# Confusion of the electrical conductivity unit in CGS.

Hello.

Electrical conductivity is written as $\frac{n_{e}e^{2}}{m_{e}\nu_{ei}}$ where $n_{e}, e, m_{e}$ and $\nu_{ei}$ are for electron density, charge, mass and electron-ion collision frequency.

According to Wikipedea, http://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity

Unit of the conductivity is same to the frequency as 1/s (s for second).

I checked this is true by substituting CGS dimensions of each parameters in the expression.

How can I interpret this? conductivity is some type of frequency? Does it make sense??

Moreover, for a given conductivity in CGS unit, how can I convert it to SI unit which is the most useful for experimentalist?

vanhees71
Gold Member
Conductivity is a kind of friction coefficient. The naive derivation goes as follows:

The electrons, making up the current, move in the conductor, undergoing collisions with the surrounding matter (e.g. in a wire with the lattice vibrations of the crystal (phonons) or defects, etc.). The (non-relativistically approximated) equation of motion for an electron is thus given by the friction force and the force due to an external electric field:
$$m\ddot{\vec{x}}=-m \gamma \dot{\vec{x}}+q \vec{E}.$$
For simplicity take a constant electric field. Then this equation of motion has the general solution for the velocity
$$\vec{v}(t)=\vec{C} \exp(-\gamma t) + \frac{q}{\gamma m} \vec{E}.$$
With an integration constant $\vec{C}$. After some time $t \gg \frac{1}{\gamma}$ a stationary state has been established:
$$\vec{v}_{\infty}=\frac{q}{\gamma m}\vec{E}.$$
The current, made up of the conduction electrons is then
$$\vec{j}=n_{\text{cond}} q \vec{v}=\frac{n_{\text{cond}} q^2}{\gamma m} \vec{E}.$$
From this it follows that the electric conductivity is given by
$$\sigma=\frac{n_{\text{cond}} q^2}{\gamma m}.$$
The units are as follows.

In the Gaussian or Heaviside Lorentz system (the latter is the best from a theoretical physics point of view since it doesn't introduce difficulties such as to make the electric and magnetic components of the one and only electromagnetic field measured in different units, which is comparable to measuring distances along the north-south directions in meters and such in east-west directions in yards, as Schwartz writes in his excellent textbook): $[n_{\text{cond}}]=1/\mathrm{cm}^3$, $[\gamma]=1/\text{s}$, $[q^2]=\text{g} \text{cm}^3/\text{s}^2$, $[m]=g$. Putting this together gives indeed $$[\sigma]=1/\text{s}$$.

The complication in the Gaussian system of units (here I use the non-rationalized units as used in Jackson; in the 3rd edition he uses SI in the early and non-rationalized Gaussian units in the later chapters) is about the units of charge, currents, etc., because it defines the unit of charge by putting Coulomb's law of two repelling static electrons in the form
$$F=\frac{q^2}{r^2}.$$
The unit of charge is thus
$$[q]=\sqrt{[F]}[r]=\sqrt{\text{g} \text{cm}/\text{s}} \; \text{cm}=\frac{\text{g} \; \text{cm}^{3/2}}{\text{s}^{1/2}}.$$
The non-integer powers of the charge one tries to avoid in the SI. Thus one introduces (indirectly via Ampere's Law for the definition of the Ampere) an extra unit for charge, the Coulomb, paying the price to introduce a (pretty artificial) conversion constant into Coulomb's force law:
$$F=\frac{q^2}{4 \pi \epsilon_0 r^2}.$$
Here, the $4 \pi$ is quite natural. It's absent in the Gaussian units, because one puts it explicitly into the inhomogeneous Maxwell equations. In the Heaviside-Lorentz (rationalized Gaussian) units one leave them out in the Maxwell equations, and then it also appears in Coulomb's Law. The "dieelectric constant of the vacuum", $\epsilon_0$, is nothing but an artificial conversion factor for the units. Obviously we have
$$[q^2/\epsilon_0]=\mathrm{C}^2/[\epsilon_0]=[F]=\frac{\text{kg} \; \text{m}}{\text{s}^2}$$
and thus
$$[\epsilon_0]=\frac{\mathrm{C}^2 \text{s}^2}{\text{kg} \; \text{m}}.$$
Anyway, the formula for conductivity is the same in SI units, because there's no difference in the formula for the force on the particle for a given electric field. It's $F=q E$ in both systems of units. Only the units change. We have
$$[\sigma]=\left [\frac{n q^2}{m \gamma} \right]=\frac{\mathrm{C}^2 s}{\text{kg} \; \text{m}^3}.$$
This becomes more practical for technology when relating it to the units of electrical resistance in the macroscopic form of Ohm's Law:
$$U=R I.$$
The resistance of a in terms of conductivity is given by
$$R=\frac{L}{A \sigma},$$
where $A$ is the cross sectional area of the wire and $L$ its length. Since resistance is measured in Ohm , we have
$$[\sigma]=\frac{[L]}{[A] [R]}=\frac{1}{\text{m} \Omega}=\frac{\text{S}}{\text{m}},$$
where $1 \text{S}=1/\Omega$ is the unit Siemens, i.e., inverse Ohms.

