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This Wikipedia article gives a table of certain constants in different systems of units. I noticed that in Gaussian and electrostatic CGS the value of $\epsilon_0$ equals the dimensionless $1$.

I wondered whether anyone could provide justification for this? I was under the impression that $\epsilon_0$ is not even necessary in these units since it's essentially just a dimensional conversion factor in the SI, however I suppose it would still be possible to convert it into CGS units. How is this done?

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    $\begingroup$ Generally speaking, CGS was constructed that way. $\endgroup$
    – Jon Custer
    Commented May 28, 2020 at 15:34
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    $\begingroup$ Actually to convert from mks to cgs you set $\epsilon_0=1/4\pi$. $\endgroup$
    – my2cts
    Commented May 28, 2020 at 17:03
  • $\begingroup$ @my2cts I don't think that is correct, it is not as simple as that. The dimensions of charge are different in the two systems of units. $\epsilon_0$ is set to 1, as is stated in the Wikipedia article. $\endgroup$
    – 13509
    Commented May 28, 2020 at 17:09
  • $\begingroup$ I agree with @my2cts. I have no idea how Wikipedia can compare Coulomb’s Law in SI units, $F=q_1q_2/4\pi\epsilon_0r^2$, with Coulomb’s Law in Gaussian units, $F=q_1q_2/r_2$, and conclude that in Gaussian units $\epsilon_0$ is $1$ in any sense. To me this is patently incorrect. $\endgroup$
    – G. Smith
    Commented May 29, 2020 at 2:25
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    $\begingroup$ I am well aware that charge has different dimensions in the two systems. I find Jackson’s table as baffling as Wikipedia’s. I used to teach electromagnetism from Jackson but that was when his book used Gaussian units. From my point of view, there is no $\epsilon_0$ in Gaussian EM and it has no “value” in Gaussian units. To convert equations from SI to Gaussian one replaces $1/4\pi\epsilon_0$ with $1$. $\endgroup$
    – G. Smith
    Commented May 29, 2020 at 16:28

2 Answers 2

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Possibly the best way to see this is to consider Coulomb's Law that gives the force F (in Newtons) between two charges $Q_1, Q_2$ (in Coulomb) separated by a distance $R$ (in m): $$F = \frac{Q_1 Q_2}{4 \pi \epsilon_0 R^2}.$$ In cgs units the force is measured in dyne, $1 N = 10^5 dyne$, the unit of length is measured in cm, $1 m = 100 cm$. So if you measure force, f, in dyne and separation, r, in cm then $$10^{-5} f = \frac{Q_1 Q_2}{4 \pi \epsilon_0 (10^{-2} r)^2}.$$ You can now define a charge $q$ by $q = Q/\sqrt{10^{ +5} \times 4 \pi \epsilon_0 \times 10^{-4}}$ then Coulombs Law becomes $$f = \frac{q_1 q_2}{r^2}.$$ It follows that the relationship between charge in SI units and cgs units is $$q (Fr) \equiv 2997919999.934 \,\,\, Q (C)$$

