Derivative in cylindrical coordinates

Question

Why do we multiply a $frac{1}{r}$ factor for the  gradient unit vector in $vec{theta}$ direction? and how is the angle a vector here?

Eli · Accepted Answer

the components of the vector $vec{R}$ given with polar coordinates are:

$$vec{R}= begin{bmatrix}
   x \
   y \
 end{bmatrix}= r,begin{bmatrix}
   cos(varphi) \
   sin(varphi) \
 end{bmatrix}=r,cos(varphi)hat{e}_x+r,sin(varphi)hat{e}_y$$

with :

$$hat{vec{e}_r}=frac{1}{||frac{partial vec{R}}{partial r}||},frac{partial vec{R}}{partial r}=begin{bmatrix}
   cos(varphi) \
   sin(varphi) \
 end{bmatrix}$$

and

$$hat{vec{e}_phi}=frac{1}{||frac{partial vec{R}}{partial varphi}||},frac{partial vec{R}}{partial varphi}=begin{bmatrix}
   -sin(varphi) \
   cos(varphi) \
 end{bmatrix}$$

you can write the vector R in coordinates  $quad hat{vec{e}_r},,hat{vec{e}_phi}quad$ system

$$vec{R}=a_r,hat{vec{e}_r}+a_phi,hat{vec{e}_phi}$$

where $a_r=r$ and $a_varphi=0$

The transformation matrix between those  two coordinate system is:

$$S=left[hat{vec{e}_r},,hat{vec{e}_phi}right]$$

Buraian · Answer

The position vector (or the radius vector) is a vector R that represents the position of
points in the Euclidean space with respect to an arbitrarily selected point O, known
as the origin. We can parameterize this position vector as a function of coordinates and define basis vectors using it. In polar coordinates, it is a function of $r$ and $ theta$, where $r$ is the radial distance from a chosen origin and $theta$, is the angle which the radial distance from the arbitrarily chosen origin makes with a line.
We can derive the basis vector of this polar coordinate system by doing derivatives on this parameterized position vector. The partial derivative of this position vector with respect to $theta$ gives the local basis in $theta$ direction. The word local is used because unlike the cartesian coordinate system, the polar coordinate system has a natural basis that changes from point to point.

Working out the angular polar basis vector:

Consider taking the partial of $ R$ w.r.t $ theta$ when the radial distance is some $r$ away, you get:
$$ frac{ partial R}{partial theta} =vec{ e_{theta}}$$
But if you did this at some other $r'$ distance away with $ r'>r$, your basis vector scales up by some factor. Most precisely, the basis evaluated on the unit circle scales up linearly scales a factor of $r$ as we move outwards from the origin and hence the unit polar basis vector at some radial distance is given as:
$$ overline{vec{e_{theta}}} = frac{ e_{theta}}{r}$$
The overline stands for the vector being normalized

References:
Pavel Grinfield's Tensor Analysis book

trula · Answer

we do it since it comes out this way, if yo transform from cartesian ti polar coordinates see https://en.wikipedia.org/wiki/Polar_coordinate_system
in the grad you alway have 1/length so you need the 1/r to get the right dimension of grad.
how will you describe motion in polar coordinates if the angle has no direction?

Derivative in cylindrical coordinates

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