Can anyone provide a good explanation to converting a vector in cartesian coordinates to cylindricals?

Question

I am sorry about the perhaps trivial question, but for some reason I am really struggling to do this. I was recently taught about curvilinear coordinates, which I believe provides a system for bouncing between various coordinate systems by pretty much giving a general coordinate system. However I am struggling to actually apply what I learnt to converting a vector given in cartesian coordinates to the same vector in cylindrical coordinates. Take, for example, $$\vec{A}=10x^2\hat{i}+y^2\hat{j}+z^3\hat{k}.$$ What is a good method for obtaining this in cylindrical coordinates so that the answer is in the form $$\vec{A}=a\hat{e}_\rho+b\hat{e}_\phi+c\hat{e}_z$$ where $a,b$ and $c$ are some numbers and $\hat{e}_\rho,\hat{e}_\phi$ and $\hat{e}_z$ are base vectors in cylindrical coordinates (the $\hat{i},\hat{j},\hat{k}$ equivalent, if you will).

Please provide a basic method to do this.

Answer 1

I hadn't come across the idea of unit vectors in cylindrical coordinates before I read your question, but Wikipedia provides a good introduction.

Unit vectors in cylindrical coordinates are dependent on position, so you can't convert an arbitrary cartesian vector into cylindrical coordinates unless you interpret it, e.g., as the strength of a vector field at a point.

Suppose we have a point $(x,y,z)$ and a vector $\vec F = F_1 ~\hat i + F_2 ~\hat j + F_3 ~\hat k$ where the $F_i$ are functions of $x,y,z$. The cylindrical unit vectors at $(x,y,z)$ are: $$\begin{bmatrix}\hat{e_\rho}\\\hat{e_\phi}\end{bmatrix}=\begin{bmatrix}\cos\phi&\sin\phi\\-\sin\phi&\cos\phi\end{bmatrix}\begin{bmatrix}\hat i\\\hat j\end{bmatrix}, \hat{e_z}=\hat k$$

So we're looking for constants $c_\rho,c_\theta,c_z$ such that $c_\rho\hat{e_\rho}+c_\theta\hat{e_\theta}+c_z\hat{e_z} = F_1 \hat i + F_2 \hat j + F_3 \hat k$

$c_z=F_3$ is the easy bit. For the rest, we can equate coefficients of $\hat i,\hat j$.

Solving $\begin{bmatrix}\cos\phi&\sin\phi\\-\sin\phi&\cos\phi\end{bmatrix} \begin{bmatrix}c_\rho\\c_\theta\end{bmatrix}=\begin{bmatrix}F_1\\F_2\end{bmatrix}$ gives $\begin{bmatrix}c_\rho\\c_\theta\end{bmatrix}=\begin{bmatrix}\cos\phi&-\sin\phi\\\sin\phi&\cos\phi\end{bmatrix} \begin{bmatrix}F_1\\F_2\end{bmatrix}$

Using $\cos\phi=\dfrac{x}{\sqrt{x^2+y^2}}, \sin\phi=\dfrac{y}{\sqrt{x^2+y^2}}$ the general formula to translate $\vec F = F_1 ~\hat i + F_2 ~\hat j + F_3 ~\hat k$ into cylindrical coordinates is:

$$\vec F=(\dfrac{F_1x-F_2y}{\sqrt{x^2+y^2}})~\hat{e_\rho}+(\dfrac{F_1y+F_2x}{\sqrt{x^2+y^2}})~\hat{e_\phi}+F_3~\hat{e_z}$$

In your case, you have $\vec A = 10x^2~\hat i+y^2~\hat j+z^3~\hat k$ and you get $$\vec A=(\dfrac{10x^3-y^3}{\sqrt{x^2+y^2}})~\hat{e_\rho}+(\dfrac{10x^2y+xy^2}{\sqrt{x^2+y^2}})~\hat{e_\phi}+z^3~\hat{e_z}$$

Warning: I've just derived this and have to run off on an errand. Check it carefully!

Answer 2

Note that we have the equality

$$\begin{align} \hat xA_x+\hat yA_y+\hat zA_z=\hat \rho A_{\rho}+\hat \phi A_{\phi}+\hat zA_z \tag 1 \end{align}$$

Now, if we form inner the inner product $\hat \rho \cdot \vec A=A_{\rho}$ on both sides of $(1)$, we see that

$$\begin{align} A_{\rho}&=(\hat \rho \cdot \hat x)A_x+(\hat \rho \cdot \hat y)A_y\\\\ &\bbox[5px,border:2px solid #C0A000]{=\cos(\phi)A_x+\sin(\phi)A_y} \end{align}$$

Similarly, if we form inner the inner product $\hat \phi \cdot \vec A=A_{\phi}$, we see that

$$\begin{align} A_{\phi}&=(\hat \phi \cdot \hat x)A_x+(\hat \phi \cdot \hat y)A_y\\\\ &=\bbox[5px,border:2px solid #C0A000]{-\sin(\phi)A_x+\cos(\phi)A_y} \end{align}$$

To finish the transformation, we note that

$$A_x=10x^2=10\rho^2\cos^2(\phi)$$

and

$$A_y=y^2=\rho^2\sin^2(\phi)$$

The rest is a simple exercise of putting it all together.