2. Sect. 2.6: Conservation Theorems & Symmetry Properties Lagrange Method: A method to get the eqtns of motion. Solving them = math! n degrees of freedom.  Need n eqtns of motion (2 nd order diff. eqtns)  2n constants of integration (2n initial conditions). –Sometimes, can easily integrate the eqtns of motion. Sometimes not (except numerically). –Even when we cannot integrate the eqtns of motion, can get a lot of info about PHYSICS of motion by considering Conservation Theorems.

3. 1 st Integrals of Motion  Relations between generalized coords, generalized velocities, & time which are 1 st order diff eqtns. Of the form: f(q 1,q 2,…,q 1,q 2,.., t) = constant 1 st Integrals of Motion: Very interesting because they tell us a lot about the system physics. They come from Conservation Theorems.

4. Consider: Point masses & conservative forces: Eqtns of motion in Cartesian coords: L = T - V = (½)∑ i m i [(x i ) 2 +(y i ) 2 +(z i ) 2 ] - V(r 1,r 2, …r N ) ( d/dt)[(∂L/∂x i )] - (∂L/∂x i ) = 0 Look at: (∂L/∂x i ) = (∂T/∂x i ) - (∂V/∂x i ) = (∂T/∂x i ) = (∂/∂x i )[ (½)∑ j m j [(x j ) 2 +(y j ) 2 +(z j ) 2 ] = m i x i = p ix (x component of momentum of i th particle)  DEFINE: Generalized Momentum associated with Generalized Coord q j : p j  (∂L/∂q j )

5. Generalized Momentum  Generalized Momentum associated with (or Momentum Conjugate to) Generalized Coord q j : p j  (∂L/∂q j ) Points worth noting: If q j is not a Cartesian Coordinate, p j is NOT necessarily a linear momentum. For a velocity dependent potential U(q j,q j,t), then, even if q j is a Cartesian Coordinate, the Generalized Momentum p j is NOT the usual Mechanical Momentum (p j  m j q j )

6. Example: Velocity dependent potential with q j a Cartesian Coord, & where p j is not m j q j : Charged particles in E&M field (A,  ). Ch. 1: L  T - U = (½)∑ i [m i (r i ) 2 - q i  (r i ) + q i A(r i )  r i ]  p ix  (∂L/∂x i ) = m i x i + q i A x  m i x i

7. Ignorable (Cyclic) Coordinates Important special case! Cyclic or Ignorable Coordinates  Generalized Coordinates q j not appearing in Lagrangian L (but the generalized velocity MAY still appear in L). Lagrange’s Eqtn for a cyclic coordinate q j : (d/dt)[(∂L/∂q j )] - (∂L/∂q j ) = 0 By definition of cyclic: (∂L/∂q j ) = 0  Lagrange Eqtn: (d/dt)[(∂L/∂q j )] = 0 Momentum Conjugate p j  (∂L/∂q j )  Lagrange Eqtn for a cyclic coordinate: (dp j /dt) = 0

8.  If a Generalized Coord q j is cyclic or ignorable, the Lagrange Eqtn is (dp j /dt) = 0 where Generalized Momentum p j  (∂L/∂q j ) (dp j /dt) = 0  p j = constant (conserved)  A General Conservation Theorem: If the Generalized Coord q j is cyclic or ignorable, the corresponding Generalized (or Conjugate) Momentum, p j  (∂L/∂q j ) is conserved.

9. Note: Derivation assumes that q j is a Generalized Coord & linearly independent of all other coords. As we’ve seen, when constraints exist, the q j are not all linearly independent. –Recall hoop rolling without slipping in xy plane (Ch. 1): Generalized Coords x, θ,  Constraint Eqtn: rdθ = dx (1) L = (½)M(x) 2 + (½)I(  ) 2 L : indep of θ. BUT because of the constraint eqtn (1), the corresponding conjugate momentum (angular momentum) p θ = Mr 2 θ is NOT conserved!

