Black body radiation vs electric discharge in a gas

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Summary:
Is there any connection between temperature-induced black body radiation in a gas and the electric potential-induced particle acceleration in the gas?
Black body radiation formula contains power and exponential terms. Electric discharge in a gas results in the ion acceleration; the ion distribution comprises power and exponential terms too.
Any connection between these two phenomena (i.e. black body and potential) could be established?
 
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  • #2
BvU
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No. The phenomena and their descriptions are widely separated.
 
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  • #3
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No. The phenomena and their descriptions are widely separated.
Would you please explain more!
I know the description for both, but the formula are very similar, except that one is due to temperature and the other from electric potential; temperature cause emission and potential cause particle acceleration! In both the cases some type of energy is converted into two different phenomena: radiation/acceleration!
 
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  • #4
sophiecentaur
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Electric discharge in a gas results in the ion acceleration; the ion distribution comprises power and exponential terms too.
There are many mathematical descriptions of phenomena which share the same basic expressions. Exponential 'decay' occurs with time and with distance. A=B/C occurs everywhere etc. etc..
If you quote the particular examples that you have chosen to then we could all discuss, in particular, what you are referring to.
 
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There are many mathematical descriptions of phenomena which share the same basic expressions. Exponential 'decay' occurs with time and with distance. A=B/C occurs everywhere etc. etc..
If you quote the particular examples that you have chosen to then we could all discuss, in particular, what you are referring to.
The main point is that we have got a spectrum comprising both the power and exponential terms, with a peak at 30 keV and end point at 1.5 MeV. Is it possible to consider this as a Maxwell-Boltzmann distribution? (Particularly connect it with black body radiation)
 
  • #6
BvU
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Hi,

You are obviously not happy with the answers so far. Instead of starting an identical thread with a different distribution (Maxwell Boltzmann instead of Planck), why not clarify your orginal question, for instance by providing a link to the article about this gas discharge ion energy distribution ?
 
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  • #8
BvU
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Thanks for the link ! Unfortunately I can't access the article (and I know next to nothing about plasma physics :cry: so maybe I can't even read it).

But the crux of the MB distribution is that it maximizes a weight (entropy) for a given total energy .
Perhaps the same principles dictate a gas discharge ion energy (re-?)distribution in a plasma ?
 
  • #9
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Thanks for the link ! Unfortunately I can't access the article (and I know next to nothing about plasma physics :cry: so maybe I can't even read it).

But the crux of the MB distribution is that it maximizes a weight (entropy) for a given total energy .
Perhaps the same principles dictate a gas discharge ion energy (re-?)distribution in a plasma ?
Just want to know if possible to have a high energy MB dist. as in Fig 4 of the attached paper, and if yes how it is justified?
 

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  • #10
BvU
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Hereby I officially declare myself as unqualified to respond in this thread :smile:

But if you want my opinion: In the first link the abstract is rather ominous

Abstract
A detailed and precise study of high energy deuteron beams from a plasma focus device shows that they follow a non-power law distribution, unlike most of the other works. This is in agreement with our previous studies using a direct and unambiguous diagnostic of the activation yield-ratio technique. It is possible that the En dependence effectively terminates at some maximum deuteron energy. The experimental results show that the spectrum increases to very high energies whereas the power law fit fails.​

So at least the notion of uncharted (or poorly charted) territory emerges. From the fig 4 in the attachment :

1611766881824.png
I personally wouldn't like to claim any preferred underlying distribution.

Anyone more introduced to the subject ?

##\ ##
 
  • #11
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Hereby I officially declare myself as unqualified to respond in this thread :smile:

But if you want my opinion: In the first link the abstract is rather ominous

Abstract
A detailed and precise study of high energy deuteron beams from a plasma focus device shows that they follow a non-power law distribution, unlike most of the other works. This is in agreement with our previous studies using a direct and unambiguous diagnostic of the activation yield-ratio technique. It is possible that the En dependence effectively terminates at some maximum deuteron energy. The experimental results show that the spectrum increases to very high energies whereas the power law fit fails.​

So at least the notion of uncharted (or poorly charted) territory emerges. From the fig 4 in the attachment :

I personally wouldn't like to claim any preferred underlying distribution.

