When we use the cylindrical coordinate system $(r, \theta, z)$ where $r$ is the distance from the point in the $xy$-plane, $\theta$ is the angle with the $x$ axis and $z$ is the heigth. As can been seen in the picture
I hav a vector field described by $(0,U_{\theta}(r),U_z)$ but how can the angle differ when $r$ is always zero? When $r$ is zero, then there is no difference between different angles. Or am i wrong?
Some help would be great.
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$\begingroup$If $r = 0$, then the idea or notion of having an angle with the x-Axis is not defined.
However, depending on the context it might be expedient to simply define that the angle can take any value in that context
$\endgroup$ $\begingroup$A vector field is a function from points to vectors. At the point $(r, \theta, z)$, you've got the vector $(0, U_\theta(r), U_z)$, where it's not clear from your description whether $U_z$ is a constant or a function of some arguments. As an example, suppose that $U_theta(r) = \arctan(r)$, and $U_z$ is the constant $4$. Then your field, at location $(r, \theta, z ) = (1, \pi, 2)$ is just $$ F(1, \pi, 2) = (0, \arctan(1), 2) = (0, \frac{\pi}{4}, 2). $$
This vector, as you observe, points straight along the $z$ direction, and because $r$ is zero, the $\theta$ coordinate is irrelevant.
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