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Unit Conversion

K.E. Schmidt
Department of Physics and Astronomy
Arizona State University
Tempe, AZ 85287

1 Simplied Conversion from SI

The appendix on units in Jackson's book gives several useful tables to convert between different systems of units. Since most of physics uses either Gaussian or Heaviside Lorentz units, it is useful to be able to quickly convert SI equations to these units. Jackson's tables allow you to do this, but I use a related but quicker way to get the same results. The differences between this method and Jackson's is that I simply drop all the $ \epsilon_{0}^{}$ and $ \mu_{0}^{}$ factors and use dimensional analysis to get the correct factors of c, and I explicitly show how to determine the appropriate conversion factors for quantities not mentioned in Jackson's tables.

I divide quantities into categories. These are

For any new quantity, you need to put it into one of these categories to proceed. Note that the polarization and the magnetization are viewed as ``Sources'' not fields. They represent the currents and charge density in the material not specifically given by $ \vec{J} $ and $ \rho$.

The rules are simple. Given an equation in SI:

  1. Set $ \epsilon_{0}^{}$, $ \mu_{0}^{}$, and c to 1.
  2. Divide all ``Fields and Potentials'' by $ \sqrt{4 \pi}$.
  3. Multiply all ``Sources'' by $ \sqrt{4 \pi}$.
  4. Include the appropriate factors of 4$ \pi$ for any ``Responses.'' See below on how to do this.
  5. If you want to put back the appropriate powers of c, include appropriate powers of c for each term so that the units agree.

To convert to Heaviside-Lorentz, skip steps 2, 3 and 4.

For all equations that do not contain ``Responses,'' (which includes most of them in Jackson's book) the conversion is trivial. Response functions have the same definitions in all units. The conversion factor for them is therefore the appropriate factor of 4$ \pi$ to keep the definition invariant. Here are some examples that encompass the ``Responses'' given in the appendix of Jackson's book.

Resistance is the ratio of voltage to current,

V = IR . (1)

Applying our rules, we divide V by $ \sqrt{4 \pi}$ and multiply I by $ \sqrt{4 \pi}$. To keep the equation the same, we must divide RSI by 4$ \pi$ to get the gaussian formula. Therefore resistances in SI equations need to be divided by 4$ \pi$ when converting to Gaussian units.

Similarly, you can write

V = L$\displaystyle {\frac{dI}{dt}}$  
I = C$\displaystyle {\frac{dV}{dt}}$  
Q = CV  
$\displaystyle \vec{J} $ = $\displaystyle \sigma$$\displaystyle \vec{E} $  
$\displaystyle \vec{P} $ = $\displaystyle \chi_{e}^{}$$\displaystyle \vec{E} $  
$\displaystyle \vec{M} $ = $\displaystyle \chi_{m}^{}$$\displaystyle \vec{H} $ (2)

etc. to see that you also need to divide the LSI by 4$ \pi$, multiply CSI and $ \sigma_{SI}^{}$ by 4$ \pi$, and multiply the susceptibilities $ \chi_{e}^{}$ and $ \chi_{m}^{}$ by 4$ \pi$.

2 Examples

Here are some examples to help you.

2.1 Example 1

Jackson Eq. 9.24 in SI is the total power radiated from an oscillating electric dipole

P = $\displaystyle {\frac{c^2 Z_0 k^4}{12\pi}}$|$\displaystyle \vec{p} $|2   SI (3)

Power is a mechanical unit, and the definition of the dipole moment $ \vec{p} $ is charge times distance in both sets of units, so it is a source. Z0 is $ \sqrt{\mu_0/\epsilon_0}$ and therefore is replaced by 1. Applying our rules,

P = $\displaystyle {\frac{k^4}{3}}$|$\displaystyle \vec{p} $|2   Gaussian c=1 (4)

Since q2/r is an energy in Gaussian units, p2 has units of energy-length3. The units of the right hand side are therefore energy/length, and we need to multiply by 1 power of c to get power. The final result is

P = $\displaystyle {\frac{c k^4}{3}}$|$\displaystyle \vec{p} $|2   Gaussian (5)

2.2 Example 2

Jackson problem 5.34, the mutual inductance is

M12 = $\displaystyle \mu_{0}^{}$$\displaystyle \pi$a2$\displaystyle \int_{0}^{\infty}$dke-kRJ1(ka)   SI (6)

We divide the inductance by 4$ \pi$ and drop the $ \mu_{0}^{}$. The result is

M12 = 4$\displaystyle \pi^{2}_{}$a2$\displaystyle \int_{0}^{\infty}$dke-kRJ1(ka)   Gaussian c=1 (7)

