A CGS units confusions in plasma physics journal papers

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cmb

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Hi,

I was wondering of those who might write papers in this field.

What is the convention of units in plasma physics papers, and is this often all screwed around and you have to already understand it to understand it?

I'll give you an example, because I am looking at this paper to understand thermalisation times-
https://descanso.jpl.nasa.gov/SciTechBook/series1/Goebel_AppF_Electron.pdf

In the opening line it says-
Spitzer derived an expression for the slowing-down time of test particles (primary electrons in our case) with a velocity v = root[2Vp/m] , where eVp is the test particle energy....

So what units is that in? If it is cgs units, then is 'e' in MeV, yet the rest of the article references eV? Or maybe it is Rybergs? Or is it statcoulombs? None of those seem to work, MeV is the only unit that works for v in cm/s and Vp in eVs?

What am I missing here?

There is another paper I can link to later that also says it is cgs units, but if you feed in all the numbers in cgs and compare with the graph it shows in the paper, the axes are out by 3 OOM.

I get the impression, please do disabuse me of this idea, that much of plasma physics is 'theoretical' and the equations become sufficiently abstract that they lose all their constants and it is not possible to really know what system of units the equations represent, unless you already understand it.

Would I be overstating the case that any equation involving physical quantities should clearly state what units of measurement those physical variables use?

It is OK in most fields of physics of course, but when it comes to electricity and magnetism that plasma physics papers rel
 

jasonRF

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To me it looks like that particular equation has a typo. It appears to come from equating kinetic and potential energy: ##\frac{1}{2}m\, v^2 = e \, V_p## where ##V_p## is a voltage, so ##v = \sqrt{\frac{2 \, e\, V_p}{m}}##. If you use consistent units you should get the correct result.

Clearly that appendix is from an online book - when I glance at the plasma physics chapter it is clear that the author is using MKS for general equations, and when providing numerical equations seems to be good about specifying the units. But having said that, it is common in plasma physics to use "convenient" units when discussing a plasma. Just because an author indicates that a plasma has a "temperature" of 10 eV doesn't mean that you insert values in eV wherever you see ##k_B T##; you use the units that the author uses (in this case, ##T## in Kelvin); it is up to you to convert the convenient units to the units used when deriving the equations. Likewise, it is common to specify densities in ##cm^{-3}## even if using MKS, and the reader just needs to convert the quantity to MKS (##m^{-3}##) before plugging into the equations.

Can it be confusing at times? Sure - especially for authors that don't write well. But usually it is pretty straightforward.

Try re-working your numbers and see what you get. If you still have a problem you should probably look at the Spitzer reference, and show us more detail of your calculations.

jason
 

jasonRF

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To directly address your concerns/questions:
I get the impression, please do disabuse me of this idea, that much of plasma physics is 'theoretical' and the equations become sufficiently abstract that they lose all their constants and it is not possible to really know what system of units the equations represent, unless you already understand it.
Not true. My grad work was all in experimental plasma physics, and it was usually straightforward to calculate numbers from the theory to compare with our measurements. Are there some poorly written plasma physics papers that are not clear? Of course. But that is true in every field.

Would I be overstating the case that any equation involving physical quantities should clearly state what units of measurement those physical variables use?
Yes. If they state units after every equation it would be intolerable to read. If an author clearly indicates the units at the appropriate time in the exposition, it shouldn't need to be repeated for every equation thereafter. If an author uses the same units throughout a book, they shouldn't need to specify it all over again in every appendix.

Edit: Now I am thinking that I misunderstood you. Should it be clear what units the author is using? Of course. Does that always happen in plasma physics? No. But still seems unreasonable to expect an author to restate all of the assumptions used throughout a book in every appendix.

jason
 
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cmb

712
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To me it looks like that particular equation has a typo. It appears to come from equating kinetic and potential energy: ##\frac{1}{2}m\, v^2 = e \, V_p## where ##V_p## is a voltage, so ##v = \sqrt{\frac{2 \, e\, V_p}{m}}##. If you use consistent units you should get the correct result.
I already anticipated the absence of the 'e', but what is the 'e'? This is my confusion.

