### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# MECHANICS PHYS 505

UW

GPA 3.96

### View Full Document

## 32

## 0

## Popular in Course

## Popular in Physics 2

This 9 page Class Notes was uploaded by Dr. Simeon Wiza on Wednesday September 9, 2015. The Class Notes belongs to PHYS 505 at University of Washington taught by Staff in Fall. Since its upload, it has received 32 views. For similar materials see /class/192447/phys-505-university-of-washington in Physics 2 at University of Washington.

## Popular in Physics 2

## Reviews for MECHANICS

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/09/15

Lecture 14 Appendix A The general central force problem We can apply the ideas of Lecture 14 to the general central force problem and as promised at the beginning of the course ask which central potential can yield periodic clocklike behavior We consider a very general form for the potential k Up Zpkompzkjp e am 139 f with a gt 0 and B lt 2 8 8 gt 0 for bounded motion The question is which terms in this expression can yield periodic trajectories independently of the specific initial conditions and the speci c force constants k ie we are interested in when stable periodic behavior arises This question can be stated in terms of the integral WZEZPTX L2 d p 2n pmz pz 2EVp 14A2 where we ask under what conditions does this integral yield a rational result independent of the specific values of E Lzk We can study this problem by focusing first on circular orbits where E V p0 V39p0 0 V p0 gt 0 and pp 2 0 This is a single frequency problem with w q5 2 L2 ip Next we consider small perturbations around this solution and determine when these perturbed orbits are independent of the initial conditions and force constants and then when they yield rational values in Eq 14A2 As usual with such linearized perturbation problems we will start with a harmonic oscillator problem We can simplify the expressions by choosing a rescaled variable x Lz p setting u 1 ie rescaling all energies by the reduces mass and defining a new potential L wxxU Zj 14A3 Now we are interested in the quantity Physics 505 Lecture 14 Appendix A 1 Autumn 2005 mx Q l 14A4 xmm 2E w which is the half period of a particle described by the 1D effective Hamiltonian x2 h7wxE 14A5 By construction we are near a circular orbit at x0 2 L2 p0 where x x02 wxwxow xo 2 14A6 Thus the frequency of the perturbed radial motion is 7T 7T CD 14A7 and we want to know when this quantity is independent of initial conditions and force constants Using the expression for the potential in Eq 1 4A 3 we have Physics 505 Lecture 14 Appendix A 2 Autumn 2005 x0 x0 L x U i 0L2 3U g izU g x0 x0 x0 l4A8 L UZ x0 So finally we have c1 1 1 U39p0 l4A9 W 2 2w x0 5V3U po poU po and we want to know when this expression is independent of the initial conditions and the force constants and then when is it rational The dependence on the angular momentum an initial condition and the force constant appears through the dependence on p0 So we first ask how the square root in Eq l4A9 can be independent of po This can clearly only occur when the potential is just one of the terms in Eq MA 1 ie a single power of p Since in this case the above Physics 505 Lecture 14 Appendix A 3 Autumn 2005 expression is independent of the force constant and its sign we need only consider U oc p 1 2 lt a for bounded motion and nd UlmapociljUlmaa1p0672 U39 a l gt 3U pU 3a 05 1 l a 2 14A10 l 2J0 239 So a potential with a single power can produce a winding number that is independent of the initial conditions and the force constant However the result is periodic only if W above is rational Thus the allowed powers 2 lt a are l 2 7 14 a n2 2 with n an integer gtW Next we ask what happens for larger perturbations around the circular orbit Since the result in Eq l4AlO is required to be stable with respect to changes in the energy E we might as well consider large changes such that the resulting expressions are relatively simple For the case U kp z k0 gt 0 we consider the limit E gt 00 In this limit we expect that the x2 2 term in wx will dominate at the maximum turning point x gt xmax gt oo E wxmax x3quotax 2 We can scale out this behavior with the change of variables y xxmax p L x 2 L2 yxmax Wx If E a 2a x y22k 14A11 where we expect in the E gt oo limit that ymin xmin xmax gt 0 With these de nitions of the variables for the E gt oo limit and assuming that the y2 terms dominates 2 fixed xmax gt oo we have from Eq l4A4 that Physics 505 Lecture 14 Appendix A 4 Autumn 2005 WE LT d x W 2nJ2ltE w ltxgtgt 1 1 1 1 1 I s1n y 271 0 1y2 271 0 l4A12 1 7r l 27 2 4 39 From Eq l4AlO we see that this corresponds to a 2 and means that this is the only positive exponent that gives the same winding number both at the energy of the circular orbit and at