Geodesic on a cylinder - have I done this correctly?

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The discussion focuses on the derivation of the geodesic on a cylinder, specifically using the metric ds² = a²dθ² + dz². The user correctly identifies the Euler-Lagrange equation and integrates to find z' = B√(a² + z'²). The conclusion drawn is that z' must be a constant, leading to the final form of the geodesic as z = bθ + c. The user seeks confirmation on their approach and whether any mistakes were made in the derivation process.

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Geodesic on a cylinder - have I done this correctly??

Homework Statement



ds[tex]^{2}[/tex] = a[tex]^{2}[/tex]d[tex]\theta^{2}[/tex] + dz[tex]^{2}[/tex]

ds = [tex]\sqrt{a^{2}d\theta^{2} + dz^{2}}[/tex]

[tex]\int\sqrt{a^{2} + dz'^{2}}[/tex] d[tex]\theta[/tex] = Min

E-L equation

df/dz - d/d[tex]\theta[/tex](df/dz') = 0

df/dz = 0,

d/d[tex]\theta[/tex][[tex]\frac{z'}{\sqrt{a^{2} + z'^{2}}}[/tex]] = 0

Integrating gives:


[tex]\frac{z'}{\sqrt{a^{2} + z'^{2}}}[/tex] = B


z' = B[tex]\sqrt{a^{2} + z'^{2}}[/tex]

z = B [tex]\int\sqrt{a^{2} + z'^{2}}[/tex]

I am now stuck, I should be able to get to:

z = b[tex]\theta[/tex] + c (i think)

But I'm not sure how...

Have I made any mistakes??
 
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You don't have to integrate it. You just have to show that z'/sqrt(a^2+z'^2)=B means z'^2 must be a constant. Which in turn means z' is a constant.
 


ah yes, thank you so much!
 

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