Finding geodesics on a cone of infinite height

  • Thread starter Leb
  • Start date
  • #1
Leb
94
0

Homework Statement


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


The Attempt at a Solution



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

Thanks.
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,859
6,127
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?
 
  • #3
Leb
94
0
Thanks haruspex!

I just tried to write them down like this:

[itex]dx=\cos(\psi) dr - r\sin (\psi) d\psi[/itex]
[itex]dy=\sin(\psi) dr + r\cos (\psi) d\psi[/itex]
[itex]dz=\cot (\alpha) dr[/itex]
and
[itex]ds = \sqrt{(1+\cot^{2}(\alpha))dr^{2}+r^{2}d\psi^{2}}[/itex]
taking dr^{2} out of the square root and calling the constant term as k
[itex]ds = \sqrt{k+r^{2}\frac{d\psi^{2}}{dr^{2}}}dr[/itex]


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, [itex]r_{0}=const=r\cos(\frac{\psi + C}{\sqrt{k}})[/itex] I now should do it with lagrange multipliers. Will I get the same answer up to a constant ?
 
Last edited:

Related Threads on Finding geodesics on a cone of infinite height

  • Last Post
Replies
0
Views
3K
Replies
2
Views
3K
Replies
2
Views
10K
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
7
Views
3K
Replies
0
Views
1K
Replies
4
Views
606
Replies
3
Views
960
  • Last Post
Replies
10
Views
11K
Replies
5
Views
6K
Top