On surface, for which $$ds^2=du^2+dv^2$$
find the angle between lines $v=u$ and $v=-u$.
This exercise is related to the first fundamental form. I think I need to find the angle between curves with parametrization: $$u(t)=t, v(t)=t$$ and $$u(s)=s, v(s)=-s$$ therefore $$du=dt, dv=dt$$ and $$du*=ds, dv*=-ds$$
Now I can make tangent vectors to both curves $$dr=\frac{dr}{du} dt + \frac{dr}{dv} dt$$ $$dr*=\frac{dr*}{du} ds – \frac{dr*}{dv} ds$$ Am I thinking right? Unfortunately at this point I don't have more ideas. For finding the angle I can use $$cos(a)=\frac{dr .dr*}{|dr| . |dr*|}$$ And then I could use the numbers "1", "0", "1" from the first fundamental form. But I'm absolutely unsure, so I would like to ask you about it. Thank you.
Best Answer
You can parametrize your curves as $c_1(t)=(t,t)$ and $c_2(t)=(t,-t)$ (first coordinate: $u$). You are searching for the angle between the tangent vectors $v=(1,1)$ and $w=(1,-1)$ (tangent vectors along $c_1$ and $c_2$: note that they do not depend on $t$) using the metric $g=du^2+dv^2$ with components $g_{ij}(u,v)=\delta_{ij}$ for $i,j=1,2$.
Then the angle $\theta$ is s.t. $\cos\theta=\frac{\langle v,w\rangle}{\|v\| \|w\|}$,
with $\langle v,w\rangle=g_{ij}v_iw_j$ and $\|v\|=\sqrt{g_{ij}v_iv_j}$ (and similarly for $\|w\|$). We sum over repeated indices.