of a material and in that of an electromagnetic system we need consider only the latter. The conclusions drawn in § 11 evidently remain valid, so that we may start from the equation which we obtain by adding the new terms to (43). We therefore have
δ
L
+
1
ϰ
δ
Q
+
Σ
(
a
)
K
a
δ
x
a
=
−
Σ
(
a
b
)
∂
(
ψ
a
b
∗
q
a
)
∂
x
b
+
1
ϰ
Σ
(
a
b
¯
e
)
∂
∂
x
e
(
δ
Q
δ
g
a
b
,
e
δ
g
a
b
)
+
+
Σ
(
a
b
¯
)
(
∂
L
∂
g
a
b
+
1
ϰ
δ
Q
δ
g
a
b
)
δ
g
a
b
−
1
ϰ
Σ
(
a
b
¯
e
)
∂
∂
x
e
(
δ
Q
δ
g
a
b
,
e
)
δ
g
a
b
{\displaystyle {\begin{array}{c}\delta \mathrm {L} +{\dfrac {1}{\varkappa }}\delta Q+\Sigma (a)K_{a}\delta x_{a}=-\Sigma (ab){\dfrac {\partial \left(\psi _{ab}^{*}q_{a}\right)}{\partial x_{b}}}+{\dfrac {1}{\varkappa }}\Sigma ({\overline {ab}}e){\frac {\partial }{\partial x_{e}}}\left({\dfrac {\delta Q}{\delta g_{ab,e}}}\delta g_{ab}\right)+\\\\+\Sigma ({\overline {ab}})\left({\dfrac {\partial \mathrm {L} }{\partial g_{ab}}}+{\frac {1}{\varkappa }}{\frac {\delta Q}{\delta g_{ab}}}\right)\delta g_{ab}-{\frac {1}{\varkappa }}\Sigma ({\overline {ab}}e){\dfrac {\partial }{\partial x_{e}}}\left({\dfrac {\delta Q}{\delta g_{ab,e}}}\right)\delta g_{ab}\end{array}}}
(48)
When we integrate over
S
{\displaystyle S}
, the first two terms on the right hand side vanish. In the terms following them the coefficient of each
δ
g
a
b
{\displaystyle \delta g_{ab}}
must be 0, so that we find
∂
Q
∂
g
a
b
−
Σ
(
e
)
∂
∂
x
e
(
∂
Q
∂
g
a
b
,
e
)
=
−
ϰ
∂
L
∂
g
a
b
{\displaystyle {\frac {\partial Q}{\partial g_{ab}}}-\Sigma (e){\frac {\partial }{\partial x_{e}}}\left({\frac {\partial Q}{\partial g_{ab,e}}}\right)=-\varkappa {\frac {\partial \mathrm {L} }{\partial g_{ab}}}}
(49)
These are the differential equations we sought for. At the same time (48) becomes
δ
L
+
1
ϰ
δ
Q
+
Σ
(
a
)
K
a
δ
x
a
=
−
Σ
(
a
b
)
∂
(
ψ
a
b
∗
q
a
)
∂
x
b
+
1
ϰ
Σ
(
a
b
¯
e
)
∂
∂
x
c
(
∂
Q
∂
g
a
b
,
e
δ
g
a
b
)
{\displaystyle \delta \mathrm {L} +{\frac {1}{\varkappa }}\delta Q+\Sigma (a)K_{a}\delta x_{a}=-\Sigma (ab){\frac {\partial \left(\psi _{ab}^{*}q_{a}\right)}{\partial x_{b}}}+{\frac {1}{\varkappa }}\Sigma ({\overline {ab}}e){\frac {\partial }{\partial x_{c}}}\left({\frac {\partial Q}{\partial g_{ab,e}}}\delta g_{ab}\right)}
(50)
§ 15. Finally we can derive from this the equations for the momenta and the energy of the gravitation field. For this purpose we impart a virtual displacement
δ
x
c
{\displaystyle \delta x_{c}}
to this field only (comp. §§ 6 and 12). Thus we put
δ
x
a
=
0
,
q
a
=
0
{\displaystyle \delta x_{a}=0,\ q_{a}=0}
and
δ
g
a
b
=
−
g
a
b
,
c
δ
x
c
{\displaystyle \delta g_{ab}=-g_{ab,c}\delta x_{c}}
Evidently
δ
Q
=
−
∂
Q
∂
x
c
δ
x
c
{\displaystyle \delta Q=-{\frac {\partial Q}{\partial x_{c}}}\delta x_{c}}
and (comp. § 12)
δ
L
=
−
(
∂
L
∂
x
c
)
ψ
δ
x
c
{\displaystyle \delta \mathrm {L} =-\left({\frac {\partial \mathrm {L} }{\partial x_{c}}}\right)_{\psi }\delta x_{c}}
After having substituted these values in equation (50) we can deduce from it the value of
(
∂
L
∂
x
c
)
ψ
{\displaystyle \left({\dfrac {\partial \mathrm {L} }{\partial x_{c}}}\right)_{\psi }}
.
If we put
T
c
c
g
=
−
1
ϰ
Q
−
1
ϰ
Σ
(
a
b
¯
)
∂
Q
∂
g
a
b
,
c
g
a
b
,
c
{\displaystyle T_{cc}^{g}=-{\frac {1}{\varkappa }}Q-{\frac {1}{\varkappa }}\Sigma ({\overline {ab}}){\frac {\partial Q}{\partial g_{ab,c}}}g_{ab,c}}
(51)
and for
e
≠
c
{\displaystyle e\neq c}
T
e
c
g
=
−
1
ϰ
Σ
(
a
b
¯
)
∂
Q
∂
g
a
b
,
e
g
a
b
,
c
{\displaystyle T_{ec}^{g}=-{\frac {1}{\varkappa }}\Sigma ({\overline {ab}}){\frac {\partial Q}{\partial g_{ab,e}}}g_{ab,c}}
(52)
the result takes the following form