Statement of the balance of angular momentum
edit
The balance of angular momentum in an inertial frame can be expressed as:
σ
=
σ
T
{\displaystyle {\boldsymbol {\sigma }}={\boldsymbol {\sigma }}^{T}}
We assume that there are no surface couples on
∂
Ω
{\displaystyle \partial {\Omega }}
or body couples
in
Ω
{\displaystyle \Omega }
. Recall the general balance equation
d
d
t
[
∫
Ω
f
(
x
,
t
)
dV
]
=
∫
∂
Ω
f
(
x
,
t
)
[
u
n
(
x
,
t
)
−
v
(
x
,
t
)
⋅
n
(
x
,
t
)
]
dA
+
∫
∂
Ω
g
(
x
,
t
)
dA
+
∫
Ω
h
(
x
,
t
)
dV
.
{\displaystyle {\cfrac {d}{dt}}\left[\int _{\Omega }f(\mathbf {x} ,t)~{\text{dV}}\right]=\int _{\partial {\Omega }}f(\mathbf {x} ,t)[u_{n}(\mathbf {x} ,t)-\mathbf {v} (\mathbf {x} ,t)\cdot \mathbf {n} (\mathbf {x} ,t)]~{\text{dA}}+\int _{\partial {\Omega }}g(\mathbf {x} ,t)~{\text{dA}}+\int _{\Omega }h(\mathbf {x} ,t)~{\text{dV}}~.}
In this case, the physical quantity to be conserved the angular momentum
density, i.e.,
f
=
x
×
(
ρ
v
)
{\displaystyle f=\mathbf {x} \times (\rho ~\mathbf {v} )}
.
The angular momentum source at the surface is then
g
=
x
×
t
{\displaystyle g=\mathbf {x} \times \mathbf {t} }
and the angular momentum source inside the body
is
h
=
x
×
(
ρ
b
)
{\displaystyle h=\mathbf {x} \times (\rho ~\mathbf {b} )}
. The angular momentum and moments are
calculated with respect to a fixed origin. Hence we have
d
d
t
[
∫
Ω
x
×
(
ρ
v
)
dV
]
=
∫
∂
Ω
[
x
×
(
ρ
v
)
]
[
u
n
−
v
⋅
n
]
dA
+
∫
∂
Ω
x
×
t
dA
+
∫
Ω
x
×
(
ρ
b
)
dV
.
{\displaystyle {\cfrac {d}{dt}}\left[\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {v} )~{\text{dV}}\right]=\int _{\partial {\Omega }}[\mathbf {x} \times (\rho ~\mathbf {v} )][u_{n}-\mathbf {v} \cdot \mathbf {n} ]~{\text{dA}}+\int _{\partial {\Omega }}\mathbf {x} \times \mathbf {t} ~{\text{dA}}+\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {b} )~{\text{dV}}~.}
Assuming that
Ω
{\displaystyle \Omega }
is a control volume, we have
∫
Ω
x
×
[
∂
∂
t
(
ρ
v
)
]
dV
=
−
∫
∂
Ω
[
x
×
(
ρ
v
)
]
[
v
⋅
n
]
dA
+
∫
∂
Ω
x
×
t
dA
+
∫
Ω
x
×
(
ρ
b
)
dV
.
{\displaystyle \int _{\Omega }\mathbf {x} \times \left[{\cfrac {\partial }{\partial t}}(\rho ~\mathbf {v} )\right]~{\text{dV}}=-\int _{\partial {\Omega }}[\mathbf {x} \times (\rho ~\mathbf {v} )][\mathbf {v} \cdot \mathbf {n} ]~{\text{dA}}+\int _{\partial {\Omega }}\mathbf {x} \times \mathbf {t} ~{\text{dA}}+\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {b} )~{\text{dV}}~.}
Using the definition of a tensor product we can write
[
x
×
(
ρ
v
)
]
[
v
⋅
n
]
=
[
[
x
×
(
ρ
v
)
]
⊗
v
]
⋅
n
.
