PHY 711 -- Lecture notes on
PHY 711 - Lecture notes on Lagrangian for Electric and
Magnetic Fields
For simplicity, consider a Lagrangian for a single particle
having the form (in Cartesian coordinates)
L(x,y,z,[x\dot],[y\dot],[z\dot],t) º T - U. The
Euler-Lagrange equations have the form:
d dt
æ ç ç
ç è
¶L
¶
. x
ö ÷ ÷
÷ ø
-
¶L ¶x
= 0,
(1)
with similar equations for y and z. We can show that this
form is consistent with Newton's Laws if the potential function
U takes the form:
U = U0(x,y,z,t) + UEM(x,y,z,
. x
,
. y
,
. z
,t),
(2)
where UEM represents the interaction of our particle (having
charge q) with an electric field E and magnetic field
B where we can represent the fields in terms of the
scalar and vector potentials:
E = - Ñf-
1 c
¶A ¶t
and B = Ñ×A.
(3)
We must find UEM which is both consistent with the
Euler-Lagrange Eq.(1) and with the Lorentz force (written
in the x direction):
Fx = q (Ex +
1 c
(
. r
×B)ûx ) = -
¶UEM ¶x
+
d dt
æ ç ç
ç è
¶UEM
¶
. x
ö ÷ ÷
÷ ø
.
(4)
We note that the magnetic field terms can be evaluated:
. r
×(Ñ×A)ûx =
. y
æ ç
è
¶Ay ¶x
-
¶Ax ¶y
ö ÷
ø
-
. z
æ ç
è
¶Ax ¶z
-
¶Az ¶x
ö ÷
ø
.
(5)
The right hand side of Eq.(5) (with the addition and
subtraction of a convenient term) can be written:
. x
¶Ax ¶x
+
. y
¶Ay ¶x
+
. z
¶Az ¶x
¶([(r)\dot]·A)/¶x
-
. x
¶Ax ¶x
-
. y
¶Ax ¶y
-
. z
¶Ax ¶z
-d Ax/dt + ¶Ax/¶t
,
(6)
where we are assuming that Ax = Ax(x,y,z,t). Noting that Ax = ¶([(r)\dot]·A)/¶[x\dot], the
electromagnetic force can thus be written:
Fx =
-q
¶f ¶x
-
q c
¶Ax ¶t
q Ex
+
q c
æ ç ç
ç ç è
¶([(r)\dot]·A) ¶x
-
d dt
¶([(r)\dot]·A)
¶
. x
+
¶Ax ¶t
([(r)\dot]×B)ûx
ö ÷ ÷
÷ ÷ ø
.
(7)
Simplifying this equation, we obtain
Fx = -
¶ ¶x
æ ç
è
q f-
q c
. r
·A
ö ÷
ø
-
d dt
¶
¶
. x
æ ç
è
q c
. r
·A
ö ÷
ø
.
(8)
Thus, we finally have the result
UEM = q f-
q c
. r
·A.
(9)
File translated from TEX by TTH, version 2.20. On 10 Sep 1999, 18:32.