that I hope will be useful:

https://en.wikipedia.org/wiki/Plasma_diagnostics
https://www.chm.bris.ac.uk/pt/diamond/stuthesis/chapter5.htm
https://fusionforenergy.europa.eu/news/how-will-we-monitor-the-temperature-of-iter-plasma/
https://scienceinschool.org/article/2013/fusion-4/
 
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