2. 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.

3. 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)

4. 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)

5.  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)

6. 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)

7. 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₣

8. 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)