What is the function to extremise for finding geodesics on a Helicoid?

jayzhao
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Homework Statement
Given the Cartesian coordinates for a helicoid:

$$x=\rho cos\phi$$
$$y = \rho sin\phi$$
$$z=h\phi/2\pi$$

where ##\rho\in [0,\infty)##, ##\phi\in (-\infty,\infty)##, and h>0.

Set up the variational principle to search for geodesics on the helicoid. Use ##z## as the independent variable and ##\rho(z)## as the unknown function. Find the Euler equation for the ##\rho(z)## and rewrite it in the form

$$\frac{d^{2}\rho}{dz^{2}}=F(\rho,\rho')$$
Relevant Equations
Euler equations:

$$\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)$$

For ##f(x,y(x),y'(x))##

When ##f(y(x),y'(x))## only:

$$f-\frac{\partial f}{\partial y'}y'=constant$$
I've got that length of a curve on the surface is:
$$L=\int_{-\infty}^{\infty}\sqrt{1+\frac{4\pi^{2}}{h^{2}}\rho^{2}+\left(\frac{d\rho}{dz}\right)^{2}}dx$$

So the function to extremise is:
$$f(\rho,\rho')=\sqrt{1+\frac{4\pi^{2}}{h^{2}}\rho^{2}+\left(\frac{d\rho}{dz}\right)^{2}}$$
Where ##\rho'=d\rho /dz##

But I don't know how to get this in the form
$$\frac{d^{2}\rho}{dz^{2}}=F(\rho,\rho')$$
since there doesn't seem to be a second derivative in the function anywhere?
 
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jayzhao said:
Relevant Equations:: Euler equations:
$$\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)$$When ##f(y(x),y'(x))## only:$$f-\frac{\partial f}{\partial y'}y'=constant$$

Maybe they want you to use the Euler equation ##\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)## rather than the "first-integral" equation ##f-\frac{\partial f}{\partial y'}y'= \rm {const}##.
 
Last edited:
TSny said:
Maybe they want you to use the Euler equation ##\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)## rather than the "first-integral" equation ##f-\frac{\partial f}{\partial y'}y'= \rm {const}##.
Thank you! I think you're right that's what they wanted. I was getting confused because I thought the first integral equation was a "special case" of the Euler equation but it turns out they're two different things.
 
You can always differentiate the Beltrami identity to recover the Euler-Lagrange form (as long as your problem is one-dimensional).
 
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