Show that two families of curves are orthogonal (without using orthogonal trajectories) I'm reading through Hartle's General Relativity and came across this question:

Consider the following coordinate transformation from rectangular coordinates $(x,y)$, labeling points in the plane to a new set of coordinates  $(m,n)$:
  $$x = mn,$$
  $$y = (1/2)(m^2 - n^2).$$
  (c) Do the curves of constant m and constant n intersect at right angles?

I determined that the curves of constant $m$ are orthogonal trajectories to the curves of constant $n$, but the answer in the solutions manual simply states "The curves intersect at right angles because there are no cross terms $dmdn$ in the metric." I don't understand where this comes from. What does he mean? Where can I learn more about this? I imagine that my method of orthognal trajectories will get unwieldy with more variables. 
 A: Well, you can just calculate the metric:
$$\begin{aligned}
\mathrm ds^2 &= \mathrm dx^2 + \mathrm dy^2\\
 &= \mathrm d(mn)^2 + \frac12\mathrm d(m^2 - n^2)\\
 &= (m\,\mathrm dn + n\,\mathrm dm)^2 + \frac12(2m\,\mathrm dm - 2n\,\mathrm dn)^2\\
 &= m^2\,\mathrm dn^2 + 2mn\,\mathrm dm\,\mathrm dn + n^2\,\mathrm dm^2
    + (m^2\,\mathrm dm^2 + n^2\,\mathrm dn^2 - 2mn\,\mathrm dm\,\mathrm dn)\\
 &= (m^2+n^2)(\mathrm dm^2 + \mathrm dn^2)
\end{aligned}$$
As you can see, the mixed term $\mathrm dm\,\mathrm dn$ cancels out so the metric is diagonal. This means that the vectors $\partial_m$ dual to $\mathrm dm$ and $\partial_n$ dual to $\mathrm dn$ are orthogonal, since $\mathrm dm(\partial_n)=\mathrm dn(\partial_m)=0$. and thus $\mathrm dm^2(\partial_m,\partial_n) = \mathrm dm(\partial_m)\mathrm dm(\partial_n)=0$ and analogously for $\mathrm dn^2$, and thus $ds^2(\partial_m,\partial_n)=0$. If there were a term proportional to $\mathrm dm\,\mathrm dn$, then this would give $\mathrm dm(\partial_m)\mathrm dn(\partial_n)\ne 0$, and thus the vectors would not be orthogonal.
A: $$dx= ndm + m dn\:,\quad dy = mdm -ndn$$
hence
$$dx^2 + dy^2 = n^2 dm^2 + m^2 dn^2 + 2nm dn dm + m^2 dm^2 + n^2 dn^2 - 2nm dn dm  \:.$$ We conclude that:
$$ds^2 = (n^2 +m^2)(dm^2 +  dn^2)$$
and there is no $dn\: dm$ term in the metric as stated.
