Line element in cylindrical coordinates

[Catalina-]

Homework Statement

As part of a recent homework I have to convert the line element
$$
ds²=-dt²+dx²+dy²+dz²
$$
to cylindrical coordinates

Relevant Equations

The cylindrical coordinates were given by
$$
r=\sqrt{x²+y²}
$$
$$
\phi=arctan(\frac{y}{x})
$$

First I took the total derivative of these and arrived at
$$
dr=\frac{\partial r}{\partial x}dx+\frac{\partial r}{\partial y}dy \quad\rightarrow \quad r²dr=xdx+ydy
$$
$$
d\phi=\frac{\partial \phi}{\partial x}dx+\frac{\partial \phi}{\partial y}dy \quad\rightarrow \quad r²dr
\phi=-ydx+xdy
$$
After solving the system of equations I got
$$
dx= xdr-yd\phi
$$
$$
dy=ydr+xd\phi
$$
After squaring these separately and adding them I got
$$
dx²+dy²=r²dr²+r²d\phi²
$$
and therefor the line element
$$
ds²=-dt²+r²dr²+r²d\phi²+dz²
$$
However the solution is not supposed to have a r² factor with the dr² term. I have looked at it for a while now but I cant seem to find my error.

 

[pasmith]

Homework Statement: As part of a recent homework I have to convert the line element
$$
ds²=-dt²+dx²+dy²+dz²
$$
to cylindrical coordinates
Relevant Equations: The cylindrical coordinates were given by
$$
r=\sqrt{x²+y²}
$$
$$
\phi=arctan(\frac{y}{x})
$$

First I took the total derivative of these and arrived at
$$
dr=\frac{\partial r}{\partial x}dx+\frac{\partial r}{\partial y}dy \quad\rightarrow \quad r²dr=xdx+ydy
$$
$$
d\phi=\frac{\partial \phi}{\partial x}dx+\frac{\partial \phi}{\partial y}dy \quad\rightarrow \quad r²dr
\phi=-ydx+xdy
$$

You have an extra factor of $r$ on the left hand side of your result for $dr$. But since you haven't shown us how you calculated the partial derivatives, we can't tell you how it got there.

But none of this is necessary. You need to find $dx^2 + dy^2$ in terms of $dr$ and $d\phi$. The easiest way is to start from $$\left. \begin{aligned} x = r \cos \phi \\ y = r \sin \phi \end{aligned}\right\} \Rightarrow \left\{\begin{aligned} dx = \cos \phi\,dr - r\sin \phi\,d\phi \\ dy = \sin \phi \,dr + r\cos \phi\,d\phi \end{aligned}\right.$$

 

[Steve4Physics]

Hi. Welcome to PF. In addition to what @pasmith said, it may be worth noting that

$$r²dr=xdx+ydy$$

can easily be seen to be wrong on dimensional grounds. The left side has dimensions $L^3$ (length cubed) but the right hand side has dimensions $L^2$ so there's an error.

 

[Catalina-]

Thank you very much for your help @pasmith & @Steve4Physics.

You pointed me in the exact right direction, I made a silly mistake in the derivation of dr.