The Sign of the Strain Tensor Determinant and What it Means

Question

Background

I'm currently working my way through Milton's and Cherkaev's, "Which Elasticity Tensors Are Realizable?" as part of my research. I come from an electrical engineering background so I'd say my linear algebra is adequate as an engineer, but I've been teaching myself mechanical engineering, continuum mechanics, etc. this past year.

Milton goes through great efforts in the beginning of Section $2$ to talk about the extremal materials in terms of the eigenvalues of the elasticity matrix, but then shifts to discussing the eigenvalues/vectors of the strain tensor in Section $2.1$. He then says that for a strain tensor $$ \epsilon = -c \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$ the "determinant is non-negative," which is a typo because the determinant is definitely negative given $det(kA) = k^ndet(A)$ for $k \in \mathbb{R}$ and $A \in \mathbb{R}^{n\times n}$.

Besides the typo, he emphasizes that the sign of the strain tensor determinant is the "significant feature" and then begins a discussion about finding a material that supports a strain with a positive determinant.

My question is: What is the physical insight on the impact of the sign of the strain tensor's determinant and why is it the "significant feature" for Milton? I have a grasp on what a negative determinant means in general (i.e. changing the orientation), but all physical intuition goes out the window here when Milton emphasizes this feature. What is it about pure shear strain that "flips" the orientation and what orientation am I even talking about?

P.S. I've also got about a million other questions on this same paper if anyone has the time to talk directly or is willing to correspond with me about this.

Answer 1

It's been a while since I read that paper, but taking a look at it I would say that the typo is not saying "non-negative" but in saying "determinant". I suppose that the authors meant to say "trace" instead.

Let me explain why. As mentioned in another answer, the determinant of the deformation gradient represents the change in volume and it should be non-negative.

We can express the new volume differential as

$$dV' = dV(1 + \epsilon_1)(1 + \epsilon_2)(1 + \epsilon_3)\, ,$$

where $\epsilon_i$ are the eigenvalues of the strain tensor. If we neglect higher-order terms it translates to

$$dV' = dV(1 + \epsilon_1 + \epsilon_2 + \epsilon_3)\, ,$$

and $\epsilon_1 + \epsilon_2 + \epsilon_3$ represents the relative change of volume

$$\frac{dV' - dV}{dV}\, ,$$

and it is equal to the trace of the tensor.

Also, note that in the previous sentence the authors list the eigenvalues of the tensor properly, that is another indicator.

Answer 2

We know that the deformation gradient must have a positive determinant because material cannot pass through itself. The determinant equals the ratio of the deformed to the undeformed infinitesimal volume elements. If its determinant is zero, then a volume (or area) has collapsed to zero, which is also unphysical. In terms of strain, however, the significance may be different. I will take a look tomorrow if I have time.