# Review of tensors#

We present here an overview of tensors and tensor notations (also called suffix or index notation).

Definition. (from Wolfram) An nth-rank tensor in m-dimensional space is a mathematical object that has n indices and $$m^n$$ components and obeys certain transformation rules.

In three-dimensional space, i.e. $$m=3$$, we can recover some familiar objects:

$$\quad \quad n = 0$$: scalar $$a$$, a 0th order tensor with no intrinsic direction - e.g. temperature

$$\quad \quad n = 1$$: vector $$a_i$$, a 1st order tensor that has a magnitude and direction at each point in space - e.g. acceleration

$$\quad \quad n = 2$$: 2nd order tensor (dyad) $$a_{ij}$$, describes the relationship between other algebraic objects related to vector fields - e.g. stress tensor

Later we will introduce a special rank-3 tensor (triad), the Levi-Civita symbol $$\varepsilon_{ijk}$$ which therefore has three indices, $$i, j, k$$.

## Stress tensor#

To introduce the stress tensor, let us assume there is a point P in space where we want to evaluate the stress. We then draw an arbitrarily oriented surface $$S$$ with area $$\Delta S$$ around P, with the orientation of the surface being uniquely defined by the unit normal vector $$\mathbf{n} = (n_x, n_y, n_z)$$. Instead of dealing with the force $$\mathbf{F}$$ acting on the surface, we will be dealing with its density, or the stress. That is, the stress is a vector quantity defined by the limit

$\mathbf{T}_n = \lim_{\Delta S \to 0} \frac{\mathbf{F}_n}{\Delta S},$

where the limit $$\Delta S \to 0$$ means that the surface $$S$$ shrinks to point M, the point where we want to evaluate the stress. Note that the stress vector $$\mathbf{T}_n$$ (often called traction vector, hence $$T$$) depends on the orientation of the surface (and the position of the point P, of course). This is indicated by the suffix $$n$$.

Let us now place the origin of our coordinate system into point P and draw a surface $$S_x$$ around P, such that the surface is normal to the x-axis, like in the figure below. Appropriately, we denote the stress on $$S_x$$ by $$\mathbf{T}_x$$.

As shown in the figure above, the stress $$\mathbf{T}_x$$ may be represented in coordinate decomposition form:

$\mathbf{T}_x = \sigma_{xx}\mathbf{i} + \sigma_{xy}\mathbf{j} + \sigma_{xz}\mathbf{k}.$

Similarly, we can draw surfaces $$S_y$$ and $$S_z$$ and corresponding stresses $$\mathbf{T}_y$$ and $$\mathbf{T}_z$$, whose coordinate decompositions are:

$\begin{split} \mathbf{T}_y = \sigma_{yx}\mathbf{i} + \sigma_{yy}\mathbf{j} + \sigma_{yz}\mathbf{k} \\ \mathbf{T}_z = \sigma_{zx}\mathbf{i} + \sigma_{zy}\mathbf{j} + \sigma_{zz}\mathbf{k}. \end{split}$

Together, these nine components of the stress vectors form the stress tensor $$\mathbf{\sigma}$$

$\begin{split} \bar{\bar{\sigma}} = \begin{pmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} \end{pmatrix}. \end{split}$

Going back to the above figure, we see that $$\sigma_{xx}$$ acts normal to the surface $$S_x$$, while components $$\sigma_{xy}$$ and $$\sigma_{xz}$$ are parallel to $$S_x$$. We can make similar observations about the y- and z- components of the stress vector, where $$\sigma_{yy}$$ and $$\sigma_{zz}$$ will be the normal stress components. Therefore, we call the diagonal elements of the stress tensor the normal stresses and the off-diagonal elements the tangential or shear stresses. Often tangential stresses are denoted by $$\tau$$, e.g. $$\tau_{xy}$$.