{\displaystyle [\mathbf {x} \times (\rho ~\mathbf {v} )][\mathbf {v} \cdot \mathbf {n} ]=[[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]\cdot \mathbf {n} ~.}
Also,
t
=
σ
⋅
n
{\displaystyle \mathbf {t} ={\boldsymbol {\sigma }}\cdot \mathbf {n} }
. Therefore we have
∫
Ω
x
×
[
∂
∂
t
(
ρ
v
)
]
dV
=
−
∫
∂
Ω
[
[
x
×
(
ρ
v
)
]
⊗
v
]
⋅
n
dA
+
∫
∂
Ω
x
×
(
σ
⋅
n
)
dA
+
∫
Ω
x
×
(
ρ
b
)
dV
.
{\displaystyle \int _{\Omega }\mathbf {x} \times \left[{\cfrac {\partial }{\partial t}}(\rho ~\mathbf {v} )\right]~{\text{dV}}=-\int _{\partial {\Omega }}[[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]\cdot \mathbf {n} ~{\text{dA}}+\int _{\partial {\Omega }}\mathbf {x} \times ({\boldsymbol {\sigma }}\cdot \mathbf {n} )~{\text{dA}}+\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {b} )~{\text{dV}}~.}
Using the divergence theorem, we get
∫
Ω
x
×
[
∂
∂
t
(
ρ
v
)
]
dV
=
−
∫
Ω
∇
∙
[
[
x
×
(
ρ
v
)
]
⊗
v
]
dV
+
∫
∂
Ω
x
×
(
σ
⋅
n
)
dA
+
∫
Ω
x
×
(
ρ
b
)
dV
.
{\displaystyle \int _{\Omega }\mathbf {x} \times \left[{\cfrac {\partial }{\partial t}}(\rho ~\mathbf {v} )\right]~{\text{dV}}=-\int _{\Omega }{\boldsymbol {\nabla }}\bullet [[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]~{\text{dV}}+\int _{\partial {\Omega }}\mathbf {x} \times ({\boldsymbol {\sigma }}\cdot \mathbf {n} )~{\text{dA}}+\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {b} )~{\text{dV}}~.}
To convert the surface integral in the above equation into a volume
integral, it is convenient to use index notation. Thus,
[
∫
∂
Ω
x
×
(
σ
⋅
n
)
dA
]
i
=
∫
∂
Ω
e
i
j
k
x
j
σ
k
l
n
l
dA
=
∫
∂
Ω
A
i
l
n
l
dA
=
∫
∂
Ω
A
⋅
n
dA
{\displaystyle \left[\int _{\partial {\Omega }}\mathbf {x} \times ({\boldsymbol {\sigma }}\cdot \mathbf {n} )~{\text{dA}}\right]_{i}=\int _{\partial {\Omega }}e_{ijk}~x_{j}~\sigma _{kl}~n_{l}~{\text{dA}}=\int _{\partial {\Omega }}A_{il}~n_{l}~{\text{dA}}=\int _{\partial {\Omega }}{\boldsymbol {A}}\cdot \mathbf {n} ~{\text{dA}}}
where
[
]
i
{\displaystyle [~]_{i}}
represents the
i
{\displaystyle i}
-th component of the vector. Using
the divergence theorem
∫
∂
Ω
A
⋅
n
dA
=
∫
Ω
∇
∙
A
dV
=
∫
Ω
∂
A
i
l
∂
x
l
dV
=
∫
Ω
∂
∂
x
l
(
e
i
j
k
x
j
σ
k
l
)
dV
.
