On this page, we will see that rotating tensors and transforming between different base vectors are very similar operations. From linear algebra, we are familiar with the rotation of vectors by using a rotation matrix, e.g.
R = [ cos ( β ) − sin ( β ) 0 sin ( β ) cos ( β ) 0 0 0 1 ] \begin{aligned} R = \begin{bmatrix} \cos(\beta) & -\sin(\beta) & 0 \\ \sin(\beta) & \cos(\beta) & 0 \\ 0 & 0 & 1\end{bmatrix} \end{aligned} R = ⎣ ⎡ cos ( β ) sin ( β ) 0 − sin ( β ) cos ( β ) 0 0 0 1 ⎦ ⎤ to rotate a vector around the z z z rotation tensor , R = R i j e ‾ i ⊗ e ‾ j \boldsymbol{ R}=R_{ij}\underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j} R = R ij e i ⊗ e j R T = R − 1 \boldsymbol{ R}^{\mathrm{T}}=\boldsymbol{ R}^{-1} R T = R − 1
If we take the matrix product R R T RR^{\mathrm{T}} R R T [ cos ( β ) − sin ( β ) 0 sin ( β ) cos ( β ) 0 0 0 1 ] [ cos ( β ) sin ( β ) 0 − sin ( β ) cos ( β ) 0 0 0 1 ] = [ cos 2 ( β ) + sin 2 ( β ) sin ( β ) cos ( β ) − sin ( β ) cos ( β ) 0 sin ( β ) cos ( β ) − sin ( β ) cos ( β ) cos 2 ( β ) + sin 2 ( β ) 0 0 0 1 ] = [ 1 0 0 0 1 0 0 0 1 ] \begin{aligned} &\begin{bmatrix} \cos(\beta) & -\sin(\beta) & 0 \\ \sin(\beta) & \cos(\beta) & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} \cos(\beta) & \sin(\beta) & 0 \\ -\sin(\beta) & \cos(\beta) & 0 \\ 0 & 0 & 1\end{bmatrix} \\ &= \begin{bmatrix} \cos^2(\beta) + \sin^2(\beta) & \sin(\beta)\cos(\beta)-\sin(\beta)\cos(\beta) & 0 \\ \sin(\beta)\cos(\beta)-\sin(\beta)\cos(\beta) & \cos^2(\beta) + \sin^2(\beta) & 0 \\ 0 & 0 & 1\end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix} \end{aligned} ⎣ ⎡ cos ( β ) sin ( β ) 0 − sin ( β ) cos ( β ) 0 0 0 1 ⎦ ⎤ ⎣ ⎡ cos ( β ) − sin ( β ) 0 sin ( β ) cos ( β ) 0 0 0 1 ⎦ ⎤ = ⎣ ⎡ cos 2 ( β ) + sin 2 ( β ) sin ( β ) cos ( β ) − sin ( β ) cos ( β ) 0 sin ( β ) cos ( β ) − sin ( β ) cos ( β ) cos 2 ( β ) + sin 2 ( β ) 0 0 0 1 ⎦ ⎤ = ⎣ ⎡ 1 0 0 0 1 0 0 0 1 ⎦ ⎤ The rotation tensor rotates a vector, but will not change its length. To rotate a vector v ‾ \underline{\boldsymbol{ v}} v R v ‾ \boldsymbol{ R}\underline{\boldsymbol{ v}} R v v ‾ R \underline{\boldsymbol{ v}}\boldsymbol{ R} v R R i j v j R_{ij} v_j R ij v j v j R j i = R i j − 1 v j v_j R_{ji} = R^{-1}_{ij} v_j v j R ji = R ij − 1 v j R \boldsymbol{ R} R R k i R i j − 1 v j = δ k j v j = v k R_{ki}R^{-1}_{ij} v_j = \delta_{kj} v_j = v_k R ki R ij − 1 v j = δ kj v j = v k R \boldsymbol{ R} R R v ‾ \boldsymbol{ R}\underline{\boldsymbol{ v}} R v
The left illustration has two coordinate systems, e ‾ i \underline{\boldsymbol{ e}}_i e i g ‾ i \underline{\boldsymbol{ g}}_i g i v ‾ \underline{\boldsymbol{ v}} v
v ‾ = v i e e ‾ i = v i g g ‾ i \begin{aligned} \underline{\boldsymbol{ v}}=v_i^{ \mathrm{ e} } \underline{\boldsymbol{ e}}_{ i} = v_i^{ \mathrm{ g} } \underline{\boldsymbol{ g}}_i \end{aligned} v = v i e e i = v i g g i where v i e v_i^{ \mathrm{ e} } v i e v i g v_i^{ \mathrm{ g} } v i g
Assuming that we know v i e v_i^\mathrm{e} v i e v i g v_i^\mathrm{g} v i g v ‾ \underline{\boldsymbol{ v}} v β \beta β R i j R_{ij} R ij g ‾ i \underline{\boldsymbol{ g}}_i g i e ‾ i \underline{\boldsymbol{ e}}_{ i} e i Q i j e g = R i j − 1 = R i j T Q^{ \mathrm{ eg} }_{ij}=R^{-1}_{ij}=R^{\mathrm{T}}_{ij} Q ij eg = R ij − 1 = R ij T To demonstrate this, consider a vector
v ‾ \underline{\boldsymbol{ v}} v with length 1. Then
v 1 e = cos ( α e ) , v 2 e = sin ( α e ) , v 3 e = 0 v i g = Q i k e g v k e v 1 g = cos ( α e ) cos ( β ) + sin ( α e ) sin ( β ) = cos ( α e − β ) = cos ( α g ) v 2 g = − cos ( α e ) sin ( β ) + sin ( α e ) cos ( β ) = sin ( α e − β ) = sin ( α g ) v 3 g = v 3 e = 0 \begin{aligned} v_1^{ \mathrm{ e} } &= \cos(\alpha_{ \mathrm{ e} }),\,v_2^{ \mathrm{ e} }=\sin(\alpha_{ \mathrm{ e} }),\,v_3^{ \mathrm{ e} }=0\\ v_i^{ \mathrm{ g} } &= Q^{ \mathrm{ eg} }_{ik}v_k^{ \mathrm{ e} } \\ v_1^{ \mathrm{ g} } &= \cos(\alpha_{ \mathrm{ e} })\cos(\beta) + \sin(\alpha_{ \mathrm{ e} })\sin(\beta) = \cos(\alpha_{ \mathrm{ e} }-\beta) = \cos(\alpha_{ \mathrm{ g} })\\ v_2^{ \mathrm{ g} } &= -\cos(\alpha_{ \mathrm{ e} })\sin(\beta) + \sin(\alpha_{ \mathrm{ e} })\cos(\beta) = \sin(\alpha_{ \mathrm{ e} }-\beta) = \sin(\alpha_{ \mathrm{ g} })\\ v_3^{ \mathrm{ g} } &= v_3^{ \mathrm{ e} } = 0 \end{aligned} v 1 e v i g v 1 g v 2 g v 3 g = cos ( α e ) , v 2 e = sin ( α e ) , v 3 e = 0 = Q ik eg v k e = cos ( α e ) cos ( β ) + sin ( α e ) sin ( β ) = cos ( α e − β ) = cos ( α g ) = − cos ( α e ) sin ( β ) + sin ( α e ) cos ( β ) = sin ( α e − β ) = sin ( α g ) = v 3 e = 0 However, note that Q ≠ R T \boldsymbol{ Q}\neq\boldsymbol{ R}^{\mathrm{T}} Q = R T Q \boldsymbol{ Q} Q e ‾ i \underline{\boldsymbol{ e}}_{ i} e i g ‾ i \underline{\boldsymbol{ g}}_i g i
Q = Q i j e g g ‾ i ⊗ e ‾ j = ( g ‾ i ⋅ e ‾ j ) g ‾ i ⊗ e ‾ j \begin{aligned} \boldsymbol{ Q} = Q_{ij}^{ \mathrm{ eg} } \underline{\boldsymbol{ g}}_i \otimes \underline{\boldsymbol{ e}}_j = (\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j) \underline{\boldsymbol{ g}}_i \otimes \underline{\boldsymbol{ e}}_j \end{aligned} Q = Q ij eg g i ⊗ e j = ( g i ⋅ e j ) g i ⊗ e j Note that Q \boldsymbol{ Q} Q
Hence, the transformation of the vector v ‾ \underline{\boldsymbol{ v}} v e ‾ i \underline{\boldsymbol{ e}}_i e i g ‾ i \underline{\boldsymbol{ g}}_i g i
