Φ₄ = ω₁F₄₁ + ω₂F₄₂ + ω₃F₄₃

The first three quantities (φ₁, φ₂, φ₃) are the components of the

space-vector (E + [_u_, M])/√(1 - _u²_),

and further (φ₄ = _i_[_u_ E]/√(1 - _u²_).

Because F is an alternating matrix,

(49) ωΦ = ω₁ φ₁ + ω₂ Φ₂ + ω₃ Φ₃ + ω₄ Φ₄ = 0.

_i.e._ Φ is perpendicular to the vector ω; we can also write Φ₄ =

_i_[ω_{x} Φ₁ + ω_{y} Φ₂ + ω_{z} Φ₃].

I shall call the space-time vector Φ of the first kind as the _Electric

Rest Force_.[24]

Relations analogous to those holding between -ωF, E, M, U, hold amongst

-ω_f_, _e_, _m_, _u_, and in particular -ω_f_ is normal to ω. The

relation (C) can be written as

{C} ω_f_ = εωF.

The expression (ω_f_) gives four components, but the fourth can be

derived from the first three.

Let us now form the time-space vector 1st kind, ψ - _i_ω_f_*, whose

components are

ψ₁ = -_i_(ω₂ _f₃₄_ + ω₃ _f_₄₂ + ω₄ _f₂₃_)

ψ₂ = -_i_(ω₁ _f_₄₃ + ω₃ _f_₄₄ + ω₄ _f₃₁_)

ψ₃ = -_i_(ω₁ _f₂₄_ + ω₂ _f_₄₁ + ω₄ _f₁₂_)

ψ₄ = -_i_(ω₁ _f_₃₂ + ω₂ _f_₁₃ + ω₃ _f_₂₁)

Of these, the first three ψ₁, ψ₂, ψ₃, are the _x_, _y_, _z_ components

of the space-vector 51) (m - (_ue_))/√(1 - _u²_) and further (52) ψ₄ =

_i_(_u_m)/√(1 - _u²_).

The Principle of Relativity · 第 65 / 203 题