{\displaystyle \int _{\partial {\Omega }}{\boldsymbol {A}}\cdot \mathbf {n} ~{\text{dA}}=\int _{\Omega }{\boldsymbol {\nabla }}\bullet {\boldsymbol {A}}~{\text{dV}}=\int _{\Omega }{\frac {\partial A_{il}}{\partial x_{l}}}~{\text{dV}}=\int _{\Omega }{\frac {\partial }{\partial x_{l}}}(e_{ijk}~x_{j}~\sigma _{kl})~{\text{dV}}~.}
Differentiating,
∫
∂
Ω
A
⋅
n
dA
=
∫
Ω
[
e
i
j
k
δ
j
l
σ
k
l
+
e
i
j
k
x
j
∂
σ
k
l
∂
x
l
]
dV
=
∫
Ω
[
e
i
j
k
σ
k
j
+
e
i
j
k
x
j
∂
σ
k
l
∂
x
l
]
dV
=
∫
Ω
[
e
i
j
k
σ
k
j
+
e
i
j
k
x
j
[
∇
∙
σ
]
k
]
dV
.
{\displaystyle \int _{\partial {\Omega }}{\boldsymbol {A}}\cdot \mathbf {n} ~{\text{dA}}=\int _{\Omega }\left[e_{ijk}~\delta _{jl}~\sigma _{kl}+e_{ijk}~x_{j}~{\frac {\partial \sigma _{kl}}{\partial x_{l}}}\right]~{\text{dV}}=\int _{\Omega }\left[e_{ijk}~\sigma _{kj}+e_{ijk}~x_{j}~{\frac {\partial \sigma _{kl}}{\partial x_{l}}}\right]~{\text{dV}}=\int _{\Omega }\left[e_{ijk}~\sigma _{kj}+e_{ijk}~x_{j}~[{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}]_{k}\right]~{\text{dV}}~.}
Expressed in direct tensor notation,
∫
∂
Ω
A
⋅
n
dA
=
∫
Ω
[
[
E
:
σ
T
]
i
+
[
x
×
(
∇
∙
σ
)
]
i
]
dV
{\displaystyle \int _{\partial {\Omega }}{\boldsymbol {A}}\cdot \mathbf {n} ~{\text{dA}}=\int _{\Omega }\left[[{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}]_{i}+[\mathbf {x} \times ({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})]_{i}\right]~{\text{dV}}}
where
E
{\displaystyle {\mathcal {E}}}
is the third-order permutation tensor.
Therefore,
[
∫
∂
Ω
x
×
(
σ
⋅
n
)
dA
]
i
==
∫
Ω
[
[
E
:
σ
T
]
i
+
[
x
×
(
∇
∙
σ
)
]
i
]
dV
{\displaystyle \left[\int _{\partial {\Omega }}\mathbf {x} \times ({\boldsymbol {\sigma }}\cdot \mathbf {n} )~{\text{dA}}\right]_{i}==\int _{\Omega }\left[[{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}]_{i}+[\mathbf {x} \times ({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})]_{i}\right]~{\text{dV}}}
or,
∫
∂
Ω
x
×
(
σ
⋅
n
)
dA
==
∫
Ω
[
E
:
σ
T
+
x
×
(
∇
∙
σ
)
]
dV
.
{\displaystyle \int _{\partial {\Omega }}\mathbf {x} \times ({\boldsymbol {\sigma }}\cdot \mathbf {n} )~{\text{dA}}==\int _{\Omega }\left[{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}+\mathbf {x} \times ({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})\right]~{\text{dV}}~.}
The balance of angular momentum can then be written as
∫
Ω
x
×
[
∂
∂
t
(
ρ
v
)
]
dV
=
−
∫
Ω
∇
∙
[
[
x
×
(
ρ
v
)
]
⊗
v
]
dV
+
∫
Ω
[
E
:
σ
T
+
x
×
(
∇
∙
σ
)
]
dV
+
∫
Ω
x
×
(
ρ
b
)
dV
.