Q v ‾ = ( g ‾ i ⋅ e ‾ j ) ( g ‾ i ⊗ e ‾ j ) ⋅ ( v k e e ‾ k ) = ( g ‾ i ⋅ e ‾ j ) g ‾ i δ j k v k e = ( ( g ‾ i ⋅ e ‾ j ) v j e ) g ‾ i v i g = ( g ‾ i ⋅ e ‾ j ) v j e \begin{aligned} \boldsymbol{ Q}\underline{\boldsymbol{ v}} &= (\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j) (\underline{\boldsymbol{ g}}_i \otimes \underline{\boldsymbol{ e}}_j) \cdot (v_k^{ \mathrm{ e} } \underline{\boldsymbol{ e}}_k) \\ &= (\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j) \underline{\boldsymbol{ g}}_i \delta_{jk} v_k^{ \mathrm{ e} } \\ &= ((\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j) v_j^{ \mathrm{ e} }) \underline{\boldsymbol{ g}}_i \\ v_i^{ \mathrm{ g} } &= (\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j) v_j^{ \mathrm{ e} } \end{aligned} Q v v i g = ( g i ⋅ e j ) ( g i ⊗ e j ) ⋅ ( v k e e k ) = ( g i ⋅ e j ) g i δ jk v k e = (( g i ⋅ e j ) v j e ) g i = ( g i ⋅ e j ) v j e But hold on! Earlier, we said that a vector is independent of its coordinate system. This implies that Q v ‾ = v ‾ \boldsymbol{ Q}\underline{\boldsymbol{ v}}=\underline{\boldsymbol{ v}} Q v = v Q = I \boldsymbol{ Q}=\boldsymbol{ I} Q = I So let's try to do the same transformation but starting from
v ‾ \underline{\boldsymbol{ v}} v described in the
g ‾ i \underline{\boldsymbol{ g}}_i g i coordinate system.
Q v ‾ = ( g ‾ i ⋅ e ‾ j ) ( g ‾ i ⊗ e ‾ j ) ⋅ ( v k g g ‾ k ) = ( g ‾ i ⋅ e ‾ j ) g ‾ i ( e ‾ j ⋅ g ‾ k ) v k g = Q i j e g [ Q e g ] j k − 1 v k g g ‾ i = v i g g ‾ i = v ‾ \begin{aligned} \boldsymbol{ Q}\underline{\boldsymbol{ v}} &= (\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j) (\underline{\boldsymbol{ g}}_i \otimes \underline{\boldsymbol{ e}}_j) \cdot (v_k^{ \mathrm{ g} } \underline{\boldsymbol{ g}}_k) \\ &= (\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j) \underline{\boldsymbol{ g}}_i (\underline{\boldsymbol{ e}}_j \cdot \underline{\boldsymbol{ g}}_k) v_k^{ \mathrm{ g} } \\ &= Q^{ \mathrm{ eg} }_{ij} \left[Q^{ \mathrm{ eg} }\right]^{-1}_{jk} v_k^{ \mathrm{ g} } \underline{\boldsymbol{ g}}_i = v_i^{ \mathrm{ g} } \underline{\boldsymbol{ g}}_i = \underline{\boldsymbol{ v}} \end{aligned} Q v = ( g i ⋅ e j ) ( g i ⊗ e j ) ⋅ ( v k g g k ) = ( g i ⋅ e j ) g i ( e j ⋅ g k ) v k g = Q ij eg [ Q eg ] jk − 1 v k g g i = v i g g i = v Since
Q i j e g = g ‾ i ⋅ e ‾ j Q^{ \mathrm{ eg} }_{ij}=\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j Q ij eg = g i ⋅ e j and
Q j k g e = ( Q j k e g ) − 1 = e ‾ j ⋅ g ‾ k Q^{ \mathrm{ ge} }_{jk}=(Q^{ \mathrm{ eg} }_{jk})^{-1}=\underline{\boldsymbol{ e}}_j \cdot \underline{\boldsymbol{ g}}_k Q jk ge = ( Q jk eg ) − 1 = e j ⋅ g k , then
( g ‾ i ⋅ e ‾ j ) ( e ‾ j ⋅ g ‾ k ) = δ i k (\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j) (\underline{\boldsymbol{ e}}_j \cdot \underline{\boldsymbol{ g}}_k)=\delta_{ik} ( g i ⋅ e j ) ( e j ⋅ g k ) = δ ik , and we have
v i g = δ i k v k g = v i g v_i^{ \mathrm{ g} }=\delta_{ik}v_k^{ \mathrm{ g} }=v_i^{ \mathrm{ g} } v i g = δ ik v k g = v i g . So now we showed that
