Black body radiation vs electric discharge in a gas

[naviakam]

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?

 

[BvU]

No. The phenomena and their descriptions are widely separated.

 

[naviakam]

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!

 

[sophiecentaur]

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.

 

[naviakam]

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)

 

[BvU]

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 ?

 

[naviakam]

https://iopscience.iop.org/article/10.1088/0741-3335/52/8/085007

 

[BvU]

Thanks for the link ! Unfortunately I can't access the article (and I know next to nothing about plasma physics 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 ?

 

[naviakam]

Thanks for the link ! Unfortunately I can't access the article (and I know next to nothing about plasma physics 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?

 

[BvU]

Hereby I officially declare myself as unqualified to respond in this thread

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 E−n 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 ?

$\$

 

[naviakam]

Hereby I officially declare myself as unqualified to respond in this thread

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 E−n 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 :

View attachment 276934​

​

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)?

 

[naviakam]

Hereby I officially declare myself as unqualified to respond in this thread

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 E−n 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 :

View attachment 276934​

​

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!

 

[artis]

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.

 

[naviakam]

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?

 

[hutchphd]

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.

 

[naviakam]

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!

 

[hutchphd]

Please specify a direct reference so I can look.

 

[naviakam]

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)?

 

[hutchphd]

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!

??

.

 

[naviakam]

??

.

BB Rad:

 

[hutchphd]

I know it is true for blackbodies.

 

[naviakam]

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?

 

[hutchphd]

Why do you think that it is?? I honestly don't know...convince me.

 

[artis]

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/86949f14edf2debfe86adee0221d1db8e1e80c07Are you just interested in the physics of this or is there a specific application or goal for you here ?

 

[naviakam]

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:

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

 

[hutchphd]

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:
View attachment 277023
Everything looks identical, but I am not sure how physics justify it?

Can you please supply a specific reference to "the formula"? I don't seem to find one.

.

 

[naviakam]

Can you please supply a specific reference to "the formula"? I don't seem to find one.

.

 

[BvU]

And where does that peak ?

 

[naviakam]

By considering MB distribution, we have obtained a few important results which matches the experiments.
It seems that there is no such thing as MB distribution for electric discharge in a gas (where the original form of MB is for low temperature gas), right?
If there is such thing, what should be the necessary conditions or what are the problems/conflicts?

 

[artis]

@naviakam Think about what happens to a gas as you heat it up , as long as it is a gas it will follow the Boltzmann distribution curve but at some point say by introducing a strong RF field or by heating it to extreme temperatures it will become a plasma. So no more gas molecules.

I already explained why electric discharge can't produce a MB distribution within a gas. To put it simply it is because if the discharge current is low (as is in most if not all gas discharge lamps) then only a few of the outer electrons get ionized and only they participate in the light emission. This is a partial ionization state, so the bulk material stays at room or close to room temperature. Think about it , this is why fluorescent lights are on average two to three times as bright for the same power consumption compared to a equivalent color temperature incandescent. Because you don't have to heat up the whole material to get the same color temperature.

I think there could not be "such a thing" as you say. Think about it this way. A gas discharge is essentially just an electric current running through a volume of partially ionized gas , the same electric current could run through a tungsten wire. Now in the tungsten it would heat it up evenly and cause it to emit a blackbody spectrum (not perfect since even metals don't have a perfect continuum emission spectrum you can look it up) but close enough.
Now the same current through the gas doesn't produce a black body, first of all the emission spectrum is not continuous but rather with discrete peaks, second of all the material isn't heated up evenly, in fact it's physically cold and only some electrons are "hot" within the volume.
Eventually it has to do with the atomic structure of different elements , metals have higher atomic numbers and a different atomic structure within them than gases do also gases have lower atomic number on average.
The result of this is that a current through metal simply heats it up evenly while a current through gas doesn't. At some point if you would increase the discharge current to a very high value the gas instead of simply heating up would rather become fully ionized and turn into a plasma.

 

[naviakam]

@naviakam Think about what happens to a gas as you heat it up , as long as it is a gas it will follow the Boltzmann distribution curve but at some point say by introducing a strong RF field or by heating it to extreme temperatures it will become a plasma. So no more gas molecules.

I already explained why electric discharge can't produce a MB distribution within a gas. To put it simply it is because if the discharge current is low (as is in most if not all gas discharge lamps) then only a few of the outer electrons get ionized and only they participate in the light emission. This is a partial ionization state, so the bulk material stays at room or close to room temperature. Think about it , this is why fluorescent lights are on average two to three times as bright for the same power consumption compared to a equivalent color temperature incandescent. Because you don't have to heat up the whole material to get the same color temperature.

