PERMITTIVITY, PERMEABILITY AND THE SPEED OF LIGHT IN THE RECIPROCAL SYSTEM Introduction Physics textbooks do not provide a theoretical derivation of Newton‘s law of gravitation or Coulomb’s laws of electrostatics and magnetostatics; they are simply stated as empirical truths. By contrast, books on the Reciprocal System, such as Ref. [1], [2], [3], do provide a theoretical derivation of these laws. This paper will take a closer look at the terms of the electrostatic and magnetostatic equationsthe terms of the gravitational equation have already been discussed in detail, most recently in Ref. [3]. Unlike the force of gravitation the forces of electrostatics and magnetostatics can be reduced by the intervening media between the charges. The question to be answered is: what are the dimensions of all the terms of Coulomb’s laws? Permittivity Coulomb’s law of electrostatic attraction is expressed as
where Q^{1} and Q^{²} are the electric charges, r is the distance between them, F_{e} is the force, and k_{e} is the proportionality constant. In the Reciprocal System all physical quantities are expressed in terms of space and time only; there are no separate dimensions for mass or charge. So, electric charge is not taken as a fundamental entity and given an independent unit. Larson has deduced that the dimensions of force and charge are
In the gravitational expression, the second mass and the distance are considered to be dimensionless ratios. Such a procedure could be used in analyzing the electrostatic expression; however I find it more fruitful to treat the second charge and the distance as having dimensions. In this case Coulomb‘s law expressed in dimensions is
For this equation to be dimensionally correct, the dimensions of k_{e} must be
These are the dimensions of permittivity (in the Reciprocal System). However, the conventional expression for the coefficient of the law is
where Î is the permittivity, expressed in farads/meter. Thus the derivation gives a result that is the inverse of the usual coefficient. But in the Reciprocal System the farad is reducible to a length. So in the conventional units that are used, permittivity turns out to be dimensionless whereas physically it is not. It should not then be surprising that the numerical values of permittivity are inverted. Instead of saying that the permittivity of air is 1.0006 times that of free space, I would say that it is 1/1.0006 = .9994 times that of free space. This actually sounds better! (The other part of the coefficient (1/4p) of the conventional expression was put in for practical reasons having nothing to do with basic physics; there is no point to keeping it in the Reciprocal System). Of course, the end resultthe calculated forcemust be the same in both systems. Permeability Coulomb‘s law of magnetostatics is
For this equation to be dimensionally correct, the dimensions of k_{m} must be
Permittivity, Permeability, and the Speed of Light One way to confirm the identification of the dimensions of permittivity and permeability is to use them in the same expression. One such expression is Maxwell‘s famous result from electromagnetic theory:
where c is the speed of light, Î_{o} is the permittivity of free space, and Ã_{o} is the permeability of free space. In the dimensional terms of the Reciprocal System, the equation is
References 1. Dewey B. Larson, The Structure of the Physical Unieerse (Portland, Oregon: North Pacific Publishers, 1959). Note: in this, the first presentation of the Reciprocal System, the permittivity and permeability were treated as dimensionless (p. 82). 2. Dewey B. Larson, Nothing But Motion (Portland, Oregon: North Pacific Publishers, 1979). 3. Dewey B. Larson, Basic Properties of Matter (Salt Lake City, Utah: International Society of Unified Science2 1988). Note: the dimensions of permittivity are stated as [s²/t] on p. 172; the dimensions of permeability are stated as [t³ /s^{4}] on p. 222. APPENDIX: SAMPLE CALCULATIONS 1. Electrostatics What is the force exerted by a charge of one coulomb on another charge of one coulomb one km away, in air? From eq. 3, the value of the permittivity in free space is s_{nat} ²/t_{nat} = (4.558816*10^{6})²/1.520655*10^{16} = 136670.11 cm²/sec. In air, the value is .9994 times this, or 136588.11 cm²/sec. Now a coulomb is defined as the electrostatic charge which when placed at a distance of 1 meter from an equal charge of the same sign produces a repulsive force of 8.98755*10^{9} N. In spacetime terms, this force is 8.98755*10^{9} N * 10^{5} dynes/N * (7.316889*10^{6}
sec/cm²) / (3.27223*10² dynes) Then from Coulomb‘s law (for a vacuum) we have 20096664 sec/cm² = 136670.11 cm²/sec * Q² sec²/cm² / 10000 cm². Solving for Q gives 1212.6213 see/em per coulomb. So for the problem at hand we have F_{e}  136588.11*1212.6213²/100000² = 20.084604 sec/cm². Converting back to conventional units we have 20.084604 sec/cm² * 3.27223*10² dynes/7.316889*10^{6 }sec/cm²
* 10^{5} N/dynes This agrees with the value from experiment. 2. Magnetostatics What is the force exerted by a magnetic pole with a strenath of one weber against another magnetic pole of equal strength one km away, in vacuum? The value of the permeability of free space is t³/s^{4} = (1.520655*10^{16})^{³}/(4.558816*10^{6})^{4} = 8.1411073*10^{27} sec /cm^{4} . Now a weber may be defined as the strength of a magnetic pole which exerts in a vacuum a force of 63325.74 N upon another magnetic pole of the same strength one meter away. In spacetime terms this force is 63325.74 N * 10^{5} dynes/N 2 (7.316889*10^{6} sec/cm²)/(3.27223*10² dynes) = 141.59989 sec/cm . From Coulomb’s magnestatic law we get 141.59989 sec/cm² = (1/8.1411073*10 27 sec3/cm4) * M²sec^{4}/cm^{4}/10000 cm². Solving for M gives 1.0736759*10^{10} sec²/cm² per weber. So for the problem at hand we have F_{m} = (1/8·1411073*10^{27 }sec³/cm^{4}) * (1.0736759*1010^{} sec /cm ) / (100000 cm) = 1.41599890 sec/cm. Converting back to conventional units we have 1.4159989*10^{4} sec/cm² * 3.27223*10² dynes/(7.316889*10^{6} sec/cm ) * 105 N dyne = 0633257 N. This agrees with the value from experiment. SUMMARY:
