Sect. 1.5: Velocity-Dependent Potentials & the Dissipation Function Non-conservative forces? It’s still possible, in a Special Case, to use Lagrange’s Eqtns unchanged, provided a Generalized or Velocity-Dependent Potential U = U(qj,qj) exists, where the generalized forces Qj are obtained as: Qj  - (U/qj) + (d/dt)[(U/qj)] The Lagrangian is now: L  T - U & Lagrange’s Eqtns are still: (d/dt)[(L/qj)] - (L/qj) = 0 (j = 1,2,3, … n) A very important application: Electromagnetic forces on moving charges.

Sect. 1.5: Electromagnetic Force Problem Particle, mass m, charge q moving with velocity v in combined electric (E) & magnetic (B) fields. Lorentz Force (SI units!): F = q[E + (v  B)] (1) E&M results that you should know! E = E(x,y,z,t) & B = B(x,y,z,t) are derivable from a scalar potential  = (x,y,z,t) and a vector potential A = A(x,y,z,t) as: E  -  - (A/t) (2) B    A (3)

Lagrangian is: L  T - U = (½)mv2 - q + qAv Can obtain the Lorentz Force (1) from the velocity dependent potential: U  q - qAv F = - U Proof: Exercise for student! Use (1),(2),(3) together. Lagrangian is: L  T - U = (½)mv2 - q + qAv Use Cartesian coords. Lagrange Eqtn for coord x (noting v2 = (x)2 + (y)2 + (z)2 & v = x i + y j + z k) (d/dt)[(L/x)] - (L/x) = 0  mx = q[x(Ax/x) + y(Ay/x) + z(Az/x)] - q[(/x) + (dAx/dt)] (a) Note that: (dAx/dt) = vAx + (Ax/t)

 mx = - q(/x) - q(Ax/t) + q[y{(Ay/x) - (Ax/y)} + z{(Az/x)- (Ax/z)}] Using (2) & (3) this becomes: mx = q[Ex + yBz -zBy] Or: mx = q[Ex + (v  B)x] = Fx (Proven!) If some forces in the problem are conservative & some are not:  Have potential V for conservative ones & thus have the Lagrangian L  T - V for these. For non-conservative ones, still have generalized forces: Qj  ∑iFi(ri/qj)

Non-Conservative Forces L  T - V for conservative forces. Generalized forces: Qj  ∑iFi(ri/qj) for non-conservative forces. Follow derivation of Lagrange Eqtns & get: (d/dt)[(L/qj)] - (L/qj) = Qj (j = 1,2,3, ..n) Friction: A common non-conservative force. Friction (or air resistance): A common model: Components are proportional to some power of v (often the 1st power): Ffx = -kxvx (kx = const)

Frictional Forces Model for Friction (or air resistance): Ffx = -kxvx Can Include such forces in Lagrangian formalism by introducing Rayleigh’s Dissipation Function ₣ ₣  (½)∑i[kx(vix)2 + ky(viy)2 + kz(viz)2] Obtain components of the frictional force by: Ffxi  - (₣/vix), etc. Or, Ff = - v₣ Physical Interpretation of ₣ : Work done by system against friction: dWf = - Ff dr = - Ff v dt = -[kx(vix)2 + ky(viy)2 + kz(viz)2] dt = -2₣ dt  Rate of energy dissipation due to friction: (dWf /dt) = -2₣

Rayleigh’s Dissipation Function ₣ ₣  (½)∑i[kx(vix)2 + ky(viy)2 + kz(viz)2] Frictional force: Ffi = - vi ₣ Corresponding generalized force: Qj  ∑iFfi(ri/qj) = - ∑ivi₣ (ri/qj) Note that: (ri/qj) = (ri/qj) Qj = - ∑ivi₣(ri/qj) = - (₣/qj) Lagrange’s Eqtns, with frictional (dissipative) forces: (d/dt)[(L/qj)] - (L/qj) = Qj Or (d/dt)[(L/qj)] - (L/qj) + (₣/qj) = 0 (j = 1,2,3, ..n)