Material time derivatives
edit
Material time derivatives are needed for many updated Lagrangian formulations
of finite element analysis.
Recall that the motion can be expressed as
x
=
φ
(
X
,
t
)
or
X
=
φ
−
1
(
x
,
t
)
{\displaystyle \mathbf {x} ={\boldsymbol {\varphi }}(\mathbf {X} ,t)\quad {\text{or}}\qquad \mathbf {X} ={\boldsymbol {\varphi }}^{-1}(\mathbf {x} ,t)}
If we keep
X
{\displaystyle \mathbf {X} }
fixed, then the velocity is given by
V
(
X
,
t
)
=
∂
φ
∂
t
(
X
,
t
)
{\displaystyle \mathbf {V} (\mathbf {X} ,t)={\frac {\partial {\boldsymbol {\varphi }}}{\partial t}}(\mathbf {X} ,t)}
This is the material time derivative expressed in terms of
X
{\displaystyle \mathbf {X} }
.
The spatial version of the velocity is
v
(
x
,
t
)
=
V
(
φ
−
1
(
x
,
t
)
,
t
)
{\displaystyle \mathbf {v} (\mathbf {x} ,t)=\mathbf {V} ({\boldsymbol {\varphi }}^{-1}(\mathbf {x} ,t),t)}
We will use the symbol
v
{\displaystyle \mathbf {v} }
for velocity from now on by slightly abusing the
notation.
We usually think of quantities such as velocity and acceleration as spatial
quantities which are functions of
x
{\displaystyle \mathbf {x} }
(rather than material quantities which
are functions of
X
{\displaystyle \mathbf {X} }
).
Given the spatial velocity
v
(
x
,
t
)
{\displaystyle \mathbf {v} (\mathbf {x} ,t)}
, if we want to find the acceleration
we will have to consider the fact that
x
≡
x
(
X
,
t
)
{\displaystyle \mathbf {x} \equiv \mathbf {x} (\mathbf {X} ,t)}
, i.e., the
position also changes with time. We do this by using the chain rule. Thus
D
v
(
x
,
t
)
D
t
=
a
(
x
,
t
)
=
∂
v
(
x
,
t
)
∂
t
+
∂
v
(
x
,
t
)
∂
x
⋅
∂
φ
(
X
,
t
)
∂
t
=
∂
v
∂
t
+
∇
v
⋅
V
=
∂
v
∂
t
+
∇
v
⋅
v
{\displaystyle {\cfrac {D\mathbf {v} (\mathbf {x} ,t)}{Dt}}=\mathbf {a} (\mathbf {x} ,t)={\frac {\partial \mathbf {v} (\mathbf {x} ,t)}{\partial t}}+{\frac {\partial \mathbf {v} (\mathbf {x} ,t)}{\partial \mathbf {x} }}\cdot {\frac {\partial {\boldsymbol {\varphi }}(\mathbf {X} ,t)}{\partial t}}={\frac {\partial \mathbf {v} }{\partial t}}+{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {V} ={\frac {\partial \mathbf {v} }{\partial t}}+{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} }
Such a derivative is called the material time derivative expressed
in terms of
x
{\displaystyle \mathbf {x} }
. The second term in the expression is called the
convective derivative ..
Let the velocity be expressed in spatial form , i.e.,
v
(
x
,
t
)
{\displaystyle \mathbf {v} (\mathbf {x} ,t)}
.
The spatial velocity gradient tensor is given by
l
:=
∂
v
(
x
,
t
)
∂
x
=
∇
v
{\displaystyle {\boldsymbol {l}}:={\frac {\partial \mathbf {v} (\mathbf {x} ,t)}{\partial \mathbf {x} }}={\boldsymbol {\nabla }}\mathbf {v} }
The velocity gradient
l
{\displaystyle {\boldsymbol {l}}}
is a second order tensor which can expressed as
l
=
l
i
j
e
i
⊗
e
j
=
∂
v
i
∂
x
j
e
i
⊗
e
j
{\displaystyle {\boldsymbol {l}}=l_{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}={\frac {\partial v_{i}}{\partial x_{j}}}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}}
The velocity gradient is a measure of the relative velocity of two points in the current configuration.
