Precession in a central potential

noospace
Messages
69
Reaction score
0

Homework Statement



Consider the following parametrization of an orbit in polar form,

\ell u = 1 + e \cos[(\phi -\phi_0)\Gamma]

where u = 1/r.

I'm trying to find the shift in the angular position of the periapsis after one complete orbit.

The Attempt at a Solution



Choose axes so that the point of first closest approach is \phi_0.

u'(\phi) = - \Gamma e \sin[(\phi -\phi_0)\Gamma]

Setting u'(0) =0 we obtain

(\phi -\phi_0)\Gamma = n \pi where n is an integer.

So after one complete orbit I guess the shift is \Delta \phi = \phi - \phi_0 = \frac{\pi}{\Gamma}, or should that be \frac{2pi}{\Gamma}?

Thanks.
 
Physics news on Phys.org
I preffer this:

\frac{2pi}{\Gamma}-\frac{2pi}{1}

To solve this problem, you even don't need to consider the exact trajectory.
You simply need to check how much (an angle) the trajectory deviates from periodicity.
For \Gamma = 1 the trajectory is periodic and there is no shift.
 
How did you obtain 2Pi/Gamma - 2Pi?
 
On reflection I get 2\pi/\Gamma. I think the first perihelion occurs at 0, the first aphelion occurs at \pi/\Gamma and the second perihelion at 2\pi/\Gamma.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top