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Electro-Magnetic Forces and a Velocity-Dependent Potential
Review: In generalized coordinates, Newton's Second Law is:
d/dt[T/qk’] - T/qk = Qk .
IF we have a force that can be put in terms of a potential energy: Qk = -V/qk where
V = V(qk) but is NOT a function of the qk’ , then since V/qk’ = 0, we can write:
L = T - V and get:
d/dt[L/qk’] - L/qk = 0 .
Basic idea:
But what about the magnetic force, Fmag = qv  B ?
In general, we can have the Lagrangian form of Newton's Second Law
if:
d/dt[V/qk’] - V/qk = Qk
(which does occur for the normal potential energies where V is NOT a function of the qk’ so that
the V/qk’ = 0 and thus Qk = -V/qk).
Here is a review of the proof: Starting with d/dt[T/qk’] - T/qk = Qk ,
and with L = T – V, subtracting d/dt[V/qk’] - V/qk = Qk from both sides gives
d/dt[L/qk’] - L/qk = 0 .
Special case: uniform magnetic field - need: d/dt[Vm/qk’] - Vm/qk = Qk .
To address the magnetic force case specifically, let's first consider a uniform B field that does
not change in time. Then the magnetic force, F = qv  B , gives (in rectangular form)
Fx = q(y’Bz - z’By),
Fy = q(z’Bx - x’Bz),
Fz = q(x’By - y’Bx) .
Let’s just work on the Fx. We need one term that when we take the [Vm/x’] we’re left with
qyBz so that when we take the d/dt of this term, we get the qy’Bz part of Fx; this means we need
a term qx’yBz. We also need one term that when we take the - V/x we are left with the - qz’By
; this means we need a term qxz’By .
Because of the symmetries, this suggests we define a uniform magnetic potential energy, Vm, as:
Vm = q(zy’Bx + xz’By + yx’Bz) .
Check:
d/dt[Vm/y’] - Vm/y = d/dt [qzBx] – qx’Bz = qz’Bx – qx’Bz = Fy
as required (as long as dBx/dt = 0)! This also works for the other two components as well. This
gives us confidence to look more generally at electro-magnetic potentials in the hope that we can
develop more general potentials that can be made part of the Lagrangian and hence the
Hamiltonian.
Electromagnetic Fields review:
Basic laws:
Coulomb’s Law:
Biot-Savart Law:
Ept charge = (kq/r2)r and
Bsmall current length = (r2I dl  r
Gauss’ Law: ∫closed area E · dA = 4k*qenclosed and
∫closed area B · dA = 0 (no monopoles)
[Gauss’ Law for Electric Fields is equivalent to Coulomb’s Law]
Differential form: use divergence theorem: ∫ E · dA = ∫  · E dV = 4k*qenclosed = 4k ∫  dV
 · E = 4k 
Faradays’ Law:
and
Ampere’s Law:
and
·B = 0
∫closed loop E · dl = (d/dt) ∫ B · dA
∫closed loop B · dl =  I = ∫ j dA
[Ampere’s Law is equivalent to the Biot-Savart Law.]
Differential form: use Stokes’ theorem: ∫closed loop E · dl = ∫ (  E ) dA
  E = -B/t
  B =  j .
Here  is the volume charge density and j is the area current density. The  may include
individual charges and polarization charges. The j may include moving charges (regular
currents), polarization currents, magnetization currents, and convective currents.
Scalar potential: Φ
This is what we called the voltage in PHYS 251. W = ∫ F · ds = -PE and so ∫ E · ds = -Φ
and so Ex = - Φ/x and more generally, E = -Φ . For a single charge, Φ = kq/r , and more
generally Φ = k ∫ /r dV.
Vector potential: A
Since  · (  A) = 0 (for any vector, A), and since  · B = 0, we can define a vector
potential, A, where B =   A . In general, A = (∫ (j /r) dV.
From Faraday’s Law:   E = -B/t = -[  A] /t , or E = -A/t
Putting this all together:
E = -Φ - A/t
and
B = xA .
General case: Magnetic force with a not necessarily uniform magnetic field.
From E&M theory (previous page):
E = -Φ - A/t
and
B = xA
where Φ is the scalar potential (electric voltage) and A is the vector potential. The
electromagnetic force can now be written as:
F = qE + qv  B = q(-Φ - A/t) + qv  (  A) .
From the 3-D calculus, we have the identity:
a  ( b) = (a · b) - (a · )b
so
and so
v  ( A) = (v · A) - (v · )A
F = qE + qv  B = q{(-Φ - A/t) + [(v · A) - (v · )A]} .
Let's pause a bit to consider:
dA
= (A/x)dx + (A/y)dy + (A/z)dz + (A/t)dt
so
dA/dt = (A/x)x’ + (A/y)y’ + (A/z)z’ + (A/t)
so
dA/dt = (v · )A + A/t
so the two terms {- A/t - (v · )A} can be replaced with - dA/dt ; thus
F = qE + qv  B = q{(-Φ - dA/dt) + (v · A)} .
If we now try: U = qΦ - q(v · A) as our electromagnetic potential energy, then
d/dt[U/x’] - U/x = d/dt[-qAx] - {q(Φ/x) - q(v · A)/x}
= q{-(dA/dt)x - (Φ)x + (v · A)x} = Fx ,
Which matches the x component of the above expression for F ! Thus, if we define the
Lagrangian as L = T – U, we can get the correct equations of motion even with Electric and
Magnetic forces.
HOMEWORK PROBLEM #8: (9-22 in Symon)
a) Show that a uniform magnetic field B in the z-direction can be represented in cylindrical
coordinates by the vector potential: A = ½ Bρφ .
b) Write out the Lagrangian function for a particle in such a field.
c) Write down the equations of motion.
d) Find three constants of the motion.