Homework 5 # 3.31, 3.32, 3.7a

Homework 5: # 3.31, 3.32, 3.7a 3.7a Show that the angle of recoil of the target particle relative to the incident direction of the scattered particle is simply Φ = 1 (π − Θ). 2 Answer: It helps to draw a figure for this problem. I don’t yet know how to do this AT X, but I do know that in the center of mass frame both the particles in L E momentum are equal. m1 v ′1 = m2 v ′ 2 Where the prime indicates the CM frame. If you take equation (3.2) Goldstein, then its easy to understand the equation after (3.110) for the relationship of the relative speed v after the collision to the speed in the CM system. v1′ = µ m2 v= v m1 m1 + m2 Here, v is the relative speed after the collision, but as Goldstein mentions because elastic collisions conserve kinetic energy, (I’m assuming this collision is elastic even though it wasn’t explicitly stated), we have v = v0 , that is the relative speed after collision is equal to the initial velocity of the first particle in the laboratory frame ( the target particle being stationary). v ′1 = m2 v m1 + m2 0 This equation works the same way for 2 v ′ v ′2 = m1 v0 m1 + m2 From conservation of momentum, we know that the total momentum in the CM frame is equal to the incident(and thus total) momentum in the laboratory frame. (m1 + m2 )vcm = m1 v0 We see 1 vcm = m1 v0 m1 + m2 This is the same as 2v ′ v2′ = vcm If we draw both frames in the same diagram, we can see an isosceles triangle ′ where the two equal sides are 2vand vcm . Φ + Φ + Θ=π 1 Φ = (π − Θ) 2 3.31 Examine the scattering produced by a repulsive central force f + kr −3 . Show that the differential cross section is given by σ(Θ)dΘ = k (1 − x)dx 2E x2 (2 − −x)2 sin πx where x is the ratio of Θ/π and E is the energy. Answer: The differential cross section is given by Goldstein (3.93): ds       dΘ  sin Θsolve We must solve for s, and ds/dΘ. Lets for Θ(s) first, take its derivative with respect to s, and invert it to find ds/dΘ. We can solve for Θ(s) by using Goldstein (3.96): σ(Θ) = Z Θ(s) = π − 2 ∞ rm s sdr q ) − s2 E r r2 (1 − V (r) What is V (r) for our central force of f = k/r 3 ? It’s found from −dV /dr = f . V (r) = 2 k 2r2 Plug this in to Θ and we have Z sdr ∞ rm ) q 2E r r2 − (s2 + k Before taking this integral, I’d like to put it in a better form. If we look at the energy of the incoming particle, Θ(s) = π − 2 E= 1 2 mr 2m θ̇2 + s2 E k k = 2 + 2 2r2m rm 2r m where from Goldstein page 113, 2s2 E θ˙2 = mr 4m We can solve for s2 +2Ek, the term in Θ, k 2E Now we are in a better position to integrate, r2m = s2 + ∞ s 1 rm 1 π  ] = π−2s ( ) = π(1− cos−1 rm rm rm 2 m  Θ(s) = π−2 p q r  r r2 − r2 s2 + Goldstein gave us x = Θ/π, so now we have an expression for x in terms of s, lets solve for s Z ∞ rm sdr = π−2s [ x= Θ =1 − π s q s2 + k s2 = ( s2 + )(1 − x)2 2E → k 2E s2 = k (1 2E − x)2 1 − (1 − x)2 (1 − x) r k 2E s= p (2 − x)to find the cross