Derivative in cylindrical coordinates

Physics Asked on October 9, 2020

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?

3 Answers

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 :

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


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

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


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

The transformation matrix between those two coordinate system is:


Correct answer by Eli on October 9, 2020

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:

enter image description here

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


Pavel Grinfield's Tensor Analysis book

Answered by Buraian on October 9, 2020

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

Answered by trula on October 9, 2020

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