The discussion revolves around calculating the inverse of a fourth-order tensor, specifically addressing the conditions under which such an inverse exists and the implications of tensor properties like symmetry.
Some participants have provided guidance on expressing the tensor in a matrix format and applying standard inversion techniques. Others have raised questions about the implications of tensor symmetry and the conditions under which the inverse can be calculated, indicating a productive exploration of the topic.
There are discussions about the potential singularity of the matrix representation due to tensor symmetry, as well as the specific context of elasticity theory where such tensors are commonly applied.
jambaugh said:By "fourth order" I presume you mean rank 4, but that is with regard to the original vector space. A tensor is also a "Vector" in its own vector space as well as a linear operator in several settings but you seem to be implying a double contraction wherein you are mapping rank 2 tensors to rank 2 tensors.
Your task then is to write down a basis for the rank 2 tensor space upon which the rank 4 tensor acts and express that rank 4 tensor's components in that basis in the form of a matrix. You then use the standard matrix inversion techniques to find its inverse.
Example for 2 dimensional vectors you would choose as a basis for your rank 2 space say,e^{11},e^{12},e^{21},e^{22} where e^{ij}=e^i\otimes e^j. Then the rank 4 tensor has matrix form:
A=\left[\begin{array}{cccc} A_{1111} & A_{1112} & A_{1121} & A_{1122} \\<br /> A_{1211} & A_{1212} & A_{1221} & A_{1222} \\<br /> A_{2111} & A_{2112} & A_{2121} & A_{2122} \\<br /> A_{2211} & A_{2212} & A_{2221} & A_{2222}<br /> \end{array}\right]
Invert that matrix and you have the "inverse" in the same basis.
jambaugh said:By "fourth order" I presume you mean rank 4, but that is with regard to the original vector space. A tensor is also a "Vector" in its own vector space as well as a linear operator in several settings but you seem to be implying a double contraction wherein you are mapping rank 2 tensors to rank 2 tensors.
Your task then is to write down a basis for the rank 2 tensor space upon which the rank 4 tensor acts and express that rank 4 tensor's components in that basis in the form of a matrix. You then use the standard matrix inversion techniques to find its inverse.
Example for 2 dimensional vectors you would choose as a basis for your rank 2 space say,e^{11},e^{12},e^{21},e^{22} where e^{ij}=e^i\otimes e^j. Then the rank 4 tensor has matrix form:
A=\left[\begin{array}{cccc} A_{1111} & A_{1112} & A_{1121} & A_{1122} \\<br /> A_{1211} & A_{1212} & A_{1221} & A_{1222} \\<br /> A_{2111} & A_{2112} & A_{2121} & A_{2122} \\<br /> A_{2211} & A_{2212} & A_{2221} & A_{2222}<br /> \end{array}\right]
Invert that matrix and you have the "inverse" in the same basis.
jambaugh said:By "fourth order" I presume you mean rank 4, but that is with regard to the original vector space. A tensor is also a "Vector" in its own vector space as well as a linear operator in several settings but you seem to be implying a double contraction wherein you are mapping rank 2 tensors to rank 2 tensors.
Your task then is to write down a basis for the rank 2 tensor space upon which the rank 4 tensor acts and express that rank 4 tensor's components in that basis in the form of a matrix. You then use the standard matrix inversion techniques to find its inverse.
Example for 2 dimensional vectors you would choose as a basis for your rank 2 space say,e^{11},e^{12},e^{21},e^{22} where e^{ij}=e^i\otimes e^j. Then the rank 4 tensor has matrix form:
A=\left[\begin{array}{cccc} A_{1111} & A_{1112} & A_{1121} & A_{1122} \\<br /> A_{1211} & A_{1212} & A_{1221} & A_{1222} \\<br /> A_{2111} & A_{2112} & A_{2121} & A_{2122} \\<br /> A_{2211} & A_{2212} & A_{2221} & A_{2222}<br /> \end{array}\right]
Invert that matrix and you have the "inverse" in the same basis.