{\displaystyle \int _{\Omega }\mathbf {x} \times \left[{\cfrac {\partial }{\partial t}}(\rho ~\mathbf {v} )\right]~{\text{dV}}=-\int _{\Omega }{\boldsymbol {\nabla }}\bullet [[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]~{\text{dV}}+\int _{\Omega }\left[{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}+\mathbf {x} \times ({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})\right]~{\text{dV}}+\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {b} )~{\text{dV}}~.}
Since
Ω
{\displaystyle \Omega }
is an arbitrary volume, we have
x
×
[
∂
∂
t
(
ρ
v
)
]
=
−
∇
∙
[
[
x
×
(
ρ
v
)
]
⊗
v
]
+
E
:
σ
T
+
x
×
(
∇
∙
σ
)
+
x
×
(
ρ
b
)
{\displaystyle \mathbf {x} \times \left[{\cfrac {\partial }{\partial t}}(\rho ~\mathbf {v} )\right]=-{\boldsymbol {\nabla }}\bullet [[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]+{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}+\mathbf {x} \times ({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})+\mathbf {x} \times (\rho ~\mathbf {b} )}
or,
x
×
[
∂
∂
t
(
ρ
v
)
−
∇
∙
σ
−
ρ
b
]
=
−
∇
∙
[
[
x
×
(
ρ
v
)
]
⊗
v
]
+
E
:
σ
T
.
{\displaystyle {\mathbf {x} }\times {\left[{\frac {\partial }{\partial t}}(\rho ~\mathbf {v} )-{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}-\rho ~\mathbf {b} \right]}=-{\boldsymbol {\nabla }}\bullet [[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]+{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}~.}
Using the identity,
∇
∙
(
u
⊗
v
)
=
(
∇
∙
v
)
u
+
(
∇
u
)
⋅
v
{\displaystyle {\boldsymbol {\nabla }}\bullet (\mathbf {u} \otimes \mathbf {v} )=({\boldsymbol {\nabla }}\bullet \mathbf {v} )\mathbf {u} +({\boldsymbol {\nabla }}\mathbf {u} )\cdot \mathbf {v} }
we get
∇
∙
[
[
x
×
(
ρ
v
)
]
⊗
v
]
=
(
∇
∙
v
)
[
x
×
(
ρ
v
)
]
+
(
∇
[
x
×
(
ρ
v
)
]
)
⋅
v
.
{\displaystyle {\boldsymbol {\nabla }}\bullet [[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]=({\boldsymbol {\nabla }}\bullet \mathbf {v} )[\mathbf {x} \times (\rho ~\mathbf {v} )]+({\boldsymbol {\nabla }}[\mathbf {x} \times (\rho ~\mathbf {v} )])\cdot \mathbf {v} ~.}
The second term on the right can be further simplified using index
notation as follows.
[
(
∇
[
x
×
(
ρ
v
)
]
)
⋅
v
]
i
=
[
(
∇
[
ρ
(
x
×
v
)
]
)
⋅
v
]
i
=
∂
∂
x
l
(
ρ
e
i
j
k
x
j
v
k
)
v
l
=
e
i
j
k
[
∂
ρ
∂
x
l
x
j
v
k
v
l
+
ρ
∂
x
j
∂
x
l
v
k
v
l
+
ρ
x
j
∂
v
k
∂
x
l
v
l
]
=
(
e
i
j
k
x
j
v
k
)
(
∂
ρ
∂
x
l
v
l
)
+
ρ
(
e
i
j
k
δ
j
l
v
k
v
l
)
+
e
i
j
k
x
j
(
ρ
∂
v
k
∂
x
l
v
l
)
=
[
(
x
×
v
)
(
∇
ρ
⋅
v
)
+
ρ
v
×
v
+
x
×
(
ρ
∇
v
⋅
v
)
]
i
=
[
(
x
×
v
)
(
∇
ρ
⋅
v
)
+
x
×
(
ρ
∇
v
⋅
v
)
]
i
.