I think there could not be "such a thing" as you say. Think about it this way. A gas discharge is essentially just an electric current running through a volume of partially ionized gas , the same electric current could run through a tungsten wire. Now in the tungsten it would heat it up evenly and cause it to emit a blackbody spectrum (not perfect since even metals don't have a perfect continuum emission spectrum you can look it up) but close enough.
Now the same current through the gas doesn't produce a black body, first of all the emission spectrum is not continuous but rather with discrete peaks, second of all the material isn't heated up evenly, in fact it's physically cold and only some electrons are "hot" within the volume.
Eventually it has to do with the atomic structure of different elements , metals have higher atomic numbers and a different atomic structure within them than gases do also gases have lower atomic number on average.
The result of this is that a current through metal simply heats it up evenly while a current through gas doesn't. At some point if you would increase the discharge current to a very high value the gas instead of simply heating up would rather become fully ionized and turn into a plasma.

what if the gas is initially fully ionezed and then a potential is applied for as low as 50 ns time scale?

 

[artis]

So you are now asking "what happens to a plasma if a brief current pulse is applied to it"
Depends on the amperage , if low almost nothing happens apart from the plasma conducting your current, if very high then various phenomena take place like the plasma is compressed (pinched) by the magnetic field of the current and it's density and particle energy increase in other words it compreses and heats up.

Can you just explain what are you trying to understand with this ?
i suggest you read this , a very good question and answer to the problem that is relevant to you.
https://physics.stackexchange.com/q...a-black-body-radiation-curve-to-be-continuous

 

[naviakam]

So you are now asking "what happens to a plasma if a brief current pulse is applied to it"
Depends on the amperage , if low almost nothing happens apart from the plasma conducting your current, if very high then various phenomena take place like the plasma is compressed (pinched) by the magnetic field of the current and it's density and particle energy increase in other words it compreses and heats up.

Can you just explain what are you trying to understand with this ?
i suggest you read this , a very good question and answer to the problem that is relevant to you.
https://physics.stackexchange.com/q...a-black-body-radiation-curve-to-be-continuous

This is exactly what it is: a small volume heats up for a ns time scale, then the ions distribution similar to that of MB is generated!
I want to understand how sensible it is physically to have such distribution for ions at high energies which behave like a BB radiation with similar curve, total power and peak.

 

[artis]

Well optically thick plasmas do have a blackbody like spectrum , our sun is one example. Sure enough the ions as well as the electrons in a plasma each have their energy distribution curve.
Spectrum is a result of energy distribution , not sure why you think time scales have any role here.

I was objecting to your point about blackbody curves from gas discharges.
Also it would be important to point out that often the radiation generated by say electron transitions in a gas for example is not the final radiation emitted because if the medium is optically thick like a very dense gas for example or plasma for that matter the emitted photons can get scattered/absorbed and re-emitted before they get out , this broadens the total spectrum that we observe and makes the spectrum closer to a BB spectrum.

Again the gas discharge tube can be made as an example. The mercury vapor/argon discharge emits an UV peak, it is the phosphor coating on the wall that then absorbs this UV photons and re-emits them at multiple lower frequencies producing light that is more broadened in it;'s spectrum.
In fact it's still rather peaky and discrete but since our eyes are not perfect we perceive it to be similar to a BB spectrum light emitted from a heated source like a glowing filament.

 

[naviakam]

Well optically thick plasmas do have a blackbody like spectrum , our sun is one example. Sure enough the ions as well as the electrons in a plasma each have their energy distribution curve.
Spectrum is a result of energy distribution , not sure why you think time scales have any role here.

I was objecting to your point about blackbody curves from gas discharges.
Also it would be important to point out that often the radiation generated by say electron transitions in a gas for example is not the final radiation emitted because if the medium is optically thick like a very dense gas for example or plasma for that matter the emitted photons can get scattered/absorbed and re-emitted before they get out , this broadens the total spectrum that we observe and makes the spectrum closer to a BB spectrum.

Again the gas discharge tube can be made as an example. The mercury vapor/argon discharge emits an UV peak, it is the phosphor coating on the wall that then absorbs this UV photons and re-emits them at multiple lower frequencies producing light that is more broadened in it;'s spectrum.
In fact it's still rather peaky and discrete but since our eyes are not perfect we perceive it to be similar to a BB spectrum light emitted from a heated source like a glowing filament.

Yes, but even if the BB is continuous and electric discharge in the gas is discrete but follow the MB-like formula and similar results to BB could be extracted by considering it as MB dist, is there anything wrong with it? Can it be claimed that our spectrum is a Boltzmann type? Because only if it is considered as a MB dist, the results mentioned could be obtained.

 

[BvU]

This is exactly what it is: a small volume heats up for a ns time scale, then the ions distribution similar to that of MB is generated!
I want to understand how sensible it is physically to have such distribution for ions at high energies which behave like a BB radiation with similar curve, total power and peak.

