Let us now consider the space-time vector of the second kind [Φ ψ], with the components
Then the corresponding space-time vector of the first kind ω[Φ, ψ] vanishes identically owing to equations 9) and 53)
Let us now take the vector of the 1st kind
with the components
Then by applying rule (45), we have
i.e. Φ₁ψ₂ - Φ₂ψ₁ = i(ω₃Ω₄ - ω₄Ω₃) etc.
The vector Ω fulfils the relation
(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
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
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
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.
§12. The Differential Operator Lor.
with the condition that in case of a Lorentz transformation it is to be replaced by ĀSA, may be called a space-time matrix of the II kind. We have examples of this in:—
1) the alternating matrix f, which corresponds to the space-time vector of the II kind,—
2) the product fF of two such matrices, for by a transformation A, it is replaced by (A⁻¹fA·A⁻¹FA) = A⁻¹fFA,
3) further when (ω₁, ω₂, ω₃, ω₄) and (Ω₁, Ω₂, Ω₃, Ω₄) are two space-time vectors of the 1st kind, the 4 × 4 matrix with the element Shk = ωhΩk,
lastly in a multiple L of the unit matrix of 4 × 4 series in which all the elements in the principal diagonal are equal to L, and the rest are zero.
We shall have to do constantly with functions of the space-time point (x, y, z, it), and we may with advantage
employ the 1 × 4 series matrix, formed of differential symbols,—
For this matrix I shall use the shortened from “lor.”[25]
Then if S is, as in (62), a space-time matrix of the II kind, by lor S′ will be understood the 1 × 4 series matrix
where Kk = ∂S1k/∂x₁ + ∂S2k/∂x₂ + ∂S3k/∂x₃ + ∂S4h/∂x₄.
When by a Lorentz transformation A, a new reference system (x′₁ x′₂ x′₃ x₄) is introduced, we can use the operator
Then S is transformed to S′= Ā S A = | S′hk |, so by lor 'S′ is meant the 1 × 4 series matrix, whose element are
Now for the differentiation of any function of (x y z t) we have the rule ∂/∂xk′ = ∂/∂x₁ ∂x₁/∂xk′ + ∂/∂x₂ ∂x₂/∂xk′ + ∂/∂x₃ ∂x₃/∂xk′ + ∂/∂x₄ ∂x₄/∂xk′ = ∂/∂x₁ a1k + ∂/∂x₂ a2k + ∂/∂x₃ a3k + ∂/∂x₄ a4k.
so that, we have symbolically lor′ = lor A.
Therefore it follows that
i.e., lor S behaves like a space-time vector of the first kind.
If L is a multiple of the unit matrix, then by lor L will be denoted the matrix with the elements
If s is a space-time vector of the 1st kind, then
In case of a Lorentz transformation A, we have
i.e., lor s is an invariant in a Lorentz-transformation.
In all these operations the operator lor plays the part of a space-time vector of the first kind.
If f represents a space-time vector of the second kind,—lor f denotes a space-time vector of the first kind with the components
So the system of differential equations (A) can be expressed in the concise form
and the system (B) can be expressed in the form
Referring back to the definition (67) for log ṡ, we find that the combinations lor ([=(lor f)=]), and lor ([=(lor F*)]) vanish identically, when f and F* are alternating matrices. Accordingly it follows out of {A}, that
while the relation
signifies that of the four equations in {B}, only three represent independent conditions.
I shall now collect the results.
Let ω denote the space-time vector of the first kind
F the space-time vector of the second kind (M,-iE)
(M = magnetic induction, E = Electric force,
f the space-time vector of the second kind (m,-ie)
(m = magnetic force, e = Electric Induction.
s the space-time vector of the first kind (C, iρ)
(ρ = electrical space-density, C - ρu = conductivity current,
ε = dielectric constant, μ = magnetic permeability,
σ = conductivity,
then the fundamental equations for electromagnetic processes in moving bodies are[26]
ω ῶ = -1, and ωF, ωf, ωF*, ωf*, s + (ωs)ω which are space-time vectors of the first kind are all normal to ω, and for the system {B}, we have
Bearing in mind this last relation, we see that we have as many independent equations at our disposal as are necessary for determining the motion of matter as well as the vector u as a function of x, y, z, t, when proper fundamental data are given.
§ 13. The Product of the Field-vectors f F.
Finally let us enquire about the laws which lead to the determination of the vector ω as a function of (x, y, z, t.) In these investigations, the expressions which are obtained by the multiplication of two alternating matrices
are of much importance. Let us write,
Then (71) S₁₁ + S₂₂ + S₃₃ + S₄₄ = 0.
