Why is $\vec u_x \cdot \vec u_\phi =-\sin\phi $?

Question

I am learning the coordinate transform between cylindrical and cartesian,and the picture is the teaching,and i don't understand why is $\vec u_x \cdot \vec u_\phi =-\sin\phi $ ? I guess the reason of $\vec u_x \cdot \vec u_r =\cos\phi $ is $x=r\cos\phi$,but it can't explain why $\vec u_x \cdot \vec u_\phi =-\sin\phi $,can anyone teach me?

Answer 1

$ \newcommand\PD[2]{\frac{\partial#1}{\partial#2}} $

$\vec u_x$ is the vector telling you how a point $P(x, y, z)$ moves when you change $x$ a tiny bit; similarly for $\vec u_y$ and $\vec u_z$. Since the point at coordinates $(x, y, z)$ is $$ P(x, y, z) = x\vec u_x + y\vec u_y + z\vec u_z $$ and $\vec u_x, \vec u_y, \vec u_z$ are constant, we see $$ \PD Px = \vec u_x,\quad \PD Py = \vec u_y,\quad \PD Pz = \vec u_z. $$ The partial derivatives of $P$ capture the idea of moving a point a tiny bit and giving the resulting displacement vector. So now write $P$ in cylindrical coordinates: $$ P(r, \phi, z) = (r\cos\phi)\vec u_x + (r\sin\phi)\vec u_y + z\vec u_z. $$ Then $$ \vec u_\phi = \PD p\phi = -(r\sin\phi)\vec u_x + (r\cos\phi)\vec u_y, $$ so we see $$ \vec u_x\cdot\vec u_\phi = -(r\sin\phi)\vec u_x\cdot\vec u_x + (r\cos\phi)\vec u_x\cdot\vec u_y = -r\sin\phi. $$

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The coordinate basis $\vec u_r, \vec u_\phi, \vec u_z$ derived in this way does not have to consist of unit vectors, but it is often desirable to normalize and make them into unit vectors. In this particular case $$ |\vec u_\phi| = \sqrt{\vec u_\phi\cdot\vec u_\phi} = \sqrt{r^2\sin^2\phi + r^2\cos^2\phi} = \sqrt{r^2} = r, $$ the last equality following since $r$ is usually assumed to be positive. Then $$ \hat u_\phi = \frac{\vec u_\phi}{|\vec u_\phi|} = -(\sin\phi)\vec u_x + (\cos\phi)\vec u_y $$ and so $$ \vec u_x\cdot\hat u_\phi = -\sin\phi. $$

Answer 2

Start by looking at the definition of $\hat a_r$ and $\hat a_\phi$. Those are vectors of length $1$, with $\hat a_r$ along the $r$, and $\hat a_\phi$ perpendicular to $\hat a_r$, in the horizontal plane. So now you need to write these in terms of the components along the axes. For $\hat a_r$, you have $$\hat a_r=\frac x{\sqrt{x^2+y^2}}\hat a_x+\frac y{\sqrt{x^2+y^2}}\hat a_y=\cos\phi\hat a_x+\sin\phi\hat a_y$$ You can easily see that the slope is $y/x$. The perpendicular to this will have the slope $-x/y$. Using the fact that it's normalized to $1$, you get $$\hat a_\phi=\pm\sin\phi\hat a_x\mp\cos\phi\hat a_y$$ The convention about the sign is given by the cross product: $$\hat a_r\times\hat a_\phi=\hat a_z$$