Question

Find the geodesics on the cone z^2 = 2(x^2 + y^2)

Answer 1

challenge question

Answer 2

Mhm I will look into this later, I'm guessing it has to do with Euler - Lagrange equations heh

Answer 3

yes astro

let's do it in cylindrical

so we cheat right off the bat (!!) and pull out the arc length \(\Delta s^2 = \Delta r^2 + r^2 \Delta \theta^2 + \Delta z^2\), which is pretty much the hard bit

then we can re-write \( z^2 = 2(x^2 + y^2)\) as \(z^2 = 2r^2, z = \sqrt{2} r\), so \( \Delta z = \sqrt{2} \Delta r, \)

which makes \(\Delta s = \sqrt{\Delta r^2 + r^2 \Delta \theta^2 + 2 \Delta r^2} = \sqrt{3 \Delta r^2 + r^2 \Delta \theta^2}\)

so we get to pick an integration variable. not sure which \(\lim\limits_{\Delta ? \to 0} \dfrac{\Delta s}{\Delta ?} = \dfrac{ds}{d ?}\) to do, maybe try both to see what happens....

Answer 4

Option 1

\(\dfrac{\Delta s}{\Delta \theta} = \sqrt{3 \dfrac{\Delta r^2}{\Delta \theta^2} + r^2} \implies s[\theta] = \int\limits_{\theta_1}^{\theta_2} ~ \sqrt{3 r'^2 + r^2} ~ d \theta\)


E-L is

\(\dfrac{d}{d \theta} \dfrac{\partial F}{\partial r'} = \dfrac{\partial F}{\partial r}\)

Option 2

\(\dfrac{\Delta s}{\Delta r} = \sqrt{3 + r'^2 \dfrac{\Delta \theta^2}{\Delta r^2} } \implies s[\theta] = \int\limits_{r_1}^{r_2} ~ \sqrt{3 + r^2 \theta'^2} ~ d r\)


E-L is

\(\dfrac{d}{d r} \dfrac{\partial F}{\partial \theta'} = \dfrac{\partial F}{\partial \theta}\)

So Option 2 is the way to go as \(\dfrac{\partial F}{\partial \theta} = 0 \implies \dfrac{\partial F}{\partial \theta'} = const.\)

Option 1 may work OK with the First Integral: \(r' F_{r'} - F = const\) but which is often messy

Answer 5

From Option 1, the E-L is

\( \dfrac{r^2 \theta '} {\sqrt{3 + r^2 \theta'^2}} = \alpha \)

\(r^4 \theta'^2 = \alpha^2(3 + r^2 \theta'^2)\)

\( \theta'^2 = \dfrac{3 \alpha^2}{r^4 - \alpha r^2}\)

\(\theta(r) + \beta = \int \dfrac{\sqrt{3} \alpha}{\sqrt{ r^4 - \alpha^2 r^2}} ~ dr\)

\( = \int \dfrac{\sqrt{3} \alpha}{r \sqrt{ r^2 - \alpha^2 }} ~ dr\)

\(u = \sqrt{ r^2 - \alpha^2 }, du = \dfrac{r}{\sqrt{ r^2 - \alpha^2 }} dr\)

\(\implies \int ~ \dfrac{\sqrt{3} \alpha}{r \sqrt{ r^2 - \alpha^2 }} \dfrac{\sqrt{ r^2 - \alpha^2 }}{r} ~ du\)

\(= \int ~ \dfrac{\sqrt{3} \alpha}{r^2} ~ du = \int ~ \dfrac{\sqrt{3} \alpha}{u^2 + \alpha^2} ~ du\)

\(\theta + \beta = \sqrt{3} \tan^{-1} \dfrac{u}{\alpha} = \sqrt{3} \tan^{-1} \dfrac{\sqrt{ r^2 - \alpha^2 }}{\alpha} \)

\(\implies \tan \dfrac{\theta + \beta}{\sqrt{3}} = \dfrac{\sqrt{ r^2 - \alpha^2 }}{\alpha}\)

\(\implies \alpha^2 \tan^2 \dfrac{\theta + \beta}{\sqrt{3}} = r^2 - \alpha^2 \)

\(r^2 = \alpha^2 \sec^2 \dfrac{\theta + \beta}{\sqrt{3}} \)

\(\implies r \cos \dfrac{\theta + \beta}{\sqrt{3}} = const \)