1
Euler-Lagrange Equations for charged particle in a field
The Lagrangian is
L=1
2m˙
r2+q(A·˙
r−φ)
Euler Lagrange Equations are
d
dt
∂L
∂˙
r=∂L
∂r,
so calculate left and right hand sides separately:
∂L
∂r=q∂
∂r(A·˙
r)−q∂φ
∂r
d
dt
∂L
∂˙
r=m¨
r+qd
dtA
Now recall Ais the vector potential evaluated at the position of the particle rat time t. The
particle is following a trajectory r(t), so A=A(r(t), t). The total time derivative thus gives two
terms,
d
dtA=∂A
∂t +∂r
∂t ·∂
∂rA,
or, to be completely clear, for a given component Ai,
d
dtAi=∂Ai
∂t +∂rj
∂t ·∂
∂rj
Ai,
where a sum over repeated indices is implied. So, putting the Euler-Lagrange equation together for
index igives
m¨ri=−q∂
∂ri
φ−q∂Ai
∂t +qÃ∂Aj
∂ri
−∂Ai
∂rj!˙rj
=q(E+˙
r×B)i,
which is the ith component of the Lorentz force. To convince yourself that the last term in paren-
theses really turns into the ˙
r×Bterm, evaluate
(˙
r×B)=(˙
r×(∇ × A))i=²ijk ˙rj(∇ × A)k=²ijk ˙rj²k`m
∂
∂r`
Am
= (δi`δkm −δim δk`) ˙rj
∂
∂r`
Am
= ˙rj
∂
∂ri
Aj−˙rj
∂
∂rj
Ai=̶Aj
∂ri
−∂Ai
∂rj!˙rjQED
(I used the cyclic property of the Levi-Civita symbol ²ijk =²jk i =²kij , and the identity ²ijk²i`m =
δj`δk m −δjmδk `.)