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I am very confused about three measuring systems: cgs, Heavside-Lorentz (never used) and SI system. I wanted to understand, shortly, the usefulness of the measurement system (HL), because it was not familiar to me.

Into the book the Classical Electrodynamics, Jackson, 3rd edition (Appendix), there is this table 2:

$$\begin{array} {|c|c|} \hline \textbf{System} & \mathbf{\epsilon_0} & \mathbf{\mu_0} \\ \hline \textbf{Gaussian-cgs} & 1 & 1 \\ \hline \textbf{Heavside-Lorentz (HL)} & 1 & 1 \\ \hline \end{array}$$

I know that in electostatic in the SI system is $k_e=1/4\pi\epsilon_0$ and in the cgs system is: $k_e=1$ (is it my skill correct?). Therefore I have a different form between the table 2 and my skills. Infact it should be in cgs system $4\pi\epsilon_0=4\pi$ seeing the table 2.

After exists into book also the table 1, where there are, IMHO, $k_1\equiv k_e$, $k_2=k_m$ and who is $\alpha$ and $k_3$?

Table 1: $$\begin{array} {|c|c|}\hline \textbf{System} & k_1 \\ \hline \textbf{Gaussian-cgs} & 1 \\ \hline \textbf{Heavside-Lorentz (HL)} & \dfrac{1}{4\pi} \\\hline \end{array}$$

What is the reason of the use of Heavside-Lorentz system?

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  • $\begingroup$ @G.Smith Hi, I have seen that $k_3=1/\alpha$. See Appendix pag. 778 after (A.5) formula: k_1/k_2=c^2$. $\endgroup$
    – Sebastiano
    Commented Sep 10, 2019 at 21:47
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    $\begingroup$ I was the closure vote. The question has too many parts to effectively answered. It would be far better to focus on one part of the question and put some effort into clarity rather than so many poorly expressed questions. I clicked on the question because I like these unit systems and know quite a bit about them, but found that even with my knowledge and interest I could not provide an answer to the question as it is. $\endgroup$
    – Dale
    Commented Sep 10, 2019 at 22:14
  • $\begingroup$ Seems to depend on whether one equation has a factor of $4\pi$ and the other doesn't. Can you edit in the two forms of Maxwells equations? $\endgroup$
    – Kyle Kanos
    Commented Sep 10, 2019 at 22:44
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    $\begingroup$ That's a negative. I want a MathJax formula written into the post from both sources showing the formulas edited in. $\endgroup$
    – Kyle Kanos
    Commented Sep 10, 2019 at 22:58
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    $\begingroup$ The second paragraph in the Wikipedia article Lorentz-Heaviside units seems to answer you question? $\endgroup$
    – Farcher
    Commented Sep 11, 2019 at 5:32

1 Answer 1

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I am very confused about the three measuring systems: cgs, Heavside-Lorentz (first time I saw it written) and SI system.

The first thing that may be causing confusion is that there is no such thing as the cgs unit system. Cgs is a class of several unit systems that all use the centimeter, gram, and second for their mechanical units, but they differ with respect to their electrical units. Electrostatic units (esu), electromagnetic units (emu), Gaussian units, and Heaviside-Lorentz units (HL) are all distinct cgs unit systems.

The HL system is particularly nice for working with Maxwell's equations. Following the notation in your second table, Maxwell's equations can be written: $$\nabla \cdot \mathbf E = 4 \pi k_1 \rho$$ $$\nabla \cdot \mathbf B = 0$$ $$\nabla \times \mathbf E = -k_3 \frac{\partial}{\partial t} \mathbf B$$ $$\nabla \times \mathbf B = 4 \pi \alpha k_2 \mathbf J + \alpha \frac{k_2}{k_1}\frac{\partial}{\partial t}\mathbf E$$

So for HL units the above simplifies nicely to $$\nabla \cdot \mathbf E = \rho$$ $$\nabla \cdot \mathbf B = 0$$ $$c\ \nabla \times \mathbf E = -\frac{\partial}{\partial t} \mathbf B$$ $$c\ \nabla \times \mathbf B = \mathbf J + \frac{\partial}{\partial t}\mathbf E$$ Thus the main purpose/usefulness of HL units is to work with Maxwell's equations in a more simplified manner. This is similar to working with Newton's second law in SI vs. US customary units. In SI units Newton's 2nd law is $F=ma$, but in US customary units it is $F=kma$ where $k = \frac{1}{32.174}\frac{lb_{f}}{lb_{m}\ ft \ s^2}$. Choosing units consistent with Newton's laws makes using Newton's laws easier. Similarly, choosing units consistent with Maxwell's equations makes using Maxwell's equations easier.

I have seen in the Classical Electrodynamics Jackson 3rd edition (Appendix) table where cgs system is equal to Heavside-Lorentz ... and a table where this system are different.

HL units are different from Gaussian units. In both systems the value of the vacuum permittivity and permeability are set to a dimensionless 1, but the vacuum permittivity and permeability do not by themselves completely define either system. The two tables are not in conflict with each other, they are just discussing different constants. Notice the label at the top of each column, they are listing completely separate characteristics of each system.

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  • $\begingroup$ Dale hi, thank you very much for your explanation and for you answer. Than you very much for your efforts. But who is $\alpha$? $\endgroup$
    – Sebastiano
    Commented Sep 11, 2019 at 20:58
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    $\begingroup$ $\alpha$ was a column in the previous table that you had posted. $\endgroup$
    – Dale
    Commented Sep 11, 2019 at 21:40
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    $\begingroup$ @Sebastiano Down and close votes are not to stop you from asking questions, but to indicate that the questions require further work from your part. If you then proceed to give up on your questions (and going off for ways to blame other people for the work that you have to do) instead of improving it, then you're implicitly agreeing with the opinion that those questions is be closed or ranked badly. This persistent and aggressive blame game does not help. (For full clarity, I think this specific question is OK, at least in its current form.) $\endgroup$ Commented Sep 11, 2019 at 22:42
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    $\begingroup$ @Sebastiano "It is undeniable my effort also to write the tables in MathJax" -- that's the minimal bar in presentation that everybody is held to, not some heroic effort. The lesson here should be that that's the level of clarity and presentation that your future questions should have before you post them, not after you've gone ahead and annoyed people with poorly-formatted posts. Unless, of course, you're truly OK with feedback through the means of down & close votes, and you're not going to complain about them. $\endgroup$ Commented Sep 12, 2019 at 14:40
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    $\begingroup$ Also, the current second table has much less information than was in the original version. Specifically, you ask "who is $\alpha$" in the text of your question, but without the important context of the table column containing $\alpha$. Without that context people are likely to assume that it is the fine structure constant, which it is not. $\endgroup$
    – Dale
    Commented Sep 12, 2019 at 14:45

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