Computing the length of a hyperbolic circle
I have a problem with the proof of the following exercise.
Define the circle of radius $r gt 0$ around a point c $ in $ $H^2$ as
$C_r(c)= ; {x in H^2 mid d(c,x)=r}$
where d($cdot$,$cdot$) denotes the hyperbolic distance. Find the length of $C_r(c)$.
My way to solve this was as follows:
Consider the hyperboloid model.
Let ($cdot$,$cdot$) denote the usual euclidean scalar product.
Let $langle,cdot,cdotrangle$ denote the Lorentz scalar product. Let O(2,1) denote the Lorentz Group.
It is sufficient to consider circles with center c=(0,0,1). Otherwise we apply a suitable transformation T $in$ O(2,1). Then $C_r(c)$ can be written as $C_r(c)=H^2 cap E$ where $E= ; {x in H^2 mid langle,c,xrangle=-cosh(r)}$ is a plane.
Consider the line $L= ; {lambda c in H^2 mid lambda in IR}$. Define m:=$L cap E$ Then m=$lambda c$ for some $lambda in IR$. It follows
$-cosh(r)=langle,c,mrangle=langle,c,lambda crangle=lambda langle,c,crangle=-lambda$.
Then we get for x $in C_r(c)$ that
$||m-x||^2=(m-x,m-x)$
$=(lambda c-x,lambda c-x)$
$=cosh^2(r) cdot (c,c)-2 cosh(r) cdot (c,x)+(x,x)$
$=cosh^2(r) cdot langle,c,crangle-2 cosh(r) cdot langle,c,xrangle+langle,x,xrangle$
$=-cosh^2(r)+2cosh^2(r)-1=cosh^2(r)-1$
$=sinh^2(r)$.
Hence the circle $C_r(c)$ can be identified with the circle defined by the equation
$||m-x||=|sinh(r)|$ (1).
Since (1) defines a euclidean circle the length must be $2pi cdot sinh(r)$.
While the result is correct I am in doubt whether or not my arguments hold.
The problem is that this proof assumes that there is an argument that the euclidean scalar product and the Lorentz scalar product can be identified in the setup I have created. Can this somehow be proven and my proof be salvaged or should I try something completely different?
hyperbolic-geometry
asked Nov 22 '18 at 22:21
PolymorphPolymorph
1186
add a comment |
I have a problem with the proof of the following exercise.
Define the circle of radius $r gt 0$ around a point c $ in $ $H^2$ as
$C_r(c)= ; {x in H^2 mid d(c,x)=r}$
where d($cdot$,$cdot$) denotes the hyperbolic distance. Find the length of $C_r(c)$.
My way to solve this was as follows:
Consider the hyperboloid model.
Let ($cdot$,$cdot$) denote the usual euclidean scalar product.
Let $langle,cdot,cdotrangle$ denote the Lorentz scalar product. Let O(2,1) denote the Lorentz Group.
It is sufficient to consider circles with center c=(0,0,1). Otherwise we apply a suitable transformation T $in$ O(2,1). Then $C_r(c)$ can be written as $C_r(c)=H^2 cap E$ where $E= ; {x in H^2 mid langle,c,xrangle=-cosh(r)}$ is a plane.
Consider the line $L= ; {lambda c in H^2 mid lambda in IR}$. Define m:=$L cap E$ Then m=$lambda c$ for some $lambda in IR$. It follows
$-cosh(r)=langle,c,mrangle=langle,c,lambda crangle=lambda langle,c,crangle=-lambda$.
Then we get for x $in C_r(c)$ that
$||m-x||^2=(m-x,m-x)$
$=(lambda c-x,lambda c-x)$
$=cosh^2(r) cdot (c,c)-2 cosh(r) cdot (c,x)+(x,x)$
$=cosh^2(r) cdot langle,c,crangle-2 cosh(r) cdot langle,c,xrangle+langle,x,xrangle$
$=-cosh^2(r)+2cosh^2(r)-1=cosh^2(r)-1$
$=sinh^2(r)$.
Hence the circle $C_r(c)$ can be identified with the circle defined by the equation
$||m-x||=|sinh(r)|$ (1).
Since (1) defines a euclidean circle the length must be $2pi cdot sinh(r)$.
While the result is correct I am in doubt whether or not my arguments hold.
The problem is that this proof assumes that there is an argument that the euclidean scalar product and the Lorentz scalar product can be identified in the setup I have created. Can this somehow be proven and my proof be salvaged or should I try something completely different?
hyperbolic-geometry
asked Nov 22 '18 at 22:21
PolymorphPolymorph
1186
add a comment |
I have a problem with the proof of the following exercise.
