Theoretical Mechanics IPSP

The position of a particle, the direction of its motion and the angle of attack of forces are constantly changing during the motion of a particle. In Chapter 3 we explore how they are related. The calculations are feasible because the involved vector spaces also have an inner product.

Remark 2.16 The idea underlying these properties is that $\sqrt{\langle \mathbf v \mid \mathbf v \rangle}$ can be interpreted as the length of the vector $\mathbf v$.

Remark 2.17 Conjugate symmetry and linearity for the first argument imply the following relations for the second argument \begin{align*} \langle \mathbf v \mid c \odot \mathbf w \rangle &= \overline{\langle c \odot \mathbf w \mid \mathbf v \rangle} = \bar{c} \;\overline{\langle \mathbf w \mid \mathbf v \rangle} = \bar{c} \; \langle \mathbf v \mid \mathbf w \rangle \end{align*} \begin{align*} \langle \mathbf u \mid \mathbf v \oplus \mathbf w \rangle &= \overline{\langle \mathbf v \oplus \mathbf w \mid \mathbf u \rangle} = \overline{\langle \mathbf v \mid \mathbf u \rangle} + \overline{\langle \mathbf w \mid \mathbf u \rangle} = \langle \mathbf u \mid \mathbf v \rangle + \langle \mathbf u \mid \mathbf w \rangle \end{align*}

Remark 2.18 Certain properties that hold for addition and scalar multiplication do not hold for the inner product.
a) There is no inverse: The information about the direction of vectors is lost upon taking the inner product. For instance, when $\langle\mathbf{u} \mid \mathbf{v} \rangle = 0$ and $\langle \mathbf{u} \mid \mathbf{w} \rangle = 0$ then one still can not tell the result of $\langle \mathbf{v} \mid \mathbf{w} \rangle$.
b) Associativity does not hold: $\langle \mathbf{u} \mid \mathbf{v} \rangle \: \mathbf{w} \neq \mathbf{u} \: \langle \mathbf{v} \mid \mathbf{w} \rangle$.

Proof. The cosine theorem for triangles with sides of length $a$, $b$ and $c$ and angle $\theta$ opposite to $c$ states that \begin{align*} c^2 = a^2 + b^2 - 2\,a\,b\,\cos\theta \end{align*} Let now $a$, $b$, and $c$ be the length of the vectors $\mathbf a$, $\mathbf b$ and $\mathbf c = \mathbf b - \mathbf a$, as shown in Figure 2.13. Then we have \begin{align*} a^2 + b^2 - 2\,a\,b\,\cos\theta &= c^2 = \mathbf c \cdot \mathbf c = (\mathbf b - \mathbf a) \cdot (\mathbf b - \mathbf a) = (\mathbf a - \mathbf b) \cdot (\mathbf a - \mathbf b) \\ &= \mathbf a \cdot \mathbf a - 2\, \mathbf a \cdot \mathbf b + \mathbf b \cdot \mathbf b = a^2 + b^2 - 2\, \mathbf a \cdot \mathbf b \\[2mm] \Rightarrow\quad \mathbf a \cdot \mathbf b &= |\mathbf a| \, |\mathbf b| \, \cos\theta \\ \end{align*}

Remark 2.19 Theorem 2.2 entails that the inner product $\mathbf u \cdot \mathbf v$ vanishes when the vectors are orthogonal, $\theta=\pi/2$. Also in general we say that \begin{align*} \text{$\mathbf v$ and $\mathbf w$ are othogonal iff} \qquad \langle \mathbf v \mid \mathbf w \rangle = 0 \, . \end{align*}

Remark 2.20 The expression for the inner product that is provided in Theorem 2.2 does not imply that the inner product is unique. Rather it is a consequence of the cosine theorem that holds iff the geometric interpretation of the vectors applies. This is demonstrated by an example provided in Problem 2.17.

Problem 2.17: The inner product is not unique
Let $\mathbf v_1$ and $v_2$ be two non-orthogonal vectors in a two-dimensional vector space with an inner product $\langle \_ \mid \_ \rangle$, and let $\lambda_1$ and $\lambda_2$ two positive real numbers. Then the following relation defines another inner product $( \_ \mid \_ )$: \begin{align*}\label{eq:nonUniqueInnerProd} ( \mathbf a \mid \mathbf b ) = \lambda_1 \langle \mathbf a \mid \mathbf e_1 \rangle \: \langle \mathbf e_1 \mid \mathbf b \rangle + \lambda_2 \langle \mathbf a \mid \mathbf e_2 \rangle \: \langle \mathbf e_2 \mid \mathbf b \rangle \end{align*}

1. Verify that the properties a) and b) of an inner product $\langle \_ \mid \_ \rangle$ as given in Definition 2.12 are also be obeyed by $( \_ \mid \_ )$.
2. Verify that $( \mathbf a \mid \mathbf a ) \geq 0$ iff $\lambda_1$ and $\lambda_2$ two positive real numbers.
3. Verify that $( \mathbf a \mid \mathbf a ) = 0$ implies $\mathbf a = \mathbf 0$ iff the vector space is two-dimensional.

Problem 2.18: Inner products for polynomials
Let $\mathbf p = \left[\sum_{i=0}^D p_i \, x^i\right]$ and $\mathbf q = \left[\sum_{i=0}^D q_i \, x^i\right]$ be elements of the vector space of $N$-dimensional polynomials. Verify that the following rules define inner products on this space.

1. $\displaystyle \langle \mathbf p \mid \mathbf q \rangle = \sum_{i=0}^N \bar p_i \, q_i$
2. $\displaystyle \langle \mathbf p \mid \mathbf q \rangle_{[a,b]} = \displaystyle \int_a^b \mathrm{d} x \overline{\left[\sum_{i=0}^D p_i \, x^i\right]} \; \left[\sum_{i=0}^D q_i \, x^i\right]$, for $a < b \in \mathbb{R}$
3. Show that $\mathbf p = [1]$ and $\mathbf q = [x]$ are orthogonal with respect to the inner product defined in 1). Under which condition are they also orthogonal for the inner product defined in 2)?