Q v ‾ = v ‾ \boldsymbol{ Q}\underline{\boldsymbol{ v}}=\underline{\boldsymbol{ v}} Q v = v . But this would, as mentioned, imply that
Q = I \boldsymbol{ Q}=\boldsymbol{ I} Q = I . To check this, let's transform the tensor
Q \boldsymbol{ Q} Q , expressed in mixed bases, into the same bases
Q = ( g ‾ i ⋅ e ‾ j ) g ‾ i ⊗ ( Q e ‾ j ) = ( g ‾ i ⋅ e ‾ j ) g ‾ i ⊗ ( g ‾ k ⊗ e ‾ l ⋅ e ‾ j ) = ( g ‾ i ⋅ e ‾ j ) ( g ‾ k ⋅ e ‾ l ) δ l j g ‾ i ⊗ g ‾ k = ( g ‾ i ⋅ e ‾ j ) ( g ‾ k ⋅ e ‾ j ) g ‾ i ⊗ g ‾ k = Q i j e g Q j k g e g ‾ i ⊗ g ‾ k = δ i k g ‾ i ⊗ g ‾ k = I \begin{aligned} \boldsymbol{ Q} &= (\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j) \underline{\boldsymbol{ g}}_i \otimes (\boldsymbol{ Q}\underline{\boldsymbol{ e}}_j)\quad\quad\\ &= (\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j) \underline{\boldsymbol{ g}}_i \otimes ( \underline{\boldsymbol{ g}}_k \otimes \underline{\boldsymbol{ e}}_l\cdot\underline{\boldsymbol{ e}}_j) \\ &= (\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j) (\underline{\boldsymbol{ g}}_k \cdot \underline{\boldsymbol{ e}}_l) \delta_{lj} \underline{\boldsymbol{ g}}_i \otimes \underline{\boldsymbol{ g}}_k \\ &= (\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j) (\underline{\boldsymbol{ g}}_k \cdot \underline{\boldsymbol{ e}}_j) \underline{\boldsymbol{ g}}_i \otimes \underline{\boldsymbol{ g}}_k \\ &= Q^{ \mathrm{ eg} }_{ij} Q^{ \mathrm{ ge} }_{jk}\underline{\boldsymbol{ g}}_i \otimes \underline{\boldsymbol{ g}}_k = \delta_{ik}\underline{\boldsymbol{ g}}_i \otimes \underline{\boldsymbol{ g}}_k = \boldsymbol{ I} \end{aligned} Q = ( g i ⋅ e j ) g i ⊗ ( Q e j ) = ( g i ⋅ e j ) g i ⊗ ( g k ⊗ e l ⋅ e j ) = ( g i ⋅ e j ) ( g k ⋅ e l ) δ l j g i ⊗ g k = ( g i ⋅ e j ) ( g k ⋅ e j ) g i ⊗ g k = Q ij eg Q jk ge g i ⊗ g k = δ ik g i ⊗ g k = I where in the first equality we used that
e ‾ j = Q e ‾ j \underline{\boldsymbol{ e}}_j = \boldsymbol{ Q}\underline{\boldsymbol{ e}}_j e j = Q e j .
If we would like to change to coordinates for higher order tensor, we simply transform each base vector by left-multiplying by Q \boldsymbol{ Q} Q Transforming the bases of the 2nd and 4th order tensors
a \boldsymbol{ a} a and
A \textbf{\textsf{ A}} A , becomes
a i j ( Q e ‾ i ) ⊗ ( Q e ‾ j ) = Q a Q T = ( Q ⊗ ‾ Q ) : a A i j k l ( Q e ‾ i ) ⊗ ( Q e ‾ j ) ⊗ ( Q e ‾ k ) ⊗ ( Q e ‾ l ) = ( Q ⊗ ‾ Q ) : A : ( Q T ⊗ ‾ Q T ) \begin{aligned} a_{ij} (\boldsymbol{ Q}\underline{\boldsymbol{ e}}_i)\otimes(\boldsymbol{ Q}\underline{\boldsymbol{ e}}_j) = \boldsymbol{ Q}\boldsymbol{ a}\boldsymbol{ Q}^{\mathrm{T}} = (\boldsymbol{ Q}\overline{\otimes}\boldsymbol{ Q}):\boldsymbol{ a} \\ \textsf{ A}_{ ijkl} (\boldsymbol{ Q}\underline{\boldsymbol{ e}}_i)\otimes(\boldsymbol{ Q}\underline{\boldsymbol{ e}}_j)\otimes(\boldsymbol{ Q}\underline{\boldsymbol{ e}}_k)\otimes(\boldsymbol{ Q}\underline{\boldsymbol{ e}}_l) = (\boldsymbol{ Q}\overline{\otimes}\boldsymbol{ Q}):\textbf{\textsf{ A}}:(\boldsymbol{ Q}^{\mathrm{T}}\overline{\otimes}\boldsymbol{ Q}^{\mathrm{T}}) \end{aligned} a ij ( Q e i ) ⊗ ( Q e j ) = Q a Q T = ( Q ⊗ Q ) : a A ijk l ( Q e i ) ⊗ ( Q e j ) ⊗ ( Q e k ) ⊗ ( Q e l ) = ( Q ⊗ Q ) : A : ( Q T ⊗ Q T ) When actually doing the calculations however, we work in index notation. So to transform from
a i j e a_{ij}^{ \mathrm{ e} } a ij e and
A i j k l e \textsf{ A}_{ ijkl}^{ \mathrm{ e} } A ijk l e to
a i j g a_{ij}^{ \mathrm{ g} } a ij g and
A i j k l g \textsf{ A}_{ ijkl}^{ \mathrm{ g} } A ijk l g , we have
a i j g = Q i k e g Q j l e g a i j e A i j k l g = Q i m e g Q j n e g Q k o e g Q l p e g A m n o p e \begin{aligned} a_{ij}^{ \mathrm{ g} } &= Q_{ik}^{ \mathrm{ eg} } Q_{jl}^{ \mathrm{ eg} } a_{ij}^{ \mathrm{ e} } \\ \textsf{ A}_{ ijkl}^{ \mathrm{ g} } &= Q_{im}^{ \mathrm{ eg} } Q_{jn}^{ \mathrm{ eg} } Q_{ko}^{ \mathrm{ eg} } Q_{lp}^{ \mathrm{ eg} } \textsf{ A}_{ mnop}^{ \mathrm{ e} } \end{aligned} a ij g A ijk l g = Q ik eg Q j l eg a ij e = Q im eg Q jn eg Q k o eg Q lp eg A mn o p e I.e. we left-multiply each index by the transformation coefficients
Q i j e g = g ‾ i ⋅ e ‾ j Q_{ij}^{ \mathrm{ eg} }=\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j Q ij eg = g i ⋅ e j Above, we have found the following interesting facts about coordinate transformation tensor and rotation tensor.
The rotation tensor is R = R i j e ‾ i ⊗ e ‾ j \boldsymbol{ R}=R_{ij} \underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j} R = R ij e i ⊗ e j R T = R − 1 \boldsymbol{ R}^{\mathrm{T}}=\boldsymbol{ R}^{-1} R T = R − 1
The coordinate transformation tensor is Q = Q i j e g g ‾ i ⊗ e ‾ j = ( g ‾ i ⋅ e ‾ j ) g ‾ i ⊗ e ‾ j = I \boldsymbol{ Q}=Q^{ \mathrm{ eg} }_{ij} \underline{\boldsymbol{ g}}_i \otimes \underline{\boldsymbol{ e}}_j = (\underline{\boldsymbol{ g}}_i \cdot \underline{\boldsymbol{ e}}_j) \underline{\boldsymbol{ g}}_i \otimes \underline{\boldsymbol{ e}}_j = \boldsymbol{ I} Q = Q ij eg g i ⊗ e j = ( g i ⋅ e j ) g i ⊗ e j = I
If we would like to transform from basis system e ‾ i \underline{\boldsymbol{ e}}_i e i g ‾ i = R e ‾ i \underline{\boldsymbol{ g}}_i = \boldsymbol{ R}\underline{\boldsymbol{ e}}_i g i = R e i Q i j e g = R i j T Q^{ \mathrm{ eg} }_{ij} = R^{\mathrm{T}}_{ij} Q ij eg = R ij T
We have already established that to rotate a vector v ‾ \underline{\boldsymbol{ v}} v R \boldsymbol{ R} R R v ‾ \boldsymbol{ R}\underline{\boldsymbol{ v}} R v Rotating a 2nd order tensor,
a \boldsymbol{ a} a , becomes
