I assume the sphere is
x= cos(theta)sin(alpha)
Y= sin(theta)sin(alpha)
Z= cos(alpha)
For a circle theta = pi/2 can you please point me in the right direction how to implement the formula.
I guess g_ij is worked out from the sphere but am not sure how to do the rest. Thanks for any help...
Can anyone help with finding the length of a circle (theta) =pi/2 (latitude 90') on the unit sphere. I know it is related to the equation
L=
integral from 0 to T of
Sqrt(g_ij (c't,c't))
The formula is on the wikipedia page called Riemannian manifold so you can get a better idea what it...
Can someone please explain to me how this formula
integral from 0 to T of
sqrt(g_ij c'(t) c'(t))
I have seen it on wikipedia but don't know how to actually implement the formula.
Does anyone have any idea how to find the length of a circle (theta)=pi/2 on the unit sphere. I am just not 100% sure on what sort of parameterisation should be used. Thanks
If we use the mapping
(r,phi)---->(x,y)=
(2tan(r/2)cos(phi),
2tan(r/2)sin(phi))
Which metric do we have to impose on R^2 in order that the mapping preserves distance. Any help would be greatly appreciated. Thank you very much
Can someone please exPlain to me what the phrase. Which metric do we have to impose in order that the mapping preserves distance means.
The example I have is
((-),phi)--->(x,y) =
(2a tan(theta/2)cos(phi) ,
2a tan(theta/2)sin(phi)) thanks
Can anyone help in provong whether or not the algorithm
r_n+1= r_n/1+sqrt(2-r_n)
is stable. I have tried using error analysis but am struggling to get the algorithm in a form that can be easily dealt with. Thanks in advance
No. In a previous example it was in R^3 and polar coordinates were used with x=rcosx and y=rsinx and r was computed from the diagram and equaled something like 2tan(pi-theta) which could be computed from the diagram by projection onto the x-axis but I'm not sure if such a formula exists in R^4
I am computing a stereographic projection in R^4 and i think i am correct in setting
x=rcos(x)sin(y)
y=rsin(x)sin(y)
z=rcos(y)
but can't see how to compute r as I do not know to visualise it graphically as was possible in R^3, any help would be greatly appreciated
So in order to prove that it is diffeomorphic all I have to do is show that (x/x^2+y^2,y/x^2*y^2) is continuous, differentiable and state that minus the origin it is it's own inverse?
I see how it is mapping all the points into a circle centred around the origin bit I'm not sure how the polar co-ords fit in. Do I just make x=rcos(theta) and y =rsin(theta). If as you said what the mapping is doing is dividing by the square of it's distance does that mean the inverse is...
Ok, I didn't think it was defined at the origin. On order to find the inverse do I just set for example u=(x/x^2+y^2) and v=(y/x^2+y^2) and then try and get x and y in terms of u and v. Thanks