MTH3008 Lecture 2

A.J. Rushworth

Labubuuuuuuuuu

Using the content of the previous lecture, namely Suffix Notation and the Kronecker Delta, we can explore the dot product and cross product. First, the dot product…

Vector vs Suffix notation

It should always be specified what is being worked in, as the difference in notation can lead to completely different answers. For example, in vector notation we specify all values at once, so ${\delta }_{ij}=1$, but in suffix notation we evaluate sums, so ${\delta }_{ij}=3$.

This is because it’ll sum over the repeated index, for example in vector notation, ${\sum }_{j=1}^3{\delta }_{jj}={\delta }_{11}+{\delta }_{22}+{\delta }_{33}=3$ (summing $j=\mathrm{1..3}$ by our convention).

We hence normally use vector notation as an in-between to simplify suffix notation directly, such as to prove properties or expressions from “first principles”.

Using Vector Notation to Simplify Suffix Notation

As mentioned, we can use vector notation to simplify expressions in suffix notation, directly. For example, to show that ${\delta }_{ij}{\delta }_{jk}\to {\sum }_{j=1}^3{\delta }_{ij}{\delta }_{jk}={\delta }_{i1}{\delta }_{1k}+{\delta }_{i2}{\delta }_{2k}+{\delta }_{i3}{\delta }_{3k}\to \cdots \to {\delta }_{ik}$; that is, the repeated index is absorbed.

In-Lecture Example (simplifying Kronecker delta expressions)

To simplify ${\delta }_{\ell \ell }{\delta }_{mn}{\delta }_{pp}{\delta }_{nq}$, we first rearrange due to its commutativity to get ${\delta }_{\ell \ell }{\delta }_{mn}{\delta }_{nq}{\delta }_{pp}={\delta }_{\ell \ell }{\delta }_{mq}{\delta }_{pp}$ and then directly evaluate over three dimensions to get $9{\delta }_{mq}$

Weekly Problem 1.6 (simplifying to vector notation)

To simplify ${\delta }_{ij}{a}_j{b}_{\ell }{c}_k{\delta }_{i\ell }$, we rearrange using commutativity to get ${\delta }_{ij}{a}_j{\delta }_{i\ell }{b}_{\ell }{c}_k=a_i{b}_i{c}_k=(\mathbf{a}\cdot \mathbf{b})\mathbf{c}$.

This hence allows us to define the dot product as $\mathbf{a}\cdot \mathbf{b}=a_i{b}_i=a_i({\delta }_{ij}{b}_j)={\delta }_{ij}{a}_i{b}_j$, and vice versa. Next, the cross product…

The Alternating Tensor

The Alternating Tensor is defined as ${ϵ}_{ijk}=\{\begin{array}{ll}0& \text{if any of }i,j,k\text{ are equal}\\ +1& \text{if }(i,j,k)=(1,2,3),(2,3,1),\text{or }(3,1,2)\\ -1& \text{if }(i,j,k)=(1,3,2),(2,1,3),\text{or }(3,2,1)\end{array}$, i.e., 1 if $(i,j,k)$ is an even permutation of $(1,2,3)$, -1 if it’s an odd permutation, or 0 if any - note: this is given in vector notation.

This can also hence be visualised as three layers of matrices, $\left[{ϵ}_{ij1}=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& -1& 0\end{array}\right],{ϵ}_{ij2}=\left[\begin{array}{ccc}0& 0& -1\\ 0& 0& 0\\ 1& 0& 0\end{array}\right],{ϵ}_{ij3}=\left[\begin{array}{ccc}0& 1& 0\\ -1& 0& 0\\ 0& 0& 0\end{array}\right]\right]$.

Directly from the definition, we can observe the following properties:

• ${ϵ}_{ijk}$ is unchanged if indices are reordered by a cyclic permutation, i.e., ${ϵ}_{ijk}={ϵ}_{jki}={ϵ}_{kij}$.
• ${ϵ}_{ijk}$ is changed if any two of the suffices are interchanged, i.e., ${ϵ}_{ijk}=-{ϵ}_{jik}$; that is, the alternating tensor is anti-symmetric.

We can then use these properties to simplify calculations, such as ${ϵ}_{ijk}{ϵ}_{ijk}$ - which would normally be 27 terms. However, as most terms are zero then we can just focus on what is nonzero, i.e., the six cases where the tensor is either $-1$ or $1$. Hence, the answer is just $6\times 1=6$.

Relations to Other Operations

We can use this definition of the Alternating Tensor to define the cross product of two vectors, the determinant of $3\times 3$ matrices, and the scalar triple product.

We get these by substituting in the alternating tensor to allow for the $+/-$ patterns in each operation…

1. $(\mathbf{a}\times \mathbf{b}{)}_i={ϵ}_{ijk}{a}_j{b}_k$,
2. ${ϵ}_{pqr}|M|={ϵ}_{ijk}{M}_{pi}{M}_{qj}{M}_{rk}$,
3. $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})={ϵ}_{ijk}{a}_i{b}_j{c}_k$.

These are then very useful for speeding up proofs, for instance showing that $\mathbf{a}\times \mathbf{b}=-\mathbf{b}\times \mathbf{a}$, proven in this week’s problems (Weekly Problem 1.8).

