Micromechanics of composites/Proof 8

Relation between axial vector and displacement

Let $𝐮$ be a displacement field. The displacement gradient tensor is given by $\nabla 𝐮$ . Let the skew symmetric part of the displacement gradient tensor (infinitesimal rotation tensor) be

ω = \frac{1}{2} (\nabla 𝐮 - \nabla 𝐮^{T}) .

Let $θ$ be the axial vector associated with the skew symmetric tensor $ω$ . Show that

θ = \frac{1}{2} \nabla \times 𝐮 .

Proof:

The axial vector $𝐰$ of a skew-symmetric tensor $𝑾$ satisfies the condition

𝑾 \cdot 𝐚 = 𝐰 \times 𝐚

for all vectors $𝐚$ . In index notation (with respect to a Cartesian basis), we have

W_{i p} a_{p} = e_{i j k} w_{j} a_{k}

Since $e_{i j k} = - e_{i k j}$ , we can write

W_{i p} a_{p} = - e_{i k j} w_{j} a_{k} \equiv - e_{i p q} w_{q} a_{p}

or,

W_{i p} = - e_{i p q} w_{q} .

Therefore, the relation between the components of $ω$ and $θ$ is

ω_{i j} = - e_{i j k} θ_{k} .

Multiplying both sides by $e_{p i j}$ , we get

e_{p i j} ω_{i j} = - e_{p i j} e_{i j k} θ_{k} = - e_{p i j} e_{k i j} θ_{k} .

Recall the identity

e_{i j k} e_{p q k} = δ_{i p} δ_{j q} - δ_{i q} δ_{j p} .

Therefore,

e_{i j k} e_{p j k} = δ_{i p} δ_{j j} - δ_{i j} δ_{j p} = 3 δ_{i p} - δ_{i p} = 2 δ_{i p}

Using the above identity, we get

e_{p i j} ω_{i j} = - 2 δ_{p k} θ_{k} = - 2 θ_{p} .

Rearranging,

θ_{p} = - \frac{1}{2} e_{p i j} ω_{i j}

Now, the components of the tensor $ω$ with respect to a Cartesian basis are given by

ω_{i j} = \frac{1}{2} (\frac{\partial u_{i}}{\partial x_{j}} - \frac{\partial u_{j}}{\partial x_{i}})

Therefore, we may write

θ_{p} = - \frac{1}{4} e_{p i j} (\frac{\partial u_{i}}{\partial x_{j}} - \frac{\partial u_{j}}{\partial x_{i}})

Since the curl of a vector $𝐯$ can be written in index notation as

\nabla \times 𝐯 = e_{i j k} \frac{\partial u_{k}}{\partial x_{j}} 𝐞_{i}

we have

e_{p i j} \frac{\partial u_{j}}{\partial x_{i}} = [\nabla \times 𝐮]_{p} and e_{p i j} \frac{\partial u_{i}}{\partial x_{j}} = - e_{p j i} \frac{\partial u_{i}}{\partial x_{j}} = - [\nabla \times 𝐮]_{p}

where $[]_{p}$ indicates the $p$ -th component of the vector inside the square brackets.

Hence,

θ_{p} = - \frac{1}{4} (- [\nabla \times 𝐮]_{p} - [\nabla \times 𝐮]_{p}) = \frac{1}{2} [\nabla \times 𝐮]_{p} .

Therefore,

θ = \frac{1}{2} \nabla \times 𝐮 ◻

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Micromechanics of composites/Proof 8

Relation between axial vector and displacement

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