# Origin of the major symmetry property of the elasticity tensor

In linear elasticity theory the stress tensor $\sigma$ is related to the strain tensor $\epsilon$ via the elastic tensor $C$. Specifically

$$\sigma_{ij} = C_{ijkl} \epsilon_{kl}$$

Because $\sigma$ and $\epsilon$ are both symmetric second rank tensors, $\sigma_{ij} = \sigma_{ji}$ and $\epsilon_{kl} = \epsilon_{lk}$ so that $C$ has the so-called "minor symmetries":

$$C_{ijkl} = C_{jikl} = C_{ijlk}$$

I can convince myself of that, but what I'm having trouble with is the "major symmetry": $$C_{ijkl} = C_{klij}$$

This allegedly comes from symmetry of the Strain Energy Density $\psi(\epsilon)$,

$$\psi = \frac{1}{2} C_{ijkl} \epsilon_{ij} \epsilon_{kl}$$

or some property of its second partial derivatives $$C_{ijkl} = \frac{\partial^2 \psi}{\partial \epsilon_{ij}\partial \epsilon_{kl}} = \frac{\partial^2 \psi}{\partial \epsilon_{kl}\partial \epsilon_{ij}} = C_{klji}$$

To me it looks like its just switching or relabelling the indices rather than transposition. Any idea what I am missing?

** EXPANDED FOLLOWING Phoenix's POST **

Assuming a 2D solid in reduced ($IJ$) notation the strain density is

$$\psi = \epsilon_I C_{IJ} \epsilon_J$$

$$\psi = \begin{bmatrix} e & f \end{bmatrix} \begin{bmatrix} C_{11} & C_{12} \\ C_{21} & C_{22} \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix}$$

$$\psi = e[C_{11} a + C_{12} b] + f[C_{21} a + C_{22} b]$$

and the second derivatives are $$\frac{\partial^2 \psi}{\partial e \partial b} = C_{12}$$ and $$\frac{\partial^2 \psi}{\partial b \partial e} = C_{12}$$

Even in the special situation of $e=a$ and $f=b$ this still comes out as $$\frac{\partial^2 \psi}{\partial a \partial b} = C_{12} + C_{21}$$ and $$\frac{\partial^2 \psi}{\partial b \partial a} = C_{12} + C_{21}$$

i.e. nothing there requires $C$ to be symmetric.

• Your edit is unrelated to Phoenix's answer, in 2D, the multindex is I={11},{12}, or {22}, so you should have a 3x3 matrix. The multindex is irrelevant to the idea, which is that if you fix $i$,$j$,$k$, and $l$ then $C_{ijkl}=\frac{\partial^2\psi}{\partial\epsilon_{ij}\partial\epsilon_{kl}}= \frac{\partial^2\psi}{\partial\epsilon_{kl}\partial\epsilon_{ij}}=C_{klij}$. Mar 7, 2015 at 16:51
• I had the same question and I understood the answer very easily from this video, the answer is in in the part: Reduction 2. youtube.com/watch?v=IhHZ3V4uBsI Feb 1, 2022 at 9:32
• @doobs Indeed the proof above based on the strain energy density is evidently wrong! In the definition of it the possible anti symmetric part of $C$ (which is evidently admitted in the first formula you wrote) is cancelled out. So it is not possible to obtain back $C$ from $\psi$! Some physical argument needs instead. Aug 6, 2022 at 7:57

You should start with the strain energy density $\psi$, then define: $$C_{ijkl} = \frac{\partial^2 \psi}{\partial \epsilon_{ij}\partial \epsilon_{kl}},$$

and then define $$\sigma_{ij} = C_{ijkl} \epsilon_{kl}$$

The remainder of my answer will be about explaining why you have to do it that way. Firstly it is physical, there really is energy associated with strain, and if there weren't there would not be any stress. Secondly, it is linear exactly because we are considering the changes in energy due to small strains.

But let's go back to the minor symmetries. We need $C_{ijkl}=C_{jikl}$ because otherwise $\sigma_{ij}\neq\sigma_{ji}$ (and then we get arbitrarily large, hence unphysical, angular velocities for smaller and smaller regions). But the other minor symmetry is not required. If someone handed you a random rank four tensor, let's call it $B_{ijkl}$, and called it an elastic tensor and it didn't have the second minor symmetry, you can define $C_{ijkl}=(B_{ijkl}+B_{ijlk})/2$ $D_{ijkl}=(B_{ijkl}-B_{ijlk})/2$ and then $B=C+D$ but when D is contracted with a symmetric rank two tensor (like the strain tensor) it gives zero. So the part of the rank four tensor without that second minor symmetry simply doesn't contribute, as an actor it does nothing (when you think all the elasticity does is give you stress from strain). So you might as well assume your tensor has both minor symmetries because it acts like it has the second ($B$ and $C$ act the same on symmetric tensors) and it has to have the first.

Did I bring that up to be pedantic? No, I brought it up because the same thing happens if you contract the elasticity tensor with a rank four symmetric combination of the strain tensor. The part of the elasticity tensor without the major symmetry doesn't contribute to the strain energy density. So a random tensor needs the first minor symmetry to be physical. But you might as well assume it has the second minor symmetry since it doesn't affect the stress-strain relationship. And you might as well assume it has the major symmetry because the part that doesn't will not contribute to the strain energy density.

