(53) ωψ = ω₁ ψ₁ + ω₂ ψ₂ + ω₃ ψ₃ + ω₄ ψ₄ = 0
which can also be written as ψ₄ = _i_ (_u__{_x_} ψ₁ + _u__{_y_} ψ₂ +
_u__{_z_} ψ₃).
The vector ψ is perpendicular to ω; we can call it the _Magnetic
rest-force_.
Relations analogous to these hold among the quantities ωF*, M, E, _u_
and Relation (D) can be replaced by the formula
{ D } -ωF* = μψ_f_*.
We can use the relations (C) and (D) to calculate F and _f_ from Φ and ψ
we have
ωF = -Φ, ωF* = -_i_μψ, ω_f_ = -εΦ, ω_f_* = -_i_ψ.
and applying the relation (45) and (46), we have
F = [ω. Φ] + _i_μ[ω. ψ]* 55)
_f_ = ε[ω. Φ] + _i_[ω. ψ]* 56)
_i.e._
F₁₂ = (ω₁ Φ₁ - ω₂ Φ₁) + _i_μ [ω₃ Ψ₄ - ω₄ ψ₃], etc.
_f₁₂_ = ε(ω₁ Φ₂ - ω₂ φ₁) + _i_ [ω₃ ψ₄ - ω₄ ψ₃]., etc.
Let us now consider the space-time vector of the second kind [Φ ψ], with
the components
[ Φ₂ ψ₃ - Φ₃ ψ₂, Φ₃ ψ₁ - Φ₁ ψ₃, Φ₁ ψ₂ - Φ₂ ψ₁ ]
[ Φ₁ ψ₄ - Φ₄ ψ₁, Φ₂ ψ₄ - Φ₄ ψ₂, Φ₃ ψ₄ - Φ₄ ψ₃ ]
Then the corresponding space-time vector of the first kind ω[Φ, ψ]
vanishes identically owing to equations 9) and 53)
for ω[Φ.ψ] = -(ωψ)Φ + (ωΦ)ψ