# Simple spring in 3d space

1. Oct 22, 2014

### diegzumillo

1. The problem statement, all variables and given/known data
This is a pretty simple system but I'm supposed to use the methods described in Landau (equations below), which are integrables derived from energy conservation and angular momentum conservation.

2. Relevant equations
$$T=2∫_{r_{min}}^{r_{max}} \frac {dr}{\sqrt{\frac{2}{\mu}(E-\frac{k}{2}r²)-\frac{L^2}{\mu ^2r^2}}}$$
$$\Delta \phi=∫_{r_{min}}^{r_{max}} \frac {Ldr/r^2}{\sqrt{2\mu (E-\frac{k}{2}r²)-\frac{L^2}{r^2}}}$$

3. The attempt at a solution
I haven't tried the second integral yet but the first one is giving me a headache so I decided to ask for some guidance. My best solution so far (I had more, who knows which one is right, if any) is
$$T=\sqrt{\frac{\mu}{k}}coth^{-1}\sqrt{\frac{\mu E^2-L^2}{kL^2+\mu E^2}}$$
This is weird. it's the time it takes from $r_{min}$ to $r_{max}$, it should be $2\pi\sqrt{\frac{\mu }{k}}$, right? And it's not well defined for every value of E and L, here's a plot:
http://i.minus.com/ievUWK70xYyxT.png [Broken]
Makes sense, the energy cannot go below the minimum of the effective potential, which depends on the angular momentum. Well, this may be a consequence of being sleep deprived but I can't wrap my head around this.
Intuitively I know what's happening, I know this is an integrable system but it can also have a dense phase space, never closing on itself. I thought that maybe the complexity of the integral was a reflex of that. I would buy all that, the dependence of the period with the angular momentum etc. If it weren't for the fact that I can solve this system in a much simpler way by writing the Lagrangian in cartesian coordinates, where I can clearly see the frequency and period and they are much simpler than that.

So, why are these two methods giving me different results?

Last edited by a moderator: May 7, 2017
2. Oct 23, 2014

### dslowik

I still don't completely understand the energy / angular momentum relation along T=0,
but,
When L=0, you do get ur expected period, since coth(2*pi) = 1 there.
How is the rotating problem simpler in Cartesian coords?

3. Oct 23, 2014

### diegzumillo

That result is wrong! I remade the integral and got the right result.
In cartesian coordinates you can solve the system straight from the Lagrangian, the solution is simply
$x=A cos(\omega t +\phi_x)$
and equivalently for y.
So we know the period. The $\Delta \phi$ is slightly more complicated to interpretate in this particular case, usually we have a curve similar to an ellipse but it could have arms and might not even be closed, delta phi gives us this shift. In this problem the delta phi is pi. Sorry I can't give more details, I'm kind of in a hurry (got an extra 24 hours to solve the problem set :P)

4. Oct 23, 2014

### dslowik

Yes, I just got out of the shower where I also realized that result must be wrong, mainly because coth(2*pi) does not equal 1! (2*pi was close enuf to infinity for me and my calculator to conclude that odd result.) thus my struggle to make sense of the T=0 relation! And, yes, I now see the Cartesian Lagrangian would just give two separate simple harmonic oscillators with solutions as you mentioned (L&L sec 23 problem 3), and the path is closed for the space oscillator here as mentioned in L&L sec 14.