Recall that the deformation gradient is given by
F
=
∂
φ
∂
X
{\displaystyle {\boldsymbol {F}}={\frac {\partial {\boldsymbol {\varphi }}}{\partial \mathbf {X} }}}
The time derivative of
F
{\displaystyle {\boldsymbol {F}}}
(keeping
X
{\displaystyle \mathbf {X} }
fixed) is
F
˙
=
∂
∂
t
(
∂
φ
∂
X
)
=
∂
∂
X
(
∂
φ
∂
t
)
=
∂
v
∂
X
=
∇
∘
v
{\displaystyle {\dot {\boldsymbol {F}}}={\frac {\partial }{\partial t}}~\left({\frac {\partial {\boldsymbol {\varphi }}}{\partial \mathbf {X} }}\right)={\frac {\partial }{\partial \mathbf {X} }}~\left({\frac {\partial {\boldsymbol {\varphi }}}{\partial t}}\right)={\frac {\partial \mathbf {v} }{\partial \mathbf {X} }}={\boldsymbol {\nabla }}_{\circ }\mathbf {v} }
Using the chain rule
F
˙
=
∂
v
∂
x
⋅
∂
x
∂
X
=
∂
v
∂
x
⋅
∂
φ
∂
X
=
l
⋅
F
{\displaystyle {\dot {\boldsymbol {F}}}={\frac {\partial \mathbf {v} }{\partial \mathbf {x} }}\cdot {\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}={\frac {\partial \mathbf {v} }{\partial \mathbf {x} }}\cdot {\frac {\partial {\boldsymbol {\varphi }}}{\partial \mathbf {X} }}={\boldsymbol {l}}\cdot {\boldsymbol {F}}}
Form this we get the important relation
l
=
F
˙
⋅
F
−
1
.
{\displaystyle {\boldsymbol {l}}={\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}~.}
Time derivative of strain
edit
Let
d
X
1
{\displaystyle d\mathbf {X} _{1}}
and
d
X
2
{\displaystyle d\mathbf {X} _{2}}
be two infinitesimal material line segments in a
body. Then
d
x
1
=
F
⋅
d
X
1
;
d
x
2
=
F
⋅
d
X
2
{\displaystyle d\mathbf {x} _{1}={\boldsymbol {F}}\cdot d\mathbf {X} _{1}~;~~d\mathbf {x} _{2}={\boldsymbol {F}}\cdot d\mathbf {X} _{2}}
Hence,
d
x
1
⋅
d
x
2
=
(
F
⋅
d
X
1
)
⋅
(
F
⋅
d
X
2
)
=
d
X
1
⋅
(
F
T
⋅
F
)
⋅
d
X
2
=
d
X
1
⋅
C
⋅
d
X
2
=
d
X
1
⋅
(
2
E
+
1
)
⋅
d
X
2
{\displaystyle d\mathbf {x} _{1}\cdot d\mathbf {x} _{2}=({\boldsymbol {F}}\cdot d\mathbf {X} _{1})\cdot ({\boldsymbol {F}}\cdot d\mathbf {X} _{2})=d\mathbf {X} _{1}\cdot ({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}})\cdot d\mathbf {X} _{2}=d\mathbf {X} _{1}\cdot {\boldsymbol {C}}\cdot d\mathbf {X} _{2}=d\mathbf {X} _{1}\cdot (2~{\boldsymbol {E}}+{\boldsymbol {\mathit {1}}})\cdot d\mathbf {X} _{2}}
Taking the derivative with respect to
t
{\displaystyle t}
gives us
∂
∂
t
(
d
x
1
⋅
d
x
2
)
=
d
X
1
⋅
∂
C
∂
t
⋅
d
X
2
=
2
d
X
1
⋅
∂
E
∂
t
⋅
d
X
2
{\displaystyle {\frac {\partial }{\partial t}}(d\mathbf {x} _{1}\cdot d\mathbf {x} _{2})=d\mathbf {X} _{1}\cdot {\frac {\partial {\boldsymbol {C}}}{\partial t}}\cdot d\mathbf {X} _{2}=2~d\mathbf {X} _{1}\cdot {\frac {\partial {\boldsymbol {E}}}{\partial t}}\cdot d\mathbf {X} _{2}}
The material strain rate tensor is defined as
E
˙
=
∂
E
∂
t
=
1
2
∂
C
∂
t
=
1
2
C
˙
{\displaystyle {\dot {\boldsymbol {E}}}={\frac {\partial {\boldsymbol {E}}}{\partial t}}={\frac {1}{2}}~{\frac {\partial {\boldsymbol {C}}}{\partial t}}={\frac {1}{2}}~{\dot {\boldsymbol {C}}}}
Clearly,
E
˙
=
1
2
∂
∂
t
(
F
T
⋅
F
)
=
1
2
(
F
˙
T
⋅
F
+
F
T
⋅
F
˙
)
.