a i j ( R e ‾ i ) ⊗ ( R e ‾ j ) = a i j ( R k l e ‾ k ⊗ e ‾ l ⋅ e ‾ i ) ⊗ ( R m n e ‾ m ⊗ e ‾ n ⋅ e ‾ j ) = a i j R k l R m n ( e ‾ k δ l i ) ⊗ ( δ n j e ‾ m ) = R k i a i j R m j e ‾ k ⊗ e ‾ m = R k i a i j R j m T e ‾ k ⊗ e ‾ m = R a R T \begin{aligned} a_{ij} (\boldsymbol{ R}\underline{\boldsymbol{ e}}_i)\otimes(\boldsymbol{ R}\underline{\boldsymbol{ e}}_j) &= a_{ij} (R_{kl}\underline{\boldsymbol{ e}}_{ k}\otimes\underline{\boldsymbol{ e}}_{ l}\cdot\underline{\boldsymbol{ e}}_{ i}) \otimes (R_{mn}\underline{\boldsymbol{ e}}_{ m}\otimes\underline{\boldsymbol{ e}}_{ n}\cdot\underline{\boldsymbol{ e}}_{ j}) \\ &= a_{ij} R_{kl} R_{mn} (\underline{\boldsymbol{ e}}_{ k} \delta_{li}) \otimes (\delta_{nj} \underline{\boldsymbol{ e}}_{ m})\\ &= R_{ki} a_{ij} R_{mj} \underline{\boldsymbol{ e}}_{ k}\otimes\underline{\boldsymbol{ e}}_{ m} = R_{ki} a_{ij} R^{\mathrm{T}}_{jm} \underline{\boldsymbol{ e}}_{ k}\otimes\underline{\boldsymbol{ e}}_{ m} \\ &= \boldsymbol{ R} \boldsymbol{ a} \boldsymbol{ R}^{\mathrm{T}} \end{aligned} a ij ( R e i ) ⊗ ( R e j ) = a ij ( R k l e k ⊗ e l ⋅ e i ) ⊗ ( R mn e m ⊗ e n ⋅ e j ) = a ij R k l R mn ( e k δ l i ) ⊗ ( δ nj e m ) = R ki a ij R mj e k ⊗ e m = R ki a ij R jm T e k ⊗ e m = R a R T Hence, if we just consider coefficients via index expressions, we have the rotated coefficients
a k m ′ a'_{km} a km ′ as
a k m ′ = R k i a i j R j m T a'_{km} = R_{ki} a_{ij} R^{\mathrm{T}}_{jm} a km ′ = R ki a ij R jm T .
Similarly, rotating a 4th order tensor, A \textbf{\textsf{ A}} A
A i j k l ( R e ‾ i ) ⊗ ( R e ‾ j ) ⊗ ( R e ‾ k ) ⊗ ( R e ‾ l ) = R m i R n j A i j k l R k o T R l p T e ‾ m ⊗ e ‾ n ⊗ e ‾ o ⊗ e ‾ p = [ R ⊗ ‾ R ] : A : [ R T ⊗ ‾ R T ] \begin{aligned} \textsf{ A}_{ ijkl} (\boldsymbol{ R}\underline{\boldsymbol{ e}}_i)\otimes(\boldsymbol{ R}\underline{\boldsymbol{ e}}_j)\otimes(\boldsymbol{ R}\underline{\boldsymbol{ e}}_k)\otimes(\boldsymbol{ R}\underline{\boldsymbol{ e}}_l) &= R_{mi} R_{nj} \textsf{ A}_{ ijkl} R^{\mathrm{T}}_{ko} R^{\mathrm{T}}_{lp} \underline{\boldsymbol{ e}}_{ m}\otimes\underline{\boldsymbol{ e}}_{ n}\otimes\underline{\boldsymbol{ e}}_{ o}\otimes\underline{\boldsymbol{ e}}_{ p} \\ &= \left[ \boldsymbol{ R}\overline{\otimes} \boldsymbol{ R}\right] : \textbf{\textsf{ A}} : \left[\boldsymbol{ R}^{\mathrm{T}} \overline{\otimes} \boldsymbol{ R}^{\mathrm{T}}\right] \end{aligned} A ijk l ( R e i ) ⊗ ( R e j ) ⊗ ( R e k ) ⊗ ( R e l ) = R mi R nj A ijk l R k o T R lp T e m ⊗ e n ⊗ e o ⊗ e p = [ R ⊗ R ] : A : [ R T ⊗ R T ] and if we just consider coefficients via index expressions, we have the rotated coefficients
A m n o p ′ \textsf{ A}_{ mnop}' A mn o p ′ as
A m n o p ′ = R m i R n j A i j k l R k o T R l p T \textsf{ A}_{ mnop}' = R_{mi} R_{nj} \textsf{ A}_{ ijkl} R^{\mathrm{T}}_{ko} R^{\mathrm{T}}_{lp} A mn o p ′ = R mi R nj A ijk l R k o T R lp T .