Anyone more introduced to the subject ?

##\ ##
Is it physically possible at all to have MB distribution for an electric discharge in the gas (not raising the temperature), and it is similar to the figure above (peaks at 75 keV and ends at 150 keV)?
 
  • #12
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Hereby I officially declare myself as unqualified to respond in this thread :smile:

But if you want my opinion: In the first link the abstract is rather ominous

Abstract
A detailed and precise study of high energy deuteron beams from a plasma focus device shows that they follow a non-power law distribution, unlike most of the other works. This is in agreement with our previous studies using a direct and unambiguous diagnostic of the activation yield-ratio technique. It is possible that the En dependence effectively terminates at some maximum deuteron energy. The experimental results show that the spectrum increases to very high energies whereas the power law fit fails.​

So at least the notion of uncharted (or poorly charted) territory emerges. From the fig 4 in the attachment :

I personally wouldn't like to claim any preferred underlying distribution.

Anyone more introduced to the subject ?

##\ ##
" I personally wouldn't like to claim any preferred underlying distribution. "
If we can claim the related distribution then may extract some information!
 
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  • #13
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One quick note from your original post , Electric field doesn't accelerate particles in a gas , gas is an insulator. Electric field (if high enough) can cause breakdown in a gas and ionize it either partially or fully and only then liberated charges start to move with respect to the applied field.
This being said the answer to your summary would be No.

One main difference is that blackbody radiation comes from the random average thermal motion of the particles that constitute a certain volume in question. Summed up in the Boltzmann distribution.
Take on the other hand a fluorescent tube as an example, it is basically a low pressure gas through which electrical discharge happens. Without the white phosphor coating on the tube walls the discharge through the tube would give off UV rays. Yet the tube is just warm, can be touched by the bare hand.
If that tube was giving off the radiation wavelength it does as a thermal blackbody emitter , it would be so hot you could not come close to it and it would melt most likely in the process.

The reason why is because there is a very uneven distribution of particle energies within the tube. The discharge gives energy to some electrons in the gas (those that get hit by electrons from the discharge arc) , as those electrons fall back to their original energy level they give off radiation. At the same time many other electrons don't receive any energy. In a blackbody this can't happen because it's constituent particles share almost the same energy.
Like when you are heating a lump of metal soon the whole metal is at the same temperature.

This targeting of individual electrons versus heating every particle within a volume is the main difference between a blackbody emitter and a gas discharge.
Although low atomic number gases like Hydrogen would be rather poor blackbody examples even when heated conventionally and not by discharge, because of it's single electron it would still emit specific spectral lines much like during a gas discharge.
 
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  • #14
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One quick note from your original post , Electric field doesn't accelerate particles in a gas , gas is an insulator. Electric field (if high enough) can cause breakdown in a gas and ionize it either partially or fully and only then liberated charges start to move with respect to the applied field.
This being said the answer to your summary would be No.

One main difference is that blackbody radiation comes from the random average thermal motion of the particles that constitute a certain volume in question. Summed up in the Boltzmann distribution.
Take on the other hand a fluorescent tube as an example, it is basically a low pressure gas through which electrical discharge happens. Without the white phosphor coating on the tube walls the discharge through the tube would give off UV rays. Yet the tube is just warm, can be touched by the bare hand.
If that tube was giving off the radiation wavelength it does as a thermal blackbody emitter , it would be so hot you could not come close to it and it would melt most likely in the process.

The reason why is because there is a very uneven distribution of particle energies within the tube. The discharge gives energy to some electrons in the gas (those that get hit by electrons from the discharge arc) , as those electrons fall back to their original energy level they give off radiation. At the same time many other electrons don't receive any energy. In a blackbody this can't happen because it's constituent particles share almost the same energy.
Like when you are heating a lump of metal soon the whole metal is at the same temperature.

