Part 1 · Vectors

1.2 Scalar product and components

Source: Favata’s notes, appendix A.1, p. 10–11

Objectives

Read $u \cdot v$ as a length times a projected length, and get lengths and angles from it.
Work in an orthonormal basis, where a component is one scalar product.
Read and write index notation fluently: free and dummy indices, the Kronecker delta.
Recipe: project a vector on a line and on the plane normal to it.
Picture: the projection of $v$ on the line of $u$ ; components that change when the basis turns, a length that does not.
Vocabulary: scalar product, norm, versor, orthogonal, orthonormal basis, Kronecker delta, Einstein convention.

The translation space $V$ comes with a scalar product $u \cdot v$ : a number, symmetric, linear in each argument, with $u \cdot u > 0$ for $u \neq = 0$ . Everything metric in this book, from lengths to angles to the strain of a body, is built from it.

Norm, angle, orthogonality

Definition 1.2.1Norm, angle, versor, orthogonality

The scalar product induces:

the norm $∣ u ∣ := u \cdot u$ , the length of $u$ ;
the angle $ϑ \in [0, π]$ between nonzero vectors, through $u \cdot v = ∣ u ∣∣ v ∣ cos ϑ .$

A versor is a vector of norm 1. Two nonzero vectors are orthogonal if $u \cdot v = 0$ .

The picture behind the formula: drop $v$ perpendicularly onto the line of $u$ . The foot is at signed distance $∣ v ∣ cos ϑ$ from the origin, so $u \cdot v$ is $∣ u ∣$ times that signed length. Positive for an acute angle, negative for an obtuse one, zero exactly when the foot is the origin.

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Figure 1.2.1 Drag

u

and

v

. The thick green segment is the projection of

v

on the line of

u

. Make them orthogonal, then parallel: both positions snap and say so. Drag either onto the dashed unit circle to make it a versor.

Orthonormal bases and components

In dimension $n$ one can always find $n$ mutually orthogonal versors. Fix one such set as the basis ${e_{i}}$ :

Definition 1.2.2Orthonormal basis, Kronecker delta, components

$e_{i} \cdot e_{j} = δ_{ij}, δ_{ij} = {10 i = j i \neq = j (Kronecker delta) .$ The components of $u$ in this basis are $u_{i} = (u)_{i} := u \cdot e_{i}, u = u_{1} e_{1} + \dots + u_{n} e_{n} = i = 1 \sum n u_{i} e_{i} = u_{i} e_{i} .$

Why a component is a scalar productbeyond the notes

${e_{i}}$ is a basis, so $u = α_{j} e_{j}$ for some numbers $α_{j}$ (Definition 1.1.7). Dot with $e_{i}$ : $u \cdot e_{i} = α_{j} e_{j} \cdot e_{i} = α_{j} δ_{ji} = α_{i}$ .

With components, the scalar product is a sum of products. By bilinearity,

$u \cdot v = (u_{i} e_{i}) \cdot (v_{j} e_{j}) = u_{i} v_{j} δ_{ij} = u_{i} v_{i}, ∣ u ∣^{2} = u_{i} u_{i} .$

beyond the notes

In an orthonormal basis, and only there, $u \cdot v = u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3}$ . In a skewed basis the cross terms $e_{1} \cdot e_{2}$ do not vanish.

Components change when the basis turns; the scalar product, and so every length and angle, does not, because it was defined without a basis.

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Figure 1.2.2 Drag the tip of

q_{1}

to turn the orthonormal basis

{q_{1}, q_{2}}

. The components

u_{i}^{'} = u \cdot q_{i}

change;

u_{1}^{'2} + u_{2}^{'2}

stays

∣ u ∣^{2}

, the dashed circle.

Index notation

Definition 1.2.3Einstein convention

An index repeated twice in a term is summed over all its values: $u_{i} e_{i}$ means $\sum_{i = 1}^{3} u_{i} e_{i}$ , and $u_{i} v_{i}$ means $u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3}$ .

Recipe 1.2.4Reading and writing index expressionsbeyond the notes

A free index appears once in each term. An equation with a free index $i$ is three equations, one per value of $i$ . Both sides must carry the same free indices.
A dummy index appears exactly twice in a term and is summed. Its name does not matter: $u_{i} v_{i} = u_{k} v_{k}$ .
No index appears three times in a term. If one would, rename a dummy first: $(u_{i} v_{i}) (w_{i} z_{i})$ is wrong, $(u_{i} v_{i}) (w_{k} z_{k})$ is right.
Substitution by $δ$ : $δ_{ij} u_{j} = u_{i}$ . The sum over $j$ keeps only the term $j = i$ .

Handy consequences: $δ_{ii} = 1 + 1 + 1 = 3$ (in 3D), and $δ_{ij} δ_{jk} = δ_{ik}$ .

Example 1.2.5beyond the notes

$δ_{ij} u_{i} v_{j}$ : substitute $j \to i$ through $δ_{ij}$ , giving $u_{i} v_{i} = u \cdot v$ . And $(δ_{ij} - e_{i} e_{j}) v_{j} = v_{i} - e_{i} (e_{j} v_{j})$ , the components of $v - (e \cdot v) e$ : the projection below, in indices.

Projections on a line and a plane

For a versor $e$ , every vector splits into a piece along $e$ and a piece orthogonal to it:

Recipe 1.2.6Project v on the line of a versor e, and on the plane normal to itbeyond the notes

Along the line: $(e \cdot v) e$ .
On the normal plane: $v - (e \cdot v) e$ .
The two add up to $v$ and are orthogonal: $(v - (e \cdot v) e) \cdot e = e \cdot v - e \cdot v = 0$ .

If $e$ is not a versor, use $e /∣ e ∣$ : the projection of $v$ on the line of $u$ is $\frac{u \cdot v}{u \cdot u} u$ . In §2.1 Tensors as linear maps these two maps become the tensors $e \otimes e$ and $I - e \otimes e$ .

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Figure 1.2.3 Drag the versor

e

around the unit circle and

v

anywhere. The orange arrow is the projection on the line, the grey one the projection on the normal line. Put

v

on either line: one of the two pieces vanishes.

Exercises

Exercise 1.2.1beyond the notes

Expand $δ_{ij} δ_{ij}$ and $δ_{ij} δ_{jk} δ_{ki}$ .

Solution

$δ_{ij} δ_{ij} = δ_{ii} = 3$ . $δ_{ij} δ_{jk} δ_{ki} = δ_{ik} δ_{ki} = δ_{ii} = 3$ .

Exercise 1.2.2beyond the notes

Find the angle between $(1, 1, 0)$ and $(1, 0, 1)$ , and the projection of the second on the line of the first.

Solution

$u \cdot v = 1$ , $∣ u ∣ = ∣ v ∣ = 2$ , so $cos ϑ = 1/2$ and $ϑ = 6 0^{\circ}$ . Projection: $\frac{u \cdot v}{u \cdot u} u = \frac{1}{2} (1, 1, 0)$ .

Exercise 1.2.3beyond the notes

Prove the parallelogram law $∣ u + v ∣^{2} + ∣ u - v ∣^{2} = 2∣ u ∣^{2} + 2∣ v ∣^{2}$ , and say what it means for the diagonals of a parallelogram.

Solution

$∣ u \pm v ∣^{2} = ∣ u ∣^{2} \pm 2 u \cdot v + ∣ v ∣^{2}$ ; add the two. The squares of the two diagonals add up to the squares of the four sides.