aus Wikisource, der freien Quellensammlung
und erhalten dann endgültig
(8)
d
i
v
′
A
′
=
d
i
v
{
A
′
+
(
p
−
1
)
c
0
⋅
c
0
A
′
}
+
p
q
n
∂
A
′
c
0
∂
t
{\displaystyle \mathrm {div} '{\mathfrak {A}}'=\mathrm {div} \left\{{\mathfrak {A}}'+(p-1){\mathfrak {c}}_{0}\cdot {\mathfrak {c}}_{0}{\mathfrak {A}}'\right\}+pqn{\frac {\partial {\mathfrak {A}}'{\mathfrak {c}}_{0}}{\partial t}}}
wo nun
d
i
v
{\displaystyle \mathrm {div} }
t
{\displaystyle t}
Ganz analog erhalten wir auf Grund von (44)I und (136) I für einen Skalar
a
′
{\displaystyle a'}
(9)
∇
′
a
′
=
∇
a
′
+
(
p
−
1
)
c
0
⋅
c
0
∇
a
′
+
p
q
n
c
0
∂
a
′
∂
t
{\displaystyle \nabla 'a'=\nabla a'+(p-1){\mathfrak {c}}_{0}\cdot {\mathfrak {c}}_{0}\nabla a'+pqn{\mathfrak {c}}_{0}{\frac {\partial a'}{\partial t}}}
Aus (68) I folgt, da
c
0
=
k
o
n
s
t
.
{\displaystyle {\mathfrak {c}}_{0}=\mathrm {konst.} }
(10)
∇
(
c
0
B
)
=
(
c
0
∇
)
B
+
[
c
0
r
o
t
B
]
{\displaystyle \nabla \left({\mathfrak {c}}_{0}{\mathfrak {B}}\right)=\left({\mathfrak {c}}_{0}\nabla \right){\mathfrak {B}}+\left[{\mathfrak {c}}_{0}\mathrm {rot} {\mathfrak {B}}\right]}
und aus (63) I
(11)
r
o
t
[
B
c
0
]
=
(
c
0
∇
)
B
−
c
0
d
i
v
B
{\displaystyle \mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]=\left({\mathfrak {c}}_{0}\nabla \right){\mathfrak {B}}-{\mathfrak {c}}_{0}\mathrm {div} {\mathfrak {B}}}
oder
(11a)
(
c
0
∇
)
B
=
r
o
t
[
B
c
0
]
+
c
0
d
i
v
B
{\displaystyle \left({\mathfrak {c}}_{0}\nabla \right){\mathfrak {B}}=\mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]+{\mathfrak {c}}_{0}\mathrm {div} {\mathfrak {B}}}
Hieraus und aus (10) ergibt sich
(12)
∇
(
c
0
B
)
=
r
o
t
[
B
c
0
]
+
[
c
0
r
o
t
B
]
+
c
0
d
i
v
B
{\displaystyle \nabla \left({\mathfrak {c}}_{0}{\mathfrak {B}}\right)=\mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]+\left[{\mathfrak {c}}_{0}\mathrm {rot} {\mathfrak {B}}\right]+{\mathfrak {c}}_{0}\mathrm {div} {\mathfrak {B}}}
Daraus folgt
(13)
c
0
∇
(
c
0
B
)
=
d
i
v
(
c
0
⋅
c
0
B
)
=
c
0
r
o
t
[
B
c
0
]
+
d
i
v
B
{\displaystyle {\mathfrak {c}}_{0}\nabla \left({\mathfrak {c}}_{0}{\mathfrak {B}}\right)=\mathrm {div} \left({\mathfrak {c}}_{0}\cdot {\mathfrak {c}}_{0}{\mathfrak {B}}\right)={\mathfrak {c}}_{0}\mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]+\mathrm {div} {\mathfrak {B}}}
oder
(13a)
c
0
r
o
t
[
B
c
0
]
=
d
i
v
(
c
0
⋅
c
0
B
)
−
d
i
v
B
{\displaystyle {\mathfrak {c}}_{0}\mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]=\mathrm {div} \left({\mathfrak {c}}_{0}\cdot {\mathfrak {c}}_{0}{\mathfrak {B}}\right)-\mathrm {div} {\mathfrak {B}}}
Weiter ergibt (65) I
(14)
d
i
v
[
B
c
0
]
=
c
0
r
o
t
B
{\displaystyle \mathrm {div} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]={\mathfrak {c}}_{0}\mathrm {rot} {\mathfrak {B}}}
und (69) I und (12)
(15)
r
o
t
(
c
0
⋅
c
0
B
)
=
−
[
c
0
∇
(
B
c
0
)
]
=
−
[
c
0
r
o
t
[
B
c
0
]
]
−
−
[
c
0
[
c
0
r
o
t
B
]
]
=
−
[
c
0
r
o
t
[
B
c
0
]
]
−
c
0
⋅
c
0
r
o
t
B
+
r
o
t
B
{\displaystyle {\begin{aligned}\mathrm {rot} \left({\mathfrak {c}}_{0}\cdot {\mathfrak {c}}_{0}{\mathfrak {B}}\right)=&-\left[{\mathfrak {c}}_{0}\nabla \left({\mathfrak {B}}{\mathfrak {c}}_{0}\right)\right]=-\left[{\mathfrak {c}}_{0}\mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]\right]-\\-\left[{\mathfrak {c}}_{0}\left[{\mathfrak {c}}_{0}\mathrm {rot} {\mathfrak {B}}\right]\right]=&-\left[{\mathfrak {c}}_{0}\mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]\right]-{\mathfrak {c}}_{0}\cdot {\mathfrak {c}}_{0}\mathrm {rot} {\mathfrak {B}}+\mathrm {rot} {\mathfrak {B}}\end{aligned}}}