{\displaystyle {\begin{aligned}\left[({\boldsymbol {\nabla }}[\mathbf {x} \times (\rho ~\mathbf {v} )])\cdot \mathbf {v} \right]_{i}=\left[({\boldsymbol {\nabla }}[\rho ~(\mathbf {x} \times \mathbf {v} )])\cdot \mathbf {v} \right]_{i}&={\frac {\partial }{\partial x_{l}}}(\rho ~e_{ijk}~x_{j}~v_{k})~v_{l}\\&=e_{ijk}\left[{\frac {\partial \rho }{\partial x_{l}}}~x_{j}~v_{k}~v_{l}+\rho ~{\frac {\partial x_{j}}{\partial x_{l}}}~v_{k}~v_{l}+\rho ~x_{j}~{\frac {\partial v_{k}}{\partial x_{l}}}~v_{l}\right]\\&=(e_{ijk}~x_{j}~v_{k})~\left({\frac {\partial \rho }{\partial x_{l}}}~v_{l}\right)+\rho ~(e_{ijk}~\delta _{jl}~v_{k}~v_{l})+e_{ijk}~x_{j}~\left(\rho ~{\frac {\partial v_{k}}{\partial x_{l}}}~v_{l}\right)\\&=[(\mathbf {x} \times \mathbf {v} )({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )+\rho ~\mathbf {v} \times \mathbf {v} +\mathbf {x} \times (\rho ~{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} )]_{i}\\&=[(\mathbf {x} \times \mathbf {v} )({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )+\mathbf {x} \times (\rho ~{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} )]_{i}~.\end{aligned}}}
Therefore we can write
∇
∙
[
[
x
×
(
ρ
v
)
]
⊗
v
]
=
(
ρ
∇
∙
v
)
(
x
×
v
)
+
(
∇
ρ
⋅
v
)
(
x
×
v
)
+
x
×
(
ρ
∇
v
⋅
v
)
]
.
{\displaystyle {\boldsymbol {\nabla }}\bullet [[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]=(\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} )(\mathbf {x} \times ~\mathbf {v} )+({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )(\mathbf {x} \times \mathbf {v} )+\mathbf {x} \times (\rho ~{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} )]~.}
The balance of angular momentum then takes the form
x
×
[
∂
∂
t
(
ρ
v
)
−
∇
∙
σ
−
ρ
b
]
=
−
(
ρ
∇
∙
v
)
(
x
×
v
)
−
(
∇
ρ
⋅
v
)
(
x
×
v
)
−
x
×
(
ρ
∇
v
⋅
v
)
+
E
:
σ
T
{\displaystyle {\mathbf {x} }\times {\left[{\frac {\partial }{\partial t}}(\rho ~\mathbf {v} )-{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}-\rho ~\mathbf {b} \right]}=-(\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} )(\mathbf {x} \times ~\mathbf {v} )-({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )(\mathbf {x} \times \mathbf {v} )-\mathbf {x} \times (\rho ~{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} )+{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}}
or,
x
×
[
∂
∂
t
(
ρ
v
)
+
ρ
∇
v
⋅
v
−
∇
∙
σ
−
ρ
b
]
=
−
(
ρ
∇
∙
v
)
(
x
×
v
)
−
(
∇
ρ
⋅
v
)
(
x
×
v
)
+
E
:
σ
T
{\displaystyle {\mathbf {x} }\times {\left[{\frac {\partial }{\partial t}}(\rho ~\mathbf {v} )+\rho ~{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} -{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}-\rho ~\mathbf {b} \right]}=-(\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} )(\mathbf {x} \times ~\mathbf {v} )-({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )(\mathbf {x} \times \mathbf {v} )+{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}}
or,
x
×
[
ρ
∂
v
∂
t
+
∂
ρ
∂
t
v
+
ρ
∇
v
⋅
v
−
∇
∙
σ
−
ρ
b
]
=
−
(
ρ
∇
∙
v
)
(
x
×
v
)
−
(
∇
ρ
⋅
v
)
(
x
×
v
)
+
E
:
σ
T
{\displaystyle {\mathbf {x} }\times {\left[\rho {\frac {\partial \mathbf {v} }{\partial t}}+{\frac {\partial \rho }{\partial t}}~\mathbf {v} +\rho ~{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} -{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}-\rho ~\mathbf {b} \right]}=-(\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} )(\mathbf {x} \times ~\mathbf {v} )-({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )(\mathbf {x} \times \mathbf {v} )+{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}}
The material time derivative of
v
{\displaystyle \mathbf {v} }
is defined as
v
˙
=
∂
v
∂
t
+
∇
v
⋅
v
.