{\displaystyle {\dot {\boldsymbol {E}}}={\frac {1}{2}}~{\frac {\partial }{\partial t}}({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}})={\frac {1}{2}}~({\dot {\boldsymbol {F}}}^{T}\cdot {\boldsymbol {F}}+{\boldsymbol {F}}^{T}\cdot {\dot {\boldsymbol {F}}})~.}
Also,
1
2
∂
∂
t
(
d
x
1
⋅
d
x
2
)
=
d
X
1
⋅
E
˙
⋅
d
X
2
=
(
F
−
1
⋅
d
x
1
)
⋅
E
˙
⋅
(
F
−
1
⋅
d
x
2
)
=
d
x
1
⋅
(
F
−
T
⋅
E
˙
⋅
F
−
1
)
⋅
d
x
2
{\displaystyle {\frac {1}{2}}~{\frac {\partial }{\partial t}}(d\mathbf {x} _{1}\cdot d\mathbf {x} _{2})=d\mathbf {X} _{1}\cdot {\dot {\boldsymbol {E}}}\cdot d\mathbf {X} _{2}=({\boldsymbol {F}}^{-1}\cdot d\mathbf {x} _{1})\cdot {\dot {\boldsymbol {E}}}\cdot ({\boldsymbol {F}}^{-1}\cdot d\mathbf {x} _{2})=d\mathbf {x} _{1}\cdot ({\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {E}}}\cdot {\boldsymbol {F}}^{-1})\cdot d\mathbf {x} _{2}}
The spatial rate of deformation tensor or stretching tensor is
defined as
d
=
F
−
T
⋅
E
˙
⋅
F
−
1
=
1
2
F
−
T
⋅
C
˙
⋅
F
−
1
{\displaystyle {\boldsymbol {d}}={\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {E}}}\cdot {\boldsymbol {F}}^{-1}={\frac {1}{2}}~{\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {C}}}\cdot {\boldsymbol {F}}^{-1}}
In fact, we can show that
d
{\displaystyle {\boldsymbol {d}}}
is the symmetric part of the velocity
gradient, i.e.,
d
=
1
2
(
l
+
l
T
)
{\displaystyle {\boldsymbol {d}}={\frac {1}{2}}~({\boldsymbol {l}}+{\boldsymbol {l}}^{T})}
For rigid body motions we get
d
=
0
{\displaystyle {\boldsymbol {d}}={\boldsymbol {\mathit {0}}}}
.
Most of the operations above can be interpreted as push-forward
and pull-back operations. Also, time derivatives of these tensors can
be interpreted as Lie derivatives .
