1
Euler-Lagrange Equations for charged particle in a field
The Lagrangian is
L =
1 2
mṙ + q(A · ṙ − φ)
2
Euler Lagrange Equations are
d ∂L
∂L
=
,
dt ∂ ṙ
∂r
so calculate left and right hand sides separately:
∂
∂φ
∂L
= q (A · ṙ) − q
∂r
∂r
∂r
d ∂L
d
= mr̈ + q A
dt ∂ ṙ
dt
Now recall A is the vector potential evaluated at the position of the particle r at 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
∂A ∂r ∂
A =
+
· A,
dt
∂t
∂t ∂r
or, to be completely clear, for a given component Ai ,
d
∂Ai ∂rj ∂
Ai =
+
·
Ai ,
dt
∂t
∂t ∂rj
where a sum over repeated indices is implied. So, putting the Euler-Lagrange equation together for
index i gives
Ã
!
∂
∂Aj
∂Ai
∂Ai
mr̈i = −q
φ−q
+q
−
ṙj
∂ri
∂t
∂ri
∂rj
= q (E + ṙ × B)i ,
which is the ith component of the Lorentz force. To convince yourself that the last term in parentheses really turns into the ṙ × B term, evaluate
(ṙ × B) = (ṙ × (∇ × A))i = ²ijk ṙj (∇ × A)k = ²ijk ṙj ²k`m
∂
Am
∂r`
Ã
!
∂
∂
∂Ai
∂Aj
= ṙj
Aj − ṙj
Ai =
−
ṙj
∂ri
∂rj
∂ri
∂rj
∂
Am
∂r`
= (δi` δkm − δim δk` )ṙj
QED
(I used the cyclic property of the Levi-Civita symbol ²ijk = ²jki = ²kij , and the identity ²ijk ²i`m =
δj` δkm − δjm δk` .)