MTH3008 Lecture 10

Paula Lins

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This lecture continues Chapter 4, building directly on last time’s introduction of the rotation-based transformation law. We now consolidate what it means for a quantity to be a tensor, establish the quotient rule as a shortcut for proving tensor character, and introduce the symmetric/antisymmetric decomposition - a canonical way to split any second-rank tensor.

Tensors: Rank and Transformation Law

A tensor in an orthogonal coordinate system is defined by how its components transform under a coordinate rotation $L_{ij}$. The general rule is one factor of $L$ per free index:

$$T_{ij}^{\prime }=L_{im}{L}_{jn}{T}_{mn},T_{ijk}^{\prime }=L_{im}{L}_{jn}{L}_{kp}{T}_{mnp},\text{etc.}$$

On the right-hand side, the original component indices ($m,n,p,\dots$) are dummy (summed over), while the primed indices ($i,j,k,\dots$) are free - so both sides carry the same free indices and the equation balances.

Rank of a Tensor

The rank (or order) of a tensor is the number of free indices. A rank-$r$ tensor in 3D has $3^r$ components and transforms with exactly $r$ factors of the rotation matrix.

Example: Kronecker Delta is a Rank-2 Tensor

We need to verify ${\delta }_{ij}^{\prime }=L_{ik}{L}_{jm}{\delta }_{km}$. Using orthogonality of $L$:

$$L_{ik}{L}_{jm}{\delta }_{km}=L_{ik}{L}_{jk}={\delta }_{ij}.$$

Since ${\delta }_{ij}$ is defined identically in every coordinate system, ${\delta }_{ij}^{\prime }={\delta }_{ij}$, so the law holds. ​

Example: Gradient of a Vector is a Rank-2 Tensor

We want to show $\frac{\partial u_i^{\prime }}{\partial x_j^{\prime }}=L_{ik}{L}_{j\ell }\frac{\partial u_k}{\partial x_{\ell }}$.

Step 1 - Product rule:

$$\frac{\partial u_i^{\prime }}{\partial x_j^{\prime }}=\frac{\partial (L_{ik}{u}_k)}{\partial x_j^{\prime }}=L_{ik}\frac{\partial u_k}{\partial x_j^{\prime }}+u_k\underset{=0}{\underbrace{\frac{\partial L_{ik}}{\partial x_j^{\prime }}}}.$$

Step 2 - Why does $\frac{\partial L_{ik}}{\partial x_j^{\prime }}=0$?

Since $L_{ik}=\frac{\partial x_i^{\prime }}{\partial x_k}$,

$$\frac{\partial L_{ik}}{\partial x_j^{\prime }}=\frac{{\partial }^2{x}_i^{\prime }}{\partial x_j^{\prime }\partial x_k}=\frac{\partial }{\partial x_k}\left(\frac{\partial x_i^{\prime }}{\partial x_j^{\prime }}\right)=\frac{\partial }{\partial x_k}({\delta }_{ij})=0.$$

Step 3 - Chain rule:

$$\frac{\partial u_i^{\prime }}{\partial x_j^{\prime }}=L_{ik}\frac{\partial u_k}{\partial x_j^{\prime }}=L_{ik}\frac{\partial u_k}{\partial x_{\ell }}\underset{=L_{j\ell }}{\underbrace{\frac{\partial x_{\ell }}{\partial x_j^{\prime }}}}=L_{ik}{L}_{j\ell }\frac{\partial u_k}{\partial x_{\ell }}.$$ $$\boxed{\frac{\partial u_i^{\prime }}{\partial x_j^{\prime }}=L_{ik}{L}_{j\ell }\frac{\partial u_k}{\partial x_{\ell }}}$$

So $\nabla \mathbf{u}=\partial u_i/\partial x_j$ is a rank-2 tensor. ​

$L_{ij}$ constant - Cartesian coordinates only

The step $\partial L_{ik}/\partial x_j^{\prime }=0$ relies on $L$ being a constant rotation matrix. In curvilinear coordinates the analogous “transformation matrix” varies with position, so this argument completely breaks down - Chapter 5 exists precisely to handle this.

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The Quotient Rule

Quotient Rule (Lemma)

Suppose $T_{ij}$ is a quantity such that, for any vector $\mathbf{b}$, the combination $a_i=T_{ij}{b}_j$ is a vector. Then $T_{ij}$ is a rank-2 tensor.

Proof

Label the assumptions:

• (A1) $b_j=L_{mj}{b}_m^{\prime }$ (since $\mathbf{b}$ is a vector),
• (A2) $a_k=T_{kj}{b}_j$ holds in all coordinate systems,
• (A3) $a_i^{\prime }=L_{ik}{a}_k$ (since $\mathbf{a}$ is a vector).

Two expressions for $a_i^{\prime }$:

From (A3), (A2), and (A1):

$$a_i^{\prime }=L_{ik}{a}_k=L_{ik}{T}_{kj}{b}_j=L_{ik}{T}_{kj}{L}_{mj}{b}_m^{\prime }=L_{ik}{L}_{mj}{T}_{kj}{b}_m^{\prime }.$$

From (A2) in the rotated frame:

$$a_i^{\prime }=T_{im}^{\prime }{b}_m^{\prime }.$$

Subtract and use arbitrariness of $\mathbf{b}$:

$$(T_{im}^{\prime }-L_{ik}{L}_{mj}{T}_{kj})b_m^{\prime }=0.$$

Since $\mathbf{b}$ is arbitrary we may take $b_m^{\prime }\ne 0$, so:

$$\boxed{T_{im}^{\prime }=L_{ik}{L}_{mj}{T}_{kj}}$$

Confirming $T_{ij}$ is a rank-2 tensor. ​

Why "quotient"?