I'm still puzzled by the tenacity to pin an underlying distribution on such meagre data. So far I 've only seen this data

And when I try to generate the two distributions (granted: meagre attempt):

(actually $x^2\,e^{-x^2}\$ and $5.7\, x^3/(e^x - 1) \$ )
the data can't distinguish between the two (and a gaussian or Poisson, Lorentzian, Voigt, ...).

Do you happen to have a statistically useful high-volume, wide energy range, set of data to subject to a decisive fitting procedure or does this all remain conjecture ?

$\$

 

[naviakam]

I'm still puzzled by the tenacity to pin an underlying distribution on such meagre data. So far I 've only seen this data
View attachment 277126

And when I try to generate the two distributions (granted: meagre attempt):

View attachment 277127​

(actually $x^2\,e^{-x^2}\$ and $5.7\, x^3/(e^x - 1) \$ )
the data can't distinguish between the two (and a gaussian or Poisson, Lorentzian, Voigt, ...).

Do you happen to have a statistically useful high-volume, wide energy range, set of data to subject to a decisive fitting procedure or does this all remain conjecture ?

$\$

Our lecturer didn't provide us with more information but said that the data is similar to that of MB. The figure I put here was from a reference, however I couldn't find anymore.
We must see if the question in post #35 is valid.

 

[artis]

No @naviakam I'm not sure why you want that Boltzmann distribution so badly for the low pressure gas discharge, is that distribution on the FBI's most wanted list?...

Low pressure gas discharges don't produce anything resembling a BB distribution
They produce a sharp peak and that's it. The spectral lines of lower energy are then added by either a mixture of other gasses or phosphors. And yet still they don't reproduce a BB spectrum.
I think you are now mixing stuff up, before you wanted to have the distribution for ions in plasma now you want it for low pressure gas discharge.

I can't say about the plasma ions as that would need more research from my side but I can surely tell you there is no BB curve for a low pressure gas discharge.
The simple test is to ask whether the object emitting the radiation is hot or cold. low pressure gas discharge tubes are cold yet shine bright , this gives you the answer.
Thermal plasma at high energies is a different subject, there at least the energies are distributed equally and a curve energy diagram could be drawn for each of the species.

The answer to the question in post #35 is No.
You are missing a lot of energy in the single peak gas discharge. To make an object emit at a BB spectrum requires much more energy input to raise the average KE energies of the particles.

 

[naviakam]

No @naviakam I'm not sure why you want that Boltzmann distribution so badly for the low pressure gas discharge, is that distribution on the FBI's most wanted list?...

Low pressure gas discharges don't produce anything resembling a BB distribution
They produce a sharp peak and that's it. The spectral lines of lower energy are then added by either a mixture of other gasses or phosphors. And yet still they don't reproduce a BB spectrum.
I think you are now mixing stuff up, before you wanted to have the distribution for ions in plasma now you want it for low pressure gas discharge.

I can't say about the plasma ions as that would need more research from my side but I can surely tell you there is no BB curve for a low pressure gas discharge.
The simple test is to ask whether the object emitting the radiation is hot or cold. low pressure gas discharge tubes are cold yet shine bright , this gives you the answer.
Thermal plasma at high energies is a different subject, there at least the energies are distributed equally and a curve energy diagram could be drawn for each of the species.

The answer to the question in post #35 is No.

Low pressure gas discharge?!

 

[BvU]

Our lecturer didn't provide us with more information but said that the data is similar to that of MB.

Oh boy, are we dealing with a homework exercise (with a known answer $-$ thus making fools of ourselves ) or with a professor with a conjecture, who is testing the waters and let's the students do the work ?

With hindsight I'm glad I took some life insurance in post #10

Can't wait to see how this develops further -- do let us know !

$\$

 

[naviakam]

To summarize:
1. Our spectrum although not continuous but follows the MB distribution
2. Peak and total power is similar to the BB radiation, furthermore we could estimate the temperature in the gas correctly by considering MB dist
3. The spectrum is for intensity against ion energy
4. The ion energy is from keV to MeV
5. Our spectrum is obtained from applying potential in an ionized gas
6. My question was to see if it is correct to consider such spectrum as an MB distribution?

 

[naviakam]

I found this explanation in an article that probably describe what I am looking for but ask to help me to understand it:

I am trying to find the equation responsible for ion spectrum. I found this paper but don't get the point for some parts. The section I put above is everything I want to realize in simple words.
The current have magnetic field around it, but here the current is a ring , r, once contracting toward axis, z, create B in theta direction and E in z. How this is explained?

