Review of tensors

Mechanics Seismology

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.

\[\begin{split} \delta_{ij} = \begin{cases} 1 & \mbox{($i = j$)} \\ 0 & \mbox{($i \neq j$)} \end{cases} \end{split}\]

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:

\[ c_k \mathbf{\hat{e}}_k=\mathbf{c} = \mathbf{a}\times\mathbf{b} = (a_i\mathbf{\hat{e}}_i)\times (b_j\mathbf{\hat{e}}_j)=a_ib_j(\mathbf{\hat{e}}_i\times\mathbf{\hat{e}}_j)\parallel\cdot \mathbf{\hat{e}}_k \text{.}\]

\[c_k \mathbf{\hat{e}}_k\cdot\mathbf{\hat{e}}_k=c_k=a_i b_j (\mathbf{\hat{e}}_i\times\mathbf{\hat{e}}_j)\cdot\mathbf{\hat{e}}_k \text{.}\]

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 -

\[ (\boldsymbol{a} \times \boldsymbol{b} )_i = \varepsilon_{ijk} a_j b_k, \]

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.

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