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How is one meant to make sense of the terms ds, dx and dy in a metric? For example the metric for hyperbolic space is $$ds^{2} = \frac{dx^{2} +dy^{2}}{y^2}$$

Given two points in the upper half plane model of hyperbolic space, how am I to use this formula to find the distance between them?

User0112358
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2 Answers2

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Given a tangent vector $v=c_x\partial_x+c_y\partial_y\in T_pH$, the hyperbolic metric evaluates to $$(ds)^2(v)=\frac{c_x^2+c_y^2}{p_y^2}$$ To find the length of a curve joining two points, you compute the integral as you usually do, but replace the Euclidean length of the tangent vector to the curve, by the length with respect to this metric.

user2520938
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Working with the metric $$ds^2=\frac{dx^2+dy^2}{y^2}.$$ The idea is that you use this formula to calculate the length of a given path $\gamma(t)=(x(t),y(t))$. As is the case with the Euclidean metric, the length of a path, $L=\int_\gamma \,ds$, connecting two points depends on the path.

For example, consider two paths connecting the points $(0,5)$, $(4,3)$. Let $\gamma_1$ be the straight line $\gamma_1:x(t)=t, y(t)=5-t/2, 0\le t\le4$. Here $dx^2+dy^2=(5/4)dt^2$, so $$ L_1=\int_{\gamma_1}\,ds=\int_{t=0}^4\frac{\sqrt5}{10-t}\,dt=\sqrt5\ln\frac53\approx 1.14224. $$ If, instead, we consider the circular arc $\gamma_2:x=5\cos t, y=5\sin t$, $\arcsin(3/5)\le t\le\pi/2$ we get $ds=5\,dt$, and $$ L_2=\int_{\gamma_2}\,ds=\int_{t=\arcsin(3/5)}^{\pi/2}\frac{dt}{\sin t}=\ln 3\approx1.09681. $$ Thus the circular path is shorter. The heuristic reason is that the circle spends less time closer to the $x$-axis, where $1/y$ grows rapidly.

It turns out that half circles perpendicular to the $x$-axes are geodesics for this metric. Undoubtedly you noticed that $\gamma_2$ is such a half circle. Therefore $L_2$ gives the shortest distance along any path connecting these two points.

Geodesics can (at least in principle) be found by solving the related calculus of variations problem of finding the path minimizing the integral. For metrics there are differential equations describing the geodesic paths (see the other WP article). Describing those and solving them for this metric is not my cup of coffee, so you need to wait for a better answer for that.

Jyrki Lahtonen
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  • In case all this was clear to you, and you really are hoping to see an explanation/proof as to why those half-circles are geodesics for the hyperbolic metric, then I apologize for wasting your time. – Jyrki Lahtonen May 25 '17 at 14:21
  • You have not wasted my time, thank you for your answer. I do wonder how you get $dx^{2} + dy^{2} = \frac{5}{4}dt^{2}$? I know how to get that; but I do things that I am told are "bad" i.e. treating dt and dx like they mean something. I geuss my question is why aren't their genuine derivatives, like $\frac{dx}{dt}$, in the metric? When one calculates a length using path integrals you have formulae involving such derivatives and not dx, dy etc. is this just notation? or am I missing something deeper? – User0112358 May 26 '17 at 01:03
  • I don't think you missed anything essential. Undoubtedly you noticed that along a path $(x(t),y(t))$ I used $ds=dt,\sqrt{x'(t)^2+y'(t)^2}/y(t)$ - just like with Euclidean metric where we would not divide by $y(t)$. To justify this I need the path to be piecewise differentiable et cetera. This can be translated into the language of manifolds. The other answerer does a better job there. – Jyrki Lahtonen May 26 '17 at 04:36
  • thanks for your answers. I was afraid that I might need to use manifolds to make complete sense of this. One day I will do that. – User0112358 May 26 '17 at 07:44