We will now state without proof that the stress vector $$\mathbf{T}_n$$ acting on an arbitrary surface can be easily found if we know the stress tensor. This relation is:

$\mathbf{T}_n = \mathbf{n} \bar{\bar{\sigma}},$

or after the multiplication:

$\mathbf{T}_n = n_x \mathbf{T}_x + n_y \mathbf{T}_y + n_z \mathbf{T}_z.$

## Tensor indices#

Mathematics for Scientists and Engineers 2

We will most often work in three dimensions, where we can choose an orthonormal coordinate system $$(\hat{e}_1, \hat{e}_2, \hat{e}_3)$$ to describe any vector v by its three coordinates

$\mathbf{v} = (v_1, v_2, v_3) = v_1\mathbf{\hat{e}}_1 + v_2 \mathbf{\hat{e}}_2 + v_3 \mathbf{\hat{e}}_3 = \sum_{i=1}^3 v_i \mathbf{\hat{e}}_i,$

where $$i$$ is a dummy index and is allowed to take values 1, 2 and 3. We can therefore represent the vector $$\mathbf{v}$$ as $$v_i$$. Similarly, we can represent three-dimensional space by the Cartesian coordinates $$(x_1, x_2, x_3)$$, where $$\mathbf{x}$$, or equivalently $$x_i$$, is a general position vector. In this notation $$x_1$$ corresponds to $$x$$, $$x_2$$ to $$y$$ and $$x_3$$ to $$z$$.The gradient $$\nabla$$ can therefore be written in tensor notation as $$\frac{\partial}{\partial x_i}$$, or sometimes even more compactly as $$\partial_i$$.

It is convenient to introduce the Einstein summation convention here. If a dummy index is repeated twice, it is implicitly understood that this represents a sum over all possible values of that index. A dummy index cannot appear more than two times in one additive term. Therefore we can simply remove the summation symbol

$\sum_{i=1}^3 v_i \mathbf{\hat{e}}_i = v_i \mathbf{\hat{e}}_i.$

Side note: For completeness, we note that the Einstein summation convention states that the sum over an index should only be performed if that index appears once up and once down, i.e. $$v_i \hat{e}^i$$. However, this notation is often abused when dealing with flat Cartesian space to avoid confusing the up-index for the power (e.g. $$v^2$$ is not a square of $$v$$). We can forgive ourselves for this (incorrect) application of the sum convention since raising and lowering an index in flat space in Cartesian coordinates is achieved as follows:

$v^i \equiv \delta^{ij}v_j, \quad \quad v_i \equiv \delta_{ij}v^j,$

where $$\delta_{ij}$$ is the Kronecker delta and $$\delta^{ij}$$ is its inverse. We will see below that this is a trivial operation and that $$v_i = v^i$$, as Kronecker delta functions similarly to the identity matrix and its inverse is also trivially related to it.

### Tensor operations#

In order to be able to perform operations on tensors that we are probably more used to in matrix notation, it is necessary to introduce two special tensors: the Kronecker delta $$\delta_{ij}$$ and the Levi-Civita symbol $$\varepsilon_{ijk}$$.

Kronecker delta is a function of two variables which has a value of 1 if the two variables are the same, or 0 if they are not.

$$\delta_{ij} = \left{ \begin{array}\ 1 & \mbox{(\(i = j$$)} \ 0 & \mbox{($$i \neq j$$)} \end{array} \right.\)

The values can be arranged into a 2x2 matrix, which will be identical to the identity matrix. Kronecker delta can be also expressed as $$\delta_{ij}=\mathbf{\hat{e}}_i\cdot\mathbf{\hat{e}}_j$$.

The three-dimensional fully antisymmetric Levi-Civita symbol $$\varepsilon_{ijk}$$ is defined by:

$\begin{split}\varepsilon_{ijk} = \left\{ \begin{array}\\ +1 & \mbox{if (ijk) = (123), (231), (312)} \\ %\ x \in \mathbf{N}^* \\ -1 & \mbox{if (ijk) = (132), (231), (321)} \\ %\ x = 0 \\ 0 & \mbox{otherwise (i.e. any 2 are equal)} \end{array} \right. \end{split}$

That is, it is 1 if (ijk) is an even permutation of (123), -1 if (ijk) is an odd permutation of 123 and 0 if any index (ijk) is repeated. For example:

$\varepsilon_{ijk} = - \varepsilon_{ikj} = - \varepsilon_{jik} = \varepsilon_{kij}.$

