Prove that an isotropic substance has only two elastic constants

Question

I've been following the derivation from Ch. II-31 of Feynman's Lectures, and there is a bit that I don't understand. We start from the relation between the strain $T_{ij}$ and the stress $S_{ij}$, $$T_{ij}=\sum_{k,l}\gamma_{ijkl}S_{kl}.$$ Feynman's argument is as follows:

How can the components of $\gamma_{ijkl}$ be independent of the direction of the axes, as they must be if the material is isotropic? Answer: They can be independent only if they are expressible in terms of the tensor $\delta_{ij}$. There are two possible expressions, $\delta_{ij}\delta_{kl}$ and $\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}$, which have the required symmetry, so $\gamma_{ijkl}$ must be a linear combination of them: $$\gamma_{ijkl}=a(\delta_{ij}\delta_{kl})+ b(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}).$$

What is unclear to me is:

1. Why is it necessary that components of $\gamma_{ijkl}$ be expressible in terms of $\delta_{ij}$, if they are independent of the direction of the axes?
2. How can one show that there are only two combinations, $\delta_{ij}\delta_{kl}$ and $\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}$, that have the required symmetry?

Answer 1

In the context you are describing the isotropy implies that the tensor $\gamma_{ijkl}$ must be strictly invariant under arbitrary rotations of the three dimensional Euclidean space $M^i{}_j \in SO(3)$. What I mean by this is that if we perform a rotation, the new components that we get, $$\gamma'_{mnop}=\sum_{i,j,k,l}M^i{}_m M^j{}_n M^k{}_o M^l{}_p \gamma_{ijkl},$$ satisfy $$\gamma'_{mnop}=\gamma_{mnop}$$ i.e., they are the same as before. If you imagine $\gamma$ as a kind of "hyper-matrix", then after the rotation we get the same hyper-matrix (the same numbers in the same positions).

It can be proved that the only tensors (with a non-trivial number of indices) that are invariant under 3-dimensional rotations are the Euclidean metric $\delta_{ij}$ (symmetric in its lower indices) and the Levi-Civita tensor $\epsilon_{ijk}$ (antisymmetric in any pair of indices); and, of course, any linear combination or tensor product of them: $$\delta_{ij}\epsilon_{klm},\quad \delta_{ij}\delta_{kl},\quad \alpha\delta_{ij}\delta_{kl}+\beta\delta_{il}\delta_{kj}\quad...$$ where $\alpha$ and $\beta$ are numbers.

Now I invite you to construct all of these (tensor) products of $\delta_{ij}$ and $\epsilon_{ijk}$ with four indices that have the same index-exchange symmetries of $\gamma_{ijkl}$. It is not difficult to realize that there are only two possibilities: $\delta_{ij}\delta_{kl}$ and the combination $\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}$. Therefore, $\gamma_{ijkl}$ must be a linear combination of them, which is the answer to your question.

Answer 2

If you did some quantum mechanics, you've probably seen how to combine independent spins in terms of total spin with the associated Clebsch-Gordan coefficients. Without looking at the details, you can view this as: $$ j_1\otimes j_2 = \bigoplus_{j=|j_1-j_2|}^{j_1+j_2} j $$ with $j$ being an irreducible representation of spin $j$. The mathematical result can be applied more abstractly on any representations of SO(3).

In particular, your stress tensor lives in $1^{\otimes 4}$, by definition. You question is therefore equivalent to asking what are the trivial irreducible representations in this space. You just need to apply the spin composition 3 times: $$ 1\otimes 1 = 2\oplus 1\oplus 0 $$ which makes sense since and matrix $A$ can be written as: $$ A = \left(\frac{A+A^T}{2}-\frac{Tr(A)}{3}I\right)+\left(\frac{A-A^T}{2}\right)+\left(\frac{Tr(A)}{3}I\right) $$ the first term being in the set of traceless symmetric matrices, a spin 2 representation; the second term in the set of antisymmetric matrices, a spin 1 representation; and the third term in the set of scalar matrices, a spin 0 representation.

You then need only take the tensor product with itself: $$ \begin{align} 1^{\otimes 4} &= (2\oplus 1\oplus 0)^{\otimes 2} \\ &= 0\oplus0\oplus0\oplus ... \end{align} $$ with the $...$ being higher spin representations. The three terms come from one of the expansion terms of $2\otimes 2, 1\otimes 1, 0\otimes 0$.

Concretely, this means that all isotopic tensors rank 4 are of the form: $$ \gamma_{ijkl} = a\delta_{ij}\delta_{kl}+b\delta_{ik}\delta_{jl}+c\delta_{il}\delta_{jk} $$ which has the correct dimension: 3. As Gravitino pointed out, in general, you cannot express symmetric tensors in terms of $\delta$ only, you'll also need $\epsilon$. It turns out those two only suffice. You can see an example for $\epsilon$ for isotropic rank 3 tensors which are all of the form: $$ B_{ijk} \propto \epsilon_{ijk} $$

The fact that you are only interested in $b=c$ is due to the added symmetry that is imposed: $$ \gamma_{ijkl} = \gamma_{ijlk} $$ This is because $S_{kl}=S_{lk}$ ie the stress tensor is symmetric. Upon switching the last two indices, $b,c$ are switched, so for $\gamma$ to be invariant, they need to be equal. This gives the final form of: $$ \gamma_{ijkl} = a\delta_{ij}\delta_{kl}+b(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}) $$

Hope this helps.