Figure 58: Particle moving on a circle in cartesian coordinates and in k-space. Something moving
on a circle at a located position, moves around the origin in k-space, because the x- and
y-velocity keep going to zero and that’s the path taken in k-space.
So, one has to construct a Lagrangian, which is a function of velocity. Normally one guesses the
Lagrangian and looks if the right force comes out, but luckily in this case the Lagrangian was guessed
before to be
L=1
2
mv^2 −qV(r,t) +qv·A(r,t). (67)So by putting this Lagrangian into the Euler-Lagrange equations (eqn. (68)), we obtain the Lorentz
force law^6 :
d
dt(∂L
∂x ̇i)
−∂L
∂xi= 0 (68)
d
dt(
∂L
∂vx)
−∂L
∂x= 0
d
dt(
∂L
∂vx)
=d
dt
(mvx+qAx) =mdvx
dt
+qdAx
dt=m
dvx
dt+q(
dx
dt∂Ax
∂x+
dy
dt∂Ax
∂y+
dz
dt∂Ax
∂z+
∂Ax
∂t)=m
dvx
dt+q(
vx
∂Ax
∂x+vy
∂Ax
∂y+vz
∂Ax
∂z+
∂Ax
∂t)∂L
∂x=−q∂V
∂x+q(
vx
∂Ax
∂x+vy
∂Ay
∂x+vz
∂Az
∂x)m
dvx
dt
=−q(
∂V
∂x+
∂Ax
∂t)︸ ︷︷ ︸
x component of electric field+q(
vy(
∂Ay
∂x−
∂Ax
∂y)
+vz(
∂Az
∂x−
∂Ax
∂z))︸ ︷︷ ︸
x component ofvxBWith the calculation above in all three dimensions,E=−∇V −∂dtA andB=∇×A, eqn. (66) is
obtained, which proves eqn. (67) to be the right Lagrangian to solve this problem.
(^6) Der Beweis fuer die Richtigkeit der Lagrangefunktion ist nicht pruefungsrelevant, wird aber der Vollstaendigkeit halber
wiedergegeben. Man sollte aber wissen, dass genau diese Lagrangefunktion das Problem beschreibt.