oder unter Berücksichtigung von (14)
(16)
[
c
0
r
o
t
[
B
c
0
]
]
=
r
o
t
(
B
−
c
0
⋅
c
0
B
)
−
c
0
d
i
v
[
B
c
0
]
{\displaystyle \left[{\mathfrak {c}}_{0}\mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]\right]=\mathrm {rot} \left({\mathfrak {B}}-{\mathfrak {c}}_{0}\cdot {\mathfrak {c}}_{0}{\mathfrak {B}}\right)-{\mathfrak {c}}_{0}\mathrm {div} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]}
Auf Grund von (2), (3) und (46) I erhalten wir
(17)
r
o
t
′
A
′
=
∫
[
d
f
′
A
′
]
V
′
lim
V
′
=
0
=
∫
[
d
f
A
′
]
V
lim
V
=
0
+
(
p
−
1
)
∫
c
0
d
f
⋅
[
c
0
A
′
]
V
lim
V
=
0
=
r
o
t
A
′
+
(
p
−
1
)
[
c
0
∫
c
0
d
f
⋅
A
′
V
]
lim
V
=
0
{\displaystyle {\begin{aligned}\mathrm {rot} '{\mathfrak {A}}'&={\underset {\lim V'=0}{\frac {\int \left[d{\mathfrak {f}}'{\mathfrak {A}}'\right]}{V'}}}={\underset {\lim V=0}{\frac {\int \left[d{\mathfrak {f}}{\mathfrak {A}}'\right]}{V}}}+{\underset {\lim V=0}{(p-1){\frac {\int {\mathfrak {c}}_{0}d{\mathfrak {f}}\cdot \left[{\mathfrak {c}}_{0}{\mathfrak {A}}'\right]}{V}}}}\\&=\mathrm {rot} {\mathfrak {A}}'+{\underset {\lim V=0}{(p-1)\left[{\mathfrak {c}}_{0}{\frac {\int {\mathfrak {c}}_{0}d{\mathfrak {f}}\cdot {\mathfrak {A}}'}{V}}\right]}}\end{aligned}}}
oder wegen (57) I und (11a)
r
o
t
′
A
′
=
r
o
t
A
′
+
(
p
−
1
)
[
c
0
r
o
t
[
A
′
c
0
]
]
{\displaystyle \mathrm {rot} '{\mathfrak {A}}'=\mathrm {rot} {\mathfrak {A}}'+(p-1)\left[{\mathfrak {c}}_{0}\mathrm {rot} \left[{\mathfrak {A}}'{\mathfrak {c}}_{0}\right]\right]}
und wegen (16)
r
o
t
′
A
′
=
r
o
t
{
p
A
′
−
(
p
−
1
)
c
0
⋅
c
0
A
′
}
−
(
p
−
1
)
c
0
⋅
d
i
v
[
A
′
c
0
]
{\displaystyle \mathrm {rot} '{\mathfrak {A}}'=\mathrm {rot} \left\{p{\mathfrak {A}}'-(p-1){\mathfrak {c}}_{0}\cdot {\mathfrak {c}}_{0}{\mathfrak {A}}'\right\}-(p-1){\mathfrak {c}}_{0}\cdot \mathrm {div} \left[{\mathfrak {A}}'{\mathfrak {c}}_{0}\right]}
![{\displaystyle \nabla \left({\mathfrak {c}}_{0}{\mathfrak {B}}\right)=\left({\mathfrak {c}}_{0}\nabla \right){\mathfrak {B}}+\left[{\mathfrak {c}}_{0}\mathrm {rot} {\mathfrak {B}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1786f447cb9ba15e94d6658f675b7129a68b9e2b)
![{\displaystyle \mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]=\left({\mathfrak {c}}_{0}\nabla \right){\mathfrak {B}}-{\mathfrak {c}}_{0}\mathrm {div} {\mathfrak {B}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4a1501f0819bee3c8abd678073ffe1d36980554)
![{\displaystyle \left({\mathfrak {c}}_{0}\nabla \right){\mathfrak {B}}=\mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]+{\mathfrak {c}}_{0}\mathrm {div} {\mathfrak {B}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9383908f395f27f47229b09e40e2962bdf236099)
![{\displaystyle \nabla \left({\mathfrak {c}}_{0}{\mathfrak {B}}\right)=\mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]+\left[{\mathfrak {c}}_{0}\mathrm {rot} {\mathfrak {B}}\right]+{\mathfrak {c}}_{0}\mathrm {div} {\mathfrak {B}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee9218ca94a81110501a405caadf8a78dc47d1da)
