Tensors

Roughly speaking, tensors are generalisations of objects like vectors and matrices which have any number of indices, i.e. $T_{ij…k}$. The underlying object described by a tensor has to be independent of the choice of basis, meaning the components of a tensor should transform in the right way under rotations.

Consider a rotation $R$ from a basis $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ to a new one $\mathbf{e}_1’, \mathbf{e}_2’, \mathbf{e}_3’$, we have

\[R_{ij} = \mathbf{e}'_i \cdot \mathbf{e}_j\]

and the components of a position vector $\mathbf{x}$ are related by

\[x_i' = R_{ij} x_j\]

Since we are considering rigid rotations of the coordinate axes, the transformation matrix will be orthogonal such that $R^{-1} = R^\intercal$, we have

\[x_i = R_{ji} x_j' \qquad \text{and} \qquad {\partial x_i \over \partial x_j'} = R_{ji} \qquad \text{and} \qquad R_{ik} R_{jk} = R_{ki} R_{kj} = \delta_{ij}\]

With the above we can start the study of tensors of different ranks.

Definition. A rank $0$ Cartesian tensor is the set of quantities that is invariant under rotations, i.e. a scalar.

Definition. A rank $1$ Cartesian tensor is the set of quantities with components that change under rotations following the condition

\[v_i' = R_{ij} v_j \qquad \text{and} \qquad v_i = R_{ji} v_j'\]

i.e. a vector.

Proposition. The scalar product of two rank $1$ tensors is a rank $0$ tensor.

Proof.

\[\mathbf{u} \cdot \mathbf{v} = u_i' v_i' = R_{ij} u_j R_{ik} v_k = \delta_{jk} u_j v_k = u_j v_j\]

Proposition. The gradient of a rank $0$ tensor is a rank $1$ tensor.

Proof.

\[F_i' = (\nabla \phi)_i' = { \partial \phi' \over \partial x_i'} = { \partial x_j \over \partial x_i' } {\partial \phi \over \partial x_j} = R_{ij} F_j\]

Proposition. The divergence of a rank $1$ tensor is a rank $0$ tensor.

Proof.

\[\left( {\partial v_i \over \partial x_i} \right)' = {\partial v_i' \over \partial x_i'} = {\partial x_j \over \partial x_i'} {\partial R_{ik} v_k \over \partial x_j} = R_{ij} R_{ik} {\partial v_k \over \partial x_j} = \delta_{jk} {\partial v_k \over \partial x_j} = {\partial v_j \over \partial x_j}\]

Following on from rank $0$ tensor with no subscripts and rank $1$ tensor with one subscript, we can extend this idea to quantities that require two or more subscripts to identify a particular element.

Definition. A rank $2$ Cartesian tensor is the set of quantities with components that change under rotations following the condition

\[T_{ij}' = R_{ik} R_{jl} T_{kl} \qquad \text{and} \qquad T_{ij} = R_{ki} R_{lj} T_{kl}'\]

Definition. A rank $p$ Cartesian tensor is the set of quantities with components that change under rotations following the condition

\[T_{i_1 i_2 ... i_p}' = R_{i_1 j_1} R_{i_2 j_2} ... R_{i_p j_p} T_{j_1 j_2 ... j_p} \qquad \text{and} \qquad T_{i_1 i_2 ... i_p} = R_{j_1 i_1} R_{j_2 i_2} ... R_{j_p i_p} T_{j_1 j_2 ... j_p}'\]

A rank $2$ tensor is geometrically like linear map that transform one vector into another without reference to any coordinate system, and behaves in the same way under orthogonal transformations $\mathsf{T}’ = \mathsf{RTR’}$. However, not all linear maps are rank $2$ tensors. The critical idea behind tensors is that all the subscripts must refer to the same coordinate system. For example, a rank $2$ tensor in $\mathbb{R}^3$ has 9 components because the underlying coordinate system has $3$ axes, and therefore $3 \times 3$ combinations among them. In order to identify a quantity, we will need a value for each of them, i.e. $xx, xy, xz, yx, yy, yz, zx, zy, zz$ and requires two subscripts.