10. Cyclic Coords & Conservation Theorems If q j is cyclic, p j = (  L/  q j ) is conserved. In this case, p j  a 1 st integral of the motion  Can use to formally eliminate the cyclic coordinate from the problem ( reducing # of diff eqtns of motion needed to solve from n to n-1).  Routh Method: If q j is cyclic, compute p j = (  L/  q j ) (= constant). Solve for q j in terms of p j : q j = f(p j ). Everywhere q j appears in the Lagrangian L, make the replacement:  L(q j, + rest of q’s, q’s)  L´(p j,+rest of q’s, q’s) Advantage: p j = integration constant. Remaining eqtns involve only non-cyclic coords. Return to this later when discuss Hamiltonian formulation.

11. Conditions for conservation of generalized momenta are actually MORE GENERAL than those used to derive conservation theorems for many particle systems in Ch. 1. –Those in Ch. 1 needed assumptions of Newton’s 3 rd Law in weak & strong forms for internal forces. –Present case: Generalized Momentum conservation theorems result even if these 3 rd Law assumptions are violated!

12. Example: A single charged particle in an E&M field (A,  ). Assume A &  are indep of x (but DO depend on y & z, or else E = B = 0!) L = (½)m[(x) 2 + (y) 2 + (z) 2 ] - q  (y,z) + qA(y,z)  (x i + y j + z k)  p x  (∂L/∂x) = mx + qA x = constant (since (∂L/∂x) = 0) mx + qA x = Constant instead of mx. Physics: From E&M can show: qA x = x component of momentum of EM field associated with charge q.  mx + qA x = constant  x component of mech. momentum + x component of field momentum is conserved (allowing for exchange of momentum between field & particle!)

13. Generalized Momenta & Forces Consider a specific Generalized Coord q j for which infinitesimal change dq j = translation of the whole system in some direction (change of origin).  q j cannot appear in the KE T because velocities aren’t affected by the origin shift: (∂T/∂q j ) = 0 Assume also, conservative forces, so the PE V is not a function of the velocities: (∂V/∂q j ) = 0  (∂L/∂q j ) = -(∂V/∂q j )  Q j (generalized force) Also:  (∂L/∂q j )  p j = (∂T/∂q j ) Lagrange’s Eqtn for this coordinate q j : (d/dt)[(∂L/∂q j )] - (∂L/∂q j ) = 0 Combining  This becomes: (dp j /dt) - Q j = 0

14. Summary: For Generalized Coord q j for which change dq j = change of origin.  q j & for conservative forces, Lagrange’s Eqtn is: (dp j /dt) = p j = Q j (Newton’s 2 nd Law in generalized coord language) Ch. 1: Generalized Force Q j, in terms of Cartesian vector forces F i : Q j  ∑ i F i  (∂r i /∂q j ) For the special case considered, dq j = change of origin, so r i (q j ) & r i (q j +dq j ) are related as in the figure:

15. So: (dp j /dt) = p j = Q j and Q j  ∑ i F i  (∂r i /∂q j ) In this case, r i (q j ) & r i (q j +dq j ) are as in figure: By definition: (∂r i /∂q j ) = lim (dq j  0) [r i (q j +dq j ) - r i (q j )]/(dq j ) = (dq j /dq j )n = n n = unit vector in direction of system translation  Q j  ∑ i F i  (∂r i /∂q j ) = ∑ i F i  n = n  F

16. In special the case considered, Lagrange gives Newton’s 2 nd Law: (dp j /dt) = p j = Q j = n  F Generalized Force = Component of Cartesian vector force F in direction of origin shift. Also, in this case, KE: T = (½)∑ i m i (r i ) 2. Conjugate momentum: p j = (∂T/∂q j ) = ∑ i m i r i  (∂r i /∂q j ) (note that (∂r i / ∂ q j ) = ( ∂ r i / ∂ q j ) )  p j = ∑ i m i v i  (∂r i /∂q j )  n  ∑ i m i v i = n  p Generalized Momentum = Component of Cartesian vector momentum p in direction of origin shift.

