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PHYS 705: Classical Mechanics Central Force Problems I 2

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PHYS 705: Classical Mechanics Central Force Problems I 2 1 PHYS 705: Classical Mechanics Central Force Problems I 2 Two-Body Central Force Problem Historical Background: Kepler’s Laws on celestial bodies (~1605) - Johannes Kepler based his 3 laws on observational data from Tycho Brahe - Formulate his famous 3 laws: - Orbit of each planet is an ellipse with sun at one of its foci - Equal areas swept out in equal time by an orbit - The ratio t 2 R 3 is the same for all planets, where t is the period and R is the semi-major axis -All these results were obtained through amazing sheer mathematical efforts. Before we get back to this important celestial application, we will address the general problem… 3 Kepler’s Notes 4 Kepler’s Notes 5 Two-Body Central Force Problem Set up of the general problem: 2 masses interact via forces directed along the line that connects them (central force): strong form of 3rd law First Step: Central force problems can be reduced to an effective 1-body problem: Change to generalized coordinates: CM position R and relative position r, r' r 2 ' r1 r2 CM R r1 ' ' ' r r2 r 1 r1 Rr 1 r r2 r 1 ' r2 Rr 2 Instead of using r 1 & r 2 , use R & r (CM & relative position) 6 Two-Body Central Force Problem From def. of CM: MR mmmi rrr i 1 1 2 2 i r ' ' r2 m1Rr 1 m 2 Rr 2 r ' ' 1 MR m1 m 2 R m1r 1 m 2 r 2 r' ' ' 2 r' m1r 1 m 2 r 2 0 CM 1 ' ' R Combining it with r r 2 r 1 , we have 'm2 ' m 2 ' r1 r 2 rr 1 m1 m 1 m2 ' m 2 1 r1 r m1 m 1 m ' m2 ' ' 1 r r and rrr2 1 r 1 m m m1 m 2 1 2 7 Two-Body Central Force Problem Now, form the Lagrangian: ' m2 r1 r m1 m 2 TTCM T about CM ' m1 r2 r 2 2 M 2m1 ' m 2 ' m1 m 2 R r1 r 2 2 2 2 M 1 mm2 mm 2 R 21 2 2 1 r 2 2 2 2 2 mm1 2 mm 1 2 M 21 mm1 2 2 R r (express in terms of R & r only) 2 2 m1 m 2 We have a central force so that U(r) depends on the relative distance r only. 8 Two-Body Central Force Problem So, M 21 mm1 2 2 LTUR r Ur( ) 2 2 m1 m 2 Note that R does not appear in L (cyclic) so that P R is conserved, i.e., MR const CM is either stationary or moving uniformly. Pick an inertial ref. frame (CM frame) in which CM is not movingR 0 and we can ignore the 1st term in L. The result is then: 1 2 mm1 2 Same as a single particle Lr Ur( ), where 2 m1 m 2 with mass moving in U(r). 9 Reference Frames for Central Force Problems Motion in the relative coordinate frame with R (fixed), r (3 dofs) : m2 ' m2 r1 r m1 m 2 r ' r 2 ' m1 r2 r m1 m 2 m 1 CM r1 ' m1 Effective one body problem with Motion of m 1 and m 2 in CM frame reduced mass and relative position r (dist from m 1 to m 2 ). 10 Reference Frames for Central Force Problems Motion in original r 1 , r 2 space with 6 dofs (not all independent): m1 and m 2 circle each other in space as their m2 r CM move with constant CM velocity along a fixed R r direction. m1 2 r1 11 Two-Body Central Force Problem The reduced problem is a much simpler problem with 3 dofs instead of 6. Also, we often have m 1 m 2 (e.g., sun-earth) Then, mm1 2 mm 1 2 m2 mm1 2 m 1 the CM is also close to m and ' 1 r r2 12 Two-Body Central Force Problem Further reduction of the two-body central force problem: Since U(r) is central, i.e., F always directs // toward CM along r, So, we have, dL N0and N 0 or L const dt (Alternatively, as we will see later, q is cyclic !) L will points in a constant direction fixed by initial condition. Then, L rp =const, and r & p must be to L always, i.e., motion has to be planar. If L 0 (trivial case), then the trajectory is a 1D motion through the origin. 