Micromechanics of composites/Proof 1
Tensor-vector identity - 1
editProof:
Using the identity we have
Also, using the definition we have
Therefore,
Using the identity we have
Finally, using the relation , we get
Hence,
![{\displaystyle [(\mathbf {v} \bullet \mathbf {a} )({\boldsymbol {S}}\bullet \mathbf {b} )]\cdot \mathbf {n} =\mathbf {a} \cdot [\{\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )\}\cdot \mathbf {b} ]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31a344970a0b7f91380ea49a642a8e1aa83d98c9)
Proof:
Using the identity
![{\displaystyle \mathbf {n} \cdot [(\mathbf {v} \bullet \mathbf {a} )({\boldsymbol {S}}\bullet \mathbf {b} )]=\mathbf {b} \cdot [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}^{T}\cdot \mathbf {n} )]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a931704ed14f04f96186aa4ba841eef2af7a2ec1)
Also, using the definition
![{\displaystyle (\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}^{T}\cdot \mathbf {n} )=[({\boldsymbol {S}}^{T}\cdot \mathbf {n} )\otimes \mathbf {v} ]\cdot \mathbf {a} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e65e3102db7772453a9ccaac8a945abfcaf08bd)
Therefore,
![{\displaystyle \mathbf {n} \cdot [(\mathbf {v} \bullet \mathbf {a} )({\boldsymbol {S}}\bullet \mathbf {b} )]=\mathbf {b} \cdot [\{({\boldsymbol {S}}^{T}\cdot \mathbf {n} )\otimes \mathbf {v} \}\cdot \mathbf {a} ]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dcf13178977ca8923a13c89badf31e252a29070)
Using the identity
![{\displaystyle \mathbf {b} \cdot [\{({\boldsymbol {S}}^{T}\cdot \mathbf {n} )\otimes \mathbf {v} \}\cdot \mathbf {a} ]=\mathbf {a} \cdot [\{({\boldsymbol {S}}^{T}\cdot \mathbf {n} )\otimes \mathbf {v} \}^{T}\cdot \mathbf {b} ]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26578ac487fb2b6520938aa9e2e3b74095fdcbed)
Finally, using the relation
![{\displaystyle \mathbf {a} \cdot [\{({\boldsymbol {S}}^{T}\cdot \mathbf {n} )\otimes \mathbf {v} \}^{T}\cdot \mathbf {b} ]=\mathbf {a} \cdot [\{\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\cdot \mathbf {n} )\}\cdot \mathbf {b} ]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9997ab1523c6dd6b53d23974b8ea7b7a226a4953)
Hence,
![{\displaystyle {[(\mathbf {v} \bullet \mathbf {a} )({\boldsymbol {S}}\bullet \mathbf {b} )]\cdot \mathbf {n} =\mathbf {a} \cdot [\{\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )\}\cdot \mathbf {b} ]}\qquad \qquad \qquad \square }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ac235fbe48eff267f6e027fbea1dd91ea5fd1ab)