Trasformation of vectors component when changing coordinate system

[lennyleonard]

Hi everyone
I have a little problem in understanding the trasformation of vectors component when passing to a different coordinate system (abbreviated CS).
Theory says that the components of a vector in the first CS $$x$$ with component $$(V^0,V^1,...,V^n)$$ will transform changing CS according to $$V^{\alpha'}=\Lambda^{\alpha'}_{\beta}B^{\beta}$$ where the primed (') letters refer to the new CS $$x'$$ and the Einstein notation is employed.
The matrix $$\Lambda$$ is the jacobian of the application $$x'(x)$$.
The problem I have is better expressed by an example.

Les's take the coordinate trasformation from cartesian to polar. We have:
$$x=rcos\theta\; ,\;y=rsin\theta$$ and $$r=\sqrt{x^2+y^2}\;,\;\theta=arctg\frac{y}{x}$$
The jacobian we need is:

$$\left(\begin{array}{cc}\frac{\partial r}{\partial x} & \frac{\partial r}{\partial y}\\\frac{\partial \theta}{\partial x}&\frac{\partial \theta}{\partial y}\end{array}\right)$$

and so:
$$\Lambda = \left(\begin{array}{cc}cos\theta & sin\theta\\ -\frac{1}{r}sin\theta & \frac{1}{r}cos\theta\end{array}\right)$$

At this point I understand that if I have the vector $$\vec{V}=X\vec{e}_x,Y\vec{e}_y$$ and I want to know its components on the basis $$\vec{e}_r\vec{e}_{\theta}$$ all I have to do is calculate $$V^{r}=\Lambda^{r}_{\beta}B^{\beta}$$ and $$V^{\theta}=\Lambda^{\theta}_{\beta}B^{\beta}$$ where $$\beta=X,Y$$.

So I end up with: $$V^r=Xcos\theta+Ysin\theta$$ and $$V^{\theta}=-\frac{X}{r}sin\theta+\frac{Y}{r}cos\theta$$; but i know that it should also be $$\vec{V}=r\vec{e}_r+\theta\vec{e}_{\theta}$$ and so: $$V^r=r$$ and $$V^{\theta}=\theta$$ but this is not true! In particular the $$V^{\theta}$$ I obtained above gets identically zero.

where is my mistake?

Tanks a lot for your answers!

 

[tiny-tim]

welcome to pf!

hi lennyleonard! welcome to pf!

(write "/tex" not "\tex" )

Hi everyone
I have a little problem in understanding the trasformation of vectors component when passing to a different coordinate system (abbreviated CS).
Theory says that the components of a vector in the first CS $$x$$ with component $$(V^0,V^1,...,V^n)$$ will transform changing CS according to $$V^{\alpha'}=\Lambda^{\alpha'}_{\beta}B^{beta}$$ where the primed (') letters refer to the new CS $$x'$$ and the Einstein notation is employed.
The matrix $$\Lambda$$ is the jacobian of the application $$x'(x)$$.
The problem I have is better expressed by an example.

Les's take the coordinate trasformation from cartesian to polar. We have:
$$x=rcos\theta\; ,\;y=rsin\theta$$ and $$r=\sqrt{x^2+y^2}\;,\;\theta=arctg\frac{y}{x}$$
The jacobian we need is:

$$\left(\begin{array}{cc}\frac{\partial r}{\partial x} & \frac{\partial r}{\partial y}\\\frac{\partial \theta}{\partial x}&\frac{\partial \theta}{\partial y}\end{array}\right)$$

and so:


$$\Lambda = \left(\begin{array}{cc}cos\theta & sin\theta\\-\frac{1}{r}sin\theta&\frac{1}{r}cos\theta\end{array}\right)$$

At this point I understand that if I have the vector $$\vec{V}=X\vec{e}_x,Y\vec{e}_y$$ and I want to know its components on the basis $$\vec{e}_r\vec{e}_{\theta}$$ all I have to do is calculate $$V^{r}=\Lambda^{r}_{\beta}B^{beta}$$ and $$V^{\theta}=\Lambda^{\theta}_{\beta}B^{beta}$$ where $$\beta=X,Y$$.

So I end up with: $$V^r=Xcos\theta+Ysin\theta$$ and $$V^{\theta}=-\frac{X}{r}sin\theta+\frac{Y}{r}cos\theta$$; but i know that it should also be $$\vec{V}=r\vec{e}_r+\theta\vec{e}_{\theta}$$ and so: $$V^r=r$$ and $$V^{\theta}=\theta$$ but this is not true! In particular the $$V^{\theta}$$ I obtained above gets identically zero.

where is my mistake?

