D O C U M E N T 1 2 5 D E C E M B E R 1 9 2 5 1 4 7 3. The following calculation may serve to give clear and rigorous emphasis to this contradiction. The first component of the acting force per unit volume is , or, if one takes into account that and , . . . . . . (1) where , (2) are the Maxwell potentials. Through derivation it is found that in the space outside, where , expres- sion (1) vanishes. If one then assumes that in the interior of the system , then (1) vanishes there as well. Hence, at all positions . We now want to multiply the equation by a volume element dS and then integrate it over the half-space R on the positive side of the yz-plane. As a remote boundary of this half-space we imagine an infinitely large spherical surface with its center at O. One obtains in this way , where the first integral extends over the yz-plane, the second extends over the hemisphere.[2] [I notice that the equation is valid even if the charged system has a definite boundary (rotational surface), because Xn must be continuous.] When the radius r of the latter grows, E diminishes as , and H as , then the latter integral vanishes. Therefore, if the integration is expanded over the full yz-plane, then ρEx 1 c --ρ( - vyHz vzHy) – + Exdiv E Hz© ∂Hx ∂z --------- ∂Hz ∂x ---------· – ¹ § Hy© ∂Hy ∂x --------- ∂Hx ∂y ---------· – ¹ § – + = rotE 0= divH 0= ∂Xx ∂x -------- - ∂Xy ∂y -------- - ∂Xz ∂z --------, + + Xx 1 2 -- - Ex 2 Ey 2 – Ez 2) – ( 1 2 -- - Hx 2 Hy 2 – Hz 2) – ( + = Xy ExEy HxHy += Xz ExEz HxHz += ρ 0= E 1 c --[ - ν. H] + 0 = 0 ∂Xx ∂x -------- - ∂Xy ∂y -------- - ∂Xz ∂z -------- + + = x O R n σ σ 0 ³Xxdσ – ³Xndσ + = 1 r2 ---- 1 r3 ----