# Inverse of a 4th order tensor

1. Jun 12, 2007

### titous

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 $$\mathbf{A}=\alpha*\mathbf{J}+\beta*\mathbf{K}$$. 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 $$\mathbf{A}$$ when
$$\mathbf{A} = \mathbf{C}-\mathbf{B}$$
and when $$\mathbf{B}$$ and $$\mathbf{C}$$ are already defined somewhere else and don't particularly have any convenient symmetry?

Last edited: Jun 12, 2007
2. Jun 12, 2007

### Chris Hillman

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. Jun 20, 2007

### titous

i'm working in the micromechanics field rather than the mathmatics field, and thus, when i talk about a localization tensor $$\mathbf{B}$$ i mean a tensor with the following definition: $$\mathbf{\epsilon}=\mathbf{B}:\mathbf{E}$$.
basically it's a 4th order tensor that when mulitplied by a global 2nd order field ($$\mathbf{E}$$ in this case) yields a local 2nd order field ($$\mathbf{\epsilon}$$ in this case) .

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

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

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

Last edited: Jun 20, 2007
4. Jun 20, 2007

### shoehorn

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. Jun 20, 2007

### titous

Given a 4th order tensor A, i'm trying to find 1/A

6. Jun 20, 2007

### robphy

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. Jun 20, 2007

### explain

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.