4 8 2 D O C U M E N T 4 8 3 F E B R U A R Y 1 9 2 7 . For the sphere with a radius and the surface charge –e, one accordingly obtains . Both charges together then yield . In case c.) the observer will measure the same field, since, in this case, no com- pensation charges occur on his magnet whose magnetic effect would otherwise have to be taken into account. Now, as it is supposedly entirely the same whether the observer is at rest and the sphere is rotating, or the sphere is at rest and the observer is moving around the sphere, one obtains this suspicious result: if the observer in our case is moving around the system, then he measures a magnetic field [case b.], although he, at rest, can notice nei- ther an electric nor a magnetic field! This completely contradicts the information of your last letter.[7] Where is the er- ror? The mathematical calculation is surely correct, as various other people arrived at the same result for via entirely different routes. Consequently, some- thing must not have been taken into account in my details. What is it? The centrif- ugal force acting on the electrons surely wouldn’t have any substantial influence. Should the magnetic field of sphere have an influence on the charge distribution on sphere and vice versa? [That could surely only be the case if the spheres with their charges were moving with respect to the magnetic field, hence if the magnetic field were not rotating along with them.] — — — — I would be very grateful if you would indicate to me where the error lies and what, according to your view, the observer will measure in those 3 cases. Respectfully and sincerely yours, T. Schlomka HhIorizontal ( ) 1 3 -- - e ω a1 2 r3 ----- sinϑ ⋅ ⋅ ⋅ ⋅ = a2 HhII) orizontal ( 3 -- -–1 e ω a2 2 r3 ----- sinϑ ⋅ ⋅ ⋅ ⋅ = Hhorizontal e0 3 ---- - ω r3 ---- a1 2 a2 2) – sinϑ( ⋅ ⋅ = Hhorizontal e 3 -- - ω r3 ---- a1 2 a2 2)sinϑ – ( ⋅ ⋅ = Hhorizontal a1 a2