Cross product $$\boldsymbol{c}$$ of two vectors $$\boldsymbol{a}$$ and $$\boldsymbol{b}$$ in tensor notation can be expressed as:

\begin{align}\begin{aligned}\begin{split} c_k \mathbf{\hat{e}}_k=\mathbf{c} = \mathbf{a}\times\mathbf{b} = (a_i\mathbf{\hat{e}}_i)\times (b_j\mathbf{\hat{e}}_j)=a_ib_j(\mathbf{\hat{e}}_i\times\mathbf{\hat{e}}_j)\parallel\cdot \mathbf{\hat{e}}_k,\\ c_k \mathbf{\hat{e}}_k\cdot\mathbf{\hat{e}}_k=c_k=a_i b_j (\mathbf{\hat{e}}_i\times\mathbf{\hat{e}}_j)\cdot\mathbf{\hat{e}}_k.\end{split}\\The Levi-Civita symbol is then equal to \varepsilon_{ijk} = (\mathbf{\hat{e}}_i\times\mathbf{\hat{e}}_j)\cdot\mathbf{\hat{e}}_k and in general, cross product can be expressed in tensor notation as:\\ (\boldsymbol{a} \times \boldsymbol{b} )_i = \varepsilon_{ijk} a_j b_k, \end{aligned}\end{align}

where j and k are dummy indices, while i is a free index as it only appears once in the term, i.e. it is not summed over. As per the definition of tensors, the number of free indices in a term denotes the rank of that term - in this case there is only one free index so the resulting object is a rank-1 tensor, i.e. a vector.

The inner product was already shown to be simply

$\mathbf{a} \cdot \mathbf{b} = a_i b_i = b_i a_i.$

If A is an $$n \times n$$ matrix, then $$A_{ij} = (A^T)_{ji}$$, where $$A^T$$ is the transpose of A. If A is symmetric then $$A_{ij} = A_{ji}$$.

### Examples#

For demonstration purposes let us express some common vector identities in tensor notation:

$\mbox{Curl of a vector:} \quad \quad \mathbf{a} = \nabla \times \mathbf{b} \quad or \quad a_i = \varepsilon_{ijk} \frac{\partial b_k}{\partial x_j}$
$\mbox{Dot product rule:} \quad \quad \nabla \cdot (\phi \mathbf{a}) = \phi \nabla \cdot \mathbf{a} + \nabla \phi \cdot \mathbf{a} \quad \mbox{or} \quad \frac{\partial}{\partial x_i}(\phi a_i) = \phi \frac{\partial a_i}{\partial x_i} + \frac{\partial \phi}{\partial x_i}a_i$
$\mbox{Triple vector product:} \quad \quad \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c} \quad \mbox{or} \quad \varepsilon_{ijk}a_k(\varepsilon_{klm}b_lc_m) = (a_jc_j)b_i - (a_jb_j)c_i$

Here we note that each component of $$\varepsilon_{ijk}a_k(\varepsilon_{klm}b_lc_m)$$ when written in index notation is just a number, as it points to one of its elements - i.e. $$a_k$$ is the kth element of $$\mathbf{a}$$. This means that we can commute them. The LHS can then be written, for example, as $$\varepsilon_{ijk}\varepsilon_{klm}a_kb_lc_m$$, or in any other order we want. This is one of the advantages of tensor notation, as we cannot do this in matrix notation!

$\mbox{Determinant in 2D:} \quad \quad \text{det}M = \varepsilon_{ij}m_{1i}m_{2j}$

The two-dimensional Levi-Civita symbol $$\varepsilon_{ij}$$ is still an anti-symmetric tensor defined as:

$\begin{split} \varepsilon_{ij} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \end{split}$

Let us check explicitly that this is indeed correct.

$\begin{split} \varepsilon_{ij}m_{1i}m_{2j} = \sum_{i=1}^2 \sum_{j=1}^2 \varepsilon_{ij}m_{1i}m_{2j} \\ = \varepsilon_{11}m_{11}m_{21} + \varepsilon_{12}m_{11}m_{22} + \varepsilon_{21}m_{12}m_{21} + \varepsilon_{22}m_{12}m_{22} \\ = m_{11}m_{22} - m_{12}m_{21}\end{split}$

which is indeed correct.

$\mbox{Determinant in 3D:} \quad \quad \text{det}M = \varepsilon_{ijk}m_{1i}m_{2j}m_{3k}$