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[Schweizer?]-Tomonago covariant

In the scalar theory there is no difference whether μ=0 or ≠0.
However with vector theory, [?] is quite different for the
following reason.

Ψ_μ are not all independent fields. This equation helps
to eliminate one of Ψ_μ. If μ=0, under the gauge transf^[n?]

gauge-invariant quantities

The scalar & vector potentials are not physically observed
qualities. Only the field strengths are observed, significant because the
force is determined by them. (For μ≠0, the gauge
invariante does not hold.) It is customary way to
choose the right gauge at which .
(which is not in general true). Gauge transfⁿ [?]
can be made.

Suppose μ=0

In the gauge