Converting JJ Thomson's q/m Units to Modern Standards

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In summary, the original q/m measurement done by JJ Thomson in 1897 was reported in "electrostatic units" and had a value of 1.76x10^7. However, the modern accepted value is 1.75882001076(53)×10^11 C/kg. This suggests that "electrostatic units" are not the same as C/kg. Based on further research, it is believed that Thomson used the electromagnetic charge unit, the absolute Coulomb, which can be converted to modern units by multiplying by 10. The most convenient system of units for electromagnetism today is the rationalized Gaussian units.
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How to convert from the units used to report q/m by JJ Thomson to the units used in the now accepted form of q/m.
Summary: How to convert from the units used to report q/m by JJ Thomson to the units used in the now accepted form of q/m.

Referencing the below paper... The q/m measurement was first done in 1897 and reported by JJ Thomson. Numerous others measured the same in the early 1900's. The paper below gives a chronicle of those measurements which appear to converge on a value of 1.76x10^7 and they refer to units of this measurement as "electrostatic units". Today's accepted value is different by several orders of magnitudes; 1.75882001076(53)×10^11 C/kg. This leads me to believe that "electrostatic units" are not the same as C/kg but I cannot find a source that can confirm the conversion. It appears that one can just multiply by 10,000 but that is conjecture... I really would like to have a source.

https://www.cs.princeton.edu/courses/archive/fall05/frs119/papers/smith97_thomson.pfg.pdf
 
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It's obviously not in electrostatic units. In the original paper

https://doi.org/10.1080/14786449708621070
Thomson quote the value for ##m/e## of being around ##0.5 \cdot 10^{-7}## (without giving a unit; he'd not pass the intro lab in our university today ;-)). Let's see which units he most probably has used. In matters of absolute values the SI is the save ground to start from: ##m/e \simeq 5.70 \cdot 10^{-12} \text{kg} \, \text{C}^{-1}##. Now in science even in England masses were measured in grams. So we have ##m/e \simeq 5.70 \cdot 10^{-9} \text{g} \, \text{C}^{-1}##. Now there's only a discrepancy of a factor of 10.

Checking Wikipedia for historical units of electromagnetism, we find that thus he must have used the electromagnetic charge unit, the socalled absolute Coulomb. Now ##1 \; \text{C} \widehat{=}0.1 \, \text{abC}##, i.e., we have ##m/e \simeq 5.70 \cdot 10^{-8} \text{g} \, \text{abC}^{-1}=0.570 \cdot 10^{-7} \text{g} \, \text{abC}^{-1}## in accordance with Thomson's measurements.

Unfortunately the English Wikipedia has not that very useful table the German Wikipedia has:

https://de.wikipedia.org/wiki/Elektromagnetische_Maßeinheiten#Elektromagnetische_Einheiten_in_verschiedenen_Systemen

The most convenient system of units for electromagnetism, used today in theoretical high-energy physics, are the rationalized Gaussian units, also known as Heaviside-Lorentz units (usually taken in natural units, where ##\hbar=c=1##. In the letter form it's identical with the SI setting ##\mu_0=\epsilon_0=1##. If it comes to concrete numbers, the one and only system is of course the modern SI units with all the (now exactly defined!) values for the fundamental constants left intact.
 
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Makes complete sense! Thank you kindly for your effort and well written response.
 

FAQ: Converting JJ Thomson's q/m Units to Modern Standards

What are JJ Thomson's units for q/m?

JJ Thomson's units for q/m refer to the ratio of an object's charge (q) to its mass (m). It is often used in the study of electromagnetism and is measured in coulombs per kilogram (C/kg).

Why are JJ Thomson's units for q/m important?

These units are important because they help scientists understand the behavior of charged particles in electric and magnetic fields. They also play a crucial role in the study of subatomic particles and atomic structure.

How did JJ Thomson determine these units?

JJ Thomson determined these units through his famous experiments with cathode ray tubes, which led to the discovery of the electron. By measuring the deflection of the electron beam in electric and magnetic fields, he was able to calculate the ratio of charge to mass.

Are JJ Thomson's units for q/m still used today?

Yes, these units are still used in modern physics and are an important part of the International System of Units (SI). They are also commonly used in practical applications, such as in medical imaging devices.

Can JJ Thomson's units for q/m be applied to all particles?

While these units are commonly used for subatomic particles such as electrons and protons, they can also be applied to larger objects such as ions. However, for particles with extremely small or large masses, different units may be used to better represent the ratio of charge to mass.

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