§11. Typical Representations of the Fundamental Equations.
In the statement of the fundamental equations, our leading idea had been that they should retain a covariance of form, when subjected to a group of Lorentz-transformations. Now we have to deal with ponderomotive reactions and energy in the electro-magnetic field. Here from the very first there can be no doubt that the settlement of this question is in some way connected with the simplest forms which can be given to the fundamental equations, satisfying the conditions of covariance. In order to arrive at such forms, I shall first of all put the fundamental equations in a typical form which brings out clearly their covariance in case of a Lorentz-transformation. Here I am using a method of calculation, which enables us to deal in a simple manner with the space-time vectors of the 1st, and 2nd kind, and of which the rules, as far as required are given below.
A system of magnitudes ah k formed into the matrix
| a₁₁.......a1 q |
| |
| |
| |
| ap 1.....ap q |
arranged in p horizontal rows, and q vertical columns is called a p × q series-matrix, and will be denoted by the letter A.
If all the quantities ah k are multiplied by C, the resulting matrix will be denoted by CA.
If the roles of the horizontal rows and vertical columns be intercharged, we obtain a q × p series matrix, which will be known as the transposed matrix of A, and will be denoted by Ā.
Ā = | a₁₁ ...... ap 1 |
| |
| a1 q ...... ap q |
If we have a second p × q series matrix B,
B = | b₁₁ .......... b₁q |
| |
| bp 1 .......... bp q |
then A + B shall denote the p × q series matrix whose members are ah k + bh k.
2⁰ If we have two matrices
A = | a₁₁ .......... a₁q |
| |
| ap 1 .......... ap q |
B = | b₁₁ .......... b₁r |
| |
| bq 1 .......... bp r |
where the number of horizontal rows of B, is equal to the number of vertical columns of A, then by AB, the product of the matrices A and B, will be denoted the matrix
C = | c₁₁ .......... c₁r |
| |
| cpr .......... cp p |
where ch k = ah 1 b₁k + ah 2 b2 h + ... ak s bs k + ... + ak q bq h
these elements being formed by combination of the horizontal rows of A with the vertical columns of B. For such a point, the associative law (AB)S = A(BS) holds, where S is a third matrix which has got as many horizontal rows as B (or AB) has got vertical columns.
For the transposed matrix of C = BA, we have Ċ = ḂĀ
3⁰. We shall have principally to deal with matrices with at most four vertical columns and for horizontal rows.
As a unit matrix (in equations they will be known for the sake of shortness as the matrix 1) will be denoted the following matrix (4 × 4 series) with the elements.
(34) | e₁₁ e₁₂ e₁₃ e₁₄ | = | 1 0 0 0 |
| e₂₁ e₂₂ e₂₃ e₂₄ | | 0 1 0 0 |
| e₃₁ e₃₂ e₃₃ e₃₄ | | 0 0 1 0 |
| e₄₁ e₄₂ e₄₃ e₄₄ | | 0 0 0 1 |
For a 4 × 4 series-matrix, Det A shall denote the determinant formed of the 4 × 4 elements of the matrix. If det A ≠ 0, then corresponding to A there is a reciprocal matrix, which we may denote by A⁻¹ so that A⁻¹A = 1.
A matrix
f = | 0 f₁₂ f₁₃ f₁₄ |
| f₂₁ 0 f₂₃ f₂₄ |
| f₃₁ f₃₂ 0 f₃₄ |
| f₄₁ f₄₂ f₄₃ 0 |
in which the elements fulfil the relation fh k = -fh k, is called an alternating matrix. These relations say that the transposed matrix ḟ = -f. Then by f* will be the dual, alternating matrix
(35)
f* = | 0 f₃₄ f₄₂ f₂₃ |
| f₄₃ 0 f₁₄ f₃₁ |
| f₂₄ f₄₁ 0 f₁₂ |
| f₃₂ f₁₃ f₂₁ 0 |
Then (36) f* f = f₃₄ f₂₂ + f₄₂ f₃₁ + f₃₂ f₂₄
i.e. We shall have a 4 × 4 series matrix in which all the elements except those on the diagonal from left up to right down are zero, and the elements in this diagonal agree with each other, and are each equal to the above mentioned combination in (36).
The determinant of f is therefore the square of the combination, by Det½f we shall denote the expression
Det½f
= f₃₂ f₁₄ f₁₃ f₂₄ + f₂₁ f₃₄·
4⁰. A linear transformation
xh = αh1 x₁′ + αh2 x₂′ + αh3 x₃′ + αh4 x₄′ (h = 1,2,3,
which is accomplished by the matrix
A = | α₁₁, α₁₂, α₁₃, α₁₄ |
| |
| α₂₁, α₂₂, α₂₃, α₂₄ |
| |
| α₃₁, α₃₂, α₃₃, α₃₄ |
| |
| α₄₁, α₄₂, α₄₃, α₄₄ |
will be denoted as the transformation A.