If I calculate, say, velocity of a 10keV proton in what I understand are cgs units, it looks like;
v = sqrt [(2 x 4.8e-10 esu/eV x 10000eV)/(1.6e-24g)] = 2.45e9 cm/s

which is clearly going to be wrong because 'esu's don't relate to eVs

The answer is nice and simple in SI, once converted, and I have no problem with this at all, it is all completely logical
v= sqrt [(2 x 1.6e-19J/eV x 10000eV)/(1.6e-27g)] = 1.4 Mm/s

So I can now calculate this 'cgs unit' and I find it to be;
e = [(1.6e-27g)*(1.4e8cm/s)^2]/[(2 x 10000eV] = 1.6e-12 g.cm^2/eV.s^2

I have not seen "1.6e-12 g.cm^2/eV.s^2" in the description of the elementary charge in cgs units. But unless you use that for 'e' in the equation above, it doesn't give the right answer?

Why use eV at all? I thought the Rydberg the unit of charge in cgs?
 

cmb

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Edit: Now I am thinking that I misunderstood you. Should it be clear what units the author is using? Of course. Does that always happen in plasma physics? No. But still seems unreasonable to expect an author to restate all of the assumptions used throughout a book in every appendix.

jason
I guess all I am asking you to tell me is what the consistent set of units are to assume, in physics papers and documents like this, if the units are not clarified in a physics paper (because, as you say, it doesn't always happen)?
 

f95toli

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Note that you have to be careful here with what you refer to as "system of units".
A paper can be written using the SI system (aka MKS) and still give all numerical values in units of say cm, eV or whatever is convenient. All constants in the equations (e, Kb etc) will still have their SI values.

If the paper is written using the CGS system of units the constants will have different numerical values and the equations will NOT be identical to the equations you would use if the SI system was used (typically the equations differ by factors of c, c^2 and pi; you can't always figure it out from analyzing the dimensions).

Nowadays it is very rare for papers to be written using CGS unit but there are indeed fields where it is still used (e.g. magnetics)
This is not only because of convention; after the re-definition of the SI this spring there will -AFAIK- strictly speaking be no way to convert between the two , in the CGS system of units the permeability of vacuum is equal to unity, but in the new SI (new Ampere with a fixed value of e) it is is a measurable quantity so does not have a precise value. That is, stop using CGS:rolleyes:
 

cmb

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Here is an example where the content is just unclear.


There is an equation there where one term is the Debye length cubed.

I have looked at multiple other sources of the derivation of Saha and I can't find any indication of what units those are in. It does not appear to be SI units, one gets crazy answers. The implication is it is cm, but could be in inches as far as I know.
 

phyzguy

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@cmb, I think you're confused about how units work in equations. Let's take the equation from Wikipedia you mentioned (by the way, λ in this equation is the deBroglie wavelength, not the Debye length).
[tex]\frac{n_e^2}{n-n_e} = \frac{2}{\lambda^3}\frac{g_1}{g_0}\exp[\frac{-\epsilon}{k_B T}][/tex]
This equation is valid no matter what units you use, as long as you are consistent. The units on the left side are 1/length^3, and so are the units on the right side. The units inside the exponential are energy/energy, and so are dimensionless. So you could measure the n's in particles/in^3 and then as long as you use λ in inches, everything will work out. Or you could measure n in particles/cm^3 and λ in cm. Or n in particles/parsec^3 and λ in parsecs, or whatever unit you want. You could even use MKS units for n and λ, and use eV for ε and kT. Since [itex] \frac{-\epsilon}{k_B T}[/itex] is dimensionless, as long as you use the same units in the numerator and denominator, it will give the same number.
 