Recall that the push-forward of the strain tensor from the material
configuration to the spatial configuration is given by
e
=
ϕ
∗
[
E
]
=
F
−
T
⋅
E
⋅
F
−
1
{\displaystyle {\boldsymbol {e}}=\phi _{*}[{\boldsymbol {E}}]={\boldsymbol {F}}^{-T}\cdot {\boldsymbol {E}}\cdot {\boldsymbol {F}}^{-1}}
The pull-back of the spatial strain tensor to the material configuration
is given by
E
=
ϕ
∗
[
e
]
=
F
T
⋅
e
⋅
F
{\displaystyle {\boldsymbol {E}}=\phi ^{*}[{\boldsymbol {e}}]={\boldsymbol {F}}^{T}\cdot {\boldsymbol {e}}\cdot {\boldsymbol {F}}}
Therefore, the rate of deformation tensor is a push-forward of the
material strain rate tensor, i.e.,
d
=
F
−
T
⋅
E
˙
⋅
F
−
1
=
ϕ
∗
[
E
˙
]
{\displaystyle {\boldsymbol {d}}={\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {E}}}\cdot {\boldsymbol {F}}^{-1}=\phi _{*}[{\dot {\boldsymbol {E}}}]}
Similarly, the material strain rate tensor is a pull-back of the rate
of deformation tensor to the material configuration, i.e.,
E
˙
=
F
T
⋅
d
⋅
F
=
ϕ
∗
[
d
]
{\displaystyle {\dot {\boldsymbol {E}}}={\boldsymbol {F}}^{T}\cdot {\boldsymbol {d}}\cdot {\boldsymbol {F}}=\phi ^{*}[{\boldsymbol {d}}]}
Now,
E
=
ϕ
∗
[
e
]
⟹
E
˙
=
∂
∂
t
(
ϕ
∗
[
e
]
)
{\displaystyle {\boldsymbol {E}}=\phi ^{*}[{\boldsymbol {e}}]\quad \implies \quad {\dot {\boldsymbol {E}}}={\frac {\partial }{\partial t}}\left(\phi ^{*}[{\boldsymbol {e}}]\right)}
Also,
d
=
ϕ
∗
[
E
˙
]
=
ϕ
∗
[
∂
∂
t
(
ϕ
∗
[
e
]
)
]
{\displaystyle {\boldsymbol {d}}=\phi _{*}[{\dot {\boldsymbol {E}}}]=\phi _{*}\left[{\frac {\partial }{\partial t}}\left(\phi ^{*}[{\boldsymbol {e}}]\right)\right]}
Therefore the rate of deformation tensor can be obtained by first pulling
back
e
{\displaystyle {\boldsymbol {e}}}
to the reference configuration, taking a material time
derivative in that configuration, and then pushing forward the result to
the current configuration.
Such an operation is called a Lie derivative . In general, the Lie
derivative of a spatial tensor
g
{\displaystyle \mathbf {g} }
is defined as
L
ϕ
[
g
]
:=
ϕ
∗
[
∂
∂
t
(
ϕ
∗
[
g
]
)
]
.
{\displaystyle {\mathcal {L}}_{\phi }[{\boldsymbol {g}}]:=\phi _{*}\left[{\frac {\partial }{\partial t}}\left(\phi ^{*}[{\boldsymbol {g}}]\right)\right]~.}
The velocity gradient tensor can be additively decomposed into a symmetric
part and a skew part:
l
=
1
2
(
l
+
l
T
)
+
1
2
(
l
−
l
T
)
=
d
+
w
{\displaystyle {\boldsymbol {l}}={\frac {1}{2}}~({\boldsymbol {l}}+{\boldsymbol {l}}^{T})+{\frac {1}{2}}({\boldsymbol {l}}-{\boldsymbol {l}}^{T})={\boldsymbol {d}}+{\boldsymbol {w}}}
We have seen that
d
{\displaystyle {\boldsymbol {d}}}
is the rate of deformation tensor. The quantity
w
{\displaystyle {\boldsymbol {w}}}
is called the spin tensor .
Note that
d
{\displaystyle {\boldsymbol {d}}}
is symmetric while
w
{\displaystyle {\boldsymbol {w}}}
is skew symmetric, i.e.,
d
=
d
T
;
w
=
−
w
T
.
