Kepler problem in parabolic coordinates

Thread starter Kate_12
Start date Jun 18, 2021
Tags

Coordinates Kepler

Jun 18, 2021

Kate_12

Homework Statement: kepler problem
H=1/2m(px^2+py^2+pz^2)-k/(x^2+y^2+z^2)^1/2
with parabolic coordinates (a,b,c)
x=sqrt(ab)cos c
y=sqrt(ab)sin c
z=(a-b)/2
1) rewrite H as a function of new canonical variables (a,b,c, pa,pb,pc)
2) Hamilton-Jacobi equation in this coordinate system turns out to be completely separable. Using the Ansatz S=Wa(a)+Wb(b)+Wc(c)-Et, write the partial differential equation for each Wa, Wb, Wc with suitable separation constants.

Relevant Equations: Hamilton Jacobi equation

I solve (1).
But to solve (2), What should be the suitable separation constants?
I am so confused...

E=2/(m*(a+b)) * (a*(dWa/da)^2+b*(dWb/db)^2-k)+l^2/(2mab)
where l(constant) is pc since c is cyclic.

What should I do to solve the problem?

Physics news on Phys.org

Jun 20, 2021

anuttarasammyak

Gold Member

if you do not mind could you write down H of your solution 1 ?

Thread 'Please tell me where I am going wrong in this integral'

##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...

View full post »

Thread 'Number of quantum accessible states of a particle given T, N?'

It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...

View full post »