Since V is charge/distance and I is charge/time, the units of inductance are time2/distance. The right side has units of distance. Therefore, I need to divide the right side by c2 to make them agree,

M12 = $\displaystyle {\frac{4 \pi^2 a^2}{c^2}}$$\displaystyle \int_{0}^{\infty}$dke-kRJ1(ka)   Gaussian (8)

2.3 Example 3

Jackson Eq. 4.83 gives the electrostatic energy in vacuum

W = $\displaystyle {\textstyle\frac{1}{2}}$$\displaystyle \int$d3r$\displaystyle \rho$($\displaystyle \vec{r} $)$\displaystyle \Phi$($\displaystyle \vec{r} $)   SI (9)

W is mechanical, $ \rho$ is a source and $ \Phi$ is a field, so the $ \sqrt{4 \pi}$ factors cancel. The unit of q$ \Phi$ is energy, so no c factors are required,

W = $\displaystyle {\textstyle\frac{1}{2}}$$\displaystyle \int$d3r$\displaystyle \rho$($\displaystyle \vec{r} $)$\displaystyle \Phi$($\displaystyle \vec{r} $)   Gaussian (10)

2.4 Example 4

Jackson Eq. 4.70, gives the Clausius-Mosotti result for the molecular polarizability in terms of the density and permitivity.

$\displaystyle \gamma_{mol}^{}$ = $\displaystyle {\frac{3}{N}}$$\displaystyle \left(\vphantom{
\frac{ \epsilon / \epsilon_0 -1}{ \epsilon / \epsilon_0 +2 }
}\right.$$\displaystyle {\frac{\epsilon / \epsilon_0 -1}{\epsilon / \epsilon_0 +2 }}$$\displaystyle \left.\vphantom{
\frac{ \epsilon / \epsilon_0 -1}{ \epsilon / \epsilon_0 +2 }
}\right)$   SI . (11)

We need to identify the sort of quantity that $ \gamma_{mol}^{}$ is. From Eq. 4.67,

$\displaystyle \langle$$\displaystyle \vec{p}_{mol}^{}$$\displaystyle \rangle$ = $\displaystyle \epsilon_{0}^{}$$\displaystyle \gamma_{mol}^{}$($\displaystyle \vec{E} $ + $\displaystyle \vec{E}_{i}^{}$)   SI (12)

and we see that $ \gamma$ is a response. The corresponding Gaussian equation (note the usual expression for polarizability in SI units does not contain $ \epsilon_{0}^{}$ either - Jackson uses a nonstandard convention)

$\displaystyle \langle$$\displaystyle \vec{p}_{mol}^{}$$\displaystyle \rangle$ = $\displaystyle \gamma_{mol}^{}$($\displaystyle \vec{E} $ + $\displaystyle \vec{E}_{i}^{}$)   Gaussian (13)

since $ \vec{p}_{mol}^{}$ is a source (as in the example above), the electric fields are fields, the SI response $ \gamma_{mol}^{}$ must be multiplied by 4$ \pi$ to get the Gaussian equation. Eq. 4.70 therefore becomes

$\displaystyle \gamma_{mol}^{}$ = $\displaystyle {\frac{3}{4\pi N}}$$\displaystyle \left(\vphantom{
\frac{\epsilon -1}{\epsilon +2}
}\right.$$\displaystyle {\frac{\epsilon -1}{\epsilon +2}}$$\displaystyle \left.\vphantom{
\frac{\epsilon -1}{\epsilon +2}
}\right)$   Gaussian . (14)

and checking the units, the polarizabiility has units of volume and so does 1/N, so no factors of c are needed.

2.5 Example 5

Jackson Eq. 6.105 is a statement of Poynting's theorem,

$\displaystyle \int_{V}^{}$d3r$\displaystyle \vec{J} $ . $\displaystyle \vec{E} $ = - $\displaystyle \int_{V}^{}$d3r$\displaystyle \left[\vphantom{
\vec \nabla \cdot (\vec E \times \vec H) +
\vec ...
... \vec D}{\partial t}
+ \vec H \cdot \frac{\partial \vec B}{\partial t}
}\right.$$\displaystyle \vec{\nabla} $ . ($\displaystyle \vec{E} $×$\displaystyle \vec{H} $) + $\displaystyle \vec{E} $ . $\displaystyle {\frac{\partial \vec D}{\partial t}}$ + $\displaystyle \vec{H} $ . $\displaystyle {\frac{\partial \vec B}{\partial t}}$$\displaystyle \left.\vphantom{
\vec \nabla \cdot (\vec E \times \vec H) +
\vec ...
... \vec D}{\partial t}
+ \vec H \cdot \frac{\partial \vec B}{\partial t}
}\right]$   SI (15)

