
Physicists love (one of) the cgs systems
and stubbornly keep using it  even so it is "forbidden". Take the
Barrett for example, a
rather recent book  you will find cgs
units. Here we will see why the
cgs system still holds a lot of attraction, and how to convert
cgs units to SI units. 

As long as you just use the basic units length, mass and
time, it really does not matter much if you
work with the m, kg, s, i.e. in SI units, or with cm, g and s  cgs units. 


Engineers at your (future) level of
sophistication simply can do the conversions without having to be taught
and without making mistakes. 

A (big) problem, however, emerges as soon as you
add some basic unit for electricity. 


The SI system chose the electrical current
with the unit Ampere (A)  and that is all there is. Like with the
meter, you now need some arbitrary, but generally accepted reference that defines 1 A (and at the same
time gives a recipe how it will be measured). 


For the meter, we originally picked 1/10 000 000 part of
the circumference of the earth and deposited that as a PtIr piece in
Paris. Later it was replaced by something better, but the general idea is the
same. 


Likewise for the
kilogram and the
second  but what do we take
for the Ampere? 

Well, something not all that smart (from the
viewpoint of physicists and practicing engineers): 


1 Ampere
is the magnitude of a constant electrical current, which, if running through
two infinitely long parallel wires with negligeable circular cross sections
kept at a distance of 1 m in vacuum, produces a force of exactly 2
· 10 ^{–7} N per meter of wire. 


This definition also defines
charge in Coulomb, by simply equating
Charge [C] = (Ampere · Time) [As], i.e.
1 C = 1 As. 


Who needs forces between wires? We rarely do. What
we do need a lot, however, are forces
between charges. Lets see what we get for
this. 

We start from the universal relation between an
electrical field E and the force F that a charge
Q experiences in said field (with the very important corollary
that the field produced by Q is not added to the field already there! Asking
"why" leads into really deep water, cf. chapter 28 (Vol. II) in the
Feynman lectures (perfectly
understandable to undergraduates!)) 


We have F = E ·
Q, and since F and Q are already
defined, this equation defines
E. Enter a number for F and Q and you
get a number for E. You akso get a unit:
[E] = N/As 


So far we have no problem. But now we look at the
force on a (point charge) q that results from another point
charge q' by first computing
the field of one of the charges and than applying the formula from above. 

Lets take a "point charge"
q and the simple statement (from Maxwell) that the electrical
flux density P through a closed
surface around a charge is proportional to the charge inside the surface. We
take it as proportional, because the
numerical value of the proportionality constant will depend on the choice of
units, which we try to unravel. 


Take a sphere with radius r around
q and you have 





P 
= Edf 
= 4pr^{2} ·
E = g
·q 







With g =
proportionality constant and df = incremental area element. This
gives us for the numerical value of the electrical field strength at a distance
r from a point charge q 











Now we know the field at some distance
r from the charge, and therefore the force F on a
charge q'; we have 





F 
= q · q' · 
g
4p · r^{2} 







This is completely general, as long as we make no
assumptions for g. 

OK, now lets
discuss the possibilities for the value of g. 


If we use SI units, we have no choice: We already have definitions for force, charge
and the electrical field E in
SI units the numericla value of g is
determined. We simply have (writing g as
g = 1/e_{0} 





F 
= q · q' · 
1
4p · e_{0}
· r^{2} 







With e_{0} =
8,859 · 10 ^{–12} As/Vm = proportionality constant
between the charge q inside some body and the total flux P through the surface of that body, or 





P 
= Edf = 
q
e_{0} 







Forgetting for a moment that we must use SI units, we could make life a lot
easier by simply defining g := 4p, and presto, we have a Coulomb law as you find it
quite often in text books, that can be written most simply as 











It's easy to see why people like that  you simply
save a lot of boring writing. 

However, since
you already have defined lengths and forces somehow (in the
cgs system they were given
in "dyn" (1 dyn = 1 gm/s^{2} = 10^{5}
N)), you are now making a statement as to how you
measure charge. The easiest thing to do is to define charge in such
a way that we get a unit of force (= 1 dyn) if two units of charge (=
???) are one unit of distance (= 1 cm) apart. 


This means we take the numerical value of
q and q' to be q = q' = 1 [?],
and the distance r to be 1 cm. 


