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?
  • 1
    $\begingroup$ Feynman doesn't provide much help. He says "The tensor δij is often called the “Kronecker delta.” You may amuse yourself by proving that the tensor has exactly the same form if you change the coordinate system to any other rectangular one. " $\endgroup$
    – mmesser314
    May 29 at 17:38

2 Answers 2


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.


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.


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