17. In special case considered, Newton’s 2 nd Law from Lagrange: n  p = p j = Q j = n  F Suppose, in this same case, Generalized Coord q j is cyclic: Q j = - (∂V/∂q j ) = 0  p j = n  p = 0  Conservation of linear momentum: If a component of total force vanishes, n  F = 0, then corresponding component of total linear momentum n  p = constant (is conserved)

18. Angular Momentum & Torque Do a similar analysis to show that, if a cyclic coordinate q j is such that dq j corresponds to a rotation of the entire particle system, then conservation of the conjugate momentum corresponds to conservation of angular momentum. Consider a specific Generalized Coord q j for which the infinitesimal change dq j = rotation of whole system about some axis.  q j cannot appear in KE T because velocities aren’t affected by rotation: (∂T/∂q j ) = 0

19. Also, conservative forces. PE V is not a function of the velocities: (∂V/∂q j ) = 0  (∂L/∂q j ) = -(∂V/∂q j )  Q j (generalized force) Also:  (∂L/∂q j )  p j = (∂T/∂q j ) Lagrange’s Eqtn for this coordinate q j : (d/dt)[(∂L/∂q j )] - (∂L/∂q j ) = 0  This again becomes: (dp j /dt) - Q j = 0

20. Now show: If q j = rotation coord, Q j = torque component about axis of rotation & p j = angular momentum component about same axis. Generalized force is again: Q j  ∑ i F i  (∂r i /∂q j ). Now, q j is a rotation coord.  dq j = infinitesimal rotation of vector r i, keeping magnitude of vector constant (fig). Rotation axis direction is n.

21. Q j  ∑ i F i  (∂r i /∂q j ). dq j = infinitesimal rotation of r i, keeping magnitude constant. |dr i | = r i sinθdq j  |∂r i /∂q j | = r i sinθ Also (∂r i /∂q j ) is  to both r i & rotation axis direction n  (∂r i /∂q j ) = n  r i Generalized Force: Q j  ∑ i F i  (∂r i /∂q j ). = ∑ i F i  (n  r i ) = ∑ i n  (r i  F i ) = n  ∑ i (r i  F i ) = n  ∑ i N i  Q j = n  N (N  total torque about rotation axis) Generalized Force = Component of Cartesian vector torque N in rotation direction.

22. In this case, Conjugate momentum: p j = (∂T/∂q j ) = ∑ i m i v i  (∂r i /∂q j ) with (∂r i /∂q j ) = n  r i p j = ∑ i m i v i  (n  r i ) = n  ∑ i (r i  m i v i ) = n  ∑ i L i = n  L Generalized Momentum = Component of Cartesian vector angular momentum L in direction of rotation. In special case considered, Newton’s 2 nd Law (rotational motion) from Lagrange: n  L = p j = Q j = n  N

23. Suppose, in this case, Generalized Coord. q j is cyclic. Q j = - (∂V/∂q j ) = 0  p j = n  N = 0  Conservation of angular momentum: If a component of the total torque vanishes, n  N = 0, then the corresponding component of total angular momentum n  L = constant (is conserved)

24. Physics/Philosophy Physical Significance of cyclic coords (translation or rotation): If the q j corresponding to a displacement is cyclic (& thus conjugate momentum is conserved):  System translation is as if it were rigid & has no effect on the problem. That is ( we’ve shown ), if the system is invariant under translation along a given direction n, the corresponding linear momentum along that direction is conserved. Also, if q j corresponding to a rotation is cyclic (& thus conjugate angular momentum is conserved):  System rotation is as if it were rigid & has no effect on the problem. That is ( we’ve shown ), if the system is invariant under rotation about a given direction n, the corresponding angular momentum along that direction is conserved.

25. Conservation Laws & Symmetry Bottom line: The Generalized Momentum Conservation Theorems are closely connected with the Symmetry Properties of the system. –For example, spherically symmetric systems ALWAYS have all components of angular momentum conserved. –If the system has symmetry only about one axis (say, z) then angular momentum about that axis (L z ) is conserved. –If, by inspection, a system has some symmetry, can often use it to IMMEDIATELY conclude that some Generalized Momentum component is conserved.  There is an intimate connection between symmetry properties and conservation theorems. (More later on this!)