About the definition of a vector in "Mechanics I" by Akira Harashima. Do I understand this definition?

Question

I am reading "Mechanics I" (in Japanese) by Akira Harashima.

A quantity is given by three real numbers $(A_x, A_y, A_z)$ in an orthogonal coordinate system $S$, and for another orthogonal coordinate system $S'$ (assuming the origin of $S$ and the orign of $S'$ are the same), it is given by three real numbers $(A_{x'}, A_{y'}, A_{z'})$.
If $(A_x, A_y, A_z)$ and $(A_{x'}, A_{y'}, A_{z'})$ satisfy the following relations, then $(A_x, A_y, A_z)$ and $(A_{x'}, A_{y'}, A_{z'})$ represent a single vector $\mathbf{A}$.
$$ A_{x'}=l_1A_x+m_1A_y+n_1A_z\\ A_{y'}=l_2A_x+m_2A_y+n_2A_z\\ A_{z'}=l_3A_x+m_3A_y+n_3A_z,$$ where $$ l_1^2+m_1^2+n_1^2=1\\ l_2^2+m_2^2+n_2^2=1\\ l_3^2+m_3^2+n_3^2=1\\ l_2l_3+m_2m_3+n_2n_3=0\\ l_3l_1+m_3m_1+n_3n_1=0\\ l_1l_2+m_1m_2+n_1n_2=0. $$

I don't understand what the author wants to say well.

Let $\vec{r}$ be a position vector.
Suppose $\vec{r}$ is represented by $(x,y,z)$ in an orthogonal coordinate system $S$.
We consider the quantity $(x,y,-z)$.
Suppose $r$ is represented by $(x',y',z')$ in an orthogonal coordinate system $S'$.
We can write $$ x'=l_1x+m_1y+n_1z\\ y'=l_2x+m_2y+n_2z\\ z'=l_3x+m_3y+n_3z,$$ where $$ l_1^2+m_1^2+n_1^2=1\\ l_2^2+m_2^2+n_2^2=1\\ l_3^2+m_3^2+n_3^2=1\\ l_2l_3+m_2m_3+n_2n_3=0\\ l_3l_1+m_3m_1+n_3n_1=0\\ l_1l_2+m_1m_2+n_1n_2=0. $$

But in general the following relations don't hold.
$$ x'=l_1x+m_1y+n_1(-z)\\ y'=l_2x+m_2y+n_2(-z)\\ -z'=l_3x+m_3y+n_3(-z),$$

So, the quantity $(x,y,-z)$ is not a vector.

Am I right?

Answer 1

This is a truly wild definition, but I suppose I never understood the way physicists think about things...

An answer your question: no. Also it's worth noting that this passage doesn't define "a vector", it merely defines what it means for two lists of real numbers to represent the same vector. So the question cannot be whether $(x,y,-z)$ is a vector, but only whether it represents the same vector as $(x,y,z)$.

In order to determine whether $(x,y,-z)$ is the same vector as $(x,y,z)$, we need to determine whether an appropriate $S'$ exists that transforms $(x,y,z)$ into $(x,y,-z)$.

Indeed, such an $S'$ does exist. Intuitively, it is the "left-handed" coordinate system, if we assume that $S$ is the "right-handed" coordinate system. Formally, we make the following 9 substitutions for the change-of-variables parameters:

$$\begin{bmatrix} l_1&m_1&n_1 \\ l_2&m_2&n_2 \\ l_3&m_3&n_3 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&-1 \end{bmatrix}.$$

Check that these substitutions satisfy the 6 constraints on the parameters (i.e. the 6 equations after "where" in your quote). The notation is a bit awkward, but if you're doing it right the checks will be very easy.

Thus you can see that, using the book's definition, $(x,y,z)$ in the system $S$ and $(x,y,-z)$ in the system $S'$ represent the same vector.

Answer 2

This is an old-style approach typically used in physics. I shall explain its purpose.

The mathematical part is very easy. We consider a three-dimensional real vector space $V$ with an inner product $\langle -, - \rangle$. In this abstract setting we do not have coordinates for the elements of $V$. However, we can introduce such coordinates by choosing an orthonormal basis $\mathcal B = (b_1, b_2, b_3)$ for $V$ (this means that $\langle b_i, b_j \rangle = \begin{cases} 1 & i = j \\ 0 & i \ne j \end{cases}$). Then each $v \in V$ has a unique representation $$v = v_1b_1 + v_2b_2 + v_3b_3$$ with $v_i \in \mathbb R$. These numbers are the coordinates of $v$ with respect to $\mathcal B$. The orthonormal basis $\mathcal B$ will also be called a coordinate system for $V$.

