Poincaré disk construction












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$begingroup$


I am trying to understand how the Poincaré disk is constructed using the stereographic projection for the hyperboloid $x^2+y^2-z^2=-1$.



So I want to project a line from the fixed point $(0,0,-1)$ to a point on the hyperboloid.



The parametric equation of a line between $(0,0,-1)$ and a point $(x,y,z)$ is given by $vec{r}(t)=t(x,y,z)+(1-t)(0,0,-1)=(tx,ty,tz+t-1)$, where $t in [0,1]$.



Now I want to find where this line intersects the hyperboloid. So



$(tx)^2+(ty)^2-(tz+t-1)^2=-1 iff$ $(x^2+y^2-z^2-2z-1)t^2+(2z+2)t=0$



Now $x^2+y^2-z^2=-1$ because I want to find the intersection. So I get the equation



$-(2z+2)t^2+(2z+2)t=0,$



whose only solutions are $t=0$ or $t=1$.



Shouldn't I obtain an expression for $t$ in terms of $x,y,z$?






hyperbolic-geometry







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    0












    $begingroup$


    I am trying to understand how the Poincaré disk is constructed using the stereographic projection for the hyperboloid $x^2+y^2-z^2=-1$.



    So I want to project a line from the fixed point $(0,0,-1)$ to a point on the hyperboloid.



    The parametric equation of a line between $(0,0,-1)$ and a point $(x,y,z)$ is given by $vec{r}(t)=t(x,y,z)+(1-t)(0,0,-1)=(tx,ty,tz+t-1)$, where $t in [0,1]$.



    Now I want to find where this line intersects the hyperboloid. So



    $(tx)^2+(ty)^2-(tz+t-1)^2=-1 iff$ $(x^2+y^2-z^2-2z-1)t^2+(2z+2)t=0$



    Now $x^2+y^2-z^2=-1$ because I want to find the intersection. So I get the equation



    $-(2z+2)t^2+(2z+2)t=0,$



    whose only solutions are $t=0$ or $t=1$.



    Shouldn't I obtain an expression for $t$ in terms of $x,y,z$?






    hyperbolic-geometry







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am trying to understand how the Poincaré disk is constructed using the stereographic projection for the hyperboloid $x^2+y^2-z^2=-1$.



      So I want to project a line from the fixed point $(0,0,-1)$ to a point on the hyperboloid.



      The parametric equation of a line between $(0,0,-1)$ and a point $(x,y,z)$ is given by $vec{r}(t)=t(x,y,z)+(1-t)(0,0,-1)=(tx,ty,tz+t-1)$, where $t in [0,1]$.



      Now I want to find where this line intersects the hyperboloid. So



      $(tx)^2+(ty)^2-(tz+t-1)^2=-1 iff$ $(x^2+y^2-z^2-2z-1)t^2+(2z+2)t=0$



      Now $x^2+y^2-z^2=-1$ because I want to find the intersection. So I get the equation



      $-(2z+2)t^2+(2z+2)t=0,$



      whose only solutions are $t=0$ or $t=1$.



      Shouldn't I obtain an expression for $t$ in terms of $x,y,z$?






      hyperbolic-geometry







      share|cite|improve this question











      $endgroup$




      I am trying to understand how the Poincaré disk is constructed using the stereographic projection for the hyperboloid $x^2+y^2-z^2=-1$.



      So I want to project a line from the fixed point $(0,0,-1)$ to a point on the hyperboloid.



      The parametric equation of a line between $(0,0,-1)$ and a point $(x,y,z)$ is given by $vec{r}(t)=t(x,y,z)+(1-t)(0,0,-1)=(tx,ty,tz+t-1)$, where $t in [0,1]$.



      Now I want to find where this line intersects the hyperboloid. So



      $(tx)^2+(ty)^2-(tz+t-1)^2=-1 iff$ $(x^2+y^2-z^2-2z-1)t^2+(2z+2)t=0$



      Now $x^2+y^2-z^2=-1$ because I want to find the intersection. So I get the equation



      $-(2z+2)t^2+(2z+2)t=0,$



      whose only solutions are $t=0$ or $t=1$.



      Shouldn't I obtain an expression for $t$ in terms of $x,y,z$?






      hyperbolic-geometry




      hyperbolic-geometry






      share|cite|improve this question















      share|cite|improve this question













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      edited Jan 8 at 21:53









      Bernard

      120k740113




      120k740113










      asked Jan 8 at 21:51









      vladr10vladr10

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          $begingroup$

          You should be projecting from the hyperboloid (i.e., a point such that $x^2+y^2-z^2=-1$) to the plane $z=0$. What you are doing now is you are projecting the hyperboloid ($x^2+y^2-z^2=-1$) to itself ($x^2+y^2-z^2=-1$), so that is why nothing changes (for $t=1$).



          Now I want to find where this line intersects the hyperboloid => Now I want to find where this line intersects the plane $z=0$.



          The correct $t$ is given by $tz+t-1=0$.






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            You should be projecting from the hyperboloid (i.e., a point such that $x^2+y^2-z^2=-1$) to the plane $z=0$. What you are doing now is you are projecting the hyperboloid ($x^2+y^2-z^2=-1$) to itself ($x^2+y^2-z^2=-1$), so that is why nothing changes (for $t=1$).



            Now I want to find where this line intersects the hyperboloid => Now I want to find where this line intersects the plane $z=0$.



            The correct $t$ is given by $tz+t-1=0$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              You should be projecting from the hyperboloid (i.e., a point such that $x^2+y^2-z^2=-1$) to the plane $z=0$. What you are doing now is you are projecting the hyperboloid ($x^2+y^2-z^2=-1$) to itself ($x^2+y^2-z^2=-1$), so that is why nothing changes (for $t=1$).



              Now I want to find where this line intersects the hyperboloid => Now I want to find where this line intersects the plane $z=0$.



              The correct $t$ is given by $tz+t-1=0$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                You should be projecting from the hyperboloid (i.e., a point such that $x^2+y^2-z^2=-1$) to the plane $z=0$. What you are doing now is you are projecting the hyperboloid ($x^2+y^2-z^2=-1$) to itself ($x^2+y^2-z^2=-1$), so that is why nothing changes (for $t=1$).



                Now I want to find where this line intersects the hyperboloid => Now I want to find where this line intersects the plane $z=0$.



                The correct $t$ is given by $tz+t-1=0$.






                share|cite|improve this answer









                $endgroup$



                You should be projecting from the hyperboloid (i.e., a point such that $x^2+y^2-z^2=-1$) to the plane $z=0$. What you are doing now is you are projecting the hyperboloid ($x^2+y^2-z^2=-1$) to itself ($x^2+y^2-z^2=-1$), so that is why nothing changes (for $t=1$).



                Now I want to find where this line intersects the hyperboloid => Now I want to find where this line intersects the plane $z=0$.



                The correct $t$ is given by $tz+t-1=0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 9 at 1:32









                Zeno RogueZeno Rogue

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