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Suppose we are given a certain Riemannian metric $$\frac{(dx)^2 + (dy)^2}{y^2}$$

How do I calculate geodesics?

I know that we have the geodesic equation $$\ddot{x}^k(t) + \dot{x}^i(t)\dot{x}^j\Gamma_{ij}^k(x(t)) = 0$$

and that we have only two components, hence a geodesic $\gamma$ should be of the form $\gamma(t) = (x^1(t),x^2(t))$. Writing out the case $k = 1$ we get $$\ddot{x}^1(t) + (\dot{x}^1(t))^2\Gamma_{11}^1(x(t)) + 2\dot{x}^1(t)\dot{x}^2(t)\Gamma_{12}^1(x(t)) + (\dot{x}^2(t))^2\Gamma_{22}^1(x(t)) = 0$$

I also calculated $\Gamma^1_{11} = \Gamma^1_{22} = 0$ and $\Gamma^1_{12} = -1/y$. Now my problem is that I do not really know how to plug in the Christoffel symbols since they are evaluated in $x(t)$.

I know that there is a similar problem here, but I do know how to solve the resulting ODE. My problem is, that I would like to get to the initial equation.

Edit. The geodesic equation is taken from Riemannian Manifolds by John M. Lee, p. 58.

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  • $\begingroup$ Just plug $\gamma(t)=(x^1(t),x^2(t))$ in Christoffel symbols. For example, $\Gamma_{12}^1=-1/y$, so on $\gamma$ we have $\Gamma_{12}^1=-1/x^2(t)$. $\endgroup$ – Cave Johnson Dec 18 '16 at 12:31
  • $\begingroup$ Sorry but I don't quite understand what you mean :( $\gamma$ is of course not a vector: it's a curve, or a map, but $\gamma(t)$, i.e., the value of $\gamma$ at some $t$, can be considered as a vector(more precisely, a point), if this is what you are wondering. And Lee actually meant $\ddot{x}^k(t) + \dot{x}^i(t)\dot{x}^j\Gamma_{ij}^k(\gamma(t)) = 0$ in the equation, since $\gamma(t)=(x^1(t),\cdots,x^n(t))$, while $x(t)$ is usually short for the point $(x^1(t),\cdots,x^n(t))$. $\endgroup$ – Cave Johnson Dec 18 '16 at 12:44
  • $\begingroup$ @CaveJohnson Thanks, sorry for the confusion (deleted my comment because I wrote too fast). You are right. This was also my initial thought, but you will end up with $\ddot{x}^1(t) - 2\frac{\dot{x}^1(t)\dot{x}^2(t)}{x^2(t)} = 0 $ wheras for example here people.maths.ox.ac.uk/hitchin/hitchinnotes/Geometry_of_surfaces/… we just find the ODE $d/dt (x'/y^2) = 0$ which is a bit simpler to solve. Thus I thought I was wrong in the beginning. $\endgroup$ – TheGeekGreek Dec 18 '16 at 12:48
  • $\begingroup$ But in our situation we also come to solve the same ODE. Since $\Gamma_{11}^1=\Gamma_{22}^1=0$ and $\Gamma_{12}^1=-1/y$, the equation comes to $\ddot x^1(t)-2\dot x^1(t)\dot x^2(t)/x^2(t)=0$, or equivalently $\dfrac d {dt}\left(\frac{\dot x^1(t)}{(x^2(t))^2}\right)=0$, which is the same as what you mentioned, isn't it? $\endgroup$ – Cave Johnson Dec 18 '16 at 12:56
  • $\begingroup$ @CaveJohnson How to resolve $x'=cy^2$ ? $\endgroup$ – lanse7pty Dec 18 '16 at 13:03
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$x(t)$ means the same thing as $\gamma(t)$ in our situation, where $\gamma(t)=(x^1(t),\ldots,x^n(t))$, since $x(t)$ is usually short for $(x^1(t),\ldots,x^n(t))$. So we only need to plug $(x^1(t),\ldots,x^n(t))$ in the Christoffel symbol $\Gamma_{ij}^k(x^1,\ldots,x^n)$. For example, in this particular problem we have $$\Gamma_{11}^1=\Gamma_{22}^1=\Gamma_{12}^2=\Gamma_{21}^2=0,~\Gamma_{11}^2=1/y,~\Gamma_{22}^2=\Gamma_{12}^1=\Gamma_{21}^1=-1/y$$ Then plug in $x=x(t),y=y(t)$ to see that along the geodesic $\gamma$, we have $$\Gamma_{11}^1=\Gamma_{22}^1=\Gamma_{12}^2=\Gamma_{21}^2=0,~\Gamma_{11}^2=1/y(t),~\Gamma_{22}^2=\Gamma_{12}^1=\Gamma_{21}^1=-1/y(t)$$ The geodesic equation becomes \begin{align} \ddot x(t)-2\dot x(t)\dot y(t)/y(t)=0\\ \ddot y(t)+(\dot x(t))^2/y(t)-(\dot y(t))^2/y(t)=0 \end{align} The first equation can be integrated easily to $$\dot x(t)=c(y(t))^2$$ While the second equation is more complicated to deal with. However, it can be checked that $$\dfrac d{dt}\frac{\dot x^2+\dot y^2}{y^2}=\frac{2}{y^3}((\dot x\ddot x+\dot y\ddot y)y-(\dot x^2+\dot y^2)\dot y)=0$$ which leads to another integration. Now it's not hard to solve those equations. The complete solution can be found here, which is a semicircle centered on the real axis.

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  • $\begingroup$ Sorry to ask again, but I do not yet really see, why I have to calculate $d/dt ...$ in the last equation. I mean, I do not see this from the system of geodesic equations. $\endgroup$ – TheGeekGreek Dec 18 '16 at 14:28
  • $\begingroup$ @TheGeekGreek It can be infered from a geometric point of view (geodesic curvature is $0$, or tangent vector is orthogonal to its covariant derivative) that $\langle\dot\gamma(t),\dot\gamma(t)\rangle$ remains constant along the geodesic $\gamma$. This is where the calculation orignates. $\endgroup$ – Cave Johnson Dec 18 '16 at 14:44
  • $\begingroup$ Sorry, if I was not clear. This was not exactly what I meant: I mean, how does $\dfrac d{dt}\frac{\dot x^2+\dot y^2}{y^2} = 0$ helps to solve $\ddot y(t)+(\dot x(t))^2/y(t)-(\dot y(t))^2/y(t)=0$? $\endgroup$ – TheGeekGreek Dec 18 '16 at 14:54
  • $\begingroup$ @TheGeekGreek We do not need the second equation any more, since it is already contained in $\dot x=cy^2$ and $\dot x^2+\dot y^2=c'y^2$. We only need to consider the latter two equations, which is quite easy. $\endgroup$ – Cave Johnson Dec 18 '16 at 14:59

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