- 46

- 2

I am really confused about coordinate transformations right now, specifically, from cartesian to polar coordinates.

A vector in cartesian coordinates is given by ##x=x^i \partial_i## with ##\partial_x, \partial_y \in T_p \mathcal{M}## of some manifold ##\mathcal{M}## and and ##x^i## being some coefficients.

Assuming an euclidean metric ##[\delta_{ij}]=

\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)##, the norm of the vector ##x## is given by $$\sqrt{\delta(x,x)}=\sqrt{\delta_{ij}x^i x^j}=\sqrt{x^2+y^2}=:r$$.

Now, we make a transformation into polar coordinates:

##x=r \cos{\phi}##

##y=r \sin{\phi}##

Which gives us the vector ##x=y^i \partial_i## in terms of the basis vectors ##\partial_r, \partial_\phi##.

The euclidean metric in polar coordinates looks as follows:

##[\hat{\delta_{ij}}]=

\left( \begin{array}{cc} 1 & 0 \\ 0 & r^2 \end{array} \right)##

Computing the norm in the polar coordinates yields:

$$\sqrt{\delta(x,x)}=\sqrt{\hat{\delta_{ij}}y^i y^j}=\sqrt{r^2+r^2\phi^2}.$$

As ##\delta(x,x)## is a tensor relation, though, it should be coordinate independent. Furthermore, the inner product is invariant over coordinate transformations. So why is the norm different?

To add to the confusion, I tried transforming the coefficients from cartesian to polar coordinates:

$$x'^i=\frac{\partial f^i}{\partial x^j} x^j \;\;\; (1)$$

With ##\frac{\partial f^i}{\partial x^j}=\left( \begin{array}{cc} \cos{\phi} & \sin{\phi} \\ -\frac{\sin{\phi}}{r} & \frac{\cos{\phi}}{r} \end{array} \right)## being the matrix of partial derivatives.

Applying ##(1)## for the first component ##r## yields:

$$x'^1= \frac{\partial f^1}{\partial x^j} x^j= \cos{\phi}\cdot r \cos{\phi} + \sin{\phi}\cdot r \sin{\phi}=r$$

So far so good. For the second component ##\phi##, however, we get:

$$x'^2= \frac{\partial f^2}{\partial x^j} x^j=-\frac{\sin{\phi}}{r} \cdot r \cos{\phi}+\frac{\cos{\phi}}{r} \cdot r \sin{\phi}=0$$

So what's going on here? Why does any general vector ##x=(x,y)^T## have a zero second component in polar coordinates?

Help!

A vector in cartesian coordinates is given by ##x=x^i \partial_i## with ##\partial_x, \partial_y \in T_p \mathcal{M}## of some manifold ##\mathcal{M}## and and ##x^i## being some coefficients.

Assuming an euclidean metric ##[\delta_{ij}]=

\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)##, the norm of the vector ##x## is given by $$\sqrt{\delta(x,x)}=\sqrt{\delta_{ij}x^i x^j}=\sqrt{x^2+y^2}=:r$$.

Now, we make a transformation into polar coordinates:

##x=r \cos{\phi}##

##y=r \sin{\phi}##

Which gives us the vector ##x=y^i \partial_i## in terms of the basis vectors ##\partial_r, \partial_\phi##.

The euclidean metric in polar coordinates looks as follows:

##[\hat{\delta_{ij}}]=

\left( \begin{array}{cc} 1 & 0 \\ 0 & r^2 \end{array} \right)##

Computing the norm in the polar coordinates yields:

$$\sqrt{\delta(x,x)}=\sqrt{\hat{\delta_{ij}}y^i y^j}=\sqrt{r^2+r^2\phi^2}.$$

As ##\delta(x,x)## is a tensor relation, though, it should be coordinate independent. Furthermore, the inner product is invariant over coordinate transformations. So why is the norm different?

To add to the confusion, I tried transforming the coefficients from cartesian to polar coordinates:

$$x'^i=\frac{\partial f^i}{\partial x^j} x^j \;\;\; (1)$$

With ##\frac{\partial f^i}{\partial x^j}=\left( \begin{array}{cc} \cos{\phi} & \sin{\phi} \\ -\frac{\sin{\phi}}{r} & \frac{\cos{\phi}}{r} \end{array} \right)## being the matrix of partial derivatives.

Applying ##(1)## for the first component ##r## yields:

$$x'^1= \frac{\partial f^1}{\partial x^j} x^j= \cos{\phi}\cdot r \cos{\phi} + \sin{\phi}\cdot r \sin{\phi}=r$$

So far so good. For the second component ##\phi##, however, we get:

$$x'^2= \frac{\partial f^2}{\partial x^j} x^j=-\frac{\sin{\phi}}{r} \cdot r \cos{\phi}+\frac{\cos{\phi}}{r} \cdot r \sin{\phi}=0$$

So what's going on here? Why does any general vector ##x=(x,y)^T## have a zero second component in polar coordinates?

Help!

Last edited: