How to Find the Inverse of a 4th Order Tensor in the Isotropic Case

In summary, the conversation discusses a problem in micromechanics involving the inversion of a 4th order tensor, and the difficulties in doing so when the tensor is not in a convenient form. The person asking for help clarifies their meaning of a localization tensor and restates their question regarding finding the inverse of a 4th order tensor in order to solve for another tensor. However, the use of nonstandard mathematical terminology hinders their ability to receive help.
  • #1
titous
3
0
Hello,

I'm doing some early work in my PhD and I'm coding a micromechanical scheme in which I have many 4th order localization tensors. The problem I'm facing is taking an expression for a 4th order tensor, and then finding the inverse of it. I am assuming the isotropic case and I fully understand that a tensor is easily invertible if it is in the form [tex]\mathbf{A}=\alpha*\mathbf{J}+\beta*\mathbf{K}[/tex]. However, how does one invert a 4th order tensor if it is not in this convenient form?

For example: how do I find the inverse of [tex]\mathbf{A}[/tex] when
[tex]\mathbf{A} = \mathbf{C}-\mathbf{B}[/tex]
and when [tex]\mathbf{B}[/tex] and [tex]\mathbf{C}[/tex] are already defined somewhere else and don't particularly have any convenient symmetry?


Thank you for your help!
 
Last edited:
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  • #2
Request clarification

Hi, titous,

Context is everything! I'll go out on a limb here and guess you are working on something related to nonlinear elasticity. Since a fourth rank tensor is mathematically a multilinear mapping [itex]V^4 \rightarrow R[/tex], it makes no sense to speak of "inverting" a tensor. Be warned further that localization has a standard technical meaning in mathematics which I suspect you do not intend to invoke.

I guess you mean a tensor field and I guess you are talking about converting between Lagrangian and Eulerian viewpoints, but I think you will need to clarify before we can offer any advice/assistance.
 
  • #3
i'm working in the micromechanics field rather than the mathematics field, and thus, when i talk about a localization tensor [tex]\mathbf{B}[/tex] i mean a tensor with the following definition: [tex]\mathbf{\epsilon}=\mathbf{B}:\mathbf{E}[/tex].
basically it's a 4th order tensor that when mulitplied by a global 2nd order field ([tex]\mathbf{E}[/tex] in this case) yields a local 2nd order field ([tex]\mathbf{\epsilon}[/tex] in this case) .

thus, to restate my question, suppose i have the following 4th order tensor [tex]\mathbf{B}[/tex]:
[tex]\mathbf{B}=\mathbf{C^K}+\mathbf{A^-1}[/tex] (i'm trying to raise [tex]\mathbf{A}[/tex] to the -1 power but I'm not sure how to do it in latex)

and suppose that [tex]\mathbf{C^K}[/tex] is known and that [tex]\mathbf{A}[/tex] is defined as:
[tex]\mathbf{A}=\mathbf{b^f}-\mathbf{b^i}[/tex] in which both 4th order tensors [tex]\mathbf{b}[/tex] are known.

how to i go about finding [tex]\mathbf{A^-1}[/tex] (again the -1 means a "raised to") in order to find [tex]\mathbf{B}[/tex]...?
 
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  • #4
titous said:
i'm working in the micromechanics field rather than the mathematics field, and thus, when i talk about a localization tensor [tex]\mathbf{B}[/tex] i mean a tensor with the following definition: [tex]\mathbf{\epsilon}=\mathbf{B}:\mathbf{E}[/tex].
basically it's a 4th order tensor that when mulitplied by a global 2nd order field ([tex]\mathbf{E}[/tex] in this case) yields a local 2nd order field ([tex]\mathbf{\epsilon}[/tex] in this case) .

thus, to restate my question, suppose i have the following 4th order tensor [tex]\mathbf{B}[/tex]:
[tex]\mathbf{B}=\mathbf{C^K}+\mathbf{A^-1}[/tex] (i'm trying to raise [tex]\mathbf{A}[/tex] to the -1 power but I'm not sure how to do it in latex)

and suppose that [tex]\mathbf{C^K}[/tex] is known and that [tex]\mathbf{A}[/tex] is defined as:
[tex]\mathbf{A}=\mathbf{b^f}-\mathbf{b^i}[/tex] in which both 4th order tensors [tex]\mathbf{b}[/tex] are known.

how to i go about finding [tex]\mathbf{A^-1}[/tex] (again the -1 means a "raised to") in order to find [tex]\mathbf{B}[/tex]...?

I would hazard the guess that people will find this no more intelligible than your first post. Could you perhaps try to explain what you're doing using standard terminology?
 
  • #5
shoehorn said:
I would hazard the guess that people will find this no more intelligible than your first post. Could you perhaps try to explain what you're doing using standard terminology?

Given a 4th order tensor A, I'm trying to find 1/A
 
  • #6
titous said:
Given a 4th order tensor A, I'm trying to find 1/A

Can you at least point to an online source for your notation?
Or translate the rather vague "4th order tensor A" and the other tensors into a standard [abstract] index notation or its equivalent [in terms of how it maps vectors and covectors to scalars]?
If the operation (A)(1/A)=1 is legal, what kind of objects are (1/A) and 1?
 
  • #7
titous
The problem is that people here have no clue what is a "global 2nd order field" or "B:E" or "alpha * K". People here know standard mathematics, but you are using mathematical terminology that is highly nonstandard (although maybe everyone in your field uses it). So you won't be able to get any help here unless you give a conventional definition (at the level of beginner's algebra textbook) of what you mean by a 4th order tensor and by an inverse of a 4th order tensor.
 

1. What is a 4th order tensor?

A 4th order tensor is a mathematical object that has four indices and can be represented as a multi-dimensional array. It is used to describe the relationship between two sets of vectors or higher order tensors in a particular coordinate system.

2. What is the inverse of a 4th order tensor?

The inverse of a 4th order tensor is a tensor that, when multiplied with the original tensor, results in the identity tensor. It is used to undo the effect of a given tensor and can be useful in solving certain mathematical problems.

3. How is the inverse of a 4th order tensor calculated?

The inverse of a 4th order tensor is calculated using various mathematical techniques such as matrix inversion or decomposition methods. The specific method used depends on the properties of the tensor and the desired accuracy of the result.

4. What are the applications of the inverse of a 4th order tensor?

The inverse of a 4th order tensor has various applications in fields such as physics, engineering, and mathematics. It is used in solving problems related to elasticity, fluid mechanics, and structural mechanics, among others.

5. Can a 4th order tensor always have an inverse?

No, not all 4th order tensors have an inverse. The invertibility of a tensor depends on its properties and the conditions under which it is defined. Some tensors may have a unique inverse, while others may have multiple or no inverses at all.

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