(66)
|
∂
L
∂
x
1
,
∂
L
∂
x
2
,
∂
L
∂
x
3
,
∂
L
∂
x
4
|
{\displaystyle \left|{\frac {\partial L}{\partial x_{1}}},\ {\frac {\partial L}{\partial x_{2}}},\ {\frac {\partial L}{\partial x_{3}}},\ {\frac {\partial L}{\partial x_{4}}}\right|}
If
s
=
|
s
1
,
s
2
,
s
3
,
s
4
|
{\displaystyle s=\left|s_{1},\ s_{2},\ s_{3},\ s_{4}\right|}
(67)
l
o
r
s
¯
=
∂
s
1
∂
x
1
+
∂
s
2
∂
x
2
+
∂
s
3
∂
x
3
+
∂
s
4
∂
x
4
{\displaystyle lor\ {\bar {s}}={\frac {\partial s_{1}}{\partial x_{1}}}+{\frac {\partial s_{2}}{\partial x_{2}}}+{\frac {\partial s_{3}}{\partial x_{3}}}+{\frac {\partial s_{4}}{\partial x_{4}}}}
In case of a Lorentz transformation
A
{\displaystyle {\mathsf {A}}}
l
o
r
′
s
¯
′
=
(
l
o
r
A
)
(
A
¯
s
¯
)
=
l
o
r
s
¯
{\displaystyle lor'\ {\bar {s}}'=(lor\ {\mathsf {A}})({\mathsf {\bar {A}}}{\bar {s}})=lor\ {\bar {s}}}
i.e. , lor s is an invariant in a {sc|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
∂
f
12
∂
x
2
+
∂
f
13
∂
x
3
+
∂
f
14
∂
x
4
,
∂
f
21
∂
x
1
+
∂
23
∂
x
3
+
∂
24
∂
x
4
,
∂
f
31
∂
x
1
+
∂
32
∂
x
2
+
∂
34
∂
x
4
,
∂
f
41
∂
x
1
+
∂
42
∂
x
2
+
∂
43
∂
x
3
,
{\displaystyle {\begin{array}{ccccccc}&&{\frac {\partial f_{12}}{\partial x_{2}}}&+&{\frac {\partial f_{13}}{\partial x_{3}}}&+&{\frac {\partial f_{14}}{\partial x_{4}}},\\\\{\frac {\partial f_{21}}{\partial x_{1}}}&&&+&{\frac {\partial _{23}}{\partial x_{3}}}&+&{\frac {\partial _{24}}{\partial x_{4}}},\\\\{\frac {\partial f_{31}}{\partial x_{1}}}&+&{\frac {\partial _{32}}{\partial x_{2}}}&&&+&{\frac {\partial _{34}}{\partial x_{4}}},\\\\{\frac {\partial f_{41}}{\partial x_{1}}}&+&{\frac {\partial _{42}}{\partial x_{2}}}&+&{\frac {\partial _{43}}{\partial x_{3}}},\end{array}}}
So the system o£ differential equations (A) can be expressed in the concise form
{A}
l
o
r
f
=
−
s
{\displaystyle lor\ f=-s}
and the system (B) can be expressed in the form
{B}
l
o
r
F
∗
=
0
{\displaystyle lor\ F^{*}=0}
Referring back to the definition (67) for
l
o
r
s
¯
{\displaystyle lor\ {\bar {s}}}
l
o
r
(
l
o
r
f
¯
)
{\displaystyle lor({\overline {lor\ f}})}
l
o
r
(
l
o
r
F
∗
¯
)
{\displaystyle lor({\overline {lor\ F^{*}}})}
f and F* are alternating matrices. Accordingly it follows out of (A), that
(68)
∂
s
1
∂
x
1
+
∂
s
2
∂
x
2
+
∂
s
3
∂
x
3
+
∂
s
4
∂
x
4
=
0
{\displaystyle {\frac {\partial s_{1}}{\partial x_{1}}}+{\frac {\partial s_{2}}{\partial x_{2}}}+{\frac {\partial s_{3}}{\partial x_{3}}}+{\frac {\partial s_{4}}{\partial x_{4}}}=0}