cmb

712
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@cmb, I think you're confused about how units work in equations. Let's take the equation from Wikipedia you mentioned (by the way, λ in this equation is the deBroglie wavelength, not the Debye length).
[tex]\frac{n_e^2}{n-n_e} = \frac{2}{\lambda^3}\frac{g_1}{g_0}\exp[\frac{-\epsilon}{k_B T}][/tex]
This equation is valid no matter what units you use, as long as you are consistent. The units on the left side are 1/length^3, and so are the units on the right side. The units inside the exponential are energy/energy, and so are dimensionless. So you could measure the n's in particles/in^3 and then as long as you use λ in inches, everything will work out. Or you could measure n in particles/cm^3 and λ in cm. Or n in particles/parsec^3 and λ in parsecs, or whatever unit you want. You could even use MKS units for n and λ, and use eV for ε and kT. Since [itex] \frac{-\epsilon}{k_B T}[/itex] is dimensionless, as long as you use the same units in the numerator and denominator, it will give the same number.
What I posted was wrong (typo included! doh!) but also a very bad example.

The main issue for mixing measurement systems is dealing with electro-magnetic units.

In a way you have actually illustrated the point. In equations where one side is a length or density or some other such unit, yet on the other are electric or magnetic units, there is no correlation between the 'type' of unit one one side to the other. Your guidance serves to exaggerate that point rather than diminish it.

I'll give you a better example when I come across one. I had one the other day I was trying to calculate the Brillouin limit for something and the equation in the paper just didn't work out like the graphs they had put there (i.e. an equation resulting in a number density). I'll try to find the paper again. I couldn't tell if the equation given was meant to have the electron charge in Joules or ergs because whatever the permittivity units the numbers I got didn't match up to the example graphics. Sorry I will have to find it, but this is the problem with the situation, on one side of the equation there is a collection of EM units and on the other they have boiled down to lengths or number densities or something that doesn't appear connected with the units on the other side.

In another example it took me a while to figure out the equations in the paper had been written in Rydbergs rather than eVs! Not a mention of what amounts to a pretty important fact, if one is trying to get quantitative answers out of these papers!
 
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I already anticipated the absence of the 'e', but what is the 'e'? This is my confusion.

If I calculate, say, velocity of a 10keV proton in what I understand are cgs units, it looks like;
v = sqrt [(2 x 4.8e-10 esu/eV x 10000eV)/(1.6e-24g)] = 2.45e9 cm/s

which is clearly going to be wrong because 'esu's don't relate to eVs

The answer is nice and simple in SI, once converted, and I have no problem with this at all, it is all completely logical
v= sqrt [(2 x 1.6e-19J/eV x 10000eV)/(1.6e-27g)] = 1.4 Mm/s

So I can now calculate this 'cgs unit' and I find it to be;
e = [(1.6e-27g)*(1.4e8cm/s)^2]/[(2 x 10000eV] = 1.6e-12 g.cm^2/eV.s^2

I have not seen "1.6e-12 g.cm^2/eV.s^2" in the description of the elementary charge in cgs units. But unless you use that for 'e' in the equation above, it doesn't give the right answer?

Why use eV at all? I thought the Rydberg the unit of charge in cgs?
please look up "statvolt"
 

cmb

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You misunderstand. I am asking why I should pick one system of units to look at or another, when nothing is mentioned.
 
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If I calculate, say, velocity of a 10keV proton in what I understand are cgs units, it looks like;
v = sqrt [(2 x 4.8e-10 esu/eV x 10000eV)/(1.6e-24g)] = 2.45e9 cm/s

which is clearly going to be wrong because 'esu's don't relate to eVs
I believe this was simply done incorrectly.

I avoid cgs units like the plague because I always screw them up....
 

cmb

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It looks like E<>1/2 mv^2 in cgs units.

What does the equation for kinetic energy look like in cgs?
 
406
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The equation is the same but I believe
4.8e-10 esu/eV
is not correct (as described in the Quora snip).

