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I just found (or: I think that I found) the geodesics of the upper, closed half plane of $\mathbb R^2$. To verify my solution: Is it correct that the geodesics are the circles and lines which meet the unit circle at right angles?

Thanks a lot for the confirmation (or the correction, if I'm wrong...)

Edit: Here is what I have so far (this should be correct, isn't it?):

This should be correct, isn't it?

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    $\begingroup$ might you include your calculations? so it is easier for us to analyze if your solution is correct? $\endgroup$
    – tired
    Dec 13, 2016 at 12:43
  • $\begingroup$ Do you mean the poincare disk instead of upper half plane? Or maybe you mean "x-axis" instead of "unit circle". In the upper half plane, there are lots of geodesics which don't meet the unit circle at all. And lots that meet it non-orthogonally. For each point and each tangent vector at that point there should be a unique geodesic. If what you said were true, there would only be one geodesic passing through any particular point on the unit circle $\endgroup$ Dec 13, 2016 at 12:48
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    $\begingroup$ I included my calculations and edited the question. $\endgroup$
    – JohnD
    Dec 13, 2016 at 13:16

3 Answers 3

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It is not correct that the geodesics of the upper half plane $\mathbb{H}=\{(x,y)\in\mathbb{R}^2:\;y>0\}$ with metric $g$ given by $g_{xx}=g_{yy}=1/y^2$ and $g_{xy}=g_{yx}=0$ the circles and lines which meet the unit circle at right angles. Instead, they are the circles and lines that meet the $x$-axis (the line $y=0$ in $\mathbb{R}^2$) at right angles.

A curve $\gamma:I\subset\mathbb{R}\rightarrow \mathbb{H}:s\mapsto(x_1(s),x_2(s))$ is a geodesic if it satisfies the geodesic equation: \begin{equation} \frac{d^2x_k}{ds^2} + \sum_{i,j=1}^2\Gamma^k_{ij}\frac{dx_i}{ds}\frac{dx_j}{ds}=0, \end{equation} for $k=1,2$ and where $\Gamma^k_{ij}$ are the Christoffel symbols, which in this case are \begin{equation} \Gamma^1_{12}=\Gamma^1_{21}=\Gamma^2_{22}=-1/y;\;\; \Gamma^2_{11}=1/y;\;\;\; \Gamma^1_{11}=\Gamma^2_{12}=\Gamma^2_{21}=\Gamma^1_{22}=0. \end{equation} The geodesic equations are then the following system of differential equations (writing $x=x_1$ and $y=x_2$): \begin{equation} \frac{d^2x}{ds^2}-2\frac{1}{y}\frac{dx}{ds}\frac{dy}{ds}=0;\;\;\; \frac{d^2y}{ds^2}+\frac{1}{y} \left[\left(\frac{dx}{ds}\right)^2-\left(\frac{dy}{ds}\right)^2\right]=0 \end{equation} For $\frac{dx}{ds}\neq 0$ the solutions satisfy: $x^2+y^2-ax=b$, where $a$ and $b$ are constants. Those are circles with center in the line $y=0$, and thus meeting that line at right angles. The solutions for $\frac{dx}{ds}=0$ are just vertical lines.

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    $\begingroup$ should the geodesic equation be $\dfrac {d^{2}x}{ds^{2}}-\dfrac {2}{y}\dfrac {dx}{ds}\dfrac {dy}{ds}=0$; $\dfrac {dy^{2}}{ds^{2}}+\dfrac {1}{y}\left( \left( \dfrac {dx}{ds}\right) ^{2}-\left( \dfrac {dy}{ds}\right) ^{2}\right) =0$? And I find some difficulties solving it. $\endgroup$
    – user360777
    Jan 22, 2017 at 6:12
  • $\begingroup$ @user360777 It should, thanks! That was a typo, I have corrected it in the answer $\endgroup$
    – coconut
    Jan 22, 2017 at 7:38
  • $\begingroup$ @user360777 To solve it, the non-trivial case is $\dfrac{dx}{ds}\neq0$. A relatively easy approach is to take the derivative of $y\dfrac{dy}{ds}\left(\dfrac{dx}{ds}\right)^{-1}-x$ and notice that it should vanish by substitution of both differential equations. Then, $x\dfrac{dx}{ds}+y\dfrac{dy}{ds}=(constant)\cdot\dfrac{dx}{ds}$, and that's just the derivative of the equation of a circle around a point at $y=0$ $\endgroup$
    – coconut
    Jan 22, 2017 at 7:47
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    $\begingroup$ Thanks! the vanishing term should be the derivative of $y\dfrac{dy}{ds}\left(\dfrac{dx}{ds}\right)^{-1}+x$ in your answer. it's hard for many people to come up with this smart idea without knowing the geodesics before $\endgroup$
    – user360777
    Jan 22, 2017 at 15:38
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Here I provide an alternative solution based on a physicist's convention.