This targeting of individual electrons versus heating every particle within a volume is the main difference between a blackbody emitter and a gas discharge.
Although low atomic number gases like Hydrogen would be rather poor blackbody examples even when heated conventionally and not by discharge, because of it's single electron it would still emit specific spectral lines much like during a gas discharge.
The word ionized was missed: it is an ionized gas enclosed in a cylinder and then the potential is applied.
But the ion spectrum follows the MB dist. similar to the black body radiation. How this could be justified physically?
 
  • #15
hutchphd
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There are very many processes and models which give rise to curves that look like this. The M-B distribution is basically a Chi-square distribution depending on the dimensionality. Any random walk will produce a result like that in higher dimension. Also time sequences for random processes produce Poisson distributons of similar shape. The data you show is perhaps interesting but in no way definitive.
 
  • #16
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There are very many processes and models which give rise to curves that look like this. The M-B distribution is basically a Chi-square distribution depending on the dimensionality. Any random walk will produce a result like that in higher dimension. Also time sequences for random processes produce Poisson distributons of similar shape. The data you show is perhaps interesting but in no way definitive.
But similar to the blakbody, in our case the total power is proportional to the forth power of voltage and the peak is linearly proportional to the voltage!
 
  • #17
hutchphd
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Please specify a direct reference so I can look.
 
  • #18
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Please specify a direct reference so I can look.
Just want to know if MB/Poisson dist. could be considered for high electric discharge in the gas (theoretically)?
1611858491136.png
 
  • #19
hutchphd
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But similar to the blakbody, in our case the total power is proportional to the forth power of voltage and the peak is linearly proportional to the voltage!

??

.
 
  • #21
hutchphd
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I know it is true for blackbodies.
 
  • #22
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I know it is true for blackbodies.

The Maxwell-Boltzmann Equation: kinetic energy of the gas particles to the Temperature
I am trying to re-cast the MB equation to give a distribution in terms of Ion Energy rather than Temperature.
Might be possible to consider the MB equation at very high temperatures (the sort of temperatures where a gas might form a plasma), is this right at all?
 
  • #23
hutchphd
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Why do you think that it is?? I honestly don't know...convince me.
 
  • #24
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Well @naviakam in a gas discharge type plasma like those in light tubes or elsewhere you surely don't have anything close to a MB distribution , as I said you have some rather few electrons excited to high energy while the rest of the particles are "cool". I can hold a working fluorescent tube in my hand easily. This means the average particle energy/temperature is low inside the tube.

In a real plasma unlike a partially ionized gas which is the gas discharge type or ordinary gas you would have to treat the ions and electrons separately I think for a distribution since for the same temperature the lighter electrons would have higher speeds.
https://en.wikipedia.org/wiki/Plasma_diagnostics

See the various methods used to probe plasmas to determine their properties. Typically one seeks to find the electron temperature and then the ion temp can be calculated accordingly since we know the ion/proton mass.

So don't take my word for it , maybe someone will be more precise and knowledgeable in this but I think you can indeed have a ion or electron distribution although I'm not sure whether it's curve will follow or resemble that of the ideal gas one originally made by Boltzmann. But irrespective of the curve the underlying idea stays the same.
Remember that just as in a gas so too in a plasma we can never know the exact KE of any particle , so we make a measurement and then we know the average energy distribution.

The only cases where you can pretty much narrow down each particle energy to a very steep curve is when you have just a single particle type isolated in vacuum under specific conditions , like an electron beam.
Take a look at this
https://www.semanticscholar.org/pap...eyan/86949f14edf2debfe86adee0221d1db8e1e80c07


Are you just interested in the physics of this or is there a specific application or goal for you here ?
 
  • #25
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As I said the formula is much the same as MB and more than that:
1. the total power is equal to the power forth of potential
2. the peak occurs at value of potential itself
these are similar to the BB radiation:
1611927357518.png

Everything looks identical, but I am not sure how physics justify it?
 

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