Up: PHY531 Classical Electrodynamics Links
Unit Conversion
K.E. Schmidt
Department of Physics and Astronomy
Arizona State University
Tempe, AZ 85287
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
and 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
 Mechanical Units: mass, length, time.
 Fields and Potentials: ,,,,,.
 Sources: , q, , I, , .
 Responses: conductivity , Resistance R, Capacitance C,
Inductance L, Susceptibilities .
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 and .
The rules are simple. Given an equation in SI:
 Set
, , and c to 1.
 Divide all ``Fields and Potentials'' by
.
 Multiply all ``Sources'' by
.
 Include the appropriate factors of 4 for any ``Responses.'' See
below on how to do this.
 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 HeavisideLorentz, 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 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,
Applying our rules, we divide V by
and multiply
I by
. To keep the equation the same, we must divide
R_{SI} by 4 to get the gaussian formula. Therefore resistances
in SI equations need to be divided by 4 when converting to
Gaussian units.
Similarly, you can write
V 
= 
L 

I 
= 
C 

Q 
= 
CV 


= 



= 



= 

(2) 
etc. to see that you also need to divide the L_{SI} by 4,
multiply C_{SI} and
by 4, and multiply the
susceptibilities and by 4.
Here are some examples to help you.
Jackson Eq. 9.24 in SI is the total power radiated from an oscillating
electric dipole
P = ^{2} SI

(3) 
Power is a mechanical unit, and the definition of the dipole moment
is charge times distance in both sets of units, so it is a source. Z_{0}
is
and therefore is replaced by 1.
Applying our rules,
P = ^{2} Gaussian c=1

(4) 
Since q^{2}/r is an energy in Gaussian units, p^{2} has units of
energylength^{3}. 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 = ^{2} Gaussian

(5) 
Jackson problem 5.34, the mutual inductance is
M_{12} = a^{2}dke^{kR}J_{1}(ka) SI

(6) 
We divide the inductance by 4 and drop the . The result
is
M_{12} = 4a^{2}dke^{kR}J_{1}(ka) Gaussian c=1

(7) 
Since V is charge/distance and I is charge/time, the units
of inductance are time^{2}/distance. The right side has
units of distance. Therefore, I need to divide the right side
by c^{2} to make them agree,
M_{12} = dke^{kR}J_{1}(ka) Gaussian

(8) 
Jackson Eq. 4.83 gives the electrostatic energy in vacuum
W is mechanical, is a source and is a field, so
the
factors cancel.
The unit of q is energy, so no c factors are required,
W = d^{3}r()() Gaussian

(10) 
Jackson Eq. 4.70, gives the ClausiusMosotti result for the
molecular polarizability in terms of the density and permitivity.
We need to identify the sort of quantity that
is.
From Eq. 4.67,
and we see that is a response. The corresponding Gaussian
equation (note the usual expression for polarizability in SI
units does not contain
either  Jackson uses a nonstandard
convention)
since
is a source (as in the example above),
the electric fields are fields,
the SI response
must be multiplied by 4 to get the Gaussian equation.
Eq. 4.70 therefore becomes
and checking the units, the polarizabiility has units of volume and
so does 1/N, so no factors of c are needed.
Jackson Eq. 6.105 is a statement of Poynting's theorem,
d^{3}r^{ . } =  d^{3}r^{ . }(×) + ^{ . } + ^{ . } SI

(15) 
In this equation ,
, , , are all fields, and is a source.
Putting in the 4 factors,
d^{3}r^{ . } =  d^{3}r^{ . }(×) + ^{ . } + ^{ . } 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
d^{3}r^{ . } =  d^{3}rc ^{ . }(×) + ^{ . } + ^{ . } Gaussian

(17) 
The relations between B, H and M, and between E, D and P
are in SI (Jackson Eq. 4.34 and 5.81)
Recalling the and are sources, while the others
are fields, we get

= 
+ 4 Gaussian 


= 
 4 Gaussian 
(19) 
where the units of all six quantities are the same, so no factors
of c are needed.
The plasma frequency squared is defined in Jackson Eq. 7.60,
= SI .

(20) 
e is a source, and the rest are mechanical units, so
= Gaussian .

(21) 
The units of N is length^{3} while e^{2} is energylength, so
the unit of Ne^{2} are energylength^{2} or masstime^{2},
dividing by mass we see that the units are correct and no factors
of c are needed.
The skin depth is given in Jackson Eq. 5.165,
= SI .

(22) 
By looking at Ohm's law
= , we saw that
we need to multiply the response by 4. Therefore
= Gaussian c=1 .

(23) 
has units of time^{1} and so does , so the
right side has units of time while the left side has units of length.
Therefore
= Gaussian .

(24) 
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
and to get the units right. Starting from Gaussian, you
set c = 1, and include the
the inverse of the 4 and
factors in steps 2 to 4
above, and then include factors of
and 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 has units of
Kilogrammeter/(Ampere second^{3}), while the divergence of
equation shows that has units of Amperesecond/meter^{2},
and therefore
has units of Ampere^{2}second^{4}/Kilogrammeter^{3}
or Ampere^{2}second^{2}/(Joulemeter).
Similarly, has units
of
Kilogram/(Ampere second^{2}), while has units of Ampere/meter,
so that has units of Newton/Ampere^{2}. Since
= c^{2}, multiplying by the speed of light is equivalent to multiplying
by
()^{1/2}. So the easiest way to convert is to
use either
or to get the Ampere units correct,
and then include speeds of light to get the mechanical units right.
We saw in Example 1, that
P = ^{2} Gaussian

(25) 
Converting back, noting as before that is a source,
The units of are Amperesecondmeter, so the right
hand side has units of
Ampere^{2}second^{2}/meter^{2}, and the left side has units of
Kilogrammeter^{2}/second^{3}. To match the Ampere units, I multiply
the right side by .
P = ^{2} SI c = 1

(27) 
The right side now has units of Newtonsecond^{2}/meter^{2}, multiplying
by c^{3} gives power units, so the result is
P = ^{2} SI

(28) 
Coulomb's law in Gaussian units is
 F = Gaussian

(29) 
where d is the distance and  F is the magnitude of the force.
Since q_{1} and q_{2} are sources, we divide them by
to convert from Gaussian to SI.
The units of the right side are Ampere^{2}second^{2}/meter^{2}, while
the right side has units of Newton. To match the Ampere units,
I multiply the right side by ,
 F = SI c = 1

(31) 
The units of the right side are now Newtonsecond^{2}/meter^{2}, so
I need to multiply the right side by c^{2} to make the dimensions
agree,
 F = SI

(32) 
which is Coulomb's force law in SI.
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