Understanding Index Notation: Simplifying Tensor and Vector Equations

[nabber]

Hello all, long time lurker, first time poster. I don't know if I am posting this in the proper section, but I would like to ask the following:
In index notation the term σ_{ik}x_{j}n_{k} is \bf{σx}\cdot\bf{n} or \bf{xσ}\cdot\bf{n}, where $σ$ is a second order tensor and $x,n$ are vectors.

On the same note, is $\frac{\partialσ_{ik}}{\partial x_{k}}x_{j}$ equivalent to $\nabla\cdot(\bf{xσ})$ or $\nabla\cdot(\bf{σx})$ ? For some reason there is an index notation rule that eludes me.

Pardon me for the fundamendality or even stupidity of my questions!

 

[Fredrik]

I'm not familiar with how people use that dot product notation when there's more than one index on a tensor component. It looks really ambiguous to me, but maybe there's some convention that removes the ambiguity. If your book explains the notation, maybe you can tell us how.

Since $\partial_i f_i=\nabla\cdot f$, I guess $\frac{\partial\sigma_{ik}}{\partial x_k}x_j$ would have to correspond to something like $(\nabla\cdot\sigma)x$ or $x(\nabla\cdot\sigma)$ in that notation.

 

[nabber]

I totally agree, that's why I am confused! But what about my first question?

 

[Geofleur]

I agree with what Fredrik said, except that I would write the same thing as $(\nabla \cdot \mathbf{\sigma}) \otimes \mathbf{x}$. $\sigma_{ik} x_j n_k$ also looks to me like a dyadic product, so I would write it as something like $(\mathbf{\sigma} \cdot \mathbf{n}) \otimes \mathbf{x}$. As a simpler example, if we had $x_j y_i$, that would be the $ji$ component of the dyadic $\mathbf{x}\otimes\mathbf{y}$.

 

[Chestermiller]

Fredrik's final interpretation is exactly correct for cartesian coordinates. Otherwise, $\partial σ_{ik}/\partial x_k$ does not represent the components of $\vec{∇}\centerdot \vec{σ}$.

Chet

 

[Geofleur]

Very true!

 

[dextercioby]

The dyadic notation is tricky to learn to learn and use, but it's undeniably correct, because it's a statement of a relationship between tensors, not about components in a particular base.

$\vec{v}$ is a vector, $\overleftrightarrow{\sigma}$ is a dyad, typically a 2nd rank tensor. We use $\otimes$ for the dyadic (tensor) product and $.$ for the scalar (contracted tensor) product. And one puts an arrow -> over nabla, too. So:

$$ \frac{\partial \sigma_{ik}}{\partial x_k} x_j \mapsto \left(\vec{\nabla} \bullet \overleftrightarrow{\sigma}\right) \otimes \vec{x} $$.

 

[Chestermiller]

The dyadic notation is tricky to learn to learn and use, but it's undeniably correct, because it's a statement of a relationship between tensors, not about components in a particular base.

$\vec{v}$ is a vector, $\overleftrightarrow{\sigma}$ is a dyad, typically a 2nd rank tensor. We use $\otimes$ for the dyadic (tensor) product and $.$ for the scalar (contracted tensor) product. And one puts an arrow -> over nabla, too. So:

$$ \frac{\partial \sigma_{ik}}{\partial x_k} x_j \mapsto \left(\vec{\nabla} \bullet \overleftrightarrow{\sigma}\right) \otimes \vec{x} $$.

In what way does this differ materially from the sum total of what the other responders said?

 

[dextercioby]

I don't quite get your question. "The other responders" seem to agree to a posting which starts with "I'm not familiar with" and contains "it looks really ambiguous". I just thought to write something that leaves no room to debate/uncertainty.

As a further note, for a divergence of a(n Euclidean) tensor, you usually contract by the first slot of the tensor. Older books I came against called that 'left divergence'. You can also contract by the 2nd (last to the right) slot and you'll have the 'right divergence'. Dyadic notation is really old-fashioned. Even engineering schools (should) teach tensors nowadays.