section. Solving Now that we have s we need only xds/dΘ dΘ/ds and then taking the inverse, 1 k −3 dΘ ) 2 )2s + = πs(− (s2 + 2E ds 2 2 2 k dΘ −πs + π(s + = 3 (s2 + k2E ) 2 ds So 3 ) 2E π q 2E s2πk+ = (s2 + k 2E 3 k )2 2E ) 2E k k ds 2E(s2 + 2E ) = dΘ πk Putting everything in terms of x, s2 + 3 2 k (1 − x)2 k k 1 k = + = 2E 2E 2E x(2 − x) 2E x(2 − x) So now, (1−x) 3 q 2E √ q 2E(s2 + k2E ) 2 ds x(2−x) k     sin πx  = = dΘ  πk And sin thisΘmost beautiful expression becomes.. σ(Θ) = s σ(Θ) = 1 1 k 1 2E k 3 ) 2( )2 ( )( 2E sin πx π 2E k After a bit more algebra... σ(Θ) = 1−x p 2E k √ (1−x) k 2E( 2E x(2−x) sin πx 1 )3 x(2−x) 2 πk = 1 (x(2 − x)) 3 2 x(2 − x) k 1 1 1−x 2E π sin πx (x(2 − x))2 And since we know dΘ = πdx, σ(Θ)dΘ = k (1 − x)dx 2E x2 (2 − x)2 sin πx 3.32 A central force potential frequently encountered in nuclear physics is the rectangular well, defined by the potential V = 0 V = −V0 r> a r≤a Show that the scattering produced by such a potential in classical mechanics is identical with the refraction of light rays by a sphere of radius a and relative index of refraction r E + VE0 n= This equivalence demonstrates why it was possible to explain refraction phenomena both by Huygen’s waves and by Newton’s mechanical corpuscles. Show also that the differential cross section is σ(Θ) = n2 a2 (n cos Θ2 − 1)(n − cos Θ 2 ) 4 cos 2Θ (1 + n2 − 2n cos Θ2 )2 What is the total cross section? 4 Answer: Ignoring the first part of the problem, and just solving for the differential cross section, sds sin ΘdΘ If the scattering is the same as light refracted from a sphere, then putting our total angle scattered, Θ, in terms of the angle of incidence and transmission, σ(Θ) = Θ = 2(θ1 − θ2 ) This is because the light is refracted from its horizontal direction twice, after hitting the sphere and leaving the sphere. Where θ1 − θ2 is the angle south of east for one refraction. We know sin θ1 = s/a and using Snell’s law, we know n= sin θ1 s → sin θ2 = na sin θ2 Expressing Θ in terms of just s and a we have s s − arcsin ) a na 2 2 Now the plan is, to solve for s and then ds /dΘ and solve for the cross section via Θ = 2(arcsin σ= sds ds2 1 1 = = sin ΘdΘ 2 sin ΘdΘ 4 sin 2Θ cos Here goes. Solve for sin sin Θ 2 and cos Θ 2 Θ 2 ds2 dΘ in terms of s Θ s s s s s s = sin(arcsin −arcsin ) = sin arcsin cos arcsin −cos arcsin sin arcsin na 2 a na a na a This is s r r ) s2 a2 na s2 cos(arccos 1 − ) − cos(arccos( 1 − √ n 2 a2 a Using arcsin x = arccos 1 − x2 and sin(a − b) = sin a cos b − cos a sin b. Now we have = s s Θ = p p 2 ( n 2 a2 − s 2 ) − a2 − s 2 ) na2 Θ yields Doing the same thing for cos 2 sin 5 Θ 1 = p p 2 ( a2 − s 2 n 2 a2 − s 2 + s 2 ) na2 b + sin a sin b. Still solving for s2 in terms of cos Using cos(a − b) = cos a cos and sin’s we proceed cos s2 Θ = p p 2 (n2 a2 − s2 − 2 n2 a2 − s2 a2 − s2 + a2 − s2 ) n 2 a4 sin2 This is sin2 Note that s2 2 2s4 2s2 (n + 1) − 2 4 − p p n 2 a2 n a