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Pre-Lecture Notes from University Notes

• First, recapped the last lecture:
  • Assume all vectors are three dimensional.
  • Suffix Notation (definition, nomenclature, conversion steps, and examples).
  • Kronecker Delta (definition and key property)
• Using the Kronecker delta in suffix notation, we can say that $\mathbf{a}\cdot \mathbf{b}=a_i{b}_i=a_i({\delta }_{ij}{b}_j)={\delta }_{ij}{a}_i{b}_j$. Naturally, this can be checked backwards using vector notation, too.
  • Note that we have to be careful with the context for vector and suffix notation: ${\delta }_{ij}$ is given in vector notation, where we specify all the values at once, but ${\delta }_{ii}$ in suffix notation is a sum.
  • This is because it’ll sum over the repeated index, for example in vector notation, ${\sum }_{j=1}^3{\delta }_{jj}={\delta }_{11}+{\delta }_{22}+{\delta }_{33}=3$ (summing $j=\mathrm{1..3}$ by our convention).
  • We can also hence simplify expressions with the function by using the vector conversion, like ${\delta }_{ij}{\delta }_{jk}\to {\sum }_{j=1}^3{\delta }_{ij}{\delta }_{jk}={\delta }_{i1}{\delta }_{1k}+{\delta }_{i2}{\delta }_{2k}+{\delta }_{i3}{\delta }_{3k}\to \cdots \to {\delta }_{ik}$; that is, the repeated index is absorbed.
  • Other expressions could be simplified, like ${\delta }_{\ell \ell }{\delta }_{mn}{\delta }_{pp}{\delta }_{nq}$ or ${\delta }_{ij}{a}_j{b}_{\ell }{c}_k{\delta }_{i\ell }$.
• Hence, the Kronecker delta (substitution tensor) can be used to define the dot product. Similarly, the alternating tensor can define the cross product.
  • Alternative tensor ${ϵ}_{ijk}$ is defined by ${ϵ}_{ijk}=\{\begin{array}{ll}0& \text{if any of }i,j,k\text{ are equal}\\ +1& \text{if }(i,j,k)=(1,2,3),(2,3,1),\text{or }(3,1,2)\\ -1& \text{if }(i,j,k)=(1,3,2),(2,1,3),\text{or }(3,2,1)\end{array}$, i.e., 1 if $(i,j,k)$ is an even permutation of $(1,2,3)$, -1 if it’s an odd permutation, or 0 if any index is repeated.
  • This can be visualised as three layers of matrices, $\left[{ϵ}_{ij1}=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& -1& 0\end{array}\right],{ϵ}_{ij2}=\left[\begin{array}{ccc}0& 0& -1\\ 0& 0& 0\\ 1& 0& 0\end{array}\right],{ϵ}_{ij3}=\left[\begin{array}{ccc}0& 1& 0\\ -1& 0& 0\\ 0& 0& 0\end{array}\right]\right]$.
  • As for the Kronecker delta, we must be careful to distinguish suffix and vector notation.
  • Due to all of this, in vector notation, it has the properties that:
    • ${ϵ}_{ijk}$ is unchanged if indices are reordered by a cyclic permutation, i.e., ${ϵ}_{ijk}={ϵ}_{jki}={ϵ}_{kij}$.
    • ${ϵ}_{ijk}$ is changed if any two of the suffices are interchanged, i.e., ${ϵ}_{ijk}=-{ϵ}_{jik}$; that is, the alternating tensor is anti-symmetric.
  • Similar to before, we can evaluate expressions involving this tensor by converting to vector notation: ${ϵ}_{ijk}{ϵ}_{ijk}={\sum }_{i=1}^3{\sum }_{j=1}^3{\sum }_{k=1}^3{ϵ}_{ijk}{ϵ}_{ijk}={\sum }_{i=1}^3{\sum }_{j=1}^3{\sum }_{k=1}^3{ϵ}_{ijk}^2$, which gives a total of $3\times 3\times 3=27$ terms, where only six are actually nonzero. After trivially calculating this, you hence get 6.
• Alternating tensors are naturally related to:
  • Cross product of two vectors: $\mathbf{a}\times \mathbf{b}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|=\left|\begin{array}{cc}{a}_2& a_3\\ b_2& b_3\end{array}\right|\mathbf{i}-\left|\begin{array}{cc}{a}_1& a_3\\ b_1& b_3\end{array}\right|\mathbf{j}+\left|\begin{array}{cc}{a}_1& a_2\\ b_1& b_2\end{array}\right|\mathbf{k}$.
    • Similar to the dot product, we can write $(\mathbf{a}\times \mathbf{b}{)}_i={ϵ}_{ijk}{a}_j{b}_k$ - the former meaning the $i$th component of the vector $\mathbf{a}\times \mathbf{b}$, and the latter where $j$ and $k$ are dummy indices indicating sums, and $i$ is a free index indicating the coordinate considered.
    • As always, this can be found by converting to vector notation, and can hence use the notation to simplify/evaluate expressions like $\mathbf{a}\times \mathbf{b}=-\mathbf{b}\times \mathbf{a}$.
  • Determinant of $3\times 3$ matrices.
    • Using the alternating tensor, we can calculate matrix determinants, again provable using vector notation: $|M|={ϵ}_{ijk}{M}_{1i}{M}_{2j}{M}_{3k}$ using the rows, or $|M|={ϵ}_{ijk}{M}_{i1}{M}_{j2}{M}_{k3}$ using the columns.
    • This is importantly related to the formula ${ϵ}_{pqr}|M|={ϵ}_{ijk}{M}_{pi}{M}_{qj}{M}_{rk}$.
  • The scalar triple product: $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$.
    • Re-writing the product in suffix notation, $(\mathbf{b}\times \mathbf{c}{)}_i={ϵ}_{ijk}{b}_j{c}_k$, we can simplify it to instead be ${ϵ}_{ijk}{a}_i{b}_j{c}_k$, which could also be used to prove properties of the scalar triple product (using ${ϵ}_{ijk}={ϵ}_{kij}$).
• All then practised in this week’s problems.