0

The metric tensor hyperbolic is given by $$g=4\frac{\mathrm d x^2+\mathrm d y^2}{(1-x^2-y^2)^2}.$$ I have to right it in polar. I know that the euclidien metric $\mathrm d x^2+\mathrm d y^2$ is given by $\mathrm d \rho^2+\rho^2\mathrm d \theta^2$, and thus, $$g=4\frac{\mathrm d \rho^2+\rho^2\mathrm d \theta^2}{(1-\rho^2)^2},$$ but in my solution it's written:

Since $\mathrm d x^2+\mathrm d y^2$ is given by $\mathrm d \rho^2+\rho^2\mathrm d \theta^2$, we get that $$g=\mathrm d r^2+\frac{\rho^2}{(1-\rho^2)}$$

(I think they forgot $\mathrm d \theta^2$ or $\mathrm d \rho^2$ after $\frac{\rho^2}{(1-\rho^2)^2}$, but since I don't understand the formula, I don't know what it has to be). Therefore, $$g=\mathrm d r^2+\sinh^2(r)\mathrm d \theta^2.$$

Ok, I don't understand anything of this argument. Can someone give more detail ? And why they used $\rho$ and $r$ ? Is it the same thing ? Thank you.

user330587
  • 1,624

1 Answers1

3

From $g= 4{ d\rho ^2+ \rho ^2 d \theta ^2 \over (1-\rho^2)^2}$ Let $R = \int _0^ \rho 2{1 \over (1-t^2)} dt= 2$artanh$(\rho)$, then $g=d R^2+\sinh^2 R d\theta ^2$, as ${2\tanh R/2\over (1-\tanh^2 R/2) }= \sinh R$ . Now $g=dR^2+ \sinh^2R d\theta ^2$ proves that $g$ is rotationnaly invariant and that $R$ is the distance to the origin, so these are polar coordinates.

Didier
  • 19,132
Thomas
  • 7,470
  • Thank you for the answer. Why do you set $R$ like this ? God told you ? :-) – user330587 Jun 25 '16 at 10:24
  • Not exactly (I did not gave Him my phone number). $R$ represents the length of the curve $\theta =cte, t\in [0,\rho]$, and this length is expected to be the distance to the origin, as this is the case in polar coordinates. – Thomas Jun 25 '16 at 11:43
  • great. Thank you very much, I think it's fine :-) – user330587 Jun 25 '16 at 11:49