In this equation $ \vec{E} $, $ \vec{D} $, $ \vec{B} $, $ \vec{H} $, are all fields, and $ \vec{J} $ is a source. Putting in the 4$ \pi$ factors,

$\displaystyle \int_{V}^{}$d3r$\displaystyle \vec{J} $ . $\displaystyle \vec{E} $ = - $\displaystyle {\frac{1}{4\pi}}$$\displaystyle \int_{V}^{}$d3r$\displaystyle \left[\vphantom{
\vec \nabla \cdot (\vec E \times \vec H) +
\vec ...
... \vec D}{\partial t}
+ \vec H \cdot \frac{\partial \vec B}{\partial t}
}\right.$$\displaystyle \vec{\nabla} $ . ($\displaystyle \vec{E} $×$\displaystyle \vec{H} $) + $\displaystyle \vec{E} $ . $\displaystyle {\frac{\partial \vec D}{\partial t}}$ + $\displaystyle \vec{H} $ . $\displaystyle {\frac{\partial \vec B}{\partial t}}$$\displaystyle \left.\vphantom{
\vec \nabla \cdot (\vec E \times \vec H) +
\vec ...
... \vec D}{\partial t}
+ \vec H \cdot \frac{\partial \vec B}{\partial t}
}\right]$   Gaussian c=1 (16)

The left side has units of Power. The first term on the right has units of energy/length, so it needs to be multiplied by a factor of c, the last two terms have units of energy/time, so they are fine. The result is

$\displaystyle \int_{V}^{}$d3r$\displaystyle \vec{J} $ . $\displaystyle \vec{E} $ = - $\displaystyle {\frac{1}{4\pi}}$$\displaystyle \int_{V}^{}$d3r$\displaystyle \left[\vphantom{
c \vec \nabla \cdot (\vec E \times \vec H) +
\ve...
... \vec D}{\partial t}
+ \vec H \cdot \frac{\partial \vec B}{\partial t}
}\right.$c $\displaystyle \vec{\nabla} $ . ($\displaystyle \vec{E} $×$\displaystyle \vec{H} $) + $\displaystyle \vec{E} $ . $\displaystyle {\frac{\partial \vec D}{\partial t}}$ + $\displaystyle \vec{H} $ . $\displaystyle {\frac{\partial \vec B}{\partial t}}$$\displaystyle \left.\vphantom{
c \vec \nabla \cdot (\vec E \times \vec H) +
\ve...
... \vec D}{\partial t}
+ \vec H \cdot \frac{\partial \vec B}{\partial t}
}\right]$   Gaussian (17)

2.6 Example 6

The relations between B, H and M, and between E, D and P are in SI (Jackson Eq. 4.34 and 5.81)
$\displaystyle \vec{D} $ = $\displaystyle \epsilon_{0}^{}$$\displaystyle \vec{E} $ + $\displaystyle \vec{P} $   SI  
$\displaystyle \vec{H} $ = $\displaystyle {\frac{1}{\mu_0}}$$\displaystyle \vec{B} $ - $\displaystyle \vec{M} $   SI (18)

Recalling the $ \vec{P} $ and $ \vec{M} $ are sources, while the others are fields, we get
$\displaystyle \vec{D} $ = $\displaystyle \vec{E} $ + 4$\displaystyle \pi$$\displaystyle \vec{P} $   Gaussian  
$\displaystyle \vec{H} $ = $\displaystyle \vec{B} $ - 4$\displaystyle \pi$$\displaystyle \vec{M} $   Gaussian (19)

where the units of all six quantities are the same, so no factors of c are needed.

2.7 Example 7

The plasma frequency squared is defined in Jackson Eq. 7.60,

$\displaystyle \omega_{p}^{2}$ = $\displaystyle {\frac{N Z e^2}{\epsilon_0 m}}$   SI . (20)

e is a source, and the rest are mechanical units, so

$\displaystyle \omega_{p}^{2}$ = $\displaystyle {\frac{4 \pi N Z e^2}{m}}$   Gaussian . (21)

The units of N is length-3 while e2 is energy-length, so the unit of Ne2 are energy-length-2 or mass-time-2, dividing by mass we see that the units are correct and no factors of c are needed.

2.8 Example 8

The skin depth is given in Jackson Eq. 5.165,

$\displaystyle \delta$ = $\displaystyle \sqrt{\frac{2}{\mu \sigma \omega}}$   SI . (22)

By looking at Ohm's law $ \vec{J} $ = $ \sigma$$ \vec{E} $, we saw that we need to multiply the response $ \sigma$ by 4$ \pi$. Therefore

$\displaystyle \delta$ = $\displaystyle \sqrt{\frac{1}{2 \pi \mu \sigma \omega}}$   Gaussian c=1 . (23)

$ \sigma$ has units of time-1 and so does $ \omega$, so the right side has units of time while the left side has units of length. Therefore

$\displaystyle \delta$ = $\displaystyle \sqrt{\frac{c^2}{2 \pi \mu \sigma \omega}}$   Gaussian . (24)

3 Conversion From Gaussian or Heaviside-Lorentz to SI

The same technique can be used to convert to SI if that is your desire. Presumably if you are converting into SI, you know the units of your quantities in SI. Starting from Heaviside Lorentz, set c = 1 and include factors of $ \epsilon_{0}^{}$ and $ \mu_{0}^{}$ to get the units right. Starting from Gaussian, you set c = 1, and include the the inverse of the 4$ \pi$ and $ \sqrt{4 \pi}$ factors in steps 2 to 4 above, and then include factors of $ \epsilon_{0}^{}$ and $ \mu_{0}^{}$ to get the units right. It's painful for me to fool around with the SI units, so I won't give many examples. The units of various quantities can be found from the fundamental equations. Recall that along with the mass length and time units, SI adds an extra current unit, the Ampere, along with an extra conversion factor. The Lorentz force law shows that the electric field $ \vec{E} $ has units of Kilogram-meter/(Ampere second3), while the divergence of $ \vec{D} $ equation shows that $ \vec{D} $ has units of Ampere-second/meter2, and therefore $ \epsilon_{0}^{}$ has units of Ampere2-second4/Kilogram-meter3 or Ampere2-second2/(Joule-meter). Similarly, $ \vec{B} $ has units of Kilogram/(Ampere second2), while $ \vec{H} $ has units of Ampere/meter, so that $ \mu_{0}^{}$ has units of Newton/Ampere2. Since $ \epsilon_{0}^{}$$ \mu_{0}^{}$ = c-2, multiplying by the speed of light is equivalent to multiplying by ($ \epsilon_{0}^{}$$ \mu_{0}^{}$)-1/2. So the easiest way to convert is to use either $ \epsilon_{0}^{}$ or $ \mu_{0}^{}$ to get the Ampere units correct, and then include speeds of light to get the mechanical units right.

3.1 SI Example 1

We saw in Example 1, that

P = $\displaystyle {\frac{c k^4}{3}}$|$\displaystyle \vec{p} $|2   Gaussian (25)

Converting back, noting as before that $ \vec{p} $ is a source,

P = $\displaystyle {\frac{k^4}{12 \pi}}$|$\displaystyle \vec{p} $|2   SI $\displaystyle \epsilon_{0}^{}$ = $\displaystyle \mu_{0}^{}$ = 1 (26)

The units of $ \vec{p} $ are Ampere-second-meter, so the right hand side has units of Ampere2-second2/meter2, and the left side has units of Kilogram-meter2/second3. To match the Ampere units, I multiply the right side by $ \mu_{0}^{}$.

P = $\displaystyle {\frac{k^4 \mu_0}{12 \pi}}$|$\displaystyle \vec{p} $|2   SI c = 1 (27)

The right side now has units of Newton-second2/meter2, multiplying by c3 gives power units, so the result is

P = $\displaystyle {\frac{k^4 \mu_0 c^3}{12 \pi}}$|$\displaystyle \vec{p} $|2   SI (28)

3.2 SI Example 2

Coulomb's law in Gaussian units is

| F| = $\displaystyle {\frac{q_1 q_2}{d^2}}$ Gaussian (29)

where d is the distance and | F| is the magnitude of the force. Since q1 and q2 are sources, we divide them by $ \sqrt{4 \pi}$ to convert from Gaussian to SI.

| F| = $\displaystyle {\frac{q_1 q_2}{4 \pi d^2}}$ SI $\displaystyle \epsilon_{0}^{}$ = $\displaystyle \mu_{0}^{}$ = 1 (30)

The units of the right side are Ampere2-second2/meter2, while the right side has units of Newton. To match the Ampere units, I multiply the right side by $ \mu_{0}^{}$,

| F| = $\displaystyle {\frac{\mu_0}{4\pi}}$$\displaystyle {\frac{q_1 q_2}{d^2}}$ SI c = 1 (31)

The units of the right side are now Newton-second2/meter2, so I need to multiply the right side by c2 to make the dimensions agree,

| F| = $\displaystyle {\frac{\mu_0 c^2}{4\pi}}$$\displaystyle {\frac{q_1 q_2}{d^2}}$ SI (32)

which is Coulomb's force law in SI.


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