Let $\gamma$ be a unit-speed curve on the helicoid $$\sigma (u,v)=(u\cos v, u\sin v, v)$$ I have shown that $$\dot u^2+(1+u^2)\dot v^2=1$$ and that if $\gamma$ is a geodesic on $\sigma$ then $$\dot v=\frac{a}{1+u^2}$$ where $a$ is a constant.

Could you give me a hint how we could find the geodesics corresponding to $a = 0$ and $a = 1$ ?

  • 1
    $\begingroup$ you have a coupled set of odes that you should solve. Taking this on face value for $a=0$ you have $\dot{v} = 0$ so $\dot{u}^2 = 1$ do the other one? $\endgroup$ – Chinny84 Jan 8 '16 at 11:45
  • $\begingroup$ So, for $a=0$ we have $\dot v=0 \Rightarrow v=c_1$and $\dot u^2=1 \Rightarrow \dot u=\pm 1\Rightarrow u=\pm t+c_2$, where $c_1,c_2$ are constants, right? What kind of geodesics are they? Are these lines? @Chinny84 $\endgroup$ – Mary Star Jan 8 '16 at 14:05
  • $\begingroup$ For $a=1$ we have $\dot v=\frac{1}{1+u^2}$, so $$\dot u^2+(1+u^2)\frac{1}{(1+u^2)^2}=1 \Rightarrow \dot u^2+\frac{1}{1+u^2}=1 \Rightarrow \dot u^2=1-\frac{1}{1+u^2} \Rightarrow \dot u^2=\frac{u^2}{1+u^2} \\ \Rightarrow \dot u=\pm \sqrt{\frac{u^2}{1+u^2}} \Rightarrow \frac{du}{dt} =\pm \sqrt{\frac{u^2}{1+u^2}} \Rightarrow \frac{\sqrt{1+u^2}}{u} du=\pm dt \\ \Rightarrow \int \frac{\sqrt{1+u^2}}{u} du=\pm \int dt$$ $$u=\sinh x \Rightarrow du=\cosh x dx$$ $1+u^2=1+\sinh^2 x=\cosh^2 x \Rightarrow \sqrt{1+u^2}=\sqrt{\cosh^2 x}=\cosh x$ $\endgroup$ – Mary Star Jan 8 '16 at 14:05
  • $\begingroup$ Therefore $$\int \frac{\sqrt{1+u^2}}{u} du=\int \frac{1}{\sinh x} dx=\int \frac{\sinh x}{\sinh^2 x}dx=\int \frac{\sinh x}{\cosh^2 x-1}dx$$ We set $w=\cosh x \Rightarrow dw=\sinh x \ dx$, so $$\int \frac{\sinh x}{\cosh^2 x-1}dx=\int \frac{1}{w^2-1}dw$$ We have that $\frac{1}{w^2-1}=\frac{A}{w-1}+\frac{B}{w+1}=\frac{w(A+B)+(A-B)}{w^2-1} \Rightarrow A+B=0, A-B=1 \Rightarrow A=-B=\frac{1}{2}$ $\endgroup$ – Mary Star Jan 8 '16 at 14:05
  • $\begingroup$ So, $$\int \frac{1}{w^2-1}dw=\frac{1}{2}\int \frac{1}{w-1}dw-\frac{1}{2}\int\frac{1}{w+1}dx \\ =\frac{1}{2}\left (\ln |w-1|-\ln |w+1|\right )+c \\ =\frac{1}{2}\ln \frac{w-1}{w+1} +c =\frac{1}{2}\ln \frac{\cosh x-1}{\cosh x+1} +c \\ =\frac{1}{2}\ln \frac{(\cosh x-1)^2}{\cosh^2 x-1} +c=\frac{1}{2}\ln \frac{(\cosh x-1)^2}{\sinh^2 x} +c \\ =\ln \frac{\cosh x-1}{\sinh x} +c=\ln \frac{\sqrt{u^2+1}-1}{u} +c$$ Therefore, $$\int \frac{\sqrt{1+u^2}}{u} du=\pm \int dt \Rightarrow \ln \frac{\sqrt{u^2+1}-1}{u} +c=\pm t$$ Is everything correct? What kind of geodesics are these? @Chinny84 $\endgroup$ – Mary Star Jan 8 '16 at 14:06

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