In the discussion of differential operators, we only define the gradient of scalar field. With tensors, we can define the gradient of vector field (and generalized to any order).

Proposition. The gradient of a rank $1$ tensor is a rank $2$ tensor.

Proof.

\[T_{ij}' = { \partial v_i' \over \partial x_j'} = R_{ik} { \partial x_l \over \partial x_j'} { \partial v_k \over \partial x_l } = R_{ik} R_{jl} T_{kl}\]

Algebra of Tensors

Proposition. The sum and difference of two tensors of the same rank is also a tensor. Multiplying a tensor by a constant is also a tensor.

Proof.

It follows immediately from the linearity of a rotation of coordinates.

Proposition. If $T_{i_1 i_2 … i_p}$ are the components of a tensor, then so as the set of quantities formed by interchanging the order of indices, e.g. $T_{i_2 i_1 … i_p}$.

Definition. Suppose $S$ is a tensor of rank $p$ and $T$ is a tensor of ranks $q$. The outer/tensor product of them, denoted by $S \otimes T$, is defined by

\[(S \otimes T)_{i_1 i_2 ... i_p j_1 j_2 ... j_q} = S_{i_1 i_2 ... i_p} T_{j_1 j_2 ... j_q}\]

Proposition. The outer product of two tensors of rank $p$ and $q$ respectively is also a tensor of rank $p + q$.

Proof.

Using outer product of two first-rank tensors $\mathbf{u}$ and $\mathbf{v}$ as example, we have

\[T_{ij}' = u_i' v_j' = R_{ik} R_{jl} u_k v_l = R_{ik} R_{jl} T_{kl}\]

Proposition. Suppose $S$ is a tensor $S$ of rank $p$. Then contraction on two indices using $\delta_{ij}$ results with a tensor $T$ of rank $p - 2$, i.e.

\[T_{k_1 ... k_{p-2}} = \delta_{ij} S_{ij k_1 ... k_{p-2}}\]

Proof.

\[\begin{align*} T_{i i k_1 .. k_{p-2}}' &= R_{im} R_{in} R_{k_1 l_1} ... R_{k_{p-2} l_{p-2}} T_{m n l_1 ... l_{p-2}} \\ &= \delta_{mn} R_{k_1 l_1} ... R_{k_{p-2} l_{p-2}} T_{m n l_1 ... l_{p-2}} \\ &= R_{k_1 l_1} ... R_{k_{p-2} l_{p-2}} T_{m m l_1 ... l_{p-2}} \end{align*}\]

For example, a second-rank tensor can be contracted to a scalar, which we call the trace, i.e. $\text{Tr}\, T = T_{ii}$. Another example is the scalar product of two vectors, which we first have the outer product of two vectors and contracting it to produce a scalar (rank $1 + 1 - 2$).

Quotient Law

Proposition. [Quotient Law] Suppose that $T$ and $U$ are tensors. If

\[S_{i_1 ... k ... i_p} T_{j_1 ... k ... j_q} = U_{i_1 ... i_p j_1 ... j_q}\]

holds for all rotations then $S$ is a tensor.

Proof.

We use $p = q = 2$ as example to demonstrate the principle. Given that $S_{ik} T_{jk} = U_{ij}$, then

\[\begin{align*} S_{ik}' T_{jk}' = U_{ij}' &= R_{il} R_{jm} U_{lm} \\ &= R_{il} R_{jm} S_{ln} T_{mn} \\ &= R_{il} R_{jm} S_{ln} R_{om} R_{kn} T_{ok}' \\ &= R_{il} R_{kn} S_{ln} T_{jk}' \end{align*}\]

Therefore, we have

\[(S_{ik}' - R_{il} R_{kn} S_{ln}) T_{jk}' = 0 \quad \implies \quad S_{ik}' = R_{il} R_{kn} S_{ln}\]

Similar arguments can be applied for the case $S_{ik} T_{kj} = U_{ij}$.

Quotient law provides a convenient way to test wheater a given quantity is a tensor. For example, if we want to check if $T_{i_1 i_2 … i_{p+q}}$ is a tensor, we can contract it with a rank $q$ tensor and check if the result is a rank $p$ tensor.