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    $\begingroup$ Thank you for this answer. The one thing I would comment is that $q \neq 2997919999.934Q$ since $q$ and $Q$ have different dimensions in the two systems of units; this is the reason why the physical law looks different in both systems of units (which is generally unusual, but a thing in electromagnetism!). It is better to state $q \leftrightarrow 2997919999.934Q$. $\endgroup$
    – 13509
    Commented May 28, 2020 at 17:23
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    $\begingroup$ @JamesWirth Well, yes I agree with you it is a conversion factor, I have changed $q = ....$ to $q \equiv ...$ $\endgroup$
    – jim
    Commented May 28, 2020 at 17:38
  • $\begingroup$ Why is your constant so different from the speed of light $c=299\,792\,458\,\mathrm{m/s}$? $\endgroup$
    – Ruslan
    Commented Jul 27, 2020 at 21:51
  • $\begingroup$ @Ruslan The constant relating charge measured in $Fr$ to charge measured in $C$ doesn't have units of $m/s$. That the value has the same numbers as the the speed of light is because $\mu_0 \approx 4 \pi \times 10^{-7} H m^{-1}$ (at one time $\mu_0$ was defined as exactly $4 \pi \times 10^{-7} H m^{-1}$ but this changed in CODATA 2018), so that $4 \pi \epsilon_0$ is conveniently related to the (SI) value of $c$. $\endgroup$
    – jim
    Commented Jul 29, 2020 at 9:32
  • $\begingroup$ @Ruslan in fact the definition of $\mu_0$ was changed in 2019, so that every book on electromagnetism where you see the statement $\mu_0 = 4 \pi \times 10^{-7} H m^{-1}$ are now technically incorrect. $\endgroup$
    – jim
    Commented Jul 29, 2020 at 9:40
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The value of $\epsilon_o$ alone is meaningless, but the question becomes meaningful if examined in a wider perspective.

For historical reasons electromagnetism was developed following various formal schemes (characterized by different definitions of some quantities). Consequently many mathematical formulae, dimensions of quantities and systems of units don't match. This may be a bit annoying when looking at different textbooks and when we need to compare units with not equivalent dimensions.

This well known formal problem can be solved by building an "ancestor" system having three arbitrary constants, as shown by Wikipedia and Jackson. Each formal scheme is uniquely determined by a suitable set of constants, that enable us to write all electromagnetic relations in every form, as well as to find all dimensional relationships (allowing an automatic conversion of units belonging to totally different systems).

A good choise of the parameters is tabulated below.\ Here the three constants $k_o,\epsilon_o,\mu_o $ are indipendent, instead $\chi = c\sqrt{(\epsilon_o \mu_o)}$ and $\xi = \frac{1}{(\epsilon_o k_o)} $ are two useful dependent parameters.

\begin{array}{|l||c|c|c||c|c|c|} \hline \text{Formal system} \rule{0pt}{14pt} \rule[-8pt]{0pt}{12pt} & \quad k_o\quad & \quad k_o\epsilon_o \quad & \quad \epsilon_o\mu_o \quad & \ \chi \ & \xi \\ \hline \text{BIPM - SI} \rule{0pt}{18pt} & k_o & \frac{1}{4\pi} & \frac{1}{c^2} & 1 & 4\pi \\ \text{ES abs} \rule{0pt}{14pt} & 1 & 1 & \frac{1}{c^2} & 1 & 1 \\ \text{EM abs} \rule{0pt}{14pt} & c^2 & 1 & \frac{1}{c^2} & 1 & 1 \\ \text{Gauss} \rule{0pt}{14pt} & 1 & 1 & 1 & c & 1 \\ \text{Heaviside} \rule{0pt}{14pt} \rule[-10pt]{0pt}{10pt} & \frac{1}{4\pi} & \frac{1}{4\pi} & 1 & c & 4\pi \\ \hline \end{array}

So e.g. Maxwell's equations can be written in a general form, which easily includes all schemes as special cases.

\begin{cases} \begin{aligned} & \nabla\cdot\boldsymbol{D} = \frac{4\pi}{\xi} \rho && \boldsymbol{D} = \epsilon_o\boldsymbol{E} + \frac{4\pi}{\xi}\boldsymbol{P} \\ & \nabla\cdot\boldsymbol{B} = 0 \\ & \nabla\times\boldsymbol{E} = -\frac{1}{\chi}\frac{\partial\boldsymbol{B}}{\partial t} \\ & \nabla\times\boldsymbol{H} = \frac{4\pi}{\xi\chi}\boldsymbol{J} + \frac{1}{\chi}\,\frac{\partial\boldsymbol{D}}{\partial t} \qquad && \boldsymbol{H} = \frac{\boldsymbol{B}}{\mu_o} - \frac{4\pi}{\xi}\boldsymbol{M} \\ \end{aligned} \end{cases}

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