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The derivation of the lorentz force on a charged particle moving in an electromagnetic field using euler-lagrange equations. The lagrangian is given, and the euler-lagrange equations are calculated to obtain the ith component of the lorentz force. The document also includes the explanation of the total time derivative and the use of the levi-civita symbol.
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Euler-Lagrange Equations for charged particle in a field
The Lagrangian is
L =
mr˙^2 + q(A · r˙ − φ)
Euler Lagrange Equations are
d dt
∂ r˙
∂r
so calculate left and right hand sides separately:
∂r = q
∂r (A · r˙) − q ∂φ ∂r d dt
∂ r˙ = m¨r + q d 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 dt
∂t
∂r ∂t
∂r
or, to be completely clear, for a given component Ai,
d dt Ai = ∂Ai ∂t
∂rj ∂t
∂rj Ai,
where a sum over repeated indices is implied. So, putting the Euler-Lagrange equation together for index i gives
m¨ri = −q
∂ri φ − q ∂Ai ∂t
( ∂Aj ∂ri
∂Ai ∂rj
) r ˙j
= 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 × B term, evaluate
(˙r × B) = (˙r × (∇ × A))i = ≤ijk r˙j (∇ × A)k = ≤ijk r˙j ≤k`m
∂r`
Am
= (δiδkm − δimδk
) ˙rj
∂r` Am
= r˙j
∂ri Aj − r˙j
∂rj Ai =
( ∂Aj ∂ri
∂Ai ∂rj
) r ˙j QED
(I used the cyclic property of the Levi-Civita symbol ≤ijk = ≤jki = ≤kij , and the identity ≤ijk≤im = δj
δkm − δjmδk`.)