Tanks a lot for your answers!


PS: I'll be extremely thankful if someone tells me how i can show my latex lines !

"but i know that it should also be $$\vec{V}=r\vec{e}_r+\theta\vec{e}_{\theta}$$ "

no … why?

 

[lennyleonard]

Thank you for your disponibility tiny-tim!

Well..i'd say that on the $$\vec{e}_r\;\vec{e}_{\theta}$$ basis to identify a point $$P$$in the plane i have to state its distance from the origin $$r$$ and the angle $$\theta$$between $$\vec{e}_r$$ and the x-axis..it is right isn't it?
This to get $$\vec{P}=r\vec{e}_r+\theta\vec{e}_{\theta}$$

And why my $$V^{\theta}=-\frac{X}{r}sin\theta+\frac{Y}{r}cos\theta$$ is always zero?!?


I've been melted my brains on this problem, so i could be mistaking all along!
I really thank you for the help you're giving me!

(I fixed the whole latex thing! :-))

 

[Fredrik]

(I fixed the whole latex thing! :-))

Almost. One expression in the first post is messed up because vBulletin inserts an extra space every 50 characters (if your text doesn't include any spaces). I don't know if that will be fixed, but at least for now, it's a good idea to throw in some spaces here and there in your code, e.g. before + and - signs. I also recommend that you use itex tags instead of tex when you want your LaTeX expression to appear on the same line as normal text.

 

[tiny-tim]

hi lennyleonard!

…… to identify a point $$P$$in the plane …

but you're not trying to identify a point P, you're trying to identify a vector PX starting at P

And why my $$V^{\theta}=-\frac{X}{r}sin\theta+\frac{Y}{r}cos\theta$$ is always zero?!?

because that's the theta coordinate of the unit vector PX in the direction OP, which of course is 0

(I fixed the whole latex thing! :-))

mmm … what about that cos sin -sin cos matrix ?

 

[lennyleonard]

..could you explain that a little deeper tiny-tim?? I'm starting to see the light!

(thanks Fredrik for your advice, now it should all be ok )

 

[tiny-tim]

..could you explain that a little deeper tiny-tim?? I'm starting to see the light!

(thanks Fredrik for your advice, now it should all be ok )

hi lennyleonard!

vectors in ordinary Cartesian coordinates are the same everywhere, eg 3i + 2j is the same direction no matter where you start

but vectors in polar coordinates depend on where you start, eg obviously e_r always points away from the origin

at each point P, the directions of e_r and e_θ depend on where P is

your original vector x starting at P needs to be expressed in terms of the local e_r and e_θ

you keep trying to find the polar coordinates of the unit vector e_r, which obviously are just … e_r !

 

[lennyleonard]

let me see if I finally got this straight..
if have a vector $\vec{V}$ in cartesian coordinates its direction will be parallel with the basis vector $\vec{e}_r$ of the polar coordinates?
So basically this change of basis is like rotating the coordinate system!

 

[tiny-tim]

if have a vector $\vec{V}$ in cartesian coordinates its direction will be parallel with the basis vector $\vec{e}_r$ of the polar coordinates?

not following you , but …

So basically this change of basis is like rotating the coordinate system!

yes, it is exactly rotating the coordinate system!

 

[Fredrik]

let me see if I finally got this straight..
if have a vector $\vec{V}$ in cartesian coordinates

$\vec V$ is a vector if and only if it's a member of a vector space. Coordinates don't have anything to do with that.

its direction will be parallel with the basis vector $\vec{e}_r$ of the polar coordinates?

Yes, if you're referring to the $\vec{e}_r$ at $\vec V$. (The polar coordinate system associates a basis with each point in the plane. It's not the same basis everywhere).

So basically this change of basis is like rotating the coordinate system!

If you switch from one orthonormal basis to another (with the same orientation), the components of the vector changes the same way as if you had just rotated the vector. (The basis change isn't like a change to another cartesian coordinate system. It is a change to another cartesian coordinate system).

 

[lennyleonard]

Thanks a lot to both of you tiny-tim and fredrik!

Boy, its'amazing how simple it is once it's all clear !

(to fredrick: the original phrase was "If I have a vector $\vec{V}$ with cartesian coordinate $X$ and $Y[/itext],..." but I messed that up by editing the post! <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f61b.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":-p" title="Stick Out Tongue :-p" data-smilie="7"data-shortname=":-p" />)<br /> <br /> Thanks one more time gentlemen!$