In the literature, there are different definitions for divergence and curl for second order and higher tensor fields. With the introduction of 3rd order Tensors, #205, we need to define this clearly. This issue is to get an overview over different sources with the goal to make a decision for which definition should be used.
Let's denote a general second-order tensor as $\boldsymbol{S}$ for the discussion. Note that we always assume an orthonormal, right-handed Cartesian coordinate system.
Just comment below the additional definitions and references, and I'll try to keep the tables updated (ping me on Slack if I forget)
Gradient
AFAIK, this is not problematic (correct me if I'm wrong (see below and [5]). To my knowledge, there are just different notations (which can be confusing in itself), i.e.
$$\begin{align*}
\mathrm{grad}(\boldsymbol{v}) &= \nabla \boldsymbol{v} = \boldsymbol{v} \otimes \nabla = \frac{\partial v_i}{x_j} \boldsymbol{e}_i\otimes\boldsymbol{e}_j \\\
\mathrm{grad}(\boldsymbol{S}) &= \nabla \boldsymbol{S} = \boldsymbol{S} \otimes \nabla = \frac{\partial S_{ij}}{x_k} \boldsymbol{e}_i\otimes\boldsymbol{e}_j\otimes\boldsymbol{e}_k
\end{align*}$$
Reddy [5] does actually define the gradient as $\nabla \boldsymbol{v} = \nabla\otimes \boldsymbol{v} = \frac{\partial v_j}{x_i} \boldsymbol{e}_i\otimes\boldsymbol{e}_j$, which is the transpose. And even more, the derivative $\frac{\partial \boldsymbol{v}}{\boldsymbol{x}} = \frac{\partial v_j}{x_i} \boldsymbol{e}_i\otimes\boldsymbol{e}_j$ follows this 👀
Divergence
| Tensor form |
Index form |
Sources |
Comment |
| $\nabla \cdot \boldsymbol{S} $ |
$d_i = \frac{\partial S_{ji}}{x_j}$ |
[1], [5] (2.5.21) |
|
| $\boldsymbol{S} \cdot \nabla $ |
$d_i = \frac{\partial S_{ij}}{x_j}$ |
[2] (2.134), [3] (2.3.11), [4] (2.112) |
Common in mechanics? |
Curl
Here it is important that our definition fulfills $\mathrm{grad}(\mathrm{curl}(\boldsymbol{S}))=\boldsymbol{0}$. There exist definitions in the literature that don't. As a precursor, we haven't defined the cross-product for 2nd-order tensors, so for the discussion, let's define the cross product with a vector $\boldsymbol{v}$ as
$$\begin{align*}
\boldsymbol{S}\times\boldsymbol{v} &= S_{ij} v_k \boldsymbol{e}_i \otimes \boldsymbol{e}_j \times \boldsymbol{e}_k = S_{ij} v_k \boldsymbol{e}_i \otimes [\varepsilon_{jkm} \boldsymbol{e}_m] = S_{ij} v_k \varepsilon_{jkm} \boldsymbol{e}_i \otimes \boldsymbol{e}_m
\\\
\boldsymbol{v}\times \boldsymbol{S} &= v_i S_{jk} \boldsymbol{e}_i \times \boldsymbol{e}_j \otimes \boldsymbol{e}_k = v_i S_{jk} [\varepsilon_{ijm} \boldsymbol{e}_m] \otimes \boldsymbol{e}_k = v_i S_{jk} \varepsilon_{ijm} \boldsymbol{e}_m \otimes \boldsymbol{e}_k
\end{align*}$$
| Tensor form |
Index form |
Sources |
Comment |
| $- \boldsymbol{S} \times \nabla $ |
$d_{ij} = \varepsilon_{opj}\frac{\partial S_{ip}}{x_o}$ |
[3] (2.3.18) |
|
| $\nabla \times \boldsymbol{S}$ |
$d_{ij} = \varepsilon_{opi}\frac{\partial S_{jp}}{x_o}$ |
[1] |
From the definition of the cross-product, this should be $\nabla \times \boldsymbol{S}^\mathrm{T}$
|
where $\varepsilon_{ijk}$ is the Levi-Civita symbol.
Sources
[1] https://en.wikipedia.org/wiki/Tensor_derivative_(continuum_mechanics)
[2] Bonet and Wood (2008)
[3] Rubin (2000)
[4] Itskov (2015)
[5] Reddy (2008) (Reddy, J. N. (2008). An introduction to continuum mechanics: with applications. Cambridge University Press.)
