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 since the action of $$K$$, $$\overline{\mathfrak{S}}$$, $$A$$ and $$B$$ on $$\overline{\mathfrak{S}}$$, $$A$$ and $$B$$, has to be considered each time. It is now

so that by the aforementioned expression it only remains

Those forces represented by the first two members would also exist, when $$\overline{\mathfrak{S}}=0$$, and the last two members are independent of the charged body K. An action of K exerted on the conductor as such, doesn't exist.

Besides, in each of the four members (30), the part that depends on $$\mathfrak{p}$$ is of second order. We already know this from $$(K, B) + (B, B)$$, since this represents an electrostatic effect. $$(A, \overline{\mathfrak{S}})$$ and $$(\overline{\mathfrak{S}}, \overline{\mathfrak{S}})$$, however, represent forces acting on a current, in which the mean electric density is zero. As it can be seen from (Va), such forces are determined by the value of $$\mathfrak{H}$$, which belongs to the acting system. Inasmuch as $$\mathfrak{H}$$ (that belongs to $$\overline{\mathfrak{S}}$$) depends on $$\mathfrak{p}$$, it is of second order (§ 25), and the compensation charge A only produces by its velocity $$\mathfrak{p}$$ a magnetic force of second order, since its density already contains the factor $$\mathfrak{p}/V$$.

Electrodynamic actions.
§ 27. The question as to how these effects are influenced by earth's motion, can now easily be answered. If we denote the currents in two conductors by $$\overline{\mathfrak{S}}$$ and $$\overline{\mathfrak{S}'}$$, and the corresponding compensation charges by A and $$A'$$, then the action exerted on the second conductor is

in which the last two terms are mutually canceled. That $$(A, \overline{\mathfrak{S}'})$$ and the $$\mathfrak{p}$$-dependent part $$(\overline{\mathfrak{S}}, \overline{\mathfrak{S}'})$$ are of order $$\mathfrak{p}^{2}/V^{2}$$, follows from considerations such as those communicated above.