Define the circle of radius $r gt 0$ around a point c $ in $ $H^2$ as
$C_r(c)= ; {x in H^2 mid d(c,x)=r}$
where d($cdot$,$cdot$) denotes the hyperbolic distance. Find the length of $C_r(c)$.
My way to solve this was as follows:
Consider the hyperboloid model.
Let ($cdot$,$cdot$) denote the usual euclidean scalar product.
Let $langle,cdot,cdotrangle$ denote the Lorentz scalar product. Let O(2,1) denote the Lorentz Group.
It is sufficient to consider circles with center c=(0,0,1). Otherwise we apply a suitable transformation T $in$ O(2,1). Then $C_r(c)$ can be written as $C_r(c)=H^2 cap E$ where $E= ; {x in H^2 mid langle,c,xrangle=-cosh(r)}$ is a plane.
Consider the line $L= ; {lambda c in H^2 mid lambda in IR}$. Define m:=$L cap E$ Then m=$lambda c$ for some $lambda in IR$. It follows
$-cosh(r)=langle,c,mrangle=langle,c,lambda crangle=lambda langle,c,crangle=-lambda$.
Then we get for x $in C_r(c)$ that
$||m-x||^2=(m-x,m-x)$
$=(lambda c-x,lambda c-x)$
$=cosh^2(r) cdot (c,c)-2 cosh(r) cdot (c,x)+(x,x)$
$=cosh^2(r) cdot langle,c,crangle-2 cosh(r) cdot langle,c,xrangle+langle,x,xrangle$
$=-cosh^2(r)+2cosh^2(r)-1=cosh^2(r)-1$
$=sinh^2(r)$.
Hence the circle $C_r(c)$ can be identified with the circle defined by the equation
$||m-x||=|sinh(r)|$ (1).
Since (1) defines a euclidean circle the length must be $2pi cdot sinh(r)$.
While the result is correct I am in doubt whether or not my arguments hold.
The problem is that this proof assumes that there is an argument that the euclidean scalar product and the Lorentz scalar product can be identified in the setup I have created. Can this somehow be proven and my proof be salvaged or should I try something completely different?
hyperbolic-geometry
asked Nov 22 '18 at 22:21
PolymorphPolymorph
1186
I have a problem with the proof of the following exercise.
Define the circle of radius $r gt 0$ around a point c $ in $ $H^2$ as
$C_r(c)= ; {x in H^2 mid d(c,x)=r}$
where d($cdot$,$cdot$) denotes the hyperbolic distance. Find the length of $C_r(c)$.
My way to solve this was as follows:
Consider the hyperboloid model.
Let ($cdot$,$cdot$) denote the usual euclidean scalar product.
Let $langle,cdot,cdotrangle$ denote the Lorentz scalar product. Let O(2,1) denote the Lorentz Group.
It is sufficient to consider circles with center c=(0,0,1). Otherwise we apply a suitable transformation T $in$ O(2,1). Then $C_r(c)$ can be written as $C_r(c)=H^2 cap E$ where $E= ; {x in H^2 mid langle,c,xrangle=-cosh(r)}$ is a plane.
Consider the line $L= ; {lambda c in H^2 mid lambda in IR}$. Define m:=$L cap E$ Then m=$lambda c$ for some $lambda in IR$. It follows
$-cosh(r)=langle,c,mrangle=langle,c,lambda crangle=lambda langle,c,crangle=-lambda$.
Then we get for x $in C_r(c)$ that
$||m-x||^2=(m-x,m-x)$
$=(lambda c-x,lambda c-x)$
$=cosh^2(r) cdot (c,c)-2 cosh(r) cdot (c,x)+(x,x)$
$=cosh^2(r) cdot langle,c,crangle-2 cosh(r) cdot langle,c,xrangle+langle,x,xrangle$
$=-cosh^2(r)+2cosh^2(r)-1=cosh^2(r)-1$
$=sinh^2(r)$.
Hence the circle $C_r(c)$ can be identified with the circle defined by the equation
$||m-x||=|sinh(r)|$ (1).
Since (1) defines a euclidean circle the length must be $2pi cdot sinh(r)$.
While the result is correct I am in doubt whether or not my arguments hold.
The problem is that this proof assumes that there is an argument that the euclidean scalar product and the Lorentz scalar product can be identified in the setup I have created. Can this somehow be proven and my proof be salvaged or should I try something completely different?
hyperbolic-geometry
hyperbolic-geometry
asked Nov 22 '18 at 22:21
PolymorphPolymorph
1186
asked Nov 22 '18 at 22:21
PolymorphPolymorph
1186
asked Nov 22 '18 at 22:21
PolymorphPolymorph
1186
asked Nov 22 '18 at 22:21
PolymorphPolymorph
1186
asked Nov 22 '18 at 22:21
PolymorphPolymorph
1186
1186
add a comment |
add a comment |
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