Finding geodesics on a cone of infinite height

Leb
Messages
87
Reaction score
0

Homework Statement


Find the geodesics on a cone of infinite height, x^{2}+y^{2} = \tan{\alpha}^{2}z^{2} using polar coordinates (x,y,z)=(r\cos{\psi},r\sin{\psi},z) with z=r\tan(\alpha)
HW.png


The Attempt at a Solution



I am not sure with how should I expres the element dz^{2} ? When it is a function of α (My calculus was always weak especially stuff with creating a derivative by dividing...)

Thanks.
 
Physics news on Phys.org
Leb said:
with ##z=r\tan(\alpha)##
The question is wrong there. They mean ##z=r\cot(\alpha)##
α is a constant, r2 = x2+y2. You can obtain an expression for dz in terms of x, y, z, dx and dy. Does that help?
 
Thanks haruspex!

I just tried to write them down like this:

dx=\cos(\psi) dr - r\sin (\psi) d\psi
dy=\sin(\psi) dr + r\cos (\psi) d\psi
dz=\cot (\alpha) dr
and
ds = \sqrt{(1+\cot^{2}(\alpha))dr^{2}+r^{2}d\psi^{2}}
taking dr^{2} out of the square root and calling the constant term as k
ds = \sqrt{k+r^{2}\frac{d\psi^{2}}{dr^{2}}}dr And now to integrate with limits from zero to infinity ? (Does not matter since we are looking for L (Lagrangian), right ?)

Update
OK, so I think I have found a solution, r_{0}=const=r\cos(\frac{\psi + C}{\sqrt{k}}) I now should do it with lagrange multipliers. Will I get the same answer up to a constant ?
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top