{\displaystyle {\boldsymbol {d}}={\boldsymbol {d}}^{T}~;~~{\boldsymbol {w}}=-{\boldsymbol {w}}^{T}~.}
So see why
w
{\displaystyle {\boldsymbol {w}}}
is called a "spin", recall that
l
=
F
˙
⋅
F
−
1
{\displaystyle {\boldsymbol {l}}={\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}}
Therefore,
w
=
1
2
(
F
˙
⋅
F
−
1
−
F
−
T
⋅
F
˙
T
)
{\displaystyle {\boldsymbol {w}}={\frac {1}{2}}({\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}-{\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {F}}}^{T})}
Also,
F
=
R
⋅
U
⟹
F
˙
=
R
˙
⋅
U
+
R
⋅
U
˙
{\displaystyle {\boldsymbol {F}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}\quad \implies \quad {\dot {\boldsymbol {F}}}={\dot {\boldsymbol {R}}}\cdot {\boldsymbol {U}}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {U}}}}
Therefore,
F
˙
⋅
F
−
1
=
(
R
˙
⋅
U
+
R
⋅
U
˙
)
⋅
(
U
−
1
⋅
R
T
)
=
R
˙
⋅
R
T
+
R
⋅
U
˙
⋅
U
−
1
⋅
R
T
{\displaystyle {\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}=({\dot {\boldsymbol {R}}}\cdot {\boldsymbol {U}}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {U}}})\cdot ({\boldsymbol {U}}^{-1}\cdot {\boldsymbol {R}}^{T})={\dot {\boldsymbol {R}}}\cdot {\boldsymbol {R}}^{T}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {U}}}\cdot {\boldsymbol {U}}^{-1}\cdot {\boldsymbol {R}}^{T}}
and
F
−
T
⋅
F
˙
T
=
(
R
⋅
U
−
1
)
⋅
(
U
⋅
R
˙
T
+
U
˙
⋅
R
T
)
=
R
⋅
R
˙
T
+
R
⋅
U
−
1
⋅
U
˙
⋅
R
T
{\displaystyle {\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {F}}}^{T}=({\boldsymbol {R}}\cdot {\boldsymbol {U}}^{-1})\cdot ({\boldsymbol {U}}\cdot {\dot {\boldsymbol {R}}}^{T}+{\dot {\boldsymbol {U}}}\cdot {\boldsymbol {R}}^{T})={\boldsymbol {R}}\cdot {\dot {\boldsymbol {R}}}^{T}+{\boldsymbol {R}}\cdot {\boldsymbol {U}}^{-1}\cdot {\dot {\boldsymbol {U}}}\cdot {\boldsymbol {R}}^{T}}
So we have
w
=
1
2
(
R
˙
⋅
R
T
+
R
⋅
U
˙
⋅
U
−
1
⋅
R
T
−
R
⋅
R
˙
T
−
R
⋅
U
−
1
⋅
U
˙
⋅
R
T
)
{\displaystyle {\boldsymbol {w}}={\frac {1}{2}}~({\dot {\boldsymbol {R}}}\cdot {\boldsymbol {R}}^{T}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {U}}}\cdot {\boldsymbol {U}}^{-1}\cdot {\boldsymbol {R}}^{T}-{\boldsymbol {R}}\cdot {\dot {\boldsymbol {R}}}^{T}-{\boldsymbol {R}}\cdot {\boldsymbol {U}}^{-1}\cdot {\dot {\boldsymbol {U}}}\cdot {\boldsymbol {R}}^{T})}
Now
R
⋅
R
T
=
1
⟹
R
˙
⋅
R
T
+
R
⋅
R
˙
T
=
0
{\displaystyle {\boldsymbol {R}}\cdot {\boldsymbol {R}}^{T}={\boldsymbol {\mathit {1}}}\quad \implies \quad {\dot {\boldsymbol {R}}}\cdot {\boldsymbol {R}}^{T}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {R}}}^{T}={\boldsymbol {\mathit {0}}}}
Therefore
w
=
R
˙
⋅
R
T
+
1
2
R
⋅
(
U
˙
⋅
U
−
1
−
U
−
1
⋅
U
˙
)
⋅
R
T
{\displaystyle {\boldsymbol {w}}={\dot {\boldsymbol {R}}}\cdot {\boldsymbol {R}}^{T}+{\frac {1}{2}}~{\boldsymbol {R}}\cdot ({\dot {\boldsymbol {U}}}\cdot {\boldsymbol {U}}^{-1}-{\boldsymbol {U}}^{-1}\cdot {\dot {\boldsymbol {U}}})\cdot {\boldsymbol {R}}^{T}}
The second term above is invariant for rigid body motions and zero for an
uniaxial stretch. Hence, we are left with just a rotation term. This is
why the quantity
w
{\displaystyle {\boldsymbol {w}}}
is called a spin.