Symmetry and Anti-Symmetry

Definition. A tensor is symmetric with respect to its first two subscripts if $T_{i_2 i_1 … i_p} = T_{i_1 i_2 … i_p}$.

Definition. A tensor is antisymmetric with respect to its first two subscripts if $T_{i_2 i_1 … i_p} = -T_{i_1 i_2 … i_p}$

Definition. A tensor is totally antisymmetric if it is antisymmetric in all pairs of indices.

Proposition. Any tensor can always be written as the sum of a symmetric tensor $S_{i_1 i_2 … i_p}$ and an antisymmetric tensor $A_{i_1 i_2 … i_p}$, i.e.

\[T_{i_1 i_2 ... i_p} = {1 \over 2} (T_{i_1 i_2 ... i_p} + T_{i_2 i_1 ... i_p}) + {1 \over 2} (T_{i_1 i_2 ... i_p} - T_{i_2 i_1 ... i_p}) = S_{i_1 i_2 ... i_p} + A_{i_1 i_2 ... i_p}\]

Invariant Tensors

There are two important invariant tensors in $\mathbb{R}^n$.

Definition. A tensor is invariant under a given rotation $R$ if

\[T_{i_1 ... i_p}' = R_{i_1 j_1} ... R_{i_p j_p} T_{j_1 ... j_p} = T_{i_1 ... i_p}\]

Definition. A tensor is isotropic if it is invariant under all rotations $R$.

Obviously, all tensors of rank $0$ are isotropic.

Proposition. The Kronecker delta $\delta_{ij}$ is a rank $2$ isotropic tensor.

Proof.

\[\delta_{ij}' = R_{ik} R_{jl} \delta_{kl} = R_{ik} R_{jk} = \delta_{ij}\]

Proposition. The Levi-Civita Symbol $\varepsilon_{i_i i_2 … i_p}$ is a rank $p$ totally anti-symmetric tensor. It is invariant when $R \in SO(p)$ but not $R \in O(p)$.

Proof.

For a $p \times p$ matrix, we have

\[(\det A) \varepsilon_{i_1 i_2 ... i_p} = A_{i_1 j_1} A_{i_2 j_2} ... A_{i_p j_p} \varepsilon_{j_1 j_2 ... j_n}\]

It can be verified this with the use of Laplace expansion,

  • when any two of the indices take the same value, the R.H.S. is the determinant of a matrix with a repeated row, so both side is $0$;

  • when the indices are of even permutation, the R.H.S. is the determinant of the matrix in Laplace expansion form;

  • when the indices are of odd permutation, the R.H.S. is the determinant of the same matrix will two rows interchanged, and both side is $- \det A$.

Hence,

\[\varepsilon_{i_1 i_2 ... i_p}' = R_{i_1 j_1} R_{i_2 j_2} ... R_{i_n j_n} \varepsilon_{j_1 j_2 ... j_n} = \det R \, \varepsilon_{i_1 i_2 ... i_p}\]

and whether $\varepsilon_{i_1 i_2 … i_p}$ is invariant depends on $\det R = \pm 1$.

Proposition. In $\mathbb{R}^3$, $T_{ij} = \alpha \delta_{ij}$ is the only rank $2$ non-zero isotropic tensor.

Proof.

Consider the rotation by $2\pi/3$ about $(1, 1, 1)$, we have $(x, y, z) \mapsto (y, z, x)$ and therefore $R_{13} = R_{21} = R_{32} = 1$ and all other $R_{ij} = 0$. This requires

  • $T_{11} = T_{22} = T_{33}$;

  • $T_{12} = T_{23} = T_{31}$;

  • $T_{13} = T_{21} = T_{32}$.

Consider another rotation by $\pi/2$ about the $z$-axis, we have $(x, y, z) \mapsto (y, -x, z)$ and therefore $R_{12} = -1$, $R_{21} = R_{33} = 1$ and all other $R_{ij} = 0$. This requires

  • $T_{13} = (-1)(1)T_{23}$;

  • $T_{23} = (1)(1)T_{13}$.