13 Two-Body Central Force Problem With all these considerations, it is natural to use spherical coordinates and to orient the polar axis with L. Then, we can analyze the problem entirely in a 2D polar plane L. Lastly, using the fact that U(r)= U(r) depends only on the magnitude of the distance, we can show that the Lagrangian is effectively one-dimensional. To see that, start with L in r , q : m L rr2 2q 2 Ur( ) (we will simply call m= from now on) 2 ˆ (Recall, vr r rˆ r q θ in polar coord.) 14 Two-Body Central Force Problem Since U(r) is central, q is cyclic (does not appear) in L. And, the generalized momentum with respect to q is conserved, L p mr2q l const q q Replacing q with the constant l in the Lagrangian, we then have an effectively 1D system in r only: q l mr 2 2 m2 m 2 2 m2 l L r rq Ur( ) r 2 Ur( ) 2 2 2 2mr Note: q is ignorable but it is NOT a constant. r and q change while l is fixed in time. Motion is still in 2D in the CM frame*. 15 Two-Body Central Force Problem Now, we calculate the EOM for both r and q : q dL L m 2 2 2 0 L rr q Ur( ) dt q q 2 L L 0 mr 2q q q (L is cyclic in q !) This gives: mr2q l const [One can recognize this as the constant ang momentum: l I ( mr 2 ) q .] (This is the equivalent statement on the conservation of L as before.) 16 Two-Body Central Force Problem Connecting back to Kepler’s 2nd law… rdq Note that the area swept out by r with an 1 r dq infinitesimal rotation dq is dA r2 dq 2 dA1 dq 1 And, r2 r 2q dt2 dt 2 d 2 d 1 2 From EL equation, we have mr q 0 r q 0 (*) dt dt 2 Thus, Eq (*) establishes Kepler’s 2nd Law! 17 Two-Body Central Force Problem m Now, we are back to the EOM for r: L rr2 2q 2 Ur( ) 2 r L L2 dU mr mrq r r dr d L L 2 dU 0 gives mr mrq 0 dt r r dr Combining with the q equation q l mr 2 , we have, 2 l dU mr mr 2 0 mr dr l2 dU mr mr3 dr 18 Two-Body Central Force Problem We have seen that the q equation give one constant of motion l. We can get another constant of motion E since: L 0 so that h (Jacobi Integral/energy function) is conserved. t U doesn’t depend on q j and r ( q j ) does not depend on t explicitly so that h = E total energy is conserved. L check: h q j L j q j 1 rmr qq mr2 mrr 2 2 q 2 U 2 1 1 1 l 2 mr2 mr 2q 2 U mr 2 UE 2 2 2 2mr 2 19 Two-Body Central Force Problem Here is another way to argue that E is conserved. Let start with the r EOM, dU l2 d l 2 mr 3 U 2 dr mr dr2 mr Multiplying r on both sides, 2 dm2 d l dr mrr r U 2 dt2 dr 2 mr dt 2 dm2 d l r U 2 (reverse chain rule) dt 2 dt 2 mr dm l 2 So, 2 E is conserved! r 2 Ur( ) 0 dt2 2 mr E 20 m Two-Body Central Force Problem L rr2 2q 2 Ur( ) 2 Summary: We get 2 EOMs and 2 integrals of motion (l, E) for this problem. mr2q l 2 2 dU l m2 l mr 3 Er Ur( ) dr mr 2 2mr 2 st Note: The E equation effectively is the 1 integral of the r equation. We can rewrite it explicitly as, 2 l 2 r EUr ( ) 2 m2 mr 21 Two-Body Central Force Problem Integrating once more time gives: r dr ' t With initial condition 2 2 l r0 at t=0 E Ur( ') 2 r0 m2 mr ' Similar, we can integrate the q equation and get, t ldt With initial condition q(t ) 2 q0 mr( t ) q0 at t=0 0 So, the problem is completely solved with t(r) and q (t). The problem has been reduced to a quadrature (doing the integrals). 22 Graphical Analysis of Central Force Problem Using the concept of an effective potential, one can get a useful qualitative understanding of the problem without actually integrating! Let consider the r equation: dU l 2 The last two terms combined can be mr 3 dr mr considered as an effective force f'( r ) This looks like a 1D problem: a single particle moving in 1 dimension under the influence of an effective force, dU l 2 f'( r ) dr mr 3 23 Graphical Analysis of Central Force Problem dU l 2 mr dr mr 3 Case 1 l 0 : No angular momentum, then, it is really a 1D problem.
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