Did I mention that I avoid cgs whenever possible?
 

cmb

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CGS look terrible and I also avoid them, but unfortunately physics papers don't! So unless one is happy to avoid physics papers too then it is not really an option.
 

cmb

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I tell you what, let's reverse engineering CGS units;-

A 5keV proton, for example, has a velocity given by v=sqrt(2 x (1.6x10^-19J/eV) x 10000eV/1.6x10^-27kg)) = 1Mm/s. We know that because SI works!! ;)

So this mystery CGS value X blahs/eV must be;-

X blahs/eV = (0.5 x 1.6x10-24g x (10^8cm/s)^2) / (5keV) = 1.6x10^-12 blahs/eV

blahs is in units of g.cm^2.s^2

So in CGS, it appears that q = 1.6x10^-12 g.cm^2.s^2

I can't find that stated anywhere in any texts I can find, so I presume the idea is that q = 1.6x10^-12 g.cm^2.s^2 = 4.8x10^-10 thingies/pi

So, little did we realise, the equation for kinetic energy is actually;-
E = (pi.m.v^2)/2 !

Perhaps the question of my thread is, instead, where physics papers are half-using CGS units are they liable to miss out the 'pi' occasionally, because like the example I gave a link to a few posts above that seems to be what has happened.

I am also taking away from this that if a paper has an equation showing the elementary charge divided by pi then it implies it is probably CGS units, however a lack of divided by pi might only mean the author believes anyone reading it are mind-readers.
 
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cmb

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Reading wiki's contribution to this befuddlement, it includes this salient sentence;-

"Furthermore, within CGS, there are several plausible choices of electromagnetic units, leading to different unit "sub-systems", including Gaussian units, "ESU", "EMU", and Lorentz–Heaviside units. Among these choices, Gaussian units are the most common today, and "CGS units" often used specifically refers to CGS-Gaussian units."

So, basically, it seems anyone attempting to write a scientific paper in CGS units, which fails to reveal the 'sub-system' of units it is using, is incomplete.

Can physics people, who have written papers using CGS units, please comment?
 

cmb

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Nightmare?!...

cgs_units_table.jpg
 
406
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I tell you what, let's reverse engineering CGS units;-

A 5keV proton, for example, has a velocity given by v=sqrt(2 x (1.6x10^-19J/eV) x 10000eV/1.6x10^-27kg)) = 1Mm/s. We know that because SI works!! ;)

So this mystery CGS value X blahs/eV must be;-

X blahs/eV = (0.5 x 1.6x10-24g x (10^8cm/s)^2) / (5keV) = 1.6x10^-12 blahs/eV

blahs is in units of g.cm^2.s^2

So in CGS, it appears that q = 1.6x10^-12 g.cm^2.s^2

I can't find that stated anywhere in any texts I can find, so I presume the idea is that q = 1.6x10^-12 g.cm^2.s^2 = 4.8x10^-10 thingies/pi

So, little did we realise, the equation for kinetic energy is actually;-
E = (pi.m.v^2)/2 !

Perhaps the question of my thread is, instead, where physics papers are half-using CGS units are they liable to miss out the 'pi' occasionally, because like the example I gave a link to a few posts above that seems to be what has happened.

I am also taking away from this that if a paper has an equation showing the elementary charge divided by pi then it implies it is probably CGS units, however a lack of divided by pi might only mean the author believes anyone reading it are mind-readers.
Here is your factor (which prompted my original advice).

1 statvolt = 299.792458 volts.(The conversion factor 299.792458 is simply the numerical value of the speed of light in m/s divided by 106)[1]The statvolt is also defined in the cgs system as 1 erg / esu.[1]


OK?
 

cmb

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Here is your factor (which prompted my original advice).

1 statvolt = 299.792458 volts.(The conversion factor 299.792458 is simply the numerical value of the speed of light in m/s divided by 106)[1]The statvolt is also defined in the cgs system as 1 erg / esu.[1]


OK?
Very nice, but how does that help me make intelligible sense of equation 4 in the paper I linked to?
 

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