256
ENERGY AND STRESS.
[643.
P
x
x
=
1
4
π
(
B
H
−
1
2
H
2
)
{\displaystyle P_{xx}={\frac {1}{4\pi }}({\mathfrak {BH}}-{\frac {1}{2}}{\mathfrak {H}}^{2})}
,
P
y
y
=
P
z
z
=
−
1
8
π
H
2
{\displaystyle P_{yy}=P_{zz}=-{\frac {1}{8\pi }}{\mathfrak {H}}^{2}}
,
(18)
and the tangential stresses disappear.
The stress in this case is therefore a hydrostatic pressure
1
8
π
H
2
{\displaystyle {\frac {1}{8\pi }}{\mathfrak {H}}^{2}}
, combined with a longitudinal tension
1
4
π
B
H
{\displaystyle {\frac {1}{4\pi }}{\mathfrak {BH}}}
along the lines of force.
643.] When there is no magnetization,
B
=
H
{\displaystyle {\mathfrak {B}}={\mathfrak {H}}}
, and the stress is still further simplified, being a tension along the lines of force equal to
1
8
π
H
2
{\displaystyle {\frac {1}{8\pi }}{\mathfrak {H}}^{2}}
, combined with a pressure in all directions at right angles to the lines of force, numerically equal also to
1
8
π
H
2
{\displaystyle {\frac {1}{8\pi }}{\mathfrak {H^{2}}}}
. The components of stress in this important case are
P
x
x
{\displaystyle P_{xx}}
=
1
8
π
(
α
2
−
β
2
−
γ
2
)
{\displaystyle {}={\frac {1}{8\pi }}(\alpha ^{2}-\beta ^{2}-\gamma ^{2})}
,
P
y
y
{\displaystyle P_{yy}}
=
1
8
π
(
β
2
−
γ
2
−
α
2
)
{\displaystyle {}={\frac {1}{8\pi }}(\beta ^{2}-\gamma ^{2}-\alpha ^{2})}
,
P
z
z
{\displaystyle P_{zz}}
=
1
8
π
(
γ
2
−
α
2
−
β
2
)
{\displaystyle {}={\frac {1}{8\pi }}(\gamma ^{2}-\alpha ^{2}-\beta ^{2})}
,
P
y
z
=
P
z
y
{\displaystyle P_{yz}=P_{zy}}
=
1
4
π
β
γ
{\displaystyle {}={\frac {1}{4\pi }}\beta \gamma }
,
P
z
x
=
P
x
z
{\displaystyle P_{zx}=P_{xz}}
=
1
4
π
γ
α
{\displaystyle {}={\frac {1}{4\pi }}\gamma \alpha }
,
P
x
y
=
P
y
x
{\displaystyle P_{xy}=P_{yx}}
=
1
4
π
α
β
{\displaystyle {}={\frac {1}{4\pi }}\alpha \beta }
.
The force arising from these stresses on an element of the medium referred to unit of volume is
X
{\displaystyle X}
=
d
d
x
p
x
x
+
d
d
y
p
y
x
+
d
d
z
p
z
x
{\displaystyle {}={\frac {d}{dx}}p_{xx}+{\frac {d}{dy}}p_{yx}+{\frac {d}{dz}}p_{zx}}
,
=
1
4
π
{
α
d
α
d
x
−
β
d
β
d
x
−
γ
d
γ
d
x
}
+
1
4
π
{
α
d
β
d
y
+
β
d
α
d
y
}
+
1
4
π
{
α
d
γ
d
z
−
γ
d
α
d
z
}
{\displaystyle {}={\frac {1}{4\pi }}\left\{\alpha {\frac {d\alpha }{dx}}-\beta {\frac {d\beta }{dx}}-\gamma {\frac {d\gamma }{dx}}\right\}+{\frac {1}{4\pi }}\left\{\alpha {\frac {d\beta }{dy}}+\beta {\frac {d\alpha }{dy}}\right\}+{\frac {1}{4\pi }}\left\{\alpha {\frac {d\gamma }{dz}}-\gamma {\frac {d\alpha }{dz}}\right\}}
,
=
1
4
π
α
(
d
α
d
x
+
d
β
d
y
+
d
γ
d
z
)
+
1
4
π
γ
(
d
α
d
z
−
d
γ
d
x
)
−
1
4
π
β
(
d
β
d
x
−
d
α
d
y
)
{\displaystyle {}={\frac {1}{4\pi }}\alpha \left({\frac {d\alpha }{dx}}+{\frac {d\beta }{dy}}+{\frac {d\gamma }{dz}}\right)+{\frac {1}{4\pi }}\gamma \left({\frac {d\alpha }{dz}}-{\frac {d\gamma }{dx}}\right)-{\frac {1}{4\pi }}\beta \left({\frac {d\beta }{dx}}-{\frac {d\alpha }{dy}}\right)}
.
Now
d
α
d
x
+
d
β
d
y
+
d
γ
d
z
=
4
π
m
{\displaystyle {\frac {d\alpha }{dx}}+{\frac {d\beta }{dy}}+{\frac {d\gamma }{dz}}=4\pi m}
,
d
α
d
z
−
d
γ
d
x
=
4
π
v
{\displaystyle {\frac {d\alpha }{dz}}-{\frac {d\gamma }{dx}}=4\pi v}
,
d
β
d
x
−
d
α
d
y
=
4
π
w
{\displaystyle {\frac {d\beta }{dx}}-{\frac {d\alpha }{dy}}=4\pi w}
,
where
m
{\displaystyle m}
is the density of austral magnetic matter referred to unit