But it is the strain energy density that is physical, and how it changes is what elasticity is. So you aren't really deriving these symmetries as much as saying that only the symmetric ones generate the physical things you want, energy when given strain. And a real derivation should start with strain energy density and strain, and then just define elasticity from that.

• you are right that the antisymmetric part doesn't play a role in the energy density, but what it is preventing it from playing a role in the relation between $\sigma$ and $\epsilon$? There is no a priori reason why the antisymmetric part should disappear from there too. Mar 6, 2015 at 10:17
• @Phoenix87 The energy density is the primary physical thing. Stress is rightly defined in terms of energy, not the other way around. Once you define the energy density (as a function of strain), then you get the elasticity in terms of the energy density and then the stress in terms of the elasticity (and the strain). So I'm saying the a priori reason is that energy is fundamental and that elasticity is nothing more or less then how energy depends on strain. The fact that stress depends on strain is, to me, simply a derived fact. Mar 7, 2015 at 16:59
• The comment on the second minor symmetry is very helpful and explains what is often hidden: everywhere one reads that the symmetry "comes from the symmetry of $\epsilon$"; as you prove, this is a (non-restrictive) choice. Thank you! Nov 4, 2015 at 17:45
• @Timaeus (Referring to the initial two formulae) Why is the C tensor in the latter identity the same as in the former? I mean, the latter is a physical fact, not a definition. With your approach you are imposing from scratch that C is symmetric (major symmetry) in a physical law without motivations, in my view. Aug 6, 2022 at 8:04
• It should be more honest to declare that major symmetry is a physical fact. @Poenix87 is right in my view. Aug 6, 2022 at 8:07

Since $$\epsilon$$ is a symmetric tensor, it has 6 independent components that determine it. Hence use a multi-index $$I\in\{(i,j)|1\leq i\leq j\leq 3\}$$ to denote them. The strain energy density then becomes (perhaps one has to be careful with "diagonal" terms here in order to get the right coefficients) $$\psi = C_{IJ}\epsilon_I\epsilon_J$$ where summation is intended on repeated multi-indices. It is clear that, since you assume that the order of the second derivatives w.r.t the $$\epsilon_I$$s doesn't matter, it follows that $$C_{IJ} = C_{JI},$$ for any pair of multi-indices $$I$$ and $$J$$. So, in the end, this is the same argument that shows that an Hessian matrix is symmetric if the order of the second derivatives doesn't matter.

• I guess this assumes that $\epsilon_I$ and $\epsilon_J$ are in the same basis, whereas I got the feeling there could be sort of permutation or rotation between their reference axes so that $C != C^T$ in general. Mar 6, 2015 at 1:26
• When I calculate the second derivatives it just comes out as $C_{IJ} = C_{IJ}$ regardless of the order of the partial derivatives. Mar 6, 2015 at 1:42
• Have a look at en.wikipedia.org/wiki/Hooke%27s_law#Anisotropic_materials. The argument you need is there Mar 6, 2015 at 10:16

The following is basically what the Ashcroft/Mermin says about it.

The idea is as following:

in harmonic approximation a relative displacement $u$ results in an energy

$U=- \frac 1 4 (\vec{ u }(\vec R) - \vec{ u }(\vec R ')) \mathbf{D}(\vec{ u }(\vec R ') - \vec{ u }(\vec R))$

The tensor $\mathbf D$ already has natural symmetry $D_{\mu\nu}=D_{\nu\mu}$, meaning that the interaction is symmetric. Approximating

$\vec{ u }(\vec R ') \approx \vec{ u }(\vec R )+ (\vec R -\vec R ')\nabla \vec u$

one gets

$U= \frac 1 2 \frac{\partial u_\mu}{\partial x_\sigma}R_\sigma D_{\mu\nu} R_\tau \frac{\partial u_\nu}{\partial x_\tau}=\frac 1 2 \frac{\partial u_\mu}{\partial x_\sigma}E_{\sigma\mu\tau\nu} \frac{\partial u_\nu}{\partial x_\tau}$

It is then stated that the harmonic energy must be unaffected by a rotation. This requires the energy to depend on the derivatives of the form

$\epsilon_{\sigma\mu}=\frac 1 2(\frac{\partial u_\mu}{\partial x_\sigma}+\frac{\partial u_\sigma}{\partial x_\mu})$

This is somewhat like a Lifschitz invariant, I guess. It should not be too difficult to prove that.

It is then stated that the energy can be written in the Form

22.78) $U=\frac 1 2 \epsilon_{\sigma\mu} c_{\sigma\mu\tau\nu} \epsilon_{\tau\nu}$

with

22.79) $c_{\sigma\mu\tau\nu} = \frac 1 8 (E_{\sigma\mu\tau\nu}+E_{\mu\sigma\tau\nu}+E_{\sigma\mu\nu\tau}+E_{\mu\sigma\nu\tau})$

Looking at the definition of $E_{\sigma\mu\tau\nu}$ this provides the symmetry of concern. So it originates in the requirement that the energy is invariant upon rotation.

However, right now I have some difficulties with the last step (that is step 22.78 to 22.79 in the Ashcroft/Mermin hence the numbers) as I think it produces some derivatives that are not present in the original form. If someone can easily explain why the form 22.78 requires $c$ to be of the form 22.79, I would be happy getting it explained.