Proof of the Symmetry of the Strain Tensor

In the study of continuum mechanics, the strain tensor serves as the fundamental mathematical object used to quantify the deformation of a material body. It captures how much a material is stretched, compressed, or sheared under applied loads. While the displacement field $\mathbf{u}(\mathbf{X})$ provides a complete description of the movement of every point in a body, it contains redundant information—specifically, it conflates actual shape changes (deformation) with pure movements that do not alter the body's shape (rigid body motions).

A critical property of the infinitesimal strain tensor is its symmetry, expressed as $\varepsilon_{ij} = \varepsilon_{ji}$. This is not merely a mathematical convenience; it is a physical necessity. The symmetry of the strain tensor ensures that rigid body rotations do not contribute to internal strain and that the energy stored within an elastic material is physically consistent. This article provides a rigorous derivation of this symmetry and explores its profound implications in solid mechanics.

The Mathematical Definition

To understand the origin of symmetry, we must first define the displacement gradient. Consider a continuous body in a reference configuration. Let $\mathbf{X}$ be the position of a material point, and let $\mathbf{u}(\mathbf{X})$ be the displacement of that point. The displacement gradient tensor, $\mathbf{H}$, is defined as the partial derivative of the displacement components with respect to the reference coordinates:

[
H_{ij} = \frac{\partial u_i}{\partial X_j}
]

In the regime of small deformations, the total displacement gradient can be decomposed into two distinct parts: a symmetric part and an anti-symmetric part. The symmetric part is defined as the infinitesimal strain tensor $\varepsilon_{ij}$:

[
\varepsilon_{ij} = \frac{1}{2} \left( H_{ij} + H_{ji} \right) = \frac{1}{2} \left( \frac{\partial u_i}{\partial X_j} + \frac{\partial u_j}{\partial X_i} \right)
]

While the definition itself is constructed to be symmetric, the physical validity of this construction rests on the fact that the anti-symmetric component represents rotation, not deformation.

Physical Significance of Symmetry

The requirement for $\varepsilon_{ij}$ to be symmetric is driven by three core principles in mechanics:

Separation of Deformation and Rotation: A fundamental requirement of any strain measure is that rigid body motion (translation and rotation) must result in zero strain. A pure rotation of a body changes the orientation of its points but does not change the distances between them. Mathematically, the gradient of a rotation field is purely anti-symmetric. By taking only the symmetric part of the gradient, we effectively "filter out" the rotation.
Thermodynamic Consistency: In hyperelastic materials, the strain energy density $\psi$ must be a scalar function of the state of deformation. If the strain tensor contained an anti-symmetric component, the energy would depend on the orientation of the body in space, which violates the principle of material frame indifference (objectivity).
Constitutive Relations: The linear elastic relationship (Hooke's Law) is expressed as $\sigma_{ij} = C_{ijkl} \varepsilon_{kl}$. For the stiffness tensor $C_{ijkl}$ to maintain its major and minor symmetries and to ensure that the strain energy is positive-definite, the strain tensor must be symmetric.

Formal Proof of Symmetry

1. The Geometric Derivation

The most intuitive proof of why the symmetric part is the "true" strain comes from examining the change in the distance between two neighboring points.

Consider two points in the reference configuration separated by an infinitesimal vector $d\mathbf{X}$. The square of the original distance is:
[
ds_0^2 = dX_i dX_i
]

After deformation, the new distance $ds^2$ between the points (accounting for the change in position $du_i \approx \frac{\partial u_i}{\partial X_j} dX_j$) is:
[
ds^2 = (dX_i + du_i)(dX_i + du_i) \approx dX_i dX_i + 2 dX_i du_i
]

Substituting the displacement gradient:
[
ds^2 = ds_0^2 + 2 \left( \frac{\partial u_i}{\partial X_j} \right) dX_i dX_j
]

Since $dX_i dX_j$ is a symmetric product (the order of $i$ and $j$ does not matter for the scalar product), only the symmetric part of the gradient $\frac{\partial u_i}{\partial X_j}$ contributes to the change in length. We can rewrite the term in the parentheses by adding and subtracting the transposed term:
[
\frac{\partial u_i}{\partial X_j} dX_i dX_j = \frac{1}{2} \left( \frac{\partial u_i}{\partial X_j} + \frac{\partial u_j}{\partial X_i} \right) dX_i dX_j + \frac{1}{2} \left( \frac{\partial u_i}{\partial X_j} - \frac{\partial u_j}{\partial X_i} \right) dX_i dX_j
]

The second term vanishes because it is the product of an anti-symmetric tensor and a symmetric tensor. Thus, the change in the squared length is governed strictly by:
[
ds^2 = ds_0^2 + 2 \varepsilon_{ij} dX_i dX_j
]
This proves that $\varepsilon_{ij}$ is the unique tensor that describes the change in the metric of the body.

2. Verification via Rigid Body Rotation

To confirm that the anti-symmetric part corresponds to rotation, let us assume a pure rigid body rotation where the displacement is given by $\mathbf{u} = \boldsymbol{\omega} \times \mathbf{X}$. In index notation, this is:
[
u_i = \epsilon_{ijk} \omega_k X_j
]
where $\epsilon_{ijk}$ is the Levi-Civita symbol. The displacement gradient becomes:
[
H_{ij} = \frac{\partial u_i}{\partial X_j} = \epsilon_{ikj} \omega_k
]

Because the Levi-Civita symbol is anti-symmetric with respect to its indices ($\epsilon_{ikj} = -\epsilon_{ijk}$), it follows that $H_{ij} = -H_{ji}$. When we apply the definition of the strain tensor:
[
\varepsilon_{ij} = \frac{1}{2}(H_{ij} + H_{ji}) = \frac{1}{2}(H_{ij} - H_{ij}) = 0
]
This confirms that the symmetric part of the gradient is zero during pure rotation, validating the physical necessity of the symmetry.

Illustrative Example: Uniaxial Tension

Consider a simple case of a bar undergoing uniform stretching along the $x_1$-axis. The displacement field is defined as:
[
u_1 = \alpha X_1, \quad u_2 = 0, \quad u_3 = 0
]
where $\alpha$ is a small constant representing the stretch ratio.

The displacement gradient components are:
[
H_{11} = \alpha, \quad \text{all other } H_{ij} = 0
]

The resulting strain tensor is:
[
\varepsilon_{ij} = \begin{bmatrix} \alpha & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}
]
In this case, $\varepsilon_{11} = \varepsilon_{11}$ and all off-diagonal terms are zero, satisfying $\varepsilon_{ij} = \varepsilon_{ji}$.

If we were to add a small rotation $\omega$ around the $x_3$-axis, the displacement gradient would gain anti-symmetric terms (e.g., $H_{12} = -\omega$ and $H_{21} = \omega$). However, the symmetric part $\varepsilon_{ij}$ would remain unchanged, demonstrating that the strain tensor is invariant to rigid body rotations.

Summary

The symmetry of the strain tensor is a cornerstone of continuum mechanics. We have established that:

Mathematically, the strain tensor is defined as the symmetric part of the displacement gradient.
Geometrically, the symmetric part is the only component that affects the change in distance between material points.
Physically, symmetry ensures that rigid body rotations do not induce strain and that the material's energy density remains objective.

Understanding this property is essential for anyone working in structural analysis, finite element modeling, or material science, as it ensures that the mathematical models used to simulate real-world deformations remain physically consistent.