Your force must come out to be one dyn; we have
F = 1dyn = 1 gcm/s^{2} = 1 [?]·[?]/cm^{2}
which gives our unit of charge to be [q]^{2} = g ·
cm^{3}/s^{2}, or 





[q] 
= 
g^{1/2} · cm^{3/2} · s^{–1} 






A bit strange, but who cares. You simply call it
an "electrostatic unit (esu)" (ESL
in German; for "elektrostatische Einheit"). 


Essentially, instead of defining the unit of
current (= 1 A) by some force law, you define a charge by some other (but more frequently used)
force law. Add it to your cgs basic units as the essential required
input for electricity, and you have the "electrostatic" cgs system, sometimes
called CGSF system (the "F" stands for "Franklin, [Fr]", which was the name
given to the unit of charge); i.e. 1 Fr = 1 g^{1/2} ·
cm^{3/2} · s^{–1}. 


However, nobody uses the name "Franklin" anymore, you just call it
electrostatic unit charge, or esu, or whatever. 


Look at this copy form the Feynman lectures, to
see how pissed people become with the SI convention! 










Now, and this brings in a lot of confusion,
instead of a special measure for charge
neede to make the cgs system "electric", you also could add
something else "electrical"  and out come many different kinds of cgs unit systems. 


But that is only for freaks (if you run across it
and cannot avoid it  look it up in a really good handbook). 

Here we will only give some
"translation" table, converting quantities from one system to the
other. It is more tricky than it looks like! 

Conversion of
charge. 


We must ask ourselves: How many esu per s do we have to run through our wires from
above to produce a force of 2 · 10 ^{–7} N per meter
of wire? This then must be 1 C. 

Problem: What is the force between two wires
running some current? 


You see again, why most of us prefer the
cgs system: We all know the Coulomb law by heart, but the force between
currentcarrying conductors??? 


OK, here is how you start: The Lorentz
law tells us that the force
F on some charge q moving with the speed
v in a magnetic field B is 











Next, the magnetic field around a wire with a
current I running through it is 











With c = (vacuum) speed of
light. 


But what is the resulting formula for the
twowirearrangement needed for defining
I ? 

Interestingly enough, several standard text books
on electrodynamics do not give you the
formula directly  of course, it is no big deal to derive it yourself. 


Still, it shows that the magnetic force formula is
far less important than the Coulomb law. Without going into the details, lets
just say that the factor for converting the charge from cgs to SI
and back, is c/10 or 10/c, respectively. 


Here are a few conversions: 



Quantity 
cgs 
Unit 
= 
SI 
Unit 
Conversion 
Charge

1 
cm^{3/2}·g^{1/2}·s^{1}
(esu) 
= 
3,3356·10^{10} 
A·s = C 
multiply by c/10 =
3,3356·10^{10}
c = number for vacuum speed of light in cm/s 
2,998· 10^{9} esu 
esu 
= 
1 


Elementary charge e 
4,8033 · 10^{10} 
esu 
= 
1,602 · 10^{19} 
A·s = C 

Current I 
1 
cm^{3/2}·g^{1/2}·s^{2} 
= 
3,3356 ·10^{10} 
A 
multiply with c/10 
Voltage U 
1 
cm^{1/2}·g^{1/2}·s^{1} 
= 
2,9979·10^{2} 
m^{2}·kg·s^{3}·A^{1} =
V 
multiply with 10^{8}·c 





And so on. For about 15  20 more
conversions, consult some handbook. 


The real
problem, however, is not conversion. The real problem is: The formulas are different! 


But first let's just look at our Coulomb
attraction between point charges again, and see if the formulas really
work 



cgs 
SI 
F = q ·
q'/ r^{2} 
F = q · q'
/4pe_{0}·r^{2} 
Use [esu] for
q, [cm] for r
F will be in [dyn]. 
Use [C] for q,
[m] for r
F will be in [N]. 
Check:
Calculate the force for r = 0,1 nm = 10^{8} cm =
10^{10} m, and q, q' = elementary charge 
F =

(4,803 · 10^{10})^{2}

[esu^{2}·10^{16}
·cm^{2}] 
1^{2} 
F 
= 2,307 · 10^{3}
[cm·g^{1}·s^{2}]


= 2,275 · 10^{3} dyn = 2,275 ·
10^{8} N 
F = 
= 2,275 · 10^{3} dyn = 2,275 ·
10^{8} N 

F =

(1,602
·10^{19})^{2}

[ 
C^{2}

] 
4 · 3,14 · 8,854·10^{12
} · 1^{} ·10^{20} 
A·s·V^{1}·m^{1}·m^{2} 
F =

2,566
·10^{38}

A^{2}·s^{2}

111.206^{}
·10^{32} 
A·s·V^{1}·m^{1} 
Now we need
1 V = m^{2}·kg·s^{3}·A^{1}
and get



F = 
2,307 ·10^{8} 
m·kg·s^{2}· 
or






Ok  it works well enough; just with some more
numerics for the SI system. 


