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

jayzhao
Messages
2
Reaction score
2
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?
 
Physics news on Phys.org
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).
 
  • Like
Likes vanhees71 and PhDeezNutz
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
Back
Top