n 2 a2 − s 2 a2 − s 2 n 2 a4 p p Θ n2 a2 − s2 a2 − s2 = na2 cos 2 − s2 So we have sin2 s2 2s2 2s2 s2 Θ Θ Θ = 2 2 (n2 + 1 − 2 − 2n cos + 2 ) = 2 2 (1 + n2 − 2n cos ) n a 2 2 a a n a 2 Solving for s2 s2 = n2 a2 sin2 Θ 2 1 + n2 − 2n cos Θ 2 Glad that that mess is over with, we can now do some calculus. I’m going to let q 2 equal the denominator squared. Also to save space, lets say2 Θ= Q. I like using the letter q . a2 sin Qn2 ds2 = [cos Q(1 − 2n cos Q + n2 ) − n sin2 Q] dΘ q2 ds2 n 2 a2 = sin Q[cos Q − 2n cos2 Q + n2 cos Q − n(1 − cos2 Q)] dΘ q2 Expand and collect n 2 a2 ds2 = sin Q[−n cos2 Q + cos Q + n2 cos Q − n] dΘ q2 Group it up ds2 n2 a2 = sin Q(n cos Q − 1)(n − cos Q) dΘ q2 Plug back in for Q and q 2 : 6 n2 a2 sin Θ2 (n cos Θ − 1)(n − cos ds2 2 = dΘ (1 − 2n cos 2Θ + n2 )2 Θ 2 ) Using our plan from above, σ= 1 4 sin Θ2 cos ds2 Θ dΘ 2 = 1 4 sin Θ2 cos n2 a2 sin Θ 2 (n cos Θ 2 − 1)(n − cos 2 2 (1 − 2n cos + n ) Θ 2 Θ 2 ) Θ 2 We obtain σ(Θ) = n2 a2 (n cos Θ − 1)(n − cos 1 2 Θ 4 cos 2 (1 − 2n cos 2Θ + n2 )2 Θ 2 ) The total cross section involves an algebraic intensive integral. The total cross section is given by Z Θmax σ(Θ) sin ΘdΘ σT = 2π 0 To find Θmax we look for when the cross section becomes zero. When (n cos Θ − 1) is zero, we’ll have Θmax . If s > a, its as if the incoming particle 2 misses the ‘sphere’. At s = a we have maximum Θ. So using Θmax = 2 arccos n1, we will find it easier to plug in x = cos2 Θas a substitution, to simplify our integral. Z σT = π 1 1 n (nx − 1)(n − x) a2 n2 (1 − 2nx + n2 )2 2dx where Θ Θmax 1 1 sin dΘ cos = 2 2 2 n The half angle formula, sin Θ = 2 sin Θ 2 cos 2Θ was used on the sin Θ, the negative sign switched the direction of integration, and the factor of 2 had to be thrown in to make the dx substitution. dx = − This integral is still hard to manage, so make another substitution, this time, let q equal the term in the denominator. q = 1 − 2nx + n2 → dq = −2ndx The algebra must be done carefully here. Making a partial substitution to see where to go: qmin = 1 − 2 + n2 = n2 − 1 Z σT = (n−1)2 n2 −1 2πa2 n2 (nx − 1)(n − x) q2 qmax = n2 − 2n + 1 = (n − 1)2 dq −2n 7 Z = πa2 (n−1)2 n2 −1 −n(nx − 1)(n − x) q2 dq Expanding q 2 to see what it gives so we can put the numerator in the above integral in terms of q 2 we see q 2 = n4 + 1 + 2n2 − 4n3 x − 4nx + 4n2 x2 Expanding the numerator −n(nx − 1)(n − x) = −n3 x − nx + n2 x2 + n2 If we take q 2 and subtract a n4 , subtract a 1, add a 2n2 and divide the whole thing by 4 we’ll get the above numerator. That is: q 2 − n4 + 2n2 − 1 4 Now, our integral is = q 2 − (n2 − 1)2 4 = −n(nx − 1)(n − x) 2 2 2 Z (n−1)2 q − (n − 1) dq 4q 2 σT = πa2 n2 −1 This is finally an integral that can be done by hand πa2 πa2 (n−1)2 (n2 − 1)2  (z + n2 −1 (n2 −q 21)2 4  1−  z After working 4 out the few steps of algebra, σT = Z dq = πa2 4n2 − 8n + 4 = πa2 4 n2 − 2n + 1 The total cross section is σT = πa2 8 )