So now you know: Whenever you see some expression
relating to electrical forces, energies, or the like, without an e_{0} or
µ_{0} in it, you are dealing with cgs units. And now you can convert to SI
units with ease, if they appear in some book, right? 


Wrong. Take the expression for the equilibrium
of forces in Bohr's model from page 19 of the "Barrett", for example. It says in
equations (24)
Ze^{2}/r = mv^{2}/ r.
So it must be given in cgs units (no particular statement is made). 


So there should be a conversion or unit table
somewhere, usually at the end of the books. And there it is, on p. 543, saying
"Electronic charge, e = 1,60 · 10^{–20} emu = 1,60 · 10^{19} C 

What the .... are emu's? In cross word puzzles, emus appear as
relatives of ostriches, but here it must be something else. 


Yes  an "emu" is the electromagnetic charge unit; we have the "magnetic" cgs system here, it
seems. 


Now you look up your trusty handbook, (e.g. the
"Physikalisches Taschenbuch" if you are a German) and find that the
number given for charge, as measured in cgs
magnetic units, is to be multiplied by 10 to get the SI
coulombs, indeed, and that an "emu" is 1
cm^{1/2} · g^{1/2}. 

Great  but now pluck it into the formula, and
things will be very wrong  by a factor c^{2}. The formula given
requires esu's, for emu's, there must be a c^{2}
somewhere. The "Barrett" simply got it wrong! Changing electric units
changes the formulas! 

In other words, just switching from esu or emu to
C will not do the trick and switch the resulting force from dyn to N in the
electric world! 


While this is quite trivial on the one hand (we
could have introduced conversions for the force, too), it simply means that if
you want to keep some quantities (like the force) expressed in conventional
units  length, mass and time  you must change your formulas, and complete
them with the required 4p, e_{0} (or alternatively m_{0} = 1/e_{0}c^{2}, and possibly other
adornments as well. 


And while this conversion is always possible and
not all that difficult to figure out, there is plenty
of room for confusion and mistakes  consider the "Barrett
example". 

Here is a conversion table for the formulas.
Whenever you encounter one of the quantities in the middle column in some
cgs system formula, you replace it with the expression in the right hand
column to obtain the formula in the SI system. But remember: You must also use the proper units! 





Quantity 
cgs (Gauss or
electrostatic) 
SI 
Speed of light in vacuum 
c 
(µ_{0}e_{0})^{1/2} 
Electrical field
or potential, voltage 
E
{F, V} 
[(4pe_{0})^{1/2}]·E {·F, ·V} 
Electrical flux density 
D 
[(4p/e_{0})^{1/2}]·D 
Charge
or current, current density, polarization 
q
{I, j, P} 
[1/(4pe_{0})^{1/2}]·q {·I, ·j,
·P} 
Magnetic induction
(= magnetic flux density) 
B 
[(4p/m_{0})^{1/2}]·B 
Magnetic field 
H 
[(4pm_{0})^{1/2}]·E 
Magnetisation 
M 
[(m_{0}/4p)^{1/2}]·M 
Conductivity 
s 
[1/(4pe_{0})]·s 
Dielectric constant 
e 
e/e_{0} 
Magnetic permeability 
m 
m/m_{0} 
Resistance or
Impedance 
R
{Z} 
(4pe_{0}) ·R
·{Z} 
Inductivity 
L 
(4pe_{0}) ·L 
Capacity 
C 
[1/(4pe_{0})]·C 





So here is what you do: 


Use SI
units  even if it gives you an ulcer and you grind your teeth a
lot. It will avoid many ulcers in the future. 


If you run across equations, numbers, relations,
anything where you are not sure what kind of unit system is used  be very careful! Often it is best to pluck in some
numbers and see if what you get makes any sense. Quite often, the result is so
far off anything sensible (many orders of magnitude, like a factor
c^{2}) that it just is clear that there is a confusion of
units. 


But, on occasion, it is only a factor 4p  be careful! 