By analogy with the calculus quotient rule: instead of directly transforming $T$, you infer its tensor character from a “product” $T_{ij}{b}_j$. The indispensable ingredient is that $\mathbf{b}$ is arbitrary - this is what forces the bracket to zero, rather than just yielding $b_m^{\prime }=0$.

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Symmetric and Antisymmetric Tensors

Symmetry Definitions

A second-rank tensor $T_{ij}$ is:

• Symmetric if $T_{ij}=T_{ji}$,
• Antisymmetric (skew-symmetric) if $T_{ij}=-T_{ji}$.

For higher-rank tensors, symmetry is defined with respect to a chosen pair of indices, e.g. ${\epsilon }_{ijk}=-{\epsilon }_{jik}$.

The Kronecker delta is symmetric (${\delta }_{ij}={\delta }_{ji}$); the alternating tensor ${\epsilon }_{ijk}$ is antisymmetric in any two of its indices.

Symmetry is a Physical Property

Symmetry Lemma

If $A_{ij}$ is symmetric in one Cartesian frame, it is symmetric in every Cartesian frame.

Proof: Suppose $A_{ij}=A_{ji}$. In a rotated frame:

$$A_{ij}^{\prime }=L_{ik}{L}_{jm}{A}_{km}=L_{ik}{L}_{jm}{A}_{mk}=L_{jm}{L}_{ik}{A}_{mk}=A_{ji}^{\prime }.\boxed{}$$

The identical argument (with a sign flip) proves antisymmetry is also frame-independent.

Symmetric/Antisymmetric Decomposition

Decomposition Theorem

Any rank-2 tensor decomposes uniquely as $T_{ik}=S_{ik}+A_{ik}$, where

$S_{ik}=\frac{1}{2}(T_{ik}+T_{ki})$ - the symmetric part of $T_{ik}$,

$A_{ik}=\frac{1}{2}(T_{ik}-T_{ki})$ - the antisymmetric part of $T_{ik}$.

The decomposition is verified directly: $S_{ik}+A_{ik}=\frac{1}{2}(T_{ik}+T_{ki})+\frac{1}{2}(T_{ik}-T_{ki})=T_{ik}$. Uniqueness follows because if $T_{ik}=S_{ik}+A_{ik}=S_{ik}^{\prime }+A_{ik}^{\prime }$ then $S_{ik}-S_{ik}^{\prime }=A_{ik}^{\prime }-A_{ik}$ is simultaneously symmetric and antisymmetric, hence zero.

Example: ${\epsilon }_{ijk}{T}_{jk}=0\Rightarrow T_{ij}$ Symmetric

Fix $i=1$ and expand - only terms where $(j,k)$ give nonzero ${\epsilon }_{1jk}$ contribute:

$${\epsilon }_{1jk}{T}_{jk}={\epsilon }_{123}{T}_{23}+{\epsilon }_{132}{T}_{32}=T_{23}-T_{32}=0⟹T_{23}=T_{32}.$$

Repeating for $i=2$ and $i=3$ gives $T_{12}=T_{21}$ and $T_{13}=T_{31}$, so $T_{ij}=T_{ji}$. ​

Contraction with ${\epsilon }_{ijk}$ kills the symmetric part

For any $T_{jk}$, the contraction ${\epsilon }_{ijk}{T}_{jk}$ is sensitive only to the antisymmetric part of $T_{jk}$ - the symmetric part contracts to zero by the antisymmetry of $\epsilon$. So ${\epsilon }_{ijk}{T}_{jk}=0$ says exactly that $T_{jk}$ has no antisymmetric part, i.e. it is symmetric.

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Pre-Lecture Notes from University Notes

• Tensor definition: a quantity is a tensor if each free index transforms with one factor of the rotation matrix $L_{ij}$; rank = number of free indices; $3^r$ components in 3D
• Kronecker delta: $L_{ik}{L}_{jm}{\delta }_{km}=L_{ik}{L}_{jk}={\delta }_{ij}$ - same in all frames, so it’s a rank-2 tensor ✓
• Gradient $\partial u_i/\partial x_j$: product rule → $\partial L_{ik}/\partial x_j^{\prime }=\partial ({\delta }_{ij})/\partial x_k=0$ → chain rule → $L_{ik}{L}_{j\ell }\partial u_k/\partial x_{\ell }$ ✓; key: $L$ is constant in Cartesian coordinates only
• Quotient rule: if $T_{ij}{b}_j$ is a vector for all vectors $\mathbf{b}$, then $T_{ij}$ is a tensor; proved by equating two expressions for $a_i^{\prime }$ and invoking arbitrariness of $\mathbf{b}$
• Symmetric/antisymmetric: $T_{ij}=T_{ji}$ / $T_{ij}=-T_{ji}$; symmetry is coordinate-independent (frame change preserves $A_{ij}^{\prime }=A_{ji}^{\prime }$)
• Decomposition: $T_{ik}=\frac{1}{2}(T_{ik}+T_{ki})+\frac{1}{2}(T_{ik}-T_{ki})$; the two parts are called symmetrisation and antisymmetrisation
• ${\epsilon }_{ijk}{T}_{jk}=0$ example: fixing each $i$ and reading off nonzero $\epsilon$ components forces $T_{23}=T_{32}$, $T_{12}=T_{21}$, $T_{13}=T_{31}$ - so $T_{ij}$ is symmetric; geometrically, $\epsilon$ only sees the antisymmetric part
• Next lecture: Chapter 5 - Local Coordinate Transforms: associated tensors, metric tensor $g_{jk}={\mathbf{e}}_j\cdot {\mathbf{e}}_k$, higher-order tensors in generalised coordinates