How to Find Geodesics on a Curved Surface Using Calculus of Variations

[matpo39]

hi, we have just got to the point in my physics course where Newtons laws are now longer that easy to work with anymore and we are now beginning to reformulate those using variational methods, and I am a little confused on one of the problems.
The shortest path between two points on a curved surface, such as the surface of a sphere is called a geodesic. To find a geodestic, one has first to set up an integral that gives the length of a path on the surface in question. Use sperical polar coordinates (r,theat,phi) to show that the length of a path joining two points on a shere of radius R is L=R*integral(from theta_1 to theta_2)sqrt(1+sin^2(theta)*(phi_prime(theta))^2)*d(theta)

if (theta_1,phi_1) and (theta_2,phi_2) specify two points and we assume that the path is expressed as phi=phi(theta).

I know how to do this problem if it were just (x,y) it would be L=integral(from x_1 to x_2) ds, where ds= sqrt(dx^2+dy^2)= sqrt(1+y_prime(x)^2) dx
but I am getting confused on how to implement the spherical polar cords. For example x=r*sin(phi)*cos(theta), but then i don't know what dx would be because I am not sure what to differantiat?

thanks for the help.

 

[ReyChiquito]

use chain rule...

dx=\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \phi}d\phi+\frac{\partial x}{\partial \theta}d\theta

same with dy

 

[matpo39]

ok i used the chain rule and came up with these values for my dx and dy

dx= cos(theta)sin(phi)*dr+r*cos(theta)cos(phi)*d(phi) - r*sin(theta)sin(phi)*d(theta)

dy= sin(theta)cos(phi)*dr-r*sin(theta)sin(phi)*d(phi)+r*cos(theta)cos(phi)*d(phi)

I then put those dy,dx values into the equation

ds=Sqrt(dx^2+dy^2) but i still can't get it to look like

L= R integral(from theta_1 to theta_2) (sqrt(1+sin^2(theta)*phi_prime(theta)^2)*d(theta)

also how would the dr,d(phi) work themselves out so all that would be left is the d(theta)?

 

[ReyChiquito]

Ok.. first of all, you are in 3 dimentions so the length of a curve is given by

L=\int_{t_{1}}^{t_{2}}{\sqrt{{dx}^2+{dy}^2+{dz}^2}}

now, given the spherical coordinates

x=r\cos\phi\sin\theta
y=r\sin\phi\sin\theta
z=r\cos\theta

you have that

{dx}^2+{dy}^2+{dz}^2={dr}^2+r^2\sin^2\theta{d\phi}^2+r^2{d\theta}^2

but you are over a sphere right? so dr=0 and r=R

using all this we have

L=R\int_{\theta_{1}}^{\theta_{2}}{\sqrt{\sin^2\theta{d\phi}^2+{d\theta}^2}}=R\int_{\theta_{1}}^{\theta_{2}}{\sqrt{\sin^2\theta{\left(\frac{d\phi}{d\theta}\right)}^2{d\theta}^2+{d\theta}^2}}

so

L=R\int_{\theta_{1}}^{\theta_{2}}{\sqrt{1+\sin^2\theta{\left(\frac{d\phi}{d\theta}\right)}^2}d\theta}