- #1
lennyleonard
- 21
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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 Einsten 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!
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 Einsten 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!
Last edited:
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… what about that cos sin -sin cos matrix ? 
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