Let L now denote the symmetrical combination of the indices 1, 2, 3, 4, given by
Then we shall have
In order to express in a real form, we write
Now Xx = ½[mxMx - myMy - mzMz + exEx - eyEy - ezEz]
so
These quantities[27] are all real. In the theory for bodies at rest, the combinations (Xx, Xy, Xz, Yz, Yy, Yz, Zx, Zy, Zz) are known as “Maxwell’s Stresses,” Tx, Ty, Tz are known as the Poynting’s Vector, Tt as the electromagnetic energy-density, and L as the Langrangian function.
On the other hand, by multiplying the alternating matrices of f* and F*, we obtain
and hence, we can put
where by L, we mean L-times the unit matrix, i.e. the matrix with elements
Since here SL = LS, we deduce that,
and find, since f*f = Det½f, F*F = Det½F, we arrive at the interesting
conclusion
i.e. the product of the matrix S into itself can be expressed as the multiple of a unit matrix—a matrix in which all the elements except those in the principal diagonal are zero, the elements in the principal diagonal are all equal and have the value given on the right-hand side of (79). Therefore the general relations
h, k being unequal indices in the series 1, 2, 3, 4, and
for h = 1, 2, 3, 4.
Now if instead of F, and f in the combinations (72) and (73), we introduce the electrical rest-force Φ, the magnetic rest-force ψ, and the rest-ray Ω [(55), (56) and (57)], we can pass over to the expressions,—
Here we have
The right side of (82) as well as L is an invariant in a Lorentz transformation, and the 4 × 4 element on the right side of (83) as well as Sk h represent a space time vector of the second kind. Remembering this fact, it suffices, for establishing the theorems (82) and (83) generally, to prove it for the special case ω₁ = 0, ω₂ = 0, ω₃ = 0, ω₄ = i. But for this case ω = 0, we immediately arrive at the equations (82) and (83) by means (45), (51), (60) on the one hand, and e = εE, M = μm on the other hand.
The expression on the right-hand side of (81), which equals
is >= 0, because (em = ε Φ [=ψ], (EM) = μ Φ [=ψ]; now referring back to 79), we can denote the positive square root of this expression as Det1/4 S.
Since ḟ = -f, and Ḟ = -F, we obtain for Ṡ, the transposed matrix of S, the following relations from (78),
Then is
an alternating matrix, and denotes a space-time vector of the second kind. From the expressions (83), we obtain,
from which we deduce that [see (57), (58)].
When the matter is at rest at a space-time point, ω = 0, then the equation 86) denotes the existence of the following equations
and from 83),
Now by means of a rotation of the space co-ordinate system round the null-point, we can make,
According to 71), we have
and according to 83), Tt > 0. In special cases, where ω vanishes it follows from 81) that
and if T, and one of the three magnitudes Xx, Yy, Zz are = ±Det1/4 S, the two others = -Det1/4 S. If Ω does not vanish let Ω ≠ 0, then we have in particular from 80)
and if Ω₁ = 0, Ω₂ = 0, Zz = -Tt It follows from (81), (see also 83) that
and -Zz = Tt = √(Det½ S + εμΩ₃²) > Det1/4S.
The space-time vector of the first kind
is of very great importance for which we now want to demonstrate a very important transformation
According to 78), S = L + fF, and it follows that
The symbol ‘lor’ denotes a differential process which in lor fF, operates on the one hand upon the components of f, on the other hand also upon the components of F. Accordingly lor fF can be expressed as the sum of two parts. The first part is the product of the matrices (lor f) F, lor f being regarded as a 1 × 4 series matrix. The second part is that part of lor fF, in which the diffentiations operate upon the components of F alone. From 78) we obtain
hence the second part of lor fF = -(lor F*)f* + the part of -2 lor L, in which the differentiations operate upon the components of F alone. We thus obtain
where N is the vector with the components
By using the fundamental relations A) and B), 90) is transformed into the fundamental relation
In the limitting case ε = 1, μ = 1, f = F, N vanishes identically.
Now upon the basis of the equations (55) and (56), and referring back to the expression (82) for L, and from 57) we obtain the following expressions as components of N,—
Now if we make use of (59), and denote the space-vector which has Ω₁, Ω₂, Ω₃ as the x, y, z components by the symbol W, then the third component of 92) can be expressed in the form
The round bracket denoting the scalar product of the vectors within it.
§ 14. The Ponderomotive Force.[28]
Let us now write out the relation K = lor S = -sF + N in a more practical form; we have the four equations
It is my opinion that when we calculate the ponderomotive force which acts upon a unit volume at the space-time point x, y, z, t, it has got, x, y, z components as the first three components of the space-time vector
This vector is perpendicular to ω; the law of Energy finds its expression in the fourth relation.
The establishment of this opinion is reserved for a separate tract.
In the limiting case ε = 1, μ = 1, σ = 0, the vector N = 0, S = ρω, ωK = 0, and we obtain the ordinary equations in the theory of electrons.