## You are here

# Seeking a logarithmic operator for a 4th order tensor

I don't know whether this question has an answer, but I'd like to see what you all think:

Does anyone know whether or not the following operation is meaningful, whether it is described and defined algorithmically somewhere, and / or how to do it?

ln(A*ij*) = B*km* ln(C*ijkm*)

A and B are second order tensors

C is a 4th order tensor

The left hand side involves the natural logarithm of the 2nd order tensor A, which is no problem.

The right hand side involves the natural log of the 4th order tensor C, which I have never encountered before.

I greatly appreciate any leads you can provide.

## Log of a 4th order tensor

Interesting question. If Cijkm is not singular, then it should have an eigensystem. If you can find the eigenvectors (2nd order tensors) and normalize them appropriately, you can certainy take the ln of each corresponding eigenvalue. The math for all of this is made easier if C has major and minor symmetries.

Matt Lewis

Los Alamos, New Mexico

## Cijkm is elasticity tensor

In my application, C

ijkmis the isotropic linear elasticity tensor, which comes with an associated inverse - the compliance tensor.Any recommended readings that detail how to approach spectral decomposition for order 4 tensors?

None of my textbooks seem to touch on this topic.

I did find today an article by M. Itskov, but haven't digested it yet. (

Computer Methods in Applied Mechanics and Engineering,Volume 189, Issue 2,1 September 2000,

Pages 419-438)

<A

TARGET="_blank" href="http://ad.doubleclick.net/click%3Bh=v8/3846/3/0/%2a/i%3B204245783%3B0-0%... src="http://m1.2mdn.net/1175468/Scopus08054_728x90_gif.gif" alt=""

BORDER=0></A>

## isotropic elasticity tensor eigensystem

If your elasticity tensor is isotropic, there are six 2nd order eigentensors for it. The first is the second order identity tensor, which you might want to scale by 1/SQRT(3) to normalize it. The corresponding eigenvalue is 3 x the bulk modulus. The remaining eigentensors are five orthogonal deviatoric tensors, which you can construct using Gram-Schmidt orthonormalization. Their associated eigenvalue is 2 x the shear modulus.

Another way of looking at this is to consider using projection operators. In that case, we represent the elasticity tensor as 3K Psp + 2G Pd where Psp is the spherical projection operator, Pd is the deviatoric projection operator, K is the bulk modulus, and G is the shear modulus. ln(C) is then ln(3K) Psp + ln(2G) Pd. The spherical projection operator is 1/3 i x i (the dyadic product of two second order identity tensors). The deviatoric projection operator is simply the 4th order identity tensor for symmetric second order tensors minus the spherical projection operator.

Hope this helps,

Matt

Matt Lewis

Los Alamos, New Mexico

## spectral decomposition

Perhaps this might be related:

http://www.imechanica.org/node/1091