![{\displaystyle {\mathfrak {c}}_{0}\nabla \left({\mathfrak {c}}_{0}{\mathfrak {B}}\right)=\mathrm {div} \left({\mathfrak {c}}_{0}\cdot {\mathfrak {c}}_{0}{\mathfrak {B}}\right)={\mathfrak {c}}_{0}\mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]+\mathrm {div} {\mathfrak {B}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7904f63a408a3b488b6cd1cce420e5ff24c6c2d9)
![{\displaystyle {\mathfrak {c}}_{0}\mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]=\mathrm {div} \left({\mathfrak {c}}_{0}\cdot {\mathfrak {c}}_{0}{\mathfrak {B}}\right)-\mathrm {div} {\mathfrak {B}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b195f71d023d784cf9112d5f534c3133487a4136)
![{\displaystyle \mathrm {div} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]={\mathfrak {c}}_{0}\mathrm {rot} {\mathfrak {B}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e75f63be7ca56f8660a7a556717ccc4d8ceb7b2)
![{\displaystyle {\begin{aligned}\mathrm {rot} \left({\mathfrak {c}}_{0}\cdot {\mathfrak {c}}_{0}{\mathfrak {B}}\right)=&-\left[{\mathfrak {c}}_{0}\nabla \left({\mathfrak {B}}{\mathfrak {c}}_{0}\right)\right]=-\left[{\mathfrak {c}}_{0}\mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]\right]-\\-\left[{\mathfrak {c}}_{0}\left[{\mathfrak {c}}_{0}\mathrm {rot} {\mathfrak {B}}\right]\right]=&-\left[{\mathfrak {c}}_{0}\mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]\right]-{\mathfrak {c}}_{0}\cdot {\mathfrak {c}}_{0}\mathrm {rot} {\mathfrak {B}}+\mathrm {rot} {\mathfrak {B}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0022e24c59d47e76008c2781e232e987b879103e)
![{\displaystyle \left[{\mathfrak {c}}_{0}\mathrm {rot} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]\right]=\mathrm {rot} \left({\mathfrak {B}}-{\mathfrak {c}}_{0}\cdot {\mathfrak {c}}_{0}{\mathfrak {B}}\right)-{\mathfrak {c}}_{0}\mathrm {div} \left[{\mathfrak {B}}{\mathfrak {c}}_{0}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea2ed5d01f085de74853e5602d93bce1df67a60d)
![{\displaystyle {\begin{aligned}\mathrm {rot} '{\mathfrak {A}}'&={\underset {\lim V'=0}{\frac {\int \left[d{\mathfrak {f}}'{\mathfrak {A}}'\right]}{V'}}}={\underset {\lim V=0}{\frac {\int \left[d{\mathfrak {f}}{\mathfrak {A}}'\right]}{V}}}+{\underset {\lim V=0}{(p-1){\frac {\int {\mathfrak {c}}_{0}d{\mathfrak {f}}\cdot \left[{\mathfrak {c}}_{0}{\mathfrak {A}}'\right]}{V}}}}\\&=\mathrm {rot} {\mathfrak {A}}'+{\underset {\lim V=0}{(p-1)\left[{\mathfrak {c}}_{0}{\frac {\int {\mathfrak {c}}_{0}d{\mathfrak {f}}\cdot {\mathfrak {A}}'}{V}}\right]}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f62fc081cf60ef39dbc6ec25e6f30a766567c6a)
![{\displaystyle \mathrm {rot} '{\mathfrak {A}}'=\mathrm {rot} {\mathfrak {A}}'+(p-1)\left[{\mathfrak {c}}_{0}\mathrm {rot} \left[{\mathfrak {A}}'{\mathfrak {c}}_{0}\right]\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7f78a5afcb4b90b6c6f459f122352fc25ea41d9)
![{\displaystyle \mathrm {rot} '{\mathfrak {A}}'=\mathrm {rot} \left\{p{\mathfrak {A}}'-(p-1){\mathfrak {c}}_{0}\cdot {\mathfrak {c}}_{0}{\mathfrak {A}}'\right\}-(p-1){\mathfrak {c}}_{0}\cdot \mathrm {div} \left[{\mathfrak {A}}'{\mathfrak {c}}_{0}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72209c9643a1391c89e1aa4eabf17180827b988d)
