Equivalence of Lagrange’s & Newton’s Equations Section 7.6 The Lagrangian & the Newtonian formulations of mechanics are 100% equivalent! –As we know, they have different starting points & different viewpoints (energy vs. forces). –However, their physics content & the results they obtain for the differential equations of motion are identical. Now, we prove this formally & mathematically by showing that the 2 sets of equations of motion are the same. I’ll skim the derivation, the details are in the text!

For simplicity (but not necessity!) for this derivation, lets choose the generalized coordinates to be rectangular coordinates.  Lagrange’s Equations (in the absence of constraints) are: (  L/  x i ) - (d/dt)[(  L/  x i )] = 0 (i = 1,2,3) (1) By definition, L  T - U (2) Put (2) into (1)  (  [T-U]/  x i ) -(d/dt)[  [T-U]/  x i ] = 0 (3) In rectangular coordinates, the KE is a function of the velocities only, T = T(x i ) (indep. of the x i ). For a conservative system the PE is a function of the coordinates only, U = U(x i ) (indep. of the x i ).  (  T/  x i ) = 0 ; (  U/  x i ) = 0

 Lagrange’s Equations become: (i = 1,2,3) - (  U/  x i ) = (d/dt)[  T/  x i ] (4) The system is conservative:  - (  U/  x i ) = F i  the i th force component.  (4) becomes: F i = (d/dt)[  T/  x i ] (5) In rectangular coordinates, T = (½)m[∑ i (x i ) 2 ] (i = 1,2,3) (6) From (6), (  T/  x i ) = mx i  p i = i th momentum component.  (5) becomes: F i = (dp i /dt) (Newton’s 2 nd Law!)  We have derived Newton’s 2 nd Law from Lagrange’s Equations  THE 2 FORMULATIONS ARE EQUIVALENT!

Now do the reverse: Derive Lagrange’s Eqtns from Newton’s 2 nd Law & other Newtonian concepts. A much more tedious & messy derivation! Details are here. I’ll skim in class! First, transform from rectangular coordinates to generalized coordinates: x i = x i (q j,t)  (dx i /dt) = ∑ j (  x i /  q j )(dq j /dt) +(  x i /  t) Or: x i = ∑ j (  x i /  q j )q j + (  x i /  t) (1) An identity: (  x i /  q j ) =  [dx i /dt]  [dq j /dt]  (  x i /  q j ) Define: Generalized Momentum (we will make great use of this a little later!): p j  (  T/  q j ) Momentum for q j –For example, in plane polar coordinates: T = (½)m(r 2 + r 2 θ 2 ), p r = mr (linear momentum) p  = mr 2 θ (angular momentum)

Determine a generalized force by considering the “virtual work” δW done when the path is varied by δx i : δW = ∑ i F i δx i = ∑ i ∑ j F i (  x i /  q j )δq j Or: δW  ∑ j Q j δq j Q j  Generalized force = ∑ i F i (  x i /  q j ) For a conservative system with a PE = U: Q j  - (  U/  q j ) (like F i = - (  U/  x i ) ) By definition the generalized momentum: p j = (  T/  q j )  p j = (  /  q j )[(½)m ∑ i (x i ) 2 ] = m∑ i x i (  x i /  q j ) p j = m ∑ i x i (  x i /  q j ) (1) Take the time derivative of (1): p j = (dp j /dt) = m ∑ i [x i (  x i /  q j ) + x i (d/dt)(  x i /  q j )] (2)

In the 2 nd term of (2): (d/dt)(  x i /  q j ) = ∑ k (  2 x i /  q k  q j )q k + (  2 x i /  q j  t) Put into (2) (& get a mess!) p j = m∑ i [x i (  x i /  q j )] + m∑ i ∑ k [x i (  2 x i /  q k  q j )q k ] + m∑ i [x i (  2 x i /  q j  t)] (3) In the 1 st term in (3): – Use: mx i = F i, Q j  ∑ i F i (  x i /  q j ) Q j = m∑ i [x i (  x i /  q j )] In the 2 nd & 3 rd terms in (3): – Manipulation on P 256 shows that (  T/  q j ) = m∑ i ∑ k [x i (  2 x i /  q k  q j )q k ] + m∑ i x i (  2 x i /  q j  t)]

After more manipulation, (3) becomes: (dp j /dt) = p j = Q j + (  T/  q j ) (4) Recall: p j  (  T/  q j ) Q j  - (  U/  q j ) so (4) becomes: - (  U/  q j ) = (d/dt) [(  T/  q j )] - (  T/  q j ) Recall that U = U(q j ) (independent of q j ) & define the Lagrangian: L  T - U  (d/dt)[(  L/  q j )] - (  L/  q j ) = 0 Lagrange’s Equations have been derived from Newton’s 2 nd Law!

