D O C U M E N T 4 4 2 J A N U A R Y 1 9 2 7 4 2 1 The conditions are thus: If the terrestrial magnetic field really is being generated by the rotation of elec- tric charges, then this charge cannot only be a surface charge or only a space charge, because then an electric field would be generated by this charge at the Earth’s sur- face that would exceed the actual observed electrostatic field by about 105-fold. [An observer not participating in the Earth’s rotation and one participating in the Earth’s rotation would measure an equivalent electric field in direction and magni- tude, as can be seen from a pertinent calculation by Bucherer (Phys. Zeitschr., 1906, page 256 + 257).[5] ][6] The terrestrial magnetic field can consequently only be caused either by the rotation of 2 surface charges, the positive one of which would be distributed over a spherical surface whose radius would have to be some- what smaller than the radius of the negatively charged spherical shell (theory by Sutherland),[7] or by the rotation of a positive space charge in the Earth’s interior and a negative surface charge. I haven’t calculated the first case yet for the second case, the following results: If R is the Earth’s radius, +e the total volume charge, hence –e the surface charge, β the polar distance of a location whose distance from the center of the Earth is = r [consequently for an observer on the Earth’s surface ], and if one denotes as ω the angular velocity and c as the velocity of light, then the south–north magnetic horizontal field strength generated by the rotation of these charges out- side the sphere, measured by an observer participating in the rotation, is (field of the rotating surface charge) (field of the rotating space charge), hence, for one point on the Earth’s surface, . An observer not participating in the rotation in this case [rotation of positive volume charge and, based on the absolute amount, equally large negative surface charge], , in particular on the Earth’s surface would therefore measure exactly the same magnetic field. r R = H R2 e ω sinβ 12π c r3 ------------------------------------- e ω sinβ c r -------------------------- - –e)ω ( R2 sinβ 20π c r3 ----------------------------------------- e ω sinβ 4πc r -------------------------- - + + R2 e ω sinβ 30 π c r3 ------------------------------------- = = H e ω sinβ 30π c R -------------------------- - = H e ω R2 sinβ 12π c r3 ------------------------------------- –e)⋅ ( ω R2 sinβ 20π c r3 ----------------------------------------- + eR2 ω sinβ 30 π c r3 --------------------------------- = = field strength of the surface charge © ¹ ¨ ¸ § · field strength of the space charge © ¹ ¨ ¸ § · H e--------------------------βsin⋅ω⋅ 30π c R - =
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