The spin tensor is a skew-symmetric tensor and has an associated axial vector
ω
{\displaystyle {\boldsymbol {\omega }}}
(also called the angular velocity vector) whose components are
given by
ω
=
[
w
1
w
2
w
3
]
{\displaystyle {\boldsymbol {\omega }}={\begin{bmatrix}w_{1}\\w_{2}\\w_{3}\end{bmatrix}}}
where
w
=
[
0
−
w
3
w
2
w
3
0
−
w
1
−
w
2
w
1
0
]
{\displaystyle \mathbf {w} ={\begin{bmatrix}0&-w_{3}&w_{2}\\w_{3}&0&-w_{1}\\-w_{2}&w_{1}&0\end{bmatrix}}}
The spin tensor and its associated axial vector appear in a number of modern
numerical algorithms.
Rate of change of volume
edit
Recall that
d
v
=
J
d
V
where
J
=
det
F
{\displaystyle dv=J~dV\qquad {\text{where}}~J=\det {\boldsymbol {F}}}
Therefore, taking the material time derivative of
d
v
{\displaystyle dv}
(keeping
X
{\displaystyle \mathbf {X} }
fixed),
we have
d
d
t
(
d
v
)
=
J
˙
d
V
=
J
˙
J
d
v
{\displaystyle {\cfrac {d}{dt}}(dv)={\dot {J}}~dV={\cfrac {\dot {J}}{J}}~dv}
At this stage we invoke the following result from tensor calculus:
If
A
{\displaystyle {\boldsymbol {A}}}
is an invertible tensor which depends on
t
{\displaystyle t}
then
d
d
t
(
det
A
)
=
(
det
A
)
tr
(
d
A
d
t
⋅
A
−
1
)
{\displaystyle {\cfrac {d}{dt}}(\det {\boldsymbol {A}})=(\det {\boldsymbol {A}})~{\text{tr}}\left({\cfrac {d{\boldsymbol {A}}}{dt}}\cdot {\boldsymbol {A}}^{-1}\right)}
In the case where
A
=
F
,
J
=
det
F
{\displaystyle {\boldsymbol {A}}={\boldsymbol {F}},~J=\det {\boldsymbol {F}}}
we have
d
d
t
(
J
)
=
J
tr
(
F
˙
⋅
F
−
1
)
{\displaystyle {\cfrac {d}{dt}}(J)=J~{\text{tr}}\left({\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}\right)}
or,
J
˙
=
J
tr
(
l
)
=
J
tr
(
d
)
{\displaystyle {\dot {J}}=J~{\text{tr}}({\boldsymbol {l}})=J~{\text{tr}}(\mathbf {d} )}
Therefore,
d
d
t
(
d
v
)
=
tr
(
d
)
d
v
{\displaystyle {\cfrac {d}{dt}}(dv)={\text{tr}}(\mathbf {d} )~dv}
Alternatively, we can also write
J
˙
=
1
2
J
C
−
1
:
C
˙
{\displaystyle {\dot {J}}={\frac {1}{2}}~J~{\boldsymbol {C}}^{-1}:{\dot {\boldsymbol {C}}}}
These relations are of immense use in numerical algorithms - particularly
those which involved incompressible behavior, i.e., when
J
˙
=
0
{\displaystyle {\dot {J}}=0}
.