Essence of Lagrangian Dynamics Section 7.7 More of the author’s historical discussion plus discussion which is almost more philosophical than physical. History: –Lagrange’s equations in generalized coordinates were derived 46 years before Hamilton’s Principle. Philosophy: –Lagrangian dynamics is not a new theory, but is completely equivalent to Newtonian dynamics. –Different methods, but get the same results! –Lagrange: Uses energy concepts only & not force.

The Lagrangian is a scalar (energy)  It is invariant under coordinate transformations. –Lagrange’s Equations are valid in generalized coordinates, which (may) make the equations of motion simpler. Lagrange vs. Newton (philosophy): –Newton: Emphasizes the interaction of a body with the outside world (forces). –Lagrange: Emphasizes quantities associated with the body itself (energy). –A major advantage of Lagrange: It can get equations of motion in situations (as in the examples) when getting forces would be very difficult.

Newton’s Laws: –Differential statements 100% equivalent to Hamilton’s Principle –This is an integral statement which yields Lagrange’s Eqtns. There is NO DISTINCTION between them in their description of Physical Effects. Philosophical Distinction: –Newton: Describes Nature with cause (force) & effect (motion). –Hamilton’s Principle: Describes motion as resulting from nature attempting to achieve a certain purpose (the minimization of an integral).

A Theorem About the KE Section 7.8 Tedious manipulation! A useful Theorem! The KE, in fixed rectangular coordinates (n particles) is T = (½)∑ α ∑ i m α (x α,i ) 2 (i =1,2,3; α = 1,2,3, n) (1) Consider the dependence of T on the generalized coordinates & velocities: x α,i = x α,i (q j,t)  x α,i = x α,i (q j,q j,t) Explicitly: x α,i = ∑ j (  x α,i /  q j )q j +(  x α,i /  t) (j = 1,2,3, s) So: (x α,i ) 2 = ∑ j ∑ k (  x α,i /  q j )(  x α,i /  q k )q j q k + 2∑ j (  x α,i /  q j )(  x α,i /  t)q j + (  x α,i /  t) 2 (2)

Put (2) into T, (1): T = ∑ α ∑ i,j,k (½)m α (  x α,i /  q j )(  x α,i /  q k )q j q k + ∑ α ∑ i,j m α (  x α,i /  q j )(  x α,i /  t)q j + ∑ α ∑ i (½)m α (  x α,i /  t) 2 In general write T in the form: T = ∑ j,k a j,k q j q k + ∑ j b j q j + c (3) Where: a j,k  ∑ α ∑ i (½)m α (  x α,i /  q j )(  x α,i /  q k ) b j  ∑ α ∑ i m α (  x α,i /  q j )(  x α,i /  t), c  ∑ α ∑ i (½)m α (  x α,i /  t) 2 A very important special case: When the system is Scleronomic  Time does not appear in the transformation between rectangular & generalized coordinates:  (  x α,i /  t) = 0  b j  0, c  0

 When time doesn’t appear in transformation between rectangular & generalized coordinates, the KE has the form: T = ∑ j,k a j,k q j q k (4) –That is, the KE is then a homogeneous, quadratic function of the generalized velocities. Also, take the derivative of (4) with respect to one of the generalized velocities (say q ): (  T/  q ) = ∑ k a,k q k + ∑ j a j, q j Multiply by q & sum over : ∑ q (  T/  q ) = ∑ k, a,k q k q + ∑ j, a j, q j q (5)

In the 2 terms on the right side of (5), the double sums are over “dummy” indices.  The 2 terms are the SAME!  ∑ q (  T/  q ) = 2 ∑ j,k a j,k q j q k From (4), the sum is just the KE, T!  ∑ q (  T/  q )  2T (6) –We will use this in the next section! –(6) is valid only if T is a homogeneous, quadratic function of the generalized velocities! –(6) is a special case of Euler’s Theorem: If f(y k ) is a homogeneous function of the y k, of degree n, then ∑ k y k (  f/  y k )  nf