3 6 2 D O C U M E N T 4 2 2 J A N U A R Y 1 9 2 3 = , 3) = , One could be satisfied with this result nevertheless, I would like to demonstrate that the asymmetric expression by Minkowski likewise seems to relate much more naturally to the field equations. For this purpose I first multiply the current with . Then the result is 1 μ -- - by 2bz 2) ( – 1 2μ ------b2 + 1 μ -- - bxby 1 μ -- - bxbz 0 1 μ -- - bxby – – 0 – – – 0 0 0 0 2μ ------b21 1 μ -- - ϕμαψνα ϕναψμα) + ( 0 0 0 i μ -- - eybz ezby) – ( 0 0 0 i μ -- - ezbx exbz) – ( 0 0 0 i μ -- - exby eybx) – ( i μ -- - eybz ezby) – ( – – 0 Iα ϕμα ψμα + ϕμα ψμα)Iα + ( ϕμα ψμα)∂xν + ( ∂ ------- - εϕαν 1 μ -- - ψαν + = = ∂ ∂xν ------- -– εϕμαϕνα) ( εϕνα------------ ∂ϕμα ∂xν ϕμα∂xν ∂ ------- - 1 μ -- - ψαν + + ∂ ∂xν ------- -– 1 μ -- - ψμαψνα 1 μ -- - ψνα------------ ∂ψμα ∂xν - ψμα------- ∂ ∂xν -(εϕαν) + + = ∂ ∂xν ------- -– εϕμαϕνα ε 4 -–--δμνϕαβ 2 1 4∂xμ -–----------(εϕαν ∂ 2 ) ∂ ∂xν –-------- 1 μ -- - ψμαψνα–4μδμνψαβ 1 ------ 2 1 4∂xμ -–---------- ∂ 1 μ -- - ψαν2