In the book I am studying the standard matrices for counterclockwise rotation around the $x$-axis through an angle $\alpha$ is:
$R_x = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos(\alpha) & -\sin(\alpha)\\ 0 & \sin(\alpha) & cos(\alpha) \end{bmatrix}$
I hope I can explain my question without using very much graphics:
For example, as an explanation for why $R_x$ looks like it does, we consider
$R_x$v.
The first column in $R_x$ makes sure $v_1$ stays the same throught the matrix transformation (rotation), this I understand. However, as for the explanation of the other two columns, the book says that one can view it as the 2-D case, where the unit circle is used to derive the coordnates for the standard matrix.
But this I can only imagine will work if you know how the coordinate system looks like, because the directions of the $y$- and $z$-axes can be interchanged. If we interchange the directions of the $y$- and $z$-axes, the points that the rotation travels between will still be the same, but the position of these points will be different relative to the $y$- and $z$-axes, if we interchange the directions of the $y$- and $z$-axes.
Therefore, I imagine that the rotation matrix around the $x$-axis could look in two different ways. One way would be as $R_x$ above, but also like:
$R^{´}_x = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -\sin(\alpha) & \cos(\alpha)\\ 0 & cos(\alpha) & \sin(\alpha) \end{bmatrix}$
And I feel like this makes sense since the resulting matrix will still be the same. But I also feel like then the standard matrix is not very "standard", so am I doing something wrong here?
Consider that when $\alpha =0$, you should get the identity matrix. That should at least convince you that your matrix is wrong, if not why it is wrong.
Swapping the order of $y$ and $z$ does get you a slightly different matrix:
$$R_x'(\alpha) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos(\alpha) & \sin(\alpha)\\ 0 &-\sin(\alpha) & cos(\alpha) \end{bmatrix}=R_x(-\alpha)$$
That is, if we swap $y$ and $z$, what was counterclockwise is now clockwise rotation by $\alpha$, and visa versa.
The above definition essentially defines clockwise as well as rotation. As you say, it is impossible to define clockwise around the $x$-axis unless you know whether the rotation of $\pi/4$ clockwise sends $(0,1,0)$ to $(0,0,1)$ or visa versa. So the above definition is giving you that. Alternatively, it is assuming a picture of the axes that makes it clear.