Choosing coordinate system means nothing else than choosing an orthogonal linear map $\phi : \mathbb R^3 \to V$ (this means that $\langle \phi(a), \phi(b) \rangle = \langle a, b \rangle$ for all $a, b \in \mathbb R^3$; note that the $\phi(e_i)$ form an orthonormal basis of $V$, where the $e_i$ are the standard basis vectors of $\mathbb R^3$).

If we choose another orthonormal basis $\mathcal B' = (b'_1, b'_2, b'_3)$ for $V$, we get another coordinate representation $$v = v'_1b'_1 + v'_2b'_2 + v'_3b'_3 .$$ What is the relation between the triples $v_{\mathcal B} = (v_1, v_2, v_3)$ and $v_{\mathcal B'} = (v'_1, v'_2, v'_3)$? Consider the corresponding orthogonal maps $\phi, \phi' : \mathbb R^3 \to V$. Then $(v'_1, v'_2, v'_3) = (\phi' \circ \phi^{-1})(v_1, v_2, v_3)$. The map $\phi' \circ \phi^{-1}$ is an orthogonal automorphism of $\mathbb R^3$ which can be represented by an orthonormal matrix $$ A_{\mathcal B, \mathcal B'} =\begin{pmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end{pmatrix}$$ Orthonormality means that the six equations in your question are satisfied. We then get $$\begin{pmatrix}v'_1 \\ v'_2 \\ v'_3\end{pmatrix} = A_{\mathcal B, \mathcal B'} \cdot \begin{pmatrix}v_1 \\ v_2 \\ v_3\end{pmatrix} . \tag{1}$$ This gives the three equations in your question. It is the coordinate transformation rule for vectors $v \in V$.

We now come to physics which is the more interesting part.

Physical entities like velocity, electric field, etc. live in our $3$-dimensional space, and they are observed and quantified via certain measurements. Since space does not have a distinguished coordinate system, we have to choose a coordinate system $\mathcal B$ before we make the appropriate measurements to quantify a physical entity $X$. With respect to the chosen coordinate system $\mathcal B$ our measurements produce an $n$-tuple $(x_1,\ldots,x_n)$ of real numbers describing $X$. For example, quantifying a velocity produces triples. But of course there are entities which need more than three measurement components.

Harashima seems to consider only physical entities $X$ which are quantified by triples. The measurement process associates to each coordinate system $\mathcal B$ a triple $X_{\mathcal B} = (x_1, x_2,x_3)$. But is the resulting collection $\{ X_{\mathcal B} \}$ indexed by all orthonormal bases $\mathcal B$ a vector in the above sense, i.e. does it satisfy the coordinate transformation rule $$X_{\mathcal B'} = A_{\mathcal B, \mathcal B'} \cdot X_{\mathcal B} ?$$

This is not a priori clear. Some physical entities do and are these are called vectors in physics. We could also call them vectorial entities.

Other physical entities do not satisfy the coordinate transformation rule, thus they are no vectors in the sense of physics. Well-known examples are pseudovectors (aka axial vectors) like angular velocity and angular momentum. See here.

Also have a look at Conceptual difference between Covariant and Contravariant tensors.

Answer 3

In his answer Paul Frost gave an explanation what Harashima wanted to say, but I do not think that Harashima formulated it correctly.

He considers a physical quantity and two orthonormal coordinate systems $S$ and $S′$. [Harashima writes orthogonal, but it is clear from the context that he means orthonormal.] The quantity has coordinate representations $A_S = (A_x,A_y,A_z)$ with respect to $S$ and $A_{S'} = (A'_x,A'_y,A'_z)$ with respect to $S'$. He then says that these triples represent a single vector $\mathbf A$ if there exists an orthonormal matrix $M_{S,S'}$ such that $A_{S'} = M_{S,S'} \cdot A_S$. Here $A_S, A_{S'}$ are understood as column vectors.

There are two serious problems here.

1. He does not require that $M_{S,S'}$ is the matrix which represents the coordinate transformation from $S$ to $S'$ (which was denoted by $\phi' \circ \phi^{-1}$ by Paul Frost). But if we allow an arbitrary orthonormal matrix $M_{S,S'}$ such that $A_{S'} = M_{S,S'} \cdot A_S$ (which may be distinct from that representing the coordinate transformation), we may only get a coincidental relationship between $A_S, A_{S'}$.

2. As Ted wrote in a comment, we must consider all orthonormal bases and all coordinate representations of the quantity under consideration. If any two are related via the matrix of the corresponding coordinate transformation, then they represent a single vector $\mathbf A$. Considering only a single pair $A_S$ and $A_{S'}$ as Harashima is not sufficient. The relation may again be coincidental in that case.