By the transformation A, the expression
x²₁ + x²₂ + x²₃ + x²₄ is changed into the quadratic for m ∑ αhk xh′ xk′,
where αhk = α1k α1k + α2h α2k + α3h α3k + α4h α4k are the members of a 4 × 4 series matrix which is the product of Ā A, the transposed matrix of A into A. If by the transformation, the expression is changed to
x′₁² + x₂′2 + x₃′2 + x′₄²,
we must have Ā A = 1.
A has to correspond to the following relation, if transformation (38) is to be a Lorentz-transformation. For the determinant of A) it follows out of (39) that (Det A)² = 1, or Det A = ± 1.
From the condition (39) we obtain
A⁻¹ = Ā,
i.e. the reciprocal matrix of A is equivalent to the transposed matrix of A.
For A as Lorentz transformation, we have further Det A = +1, the quantities involving the index 4 once in the subscript are purely imaginary, the other co-efficients are real, and a₄₄ > 0.
5⁰. A space time vector of the first kind[[21]] which s represented by the 1 × 4 series matrix,
(41) s = |s₁ s₂ s₃ s₄|
is to be replaced by sA in case of a Lorentz transformation
A. i.e. s′ = | s₁′ s₂′ s₃′ s₄′| = |s₁ s₂ s₃ s₄| A;
A space-time vector of the 2nd kind[[22]] with components f₂₃ ... f₃₄ shall be represented by the alternating matrix
(42) f = | 0 f₁₂ f₁₃ f₁₄ |
|f₂₁ 0 f₂₃ f₂₄ |
|f₃₁ f₃₂ 0 f₃₄ |
|f₄₁ f₄₂ f₄₃ 0 |
and is to be replaced by A⁻¹ f A in case of a Lorentz transformation [see the rules in [§ 5] (23) (24)]. Therefore referring to the expression (37), we have the identity Det½ (Ā f A) = Det A. Det½ f. Therefore Det½ f becomes an invariant in the case of a Lorentz transformation [see eq. (26) See. [§ 5]].
Looking back to (36), we have for the dual matrix (Āf*A) (A⁻¹fA) = A⁻¹f*fA = Det½ function. A⁻¹A = Det½f from which it is to be seen that the dual matrix f* behaves exactly like the primary matrix f, and is therefore a space time vector of the II kind; f* is therefore known as the dual space-time vector of f with components (f₁₄, f₂₄, f₃₄,), (f₂₃}, f₃₁, f₁₂).
6. If w and s are two space-time rectors of the 1st kind then by w ṡ (as well as by s ẇ) will be understood the combination (43) w₁ s₁ + w₂ s₂ + w₃ s₃ + w₄ s₄.
In case of a Lorentz transformation A, since (wA) (Āṡ) = w s, this expression is invariant.—If w ṡ = 0, then w and s are perpendicular to each other.
Two space-time rectors of the first kind (w, s) gives us a 2 × 4 series matrix
| w₁ w₂ w₃ w₄ |
| s₁ s₂ s₃ s₄ |
Then it follows immediately that the system of six magnitudes (44)
w₂ s₃ - w₃ s₂,
w₃ s₁ - w₁ s₃,
w₁ s₂ - w₂ s₁,
w₁ s₄ - w₄ s₁,
w₂ s₄ - w₄ s₂,
w₃ s₄ - w₄ s₃,
behaves in case of a Lorentz-transformation as a space-time vector of the II kind. The vector of the second kind with the components (44) are denoted by [w, s]. We see easily that Det½ [w, s] = 0. The dual vector of [w, s] shall be written as [w, s].
If ẇ is a space-time vector of the 1st kind, f of the second kind, w f signifies a 1 × 4 series matrix. In case of a Lorentz-transformation A, w is changed into w′ = wA, f into f′ = A⁻¹ f A,—therefore w′ f′ becomes = (wA A⁻¹ f A) = w f A i.e. w f is transformed as a space-time vector of the 1st kind.[[23]] We can verify, when w is a space-time vector of the 1st kind, f of the 2nd kind, the important identity
(45) [w, wf] + [w, wf*]* = (w] ẇ)f.
The sum of the two space time vectors of the second kind on the left side is to be understood in the sense of the addition of two alternating matrices.
For example, for ω₁ = 0, ω₂ = 0, ω₃ = 0, ω₄ = i,
ωf = | if₄₁, if₄₂, if₄₃, 0 |;
ωf* = | if₃₂, if₁₃, if₂₁, 0 |
[ω · ωf] = 0, 0, 0, f₄₁, f₄₂, f₄₃;
[ω · ωf*]* = 0, 0, 0, f₃₂, f₁₃, f₂₁.
The fact that in this special case, the relation is satisfied, suffices to establish the theorem (45) generally, for this relation has a covariant character in case of a Lorentz transformation, and is homogeneous in (ω₁, ω₂, ω₃, ω₄).