 

[naviakam]

Text format of the image above:
This model assumes an axisymmetric, cylindrical current distribution j(r, t) which is finite in thickness and initially annular in shape. The distribution is then assumed to contract rapidly to the axis in some manner. Such a time variation in the current density gives rise to both an azimuthal magnetic field By(r, t) and an axial electric field E(r, t) whose values are derived from Maxwell’s equations:

d(rB)/dr=rj(r,t)

dE/dr=dB/dt

The equation of motion in two dimensions is

mr’’ = -ez’B (r, t),

mz’’ = er’B (r, t) +eE(r, t)

where m is the ion mass, e is the ionic charge, and r’ and z’ are the velocity components.

 

[naviakam]

Maxwell equations:

Cylindrical coordinates:

 

[vanhees71]

It would help a lot to quote the paper properly. Maybe somebody can help then. For me there's not enough context to know what the question is. Is it about the calculation of the em. field? Then just use the retarded potentials. Is it about a self-consistent solution of the equation of motion of some charge-current distribution and the em. field? Then it's more complicated but perhaps doable due to high symmetry.

 

[naviakam]

It would help a lot to quote the paper properly. Maybe somebody can help then. For me there's not enough context to know what the question is. Is it about the calculation of the em. field? Then just use the retarded potentials. Is it about a self-consistent solution of the equation of motion of some charge-current distribution and the em. field? Then it's more complicated but perhaps doable due to high symmetry.

It says that current density changes (while it is a annular) leads to azimuthal B and axial E. I don't know how this happens.
Then based on this, and using Cylindrical Maxwell two equations appeared in the paper which I don't know mathematically how they are obtained.
And finally how E and B are used to obtain the energy from equation of motion presented in the paper?
Full paper attached.

 

[artis]

@naviakam what is the thing that you want to understand here ? I am asking because if it's simply ion or electron energy distribution within a plasma then why make it more complicated by introducing an electric discharge pinch device?

You might as well just analyze a plasma at thermal equilibrium which means that the temperature is similar between ions and electrons but if you only care about one species distribution then you might as well take as an example a plasma that is not in thermal equilibrium. Although I think it is easier to estimate the electron temperature distribution than the ion one.

I think you can see from the paper you attached that determining ion energies in a plasma that changes it's energy fast or basically a gas that gets ionized rapidly and compressed due to current discharge is quite complicated and there is no "plug n play" type mechanism or "probe" which you could stick in such a plasma to know the ion distribution.
Getting to this parameter requires some painfully complicated maths and analyzing , like if you have a certain plasma volume and pressure and you get say neutrons as a byproduct then you would analyze their energy and flux to then estimate the number of fusion reactions taking place that made them versus the number of ions total which would then give you a rough estimate of the ion energy distribution.
Although I'm sure members like @mfb or @vanhees71 or @BvU could explain this better

 

[naviakam]

@naviakam what is the thing that you want to understand here ? I am asking because if it's simply ion or electron energy distribution within a plasma then why make it more complicated by introducing an electric discharge pinch device?

You might as well just analyze a plasma at thermal equilibrium which means that the temperature is similar between ions and electrons but if you only care about one species distribution then you might as well take as an example a plasma that is not in thermal equilibrium. Although I think it is easier to estimate the electron temperature distribution than the ion one.

I think you can see from the paper you attached that determining ion energies in a plasma that changes it's energy fast or basically a gas that gets ionized rapidly and compressed due to current discharge is quite complicated and there is no "plug n play" type mechanism or "probe" which you could stick in such a plasma to know the ion distribution.
Getting to this parameter requires some painfully complicated maths and analyzing , like if you have a certain plasma volume and pressure and you get say neutrons as a byproduct then you would analyze their energy and flux to then estimate the number of fusion reactions taking place that made them versus the number of ions total which would then give you a rough estimate of the ion energy distribution.
Although I'm sure members like @mfb or @vanhees71 or @BvU could explain this better

Now what I need to understand is from basic physics and math:
how the current density here produces such B and E in the mentioned directions. Then how the Maxwell and coordinate gives this B and E. And finally how to use B and E to estimate energy from provided equation of motion.
Everything is available, current configuration, and formula but I need someone to explain in simple words!

 

[artis]

@naviakam So you say you just need to understand

basic physics and math:

yet you go on to describe very complicated setup and conditions which are normally analyzed by people working decades in the field or folks writing PhD papers.
And then lastly you say

but I need someone to explain in simple words!

from which I see you don't understand the concepts as they are complicated ,so how about accepting the given advice and learning the basics first?

 

[naviakam]

@naviakam So you say you just need to understand

yet you go on to describe very complicated setup and conditions which are normally analyzed by people working decades in the field or folks writing PhD papers.
And then lastly you sayfrom which I see you don't understand the concepts as they are complicated ,so how about accepting the given advice and learning the basics first?

It's for our BSc course to understand how the Maxwell in a Cylinder gives the B and E in first equations based on the assumptions made.
Then from equation of motion, calculate the energy. Seems that I should be able to understand it but couldn't do that. Then I must understand in simple words.