Elasticity, free energy of isotropic body mismatch

AI Thread Summary
The discussion focuses on reconciling the general expression for the free energy of an isotropic body in elasticity theory with the commonly accepted form. The user initially encounters an extra term, "+λ*uxx*uyy," in their calculations, which does not appear in the standard isotropic expression and raises concerns about its physical validity. Through analysis involving tensor properties and matrix representations, the user discovers that their initial interpretation of notation was incorrect, leading to confusion. They clarify that the correct expression does indeed account for the cross term, aligning with established theories. Ultimately, the user resolves their misunderstanding and confirms the consistency of the isotropic free energy expression.
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I'm looking at the free energy of a body (theory of elasticity) but I can't really square the general expression with the one usually used for isotropic bodies.

According to wikipedia (http://en.wikipedia.org/wiki/Elastic_energy), Landau & Lifgarbagez etc the general expression for the free (elasticity) energy "F" of a body is

½*Cijkl*uij*ukl (summation implied)

where Cijkl is the elastic modulus tensor and the uij's are the strain tensor, uij=½*((∂ui/∂xj)+(∂uj/∂xi))=uji (small deformations, i.e. no non-linear term).

Considering 2D, the non-zero components of Cijkl for an isotropic medium (as can be seen e.g. on the above wikipedia page) are

Cxxxx=Cyyyy=λ+2μ
Cxyxy=Cyxyx=Cxyyx=Cyxxy
Cxxyy=Cyyxx

where λ and μ are the Lame coefficients (material constants). Now, if I use these above values, I get

(½λ+2μ)*[uxx2+uyy2]+2μ*uxy2+λ*uxx*uyy .

According to Landau & Lifgarbagez, a million papers etc, the free energy of an isotropic body is

F=½λ*uii2+μ*uik2=(½λ+μ)*[uxx2+uyy2]+2μ*uxy2 .

I would expect the general expression to reduce to the isotropic one when using the isotropic elastic modulus tensor. However, I get an extra term "+λ*uxx*uyy" which does not appear in the usual isotropic expression. Moreover, this extra term seems quite unphysical too me, since it does not involve a square, and thus could probably be negative for a non-zero deformation.

It seems to me that this term should go, however, I do not see how.
 
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Could it possibly have something to do with the definitions of the axes or something? I really suck at tensors, but if I viewed Cijkluijukl as a four-component tensor and then presented it as a 4x4 matrix (I'm not entirely sure how legit this is), i.e. with the rows (columns), starting from the top (left) and designating the second (first) two indices: "xx", "xy, "yx", "yy", I find (using symmetry properties) that it is symmetric and real, and thus should be diagonalizable. I'm not entirely sure how to interpret such a diagonalization though.
 
Okay, here's a better idea. The initial expression can be written as a matrix of constants (the moduli) times a column vector with the components uxx, uxy, uyx, uyy, times a row vector with the same components (I checked that it sums up to the right thin). Since it's a scalar, it's invariant under coordinate transformations.

I used Mathematica to find the eigenvalues of the matrix (I included the factor ½): 0, μ, μ, λ+μ, then changed everything to the eigenbasis, making the matrix diagonal. Performing the multiplications now yields the scalar (expressed in a different basis, I use "1" and "2" instead of "x" and "y"; also, I assumed I still have the same symmetries, uij=uji) as

(λ+μ)*u222+2μ*u122 .

This seems better to me, even though it still doesn't quite look like the regular expression for the free energy of an isotropic body. I think I'm going to try messing around with that expression in a similar manner.
 
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Now I wrote the "canonical" isotropic expression in the same "matrix-vector-vector"-form, and for it the matrix looks different, but not diagonal.

My matrix:
½λ+μ 0 0 ½λ
0 ½μ ½μ 0
0 ½μ ½μ 0
½λ 0 0 ½λ+μ

The "standard" matrix:
½λ+μ 0 0 0
0 μ μ 0
0 μ μ 0
0 0 0 ½λ+μ

The vectors are as before.
 
Ok, actually, this was way simpler than I though. I had the right expression, I just misinterpreted others' notation.

uii2=uiiujj=(uxx+uyy)2 ,

including the cross term.
 
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