I'm struggeling to understand when we call a certain tuple a vector or first order Cartesian tensor $\mathbf v$.
Riley, Hobson and Riley (3rd) states that when you apply a rotation, the new components $v_i'$ should be obtained using $$v_i' = L_{ij}v_j,\tag{1}$$ where$L = \begin{pmatrix}\cos \theta & \sin \theta\\-\sin \theta& \cos \theta \end{pmatrix}$ is the inverse of the basis change matrix $S$ in $\mathbf e_j'= S_{ij}\mathbf e_i$.
The first example $\mathbf v = (x_2,-x_1)$ goes on as follows
Here, we have $v_1 = x_2$ and $v_2 = -x_1$, referred to the old axes. In terms of the new coordinates they will be $v_1'=x_2'$ and $v_2' = -x_1'$, i.e.\begin{align} v_1' &=x_2' = -sx_1 + cx_2 \\ \tag{2} v_2'&= -x_1 = -cx_1-sx_2, \end{align}denoting $\cos \theta$ by $c$ and $\sin \theta$ by $s$.
So far I'm following the argument. But now, it reads
Now if we start again and evaluate $v_1'$ and $v_2'$ as given by $(1)$, we find that\begin{align} v_1' &= L_{11}v_1 + L_{12}v_2 = cx_2 + s(-x_1) \\ v_2' &= L_{21}v_1 + L_{22}v_2 = -s(x_2)+c(-x_2) \tag{3} \end{align}The expressions for $v_1'$ and $v_2'$ in $(2)$ and $(3)$ are the same whatever the values of $\theta$ (i.e. all rotations) and thus by definition $(1)$ the pair $(x_2,-x_1)$ is a first-order Cartesian tensor.
What is the difference between $(2)$ and $(3)$? In the first case we compute $\mathbf v' = L\mathbf v$. What is done differently in the second case?
The second example $\mathbf v = (x_2,x_1)$ goes on to say that $(x_2,x_1)$ isn't a first-order Cartesian tensor.
$\endgroup$1 Answer
$\begingroup$Looking at the textbook I agree the example is confusing. The following is my best guess of what the author(s) intended.
We have a point in $\mathbb{R}^2$ represented by $\vec{x}=\begin{pmatrix}x_1\\x_2\end{pmatrix}$, which we can express in our new coordinates as$$\vec{x}'=\begin{pmatrix}x_1'\\x_2'\end{pmatrix}=L\vec{x}=\begin{pmatrix}cx_1+sx_2\\-sx_1+cx_2 \end{pmatrix}.$$In other words, $(x_1,x_2)$ is a vector / first-order Catesian tensor. (Edit: To clarify, we're defining $\vec{x}$ as the vector representing a point in space, so it has to transform as a vector.) We then form other arrays from the components of $\vec{x}$, (which are not defined as vectors a priori so might not transform as vectors) and ask whether they too transform as a tensor.
First we look at $(-x_2,x_1)$, which I write as $\vec{v}=\begin{pmatrix}-x_2\\x_1\end{pmatrix}$.
Then $$\vec{v}'=\begin{pmatrix}-x_2'\\x_1'\end{pmatrix}=\begin{pmatrix}c(-x_2)+sx_1\\-s(-x_2)+cx_1\end{pmatrix}=L\vec{v}$$so $\vec{v}$ transforms as a tensor. Note we didn't start out by calculating $L\vec{v}$, instead we calculated the new components of $\vec{v}'$ from $\vec{x}'=L\vec{x}$ and then noticed that $\vec{v}'=L\vec{v}$.
If instead we look at $(x_2,x_1)$ then we find that $\vec{v}'=\begin{pmatrix}x_2'\\x_1'\end{pmatrix}=\begin{pmatrix}cx_2-sx_1\\sx_2+cx_1\end{pmatrix}=\begin{pmatrix}c & -s\\s & c\end{pmatrix}\vec{v}\neq L\vec{v}$, so now we have that $\vec{v}$ does not transform as a tensor.
The confusing thing is that when you think of any simple geometric quantity (e.g. our $\vec{x}$ here which represented a position) it tends to be a vector since these are often the only things worth thinking about. Here to find an example of something that didn't transform as a vector, we had to form something quite artificial - $(x_2,x_1)$ isn't an obvious quantity: it says take your vector, express it in some coordinates, then swap its entries. In general an array formed from the components of another tensor in some order doesn't have to be a tensor itself - this is what we see here with $(x_2,x_1)$.
$\endgroup$ 6