1 person
Meir Achuz
Homework Helper
Gold Member
The units of sigma are best seen from its definition in all systems of units as ${\bf j}=\sigma{\bf E}$,
without introducing a naive (and wrong) model of current.
Then, in Gaussian units, the units of charge cancel, leaving 1/sec as the unit for sigma.
In SI units, this just gets multiplied by $4\pi\epsilon_0$, giving more complicated units, and honoring two great physicists.

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1 person
Thanks.

I still hard to accept the unit of the conductivity in CGS unit since it is physically not intuitive for me...Anyway thanks for all!

Is it right that conversion from CGS to SI is just being muliplied by $4\pi\epsilon_{0}$?

vanhees71
Gold Member
Let's see, whether I can figure out the messy SI units (just kidding).

The Coulomb force is written in both systems of units as
$$F=\frac{q_{\text{G}}^2}{r^2}=\frac{q_{\text{SI}}^2}{4 \pi \epsilon_0}.$$
So we have
$$q_{\text{SI}}^2=4 \pi \epsilon_0 q_{\text{G}}^2.$$
Now we have
$$\sigma_{\text{G}}=\frac{n q_{\text{G}}^2}{\gamma m},$$
where we have used that all quantities except the charge have the same units (apart from usually using cm and g in Gauß instead of m and kg in SI units). In the SI we have
$$\sigma_{\text{SI}}=\frac{n q_{\text{SI}}^2}{\gamma m}=4 \pi \epsilon_0 \sigma_{\text{G}}.$$

Meir Achuz
Homework Helper
Gold Member
Thanks.

I still hard to accept the unit of the conductivity in CGS unit since it is physically not intuitive for me...Anyway thanks for all!

Is it right that conversion from CGS to SI is just being muliplied by $4\pi\epsilon_{0}$?
1. Is
$$[\sigma]=\left [\frac{n q^2}{m \gamma} \right]=\frac{\mathrm{C}^2 s}{\text{kg} \; \text{m}^3}.$$
more physically intuitive?
2. Yes.

Meir Achuz
Homework Helper
Gold Member
To go from SI to the more complicated Gaussian units, let $1/4\pi\epsilon_0\rightarrow 1$.
To simplify Gaussian, by going to SI, let $1\rightarrow 1/4\pi\epsilon_0$.

vanhees71
Gold Member
It's the opposite way around: the Gaussian units are more natural and thus simpler than the SI units. Any system of units, where the electric and magnetic components of the electromagnetic field do not have the same units, are not natural at all. It's indeed as measuring distances in one direction in meters and in another in yards!

The only inestheticity of Gaussian units is the appearance of factors $4 \pi$ in the fundamental Maxwell equations. This is repaired by using Heaviside-Lorentz units, where the factors $4 \pi$ appear at the places, where they belong, as e.g., in the Coulomb-force law, and not in the fundamental equations.

Meir Achuz
Homework Helper
Gold Member
It's the opposite way around
Of course. Sometimes my humor is too subtle.