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1 5 Home exercise sheet 5.1 The central force problem and Scattering Exercise 7.1: Yukawa potential The Yukawa potential is 1. Write down the effective potential. U(r) = αe χr r (1) 2. Sketch the effective potential. 3. What is the condition for a circular orbit? Solution: 1. This is the Yukawa potential which is used in particle physics, it s also called a screened Coulomb potential. It is called that because the Coulomb potential is an example of Yukawa with the mass of the particle equaling zero. We first write the effective potential U eff (r) = l2 2mr αe χr 2 r (2) 2. To sketch the effective potential it s best to check the boundaries 1

2 3. To get circular motion we need to be at the minimum, this means that we define a new parameter x = χr U eff(r) = 0 = L2 mr + αχe χr + αe χr 2 r r 2 (3) L 2 αm = r(χr + 1)e χr (4) f(x) = x(x + 1)e x (5) The condition is that this equation must have real roots. To see if this is possible we need to check the graph for f(x) and see that this is possible when f(x) is, when x > 0, maximum and χl2 is smaller than it. mα Exercise 5.2: Spherically symmetric potential well Find the motion and characterize possible trajectories of a particle of mass m in a spherically symmetric potential well V (r) where V (r) = { V for r < R 0 for r > R (6) (V > 0)) for different values of angular momentum E and total energy E a. Find the particle velocity inside and outside the potential well. 2

3 b. What is the minimum distance to the center of the potential r min for the particle with angular momentum L and total energy E which is moving inside the well? c. What is the relationship between L and E for a particle that is permanently bound inside the well? d. What is the trajectory motion of the particle with E > 0 and angular momentum L which satisfies 2 > E (large angular momentum)? e. Describe qualitatively the trajectory motion of a particle with E > 0 and small angular momentum L2 < E. Note that even though the particle can move outside of the well (E > 0) its minimum distance to the center of the potential r min < R. This means that if the particle is moving towards the center it will traverse the well. f. Describe qualitatively (make a sketch) of trajectories of particle that is permanently bound in the potential well. Solution: It is good to always draw your effective potential on a graph (makes seeing everything easier). a. Since the Lagrangian is time independent we know that the energy is a constant of motion and so Outside E = 1 2E 2 mv2 v = m Inside E = 1 2(E + V ) 2 mv2 V v = m (7) (8) 3

4 b. Let us look at the effective potential V eff = L2 V, we know that the radius is 2mr 2 minimal when V eff = E (this is easily deduced from the graph we drew at the beginning). Assuming that E > we get L2 r min = L 2m(E + V ) (9) c. A particle that is permanently bound inside the well has an energy that wont allow it to escape meaning E < L2 d. There are two possible scenarios: The particle comes from infinity and completely misses the spherical potential. The particle starts inside the potential well and is stuck in there. e. This time we assume that the particle can enter the potential well and so it will go through the potential well and head out. Notice that since the angular momentum is conserved we get mbv out mr min v in and since v in > v out this means that b > r min which is why the particle moves towards to the center and not the other way. f. The bound particle will move between r min r R 4

5 Notice that the condition for closed orbits is given by arccos( r min R ) 2π rational number. = m where m is a Exercise 5.3: Interacting particles Two point particles of mass m 1 and m 2 interact via the central potential ( ) r 2 U(r) = U 0 log r 2 + b 2 (10) where b is a constant with dimensions of length. 1. For what values of the relative angular momentum l does a circular orbit exist? Find the radius r 0 of the circular orbit. Is it stable or unstable? 2. Suppose the orbit is nearly circular, with r = r 0 + η, where η << r 0. Find the equation for the shape η(φ) of the perturbation. 3. What is the angle φ through which periapsis changes each cycle? For which value(s) of l does the perturbed orbit not precess? Solution: 5

6 1. We first need to write the effective potential U eff (r) = l2 2µr 2 + U 0 log ( r 2 r 2 + b 2 To find the circular orbit we need to find the minimum of the potential ) (11) U (r) = 0 = l2 µr + 2rU 0b 2 (12) 3 r 2 (r 2 + b 2 ) r0 2 b 2 l 2 = (13) 2µb 2 U 0 l 2 The condition on l is born from the condition on r 0 which is r 2 0 > 0 which means that l < l c 2µb 2 U 0 (14) To check if the orbit is stable or unstable we just need to check if the point is a minimum or maximum. There are two ways: (a) Either take the second derivative and check to see if it s positive or negative. (b) Draw the effective potential and check if we need a maximum or minimum. The second way is easier so that s what we ll do. Since U eff (0 + ) = and U eff ( ) = 0, this means that if there s a circular orbit it will be stable. 2. To find the shape of η(φ) we will start in the same way that we would for a regular orbit. E = 1 ( ) 2 2 µṙ2 + U eff (r) = l2 dr + U eff(r) (15) 2µr 4 dφ The transition is done by using l = mr 2 φ. We can now plug in r = r0 + η and differentiate E with respect to φ, which will give us For our potential we get Which has the trivial solution η = β 2 η where β 2 = µr4 0 l 2 U eff(r 0 ) (16) ) β 2 = 2 (1 l2 l 2 c (17) η(φ) = A cos (βφ + δ) (18) 6

7 3. The amount by which periapsis changes each cycle is given by φ = φ n+1 φ n 2π (19) We now need to check what is φ n. Setting η = η 0 we obtain the sequence of φ values φ n = δ 0 + 2πn β (20) This we plug into φ to receive φ = 2π ( β 1 1 ) (21) If β > 1 then φ < 0 and the periapsis advances each cycle (i.e. it comes sooner with every cycle). If β < 1 then φ > 0 and the periapsis recedes. For β = 1 which means that l = µb 2 U 0 there is no precession and φ = 0. 7