In mathematical physics concerning differential geometry, e.g. general relativity, one often uses Killing vectors to help solving a geodesic. Each Killing vector $K$ gives a first integral (Einstein summation convention assumed) $$ \frac{\mathrm{d} x^i}{\mathrm{d} \lambda} K_i = \text{const.}, $$ so that one can deal with a set of $n_K$ first-order differential equations, instead of the second-order and often non-linear equations, if the number of Killing vectors is $n_K$.

In your case, all the Killing vectors in Poincaré half-plane are given by $$ T = \partial_1,\qquad D = 2\left(x\,\partial_x+y\,\partial_y\right),\qquad S = (y^2-x^2)\,\partial_x - 2xy\,\partial_y. $$ Their musical isomorphisms $g(K,\cdot)$ are $$ T^{\flat} = \frac{1}{y^2}\mathrm{d}x,\qquad D^{\flat} = \frac{2}{y^2}\left(x\,\mathrm{d}x+y\,\mathrm{d}y\right),\qquad S^{\flat} = \left(1-\frac{x^2}{y^2}\right)\,\mathrm{d}x - 2\frac{x}{y}\,\mathrm{d}y. $$ One has therefore three first integrals $$ t = \frac{\mathrm{d}x^i}{\mathrm{d}\lambda} T_i = \frac{1}{y^2} \frac{\mathrm{d}x}{\mathrm{d}\lambda},\\ d = \frac{\mathrm{d}x^i}{\mathrm{d}\lambda} D_i = \frac{2}{y^2} \left( x \frac{\mathrm{d}x}{\mathrm{d}\lambda} + y \frac{\mathrm{d}y}{\mathrm{d}\lambda}\right), \\ s = \frac{\mathrm{d}x^i}{\mathrm{d}\lambda} S_i = \left(1-\frac{x^2}{y^2}\right)\frac{\mathrm{d}x}{\mathrm{d}\lambda} - 2\frac{x}{y} \frac{\mathrm{d}y}{\mathrm{d}\lambda}. $$

In this case, which is named as a maximally symmetric space in physics, the number of equations $n_K = N(N+1)/2 = 3$ is already larger than that of variables $N = 2$. Algebraic elimination gives $$ (x^2 + y^2)t - x d - s = 0, $$ which is essentially the same as coconut's answer.

An additional advantage of this method is, Killing vectors are realisation of symmetries, so that each integration constant now has a geometric meaning. Unfortunately as a physicist I am not able to express the meaning in mathematical terms.

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It is possible to write down the solution in full parametric form. It is parametrized with the intial conditions: $x(0)=x_0$, $y(0)=y_0$, $x'(0)=u_0$, $y'(0)=v_0$ and additionally the constraint $u_0^2+v_0^2=y_0^2$ is implied.

For the circle:

$x(\lambda) = x_0+\frac{u_0 y_0\sinh\lambda}{y_0\cosh\lambda-v_0\sinh\lambda}$

$y(\lambda)=\frac{y_0^2}{|y_0\cosh\lambda-v_0\sinh\lambda|}$

For $u_0=0$ it follows that $v_0=y_0$ and the straight line is recovered from the general solution:

$x(\lambda)=x_0$

$y(\lambda)=y_0 \exp(\lambda)$

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    $\begingroup$ 1. The vertical line has already been included in the former expression with $u_0 = 0$, so that $v_0 = \pm 1$ according to your constraint. 2. It might make better sense with $y'(0) = w_0$ and the constraint $(u_0^2 - w_0^2)/y_0^2 = 1$, so that the velocity vector has length $1$ with the hyperbolic metric. This is related to your convention with $v_0^2 = 1 + w_0^2 - y_0^2$. $\endgroup$
    – cmp0xff
    Apr 20, 2021 at 10:16
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    $\begingroup$ Yes, that's correct, I edited the post accordingly. However, the metric has positive signature, so its $(u_0^2+v_0^2)/y_0^2=1$. $\endgroup$
    – p6majo
    Apr 23, 2021 at 11:49

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