arbitrarily large energy Thus we conclude that the only potential with a positive exponent that yields periodic motion independent of the initial conditions ie the value of E is the isotropic harmonic oscillator To study the negative exponent case U kp B 0 lt B lt 2 we consider the opposite limit E gt0 We have gt zkjlZ x Note that here xmin 0 implies that p gt 00 where we can consider that the orbits close Again we use y xxmax and noting that xi 2 kx ax L i we have k 3 x2 xfm 3 2 wx gt y y x L 2 2 14A13 Physics 505 Lecture 14 Appendix A Autumn 2005 Thus the winding number is given by l X alx W 2 Ea0 277 Xiquot EQO l 1 dyy m L1 dy MixW y2 2 ex12v 1 7 Z l j dzzmi H 2712 p0 J1 z l 1 dZZ 2 6 2 1 1 1 j dz Blnl 27r2 01 Z 27T2 1 lez 7r 27r2 r1 27r2 1 22 39 Zy2 dy 14A14 Again we demand that this result match that in Eq 14A 10 which yields B 0 1 The only negative power potential that yields periodic motion for all initial conditions is the lr Kepler potential As we have claimed since the beginning of the quarter only two central potentials yield periodic behaVior independent of the details of the initial conditions and independent of the force constant k U Kepler kr and UHO kr2 2 This result is the BertrandKonigs theorem To summarize for W irrational we nd quasiperiodic behaVior and the function for Physics 505 Lecture 14 Appendix A 6 Autumn 2005 p or better it s inverse p must be an in nitely valued function For periodic behavior we have W mn and the orbit function is an mvalued function of if n is even and a 2mvalued function of if n is odd You can confirm this result by considering the various orbits illustrated in the Appendix to this lecture The multi valuedness of p is given by how many times you intersect the orbit in moving from pmin to pmax at a fixed value of For the Kepler and harmonic oscillator problem m l with n even and the orbit function is singlevalued ie analytic This analyticity arises from the eXistence of another conserved quantity which allows the problem to be solved algebraically and analytically rather than via an integral Here we discuss the Kepler problem A similar strategy works for the oscillator problem Quite generally for the central force problem we have 14A15 Since the total angular momentum is conserved for a central force problem we can write d I x Edt x E For the Kepler potential we have U kr2 so that we can rewrite Eq 14A 15 as d a a d A EltprgtykEr ji xz lkrjzo 14A16 Physics 505 Lecture 14 Appendix A 7 Autumn 2005 Thus we have a new conserved vector quantity AprIZ ykr 14A17 called the LaplaceRunge Lenz or just RungeLenz vector Note that since the angular momentum is orthogonal to the radius vector we have AZzo 14A18 The RungeLenz vector lies in the plane orthogonal to the angular momentum It has a magnitude de ned by using helpful vector identities A2gxiyng AMf xZu2 pZL2 f7I2 2ukII7gtltfa uzk2 L2 2B 2k 2H p y r y 14A19 2 214142 L uzk2 21 r 2LEL2 uzkz It is now easy to nd the orbit function in terms of this conserved quantity and the angular momentum using the form of the scalar triple product symmetric under cyclic permutations FZF1 gtltI ukrZFX ukr 14A20 LZ ukr As before we choose to define cylindrical coordinates with the zaXis aligned with E and the XaXis along the projection of 1 into the x y plane Then for motion in that plane as required by angular momentum conservation Eq l4A20 reads Physics 505 Lecture 14 Appendix A 8 Autumn 2005 pAcos L ukp L2 2 14A21 Acos uk 2p an analytic orbit function We can further simplify this expression by defining 8 quot12ELik2 Auk and p39zLiukzlC so that I P p ecos 13 1 14A22 C8cos 1 p We identify 8 as the eccentricity of the corresponding orbit E lt 0 gt 8 lt1 gt an ellipse E 0 gt 8 1gt a circle and E gt 0 gt 8 gt1 gt a hyperbola More formally the 3 components of E and the 3 components of the properly normalized vector 13 JZuE are all conserved and viewed as operators provide a representation of the 6 generators of the group SO4 rotations in a Euclidean 4D space which is the true symmetry of the Kepler problem this is a formal extension of the true 3 D configuration space The extra conserved quantity in the isotropic harmonic oscillator problem is a symmetric 3x3 tensor which has 6 independent components However the trace of this tensor correctly normalized is just the Hamiltonian So besides the Hamiltonian we have 3 E plus 5 the components of the traceless symmetric tensor ie 8 conserved quantities These 8 quantities viewed as operators provide a representation of the group SU3 the true symmetry of the 3D isotopic harmonic oscillator Physics 505 Lecture 14 Appendix A 9 Autumn 2005

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I made $350 in just two days after posting my first study guide."

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.