Lagrange - Mass under potential in spherical

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Homework Statement



A particle of mass m moves in a force field whose potential in spherical coordinates is,

U = \frac{-K \cos \theta}{r^3}

where K is constant.

Identify the two constants of motion of the system.

The Attempt at a Solution



L = T - V = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2 \theta ~\dot{\phi}^2) + \frac{K \cos \theta}{r^3}

I don't see how there are two constants of motion if the Lagrangian is missing only \phi, i.e.,

\frac{ \partial L}{\partial \phi} = 0 \Rightarrow \frac{\partial L}{\partial \dot{\phi}} = constant
 
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I'm not 100% sure that this is what the questioner has in mind, but I can think of one quantity that is always a constant of motion whenever the Lagrangian has no explicit time dependence...:wink:
 
Energy function/Hamiltonian?

\frac{\partial L}{\partial t} = 0 = - \frac{dH}{dt}

So H = constant.
 
Yup.:smile:
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
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