{\displaystyle {\dot {\mathbf {v} }}={\frac {\partial \mathbf {v} }{\partial t}}+{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} ~.}
Therefore,
x
×
[
ρ
v
˙
−
∇
∙
σ
−
ρ
b
]
=
−
x
×
∂
ρ
∂
t
v
+
−
(
ρ
∇
∙
v
)
(
x
×
v
)
−
(
∇
ρ
⋅
v
)
(
x
×
v
)
+
E
:
σ
T
.
{\displaystyle {\mathbf {x} }\times {\left[\rho ~{\dot {\mathbf {v} }}-{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}-\rho ~\mathbf {b} \right]}=-\mathbf {x} \times {\cfrac {\partial \rho }{\partial t}}~\mathbf {v} +-(\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} )(\mathbf {x} \times ~\mathbf {v} )-({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )(\mathbf {x} \times \mathbf {v} )+{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}~.}
Also, from the conservation of linear momentum
ρ
v
˙
−
∇
∙
σ
−
ρ
b
=
0
.
{\displaystyle \rho ~{\dot {\mathbf {v} }}-{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}-\rho ~\mathbf {b} =0~.}
Hence,
0
=
x
×
∂
ρ
∂
t
v
+
(
ρ
∇
∙
v
)
(
x
×
v
)
+
(
∇
ρ
⋅
v
)
(
x
×
v
)
−
E
:
σ
T
=
(
∂
ρ
∂
t
+
ρ
∇
∙
v
+
∇
ρ
⋅
v
)
(
x
×
v
)
−
E
:
σ
T
.
{\displaystyle {\begin{aligned}0&=\mathbf {x} \times {\cfrac {\partial \rho }{\partial t}}~\mathbf {v} +(\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} )(\mathbf {x} \times ~\mathbf {v} )+({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )(\mathbf {x} \times \mathbf {v} )-{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}\\&=\left({\frac {\partial \rho }{\partial t}}+\rho {\boldsymbol {\nabla }}\bullet \mathbf {v} +{\boldsymbol {\nabla }}\rho \cdot \mathbf {v} \right)(\mathbf {x} \times \mathbf {v} )-{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}~.\end{aligned}}}
The material time derivative of
ρ
{\displaystyle \rho }
is defined as
ρ
˙
=
∂
ρ
∂
t
+
∇
ρ
⋅
v
.
{\displaystyle {\dot {\rho }}={\frac {\partial \rho }{\partial t}}+{\boldsymbol {\nabla }}\rho \cdot \mathbf {v} ~.}
Hence,
(
ρ
˙
+
ρ
∇
∙
v
)
(
x
×
v
)
−
E
:
σ
T
=
0
.
{\displaystyle ({\dot {\rho }}+\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} )(\mathbf {x} \times \mathbf {v} )-{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}=0~.}
From the balance of mass
ρ
˙
+
ρ
∇
∙
v
=
0
.
{\displaystyle {\dot {\rho }}+\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} =0~.}
Therefore,
E
:
σ
T
=
0
.
{\displaystyle {\mathcal {E}}:{\boldsymbol {\sigma }}^{T}=0~.}
In index notation,
e
i
j
k
σ
k
j
=
0
.
{\displaystyle e_{ijk}~\sigma _{kj}=0~.}
Expanding out, we get
σ
12
−
σ
21
=
0
;
σ
23
−
σ
32
=
0
;
σ
31
−
σ
13
=
0
.
{\displaystyle \sigma _{12}-\sigma _{21}=0~;~~\sigma _{23}-\sigma _{32}=0~;~~\sigma _{31}-\sigma _{13}=0~.}
Hence,
σ
=
σ
T
{\displaystyle {{\boldsymbol {\sigma }}={\boldsymbol {\sigma }}^{T}}}