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Derivation of Lorentz Force: Euler-Lagrange for Charged Particles, Study notes of Physics

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.

Typology: Study notes

Pre 2010

Uploaded on 03/10/2009

koofers-user-okw
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Euler-Lagrange Equations for charged particle in a field
The Lagrangian is
L=1
2m˙
r2+q(A·˙
rφ)
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
φq∂Ai
∂t +qÃAj
∂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 `.)

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1

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

∂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 + q d dt

A

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

A =

∂A

∂t

∂r ∂t

∂r

A,

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

  • q

( ∂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 = δjδkm − δjmδk`.)