Continuum mechanics/Curl of a gradient of a vector

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[edit] Curl of the gradient of a vector - 1

Let \mathbf{v} be a vector field. Show that

\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}) = 0 ~.

Proof: For a second order tensor field \boldsymbol{S}, we can define the curl as

(\boldsymbol{\nabla} \times \boldsymbol{S})\cdot\mathbf{a} = \boldsymbol{\nabla} \times (\boldsymbol{S}^T\cdot\mathbf{a})

where \mathbf{a} is an arbitrary constant vector. Substituting \boldsymbol{\nabla}\mathbf{v} into the definition, we have

[\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v})]\cdot\mathbf{a} = \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}^T\cdot\mathbf{a}) ~.

Since \mathbf{a} is constant, we may write

\boldsymbol{\nabla}\mathbf{v}^T\cdot\mathbf{a} = \boldsymbol{\nabla} (\mathbf{v}\cdot\mathbf{a}) = \boldsymbol{\nabla} \varphi

where \varphi = \mathbf{v} \cdot \mathbf{a} is a scalar. Hence,

[\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v})]\cdot\mathbf{a} = \boldsymbol{\nabla} \times (\boldsymbol{\nabla \varphi)} ~.

Since the curl of the gradient of a scalar field is zero (recall potential theory), we have

\boldsymbol{\nabla} \times (\boldsymbol{\nabla \varphi)} = \mathbf{0} ~.

Hence,

[\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v})]\cdot\mathbf{a} = \mathbf{0} \qquad\qquad \forall~~\mathbf{a} ~.

The arbitrary nature of \mathbf{a} gives us

{ \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}) = \mathbf{0} \qquad \square }

[edit] Curl of the transpose of the gradient of a vector

Let \mathbf{v} be a vector field. Show that

\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}^T) = \boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{v}) ~.

Proof:

The curl of a second order tensor field \boldsymbol{S} is defined as

(\boldsymbol{\nabla} \times \boldsymbol{S})\cdot\mathbf{a} = \boldsymbol{\nabla} \times (\boldsymbol{S}^T\cdot\mathbf{a})

where \mathbf{a} is an arbitrary constant vector. If we write the right hand side in index notation with respect to a Cartesian basis, we have

[\boldsymbol{S}^T\cdot\mathbf{a}]_k = [\mathbf{b}]_k = b_k = S_{pk}~a_p

and

[\boldsymbol{\nabla} \times \mathbf{b}]_i = e_{ijk}\frac{\partial b_k}{\partial x_j} = e_{ijk}\frac{\partial (S_{pk}~a_p)}{\partial x_j} = e_{ijk}\frac{\partial S_{pk}}{\partial x_j}~a_p = [(\boldsymbol{\nabla} \times \boldsymbol{S})]_{ip}~a_p ~.

In the above a quantity [~]_i represents the i-th component of a vector, and the quantity [~]_{ip} represents the ip-th components of a second-order tensor.

Therefore, in index notation, the curl of a second-order tensor \boldsymbol{S} can be expressed as

[\boldsymbol{\nabla} \times \boldsymbol{S}]_{ip} = e_{ijk}\frac{\partial S_{pk}}{\partial x_j} ~.

Using the above definition, we get

[\boldsymbol{\nabla} \times \boldsymbol{S}^T]_{ip} = e_{ijk}\frac{\partial S_{kp}}{\partial x_j} ~.

If \boldsymbol{S} = \boldsymbol{\nabla}\mathbf{v}, we have

[\boldsymbol{\nabla} \times \boldsymbol{\nabla}\mathbf{v}^T]_{ip} = e_{ijk}\frac{\partial }{\partial x_j} \left(\frac{\partial v_k}{\partial x_p}\right) = \frac{\partial }{\partial x_p}\left(e_{ijk}\frac{\partial v_k}{\partial x_j}\right) = \frac{\partial }{\partial x_p}\left([\boldsymbol{\nabla} \times \mathbf{v}]_i\right) = [\boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{v})]_{ip} ~.

Therefore,

{ \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}^T) = \boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{v}) \qquad \square }
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