Concluding from the above, we must have $T_{ij} = 0$ except $T_{11} = T_{22} = T_{33}$.

Proposition. In $\mathbb{R}^3$, $T_{ijk} = \beta \varepsilon_{ijk}$ is the only rank $3$ isotropic tensor.

Proof.

Similarily, rotating by $2\pi/3$ about the $(1, 1, 1)$ direction we have

  • $T_{111} = T_{222} = T_{333}$;

  • $T_{112} = T_{223} = T_{331}, …$;

  • $T_{123} = T_{231} = T_{312}, …$.

Then rotating by $\pi/2$ about the $z$-axis, we have

  • $T_{111} = (-1)(-1)(-1) T_{222} = -T_{222}$;

  • $T_{112} = (-1)(-1)(1) T_{221}$;

  • $T_{221} = (1)(1)(-1) T_{112}$;

  • $T_{123} = (-1)(1)(1) T_{213}$.

The quantity that matches all of the above these conditions is a multiple of $\varepsilon_{ijk}$.

By using the same procedure, the only rank $1$ isotropic tensor is the trivial one with all elements $0$.

The following result can be useful but we will not prove here.

Proposition. In $\mathbb{R}^3$, the only rank $4$ isotropic tensor is

\[T_{ijkl} = \alpha \delta_{ij} \delta_{kl} + \beta \delta_{ik} \delta_{jl} + \gamma \delta_{il} \delta_{jk}\]

Dual Tensors

Definition. A rank $p$ pseudotensor is a set of quantities in which its components satisfy

\[v_{i_1 i_2 ... i_p}' = \begin{cases} R_{i_1 j_1} R_{i_2 j_2} ... R_{i_p j_p} v_{j_1 j_2 ... j_p} & \text{for } |R| = 1 \\ - R_{i_1 j_1} R_{i_2 j_2} ... R_{i_p j_p} v_{j_1 j_2 ... j_p} & \text{for } |R| = -1 \\ \end{cases}\]

A rank $1$ pseudotensor is also called pseudovector. An example is the vector cross product in which it changes sign under a left-handed coordinate system.

It is important that pseudotensors are not themselves appropriate for the description of phyiscal phenomena but sometimes they are needed. For example,it can be useful to associate an antisymmetric tensor with the Levi-Civita symbol.

Definition. The dual of rank $2$ antisymmetric tensor $A_{ij}$ is the pseudovector $\mathbf{p}$ given by

\[p_i = {1 \over 2} \varepsilon_{ijk} A_{jk}\]

Proposition. $A_{ij} = \varepsilon_{ijk} p_k$.

Proof.

\[\begin{align*} \varepsilon_{ijk} p_k &= {1 \over 2} \varepsilon_{ijk} \varepsilon_{kpq} A_{pq} \\ &= {1 \over 2} (\delta_{ip} \delta_{jq} - \delta_{iq} \delta_{jp}) A_{pq} \\ &= {1 \over 2} (A_{ij} - A_{ji}) = {1 \over 2} (A_{ij} + A_{ij}) = A_{ij} \\ \end{align*}\]

The above provides a way to further decompose a rank $2$ tensor in $\mathbb{R}^3$.

Proposition. In $\mathbb{R}^3$, any rank $2$ tensor $T_{ij}$ can be decomposed as

\[T_{ij} = P_{ij} + \varepsilon_{ijk} B_k + {1 \over 3} \delta_{ij} Q\]

where $P_{ii} = 0$.

Proof.

From the above, we can decompose a rank $2$ tensor into the symmetric and anti-symmetric piece.

\[T_{ij} = S_{ij} + A_{ij}\]

Furthermore,

\[S_{ij} = P_{ij} + {Q \over 3} \delta_{ij}\]

where $Q = S_{ii}$ and $P_{ij}$ is traceless, and

\[A_{ij} = \varepsilon_{ijk} B_k\]

where $B_k$ is a pseudovector. Hence,

\[T_{ij} = P_{ij} + \varepsilon_{ijk} B_k + {1 \over 3} \delta_{ij} Q\]

References