Chapter 27
The Separation of Eigen Stresses
The transformations with which we replaced Eq. (384) with (389) and (390) for the
isotropic continuum in the last chapter was made possible by the assumed isotropy,
which allowed us to insert Eq. (387) into (386). This cannot be done using the general
case of Hooke’s tensor that we will assume now. By differentiating Eq. (384a), we
find
ρ
∂
∂t
1
2
(v i , k +v k , i ) −
1
2
(σ ri , rk +σ rk , ri ) =
1
2
( f i , k + f k , i )
(402)
as well as
ρ
∂
2
∂t 2 v i −
∂
∂t
σ ri , r =
∂
∂t
f i .
(403)
We further require Eq. (384c) and its derivative with respect to time (388),
∂
∂t
ε ik −
1
2
(v i , k + v k , i ) =
1
2
(J ik + J ki ) ,
(384c)
∂
2
∂t 2 ε ik −
∂
∂t
1
2
(v i , k + v k , i ) =
∂
∂t
1
2
(J ik + J ki ) .
(388)
We now need the general case of Hooke’s law, as well as its inverse formulation; this
means
σ ik = C ikrs ε rs ,
ε ik = S ikrs σ rs
(404)
with the tensor S reciprocal to C, thus
S · · C = 1
that is
S ikrs C rspq = δ i p δ kq .
(405)
© The Editor(s) (if applicable) and The Author(s), under exclusive
license to Springer Nature Singapore Pte Ltd. 2020
H. Günther, Elementary Approach to Special Relativity,
https://doi.org/10.1007/978-981-15-3168-2_27
293
The Separation of Eigen Stresses
The transformations with which we replaced Eq. (384) with (389) and (390) for the
isotropic continuum in the last chapter was made possible by the assumed isotropy,
which allowed us to insert Eq. (387) into (386). This cannot be done using the general
case of Hooke’s tensor that we will assume now. By differentiating Eq. (384a), we
find
ρ
∂
∂t
1
2
(v i , k +v k , i ) −
1
2
(σ ri , rk +σ rk , ri ) =
1
2
( f i , k + f k , i )
(402)
as well as
ρ
∂
2
∂t 2 v i −
∂
∂t
σ ri , r =
∂
∂t
f i .
(403)
We further require Eq. (384c) and its derivative with respect to time (388),
∂
∂t
ε ik −
1
2
(v i , k + v k , i ) =
1
2
(J ik + J ki ) ,
(384c)
∂
2
∂t 2 ε ik −
∂
∂t
1
2
(v i , k + v k , i ) =
∂
∂t
1
2
(J ik + J ki ) .
(388)
We now need the general case of Hooke’s law, as well as its inverse formulation; this
means
σ ik = C ikrs ε rs ,
ε ik = S ikrs σ rs
(404)
with the tensor S reciprocal to C, thus
S · · C = 1
that is
S ikrs C rspq = δ i p δ kq .
(405)
© The Editor(s) (if applicable) and The Author(s), under exclusive
license to Springer Nature Singapore Pte Ltd. 2020
H. Günther, Elementary Approach to Special Relativity,
https://doi.org/10.1007/978-981-15-3168-2_27
293