After these preparatory works let us engage ourselves with the equations (C,) (D,) (E) by means which the constants ε μ, σ will be introduced.
Instead of the space vector u, the velocity of matter, we shall introduce the space-time vector of the first kind ω with the components.
ω₁ = ux/√(1 - u²),
ω₂ = uy/√(1 - u²),
ω₃ = uz/√(1 - u²),
ω₄ = i/√(1 - u²).
(40) where ω₁² + ω₂² + ω₃² + ω₄² = -1 and -iω₄ > 0.
By F and f shall be understood the space time vectors of the second kind M - iE, m - ie.
In Φ = ωF, we have a space time vector of the first kind with components
Φ₁ = ω₂F₁₂ + ω₃F₁₃ + ω₄F₁₄
Φ₂ = ω₁F₂₁ + ω₃F₂₃ + ω₄F₂₄
Φ₃ = ω₁F₃₁ + ω₂F₃₂ + ω₄F₃₄
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[ωx Φ₁ + ωy Φ₂ + ωz 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 -ωf, e, m, u, and in particular -ωf is normal to ω. The relation (C) can be written as
{C} ωf = εωF.
The expression (ωf) gives four components, but the fourth can be derived from the first three.
Let us now form the time-space vector 1st kind, ψ - iωf*, whose components are
ψ₁ = -i(ω₂ f₃₄ + ω₃ f₄₂ + ω₄ f₂₃)
ψ₂ = -i(ω₁ f₄₃ + ω₃ f₄₄ + ω₄ f₃₁)
ψ₃ = -i(ω₁ f₂₄ + ω₂ f₄₁ + ω₄ f₁₂)
ψ₄ = -i(ω₁ f₃₂ + ω₂ f₁₃ + ω₃ f₂₁)
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(um)/√(1 - u²).
Among these there is the relation
(53) ωψ = ω₁ ψ₁ + ω₂ ψ₂ + ω₃ ψ₃ + ω₄ ψ₄ = 0
which can also be written as ψ₄ = i (ux ψ₁ + uy ψ₂ + uz 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* = μψf*.
We can use the relations (C) and (D) to calculate F and f from Φ and ψ we have
ωF = -Φ, ωF* = -iμψ, ωf = -εΦ, ωf* = -iψ.
and applying the relation (45) and (46), we have
F = [ω. Φ] + iμ[ω. ψ]* 55)
f = ε[ω. Φ] + i[ω. ψ]* 56)
i.e.
F₁₂ = (ω₁ Φ₁ - ω₂ Φ₁) + iμ [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
Let us now take the vector of the 1st kind
(57) Ω = iω[Φψ]*
with the components
Ω₁ = -i | ω₂ ω₃ ω₄ |
| Φ₂ Φ₃ Φ₄ |
| ψ₂ ψ₃ ψ₄ |, etc.
Then by applying rule (45), we have
(58) [Φ.ψ] = i[ωΩ]*
i.e. Φ₁ψ₂ - Φ₂ψ₁ = i(ω₃Ω₄ - ω₄Ω₃) etc.
The vector Ω fulfils the relation
(ωΩ) = ω₁Ω₁ + ω₂Ω₂ + ω₃Ω₃ + ω₄Ω₄ = 0,
(which we can write as Ω₄ = i(ωxΩ₁ + ωyΩ₂ + ωzΩ₃) and Ω is also normal to ω. In case ω = 0, we have Φ₄ = 0, ψ₄ = 0, Ω₄ = 0, and
[Ω₁, Ω₂, Ω₃ = | Φ₁ Φ₂ Φ₃ |
| ψ₁ ψ₂ ψ₃ |.
I shall call Ω, which is a space-time vector 1st kind the Rest-Ray.
As for the relation E), which introduces the conductivity σ we have -ωS = -(ω₁s₁ + ω₂s₂ + ω₃s₃ + ω₄s₄) = (- | u | Cu + ρ)/√(1 - u²) = ρ′.
This expression gives us the rest-density of electricity (see [§8] and [§4]).
Then 61) = s + (ωṡ)ω represents a space-time vector of the 1st kind, which since ωω = -1, is normal to ω, and which I may call the rest-current. Let us now conceive of the first three component of this vector as the (x-y-z) co-ordinates of the space-vector, then the component in the direction of u is
Cu - (| u | ρ′)/√(1 - u²)
= (cu - | u |ρ)/√(1 - u²)
= Ju/(1 - u²)
and the component in a perpendicular direction is Cu = Jū.
This space-vector is connected with the space-vector J = C - ρu, which we denoted in [§8] as the conduction-current.
Now by comparing with Φ = -ωF, the relation (E) can be brought into the form
{E} s + (ωṡ)ω = - σωF,
This formula contains four equations, of which the fourth follows from the first three, since this is a space-time vector which is perpendicular to ω.
Lastly, we shall transform the differential equations (A) and (B) into a typical form.