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 <br /> \left. \begin{aligned} x = r \cos \phi \\ y = r \sin \phi \end{aligned}\right\} \Rightarrow <br /> \left\{\begin{aligned} dx = \cos \phi\,dr - r\sin \phi\,d\phi \\<br /> 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.