Seminar 4:CHARGED PARTICLE IN ELECTROMAGNETIC FIELD
Introduction
Let take Lagrange's equations in the formthat follows fromD'Alembert's principle,
d
dt
@T
@ _q
j
!
@T
@q
j
= Q
j
;(1)
and suppose that the generalized force is derivable not from the potential V (q
j
) but
from a more general function U(q
j
;_q
j
),by the prescription
Q
j
=
@U
@q
j
+
d
dt
@U
@ _q
j
!
:(2)
Substituting this expression for the force into Eqs.(1) yields
d
dt
@T
@ _q
j
!
@T
@q
j
=
@U
@q
j
+
d
dt
@U
@ _q
j
!
;(3)
or
d
dt
@(T U)
@ _q
j
!
@(T U)
@q
j
= 0:(4)
It follows that these equations can be written in the form of Lagrange's equations,
d
dt
@L
@ _q
j
!
@L
@q
j
= 0;(5)
if we use as the Lagrangian the function
L = T U:(6)
The function U is called usually the velocitydependent potential (sometimes the
termgeneralized potential is also used).It can be thought that the possibility of using
such a strange potential is purely academic but this is not the case!On the contrary,
it appears that all the fundamental forces in physics can be expressed in the form
(2),for a suitably chosen potential function U.Its near practical importance relates
to the theory of an electric charge in an electromagnetic eld.As you know,a charge
q moving with the velocity v in an electromagnetic eld,containing both an electric,
E,and magnetic,B,elds,experiences a force (Note:1/c appears in Gauss system
of units,in SI system it will be absent)
F = q
h
E+(v B)
i
;(7)
which is called the Lorentz force.Both vectors E(r;t) and B(r;t) are continuous
functions of time t and position r = (x;y;z) derivable from the scalar and vector
potentials'(r;t) and B(r;t) by
E = r'
@A
@t
(8)
and
B = rA:(9)
Here,r is the dierential operator dened in Cartesian coordinates by
r
@
@x
;
@
@y
;
@
@z
!
;(10)
so that
r'=
@'
@x
^x +
@'
@y
^y +
@'
@z
^z (11)
and
rA=
^x ^y ^z
@
@x
@
@y
@
@z
A
x
A
y
A
z
;(12)
where ^x;^y and ^z are the unit vectors along x,y and z axes,respectively.
Notice that the electromagnetic eld dened by Eqs.(89) don't change when the
potentials are transformed according to:
'
0
='
@
@t
;A
0
= A+r ;(13)
where is an arbitrary function of the coordinates and time.These transformations
are known as the gauge transformations.
Problem 14.Lagrangian of Charged Particle in Electromagnetic Field
Show that the Lagrangian of a particle with the charge q moving with the velocity
v in an electromagnetic eld given by the scalar and vector potentials'and A is
L =
1
2
mv
2
q'+qA v:(14)
Solution:
Taking the vectors of the electric and magnetic elds E and Brepresented in terms
of the scalar and vector potentials in accordance with Eqs.(8) and (9),we have for
the Lorentz force
F = q
"
r'
@A
@t
+
v (rA)
#
:(15)
Let calculate,say,the xcomponent of this vector force,
F
x
= q
"
(r')
x
@A
x
@t
+
v (rA)
x
#
:(16)
Using the denitions (11) and (12),we obtain
(r')
x
=
@'
@x
(17)
and
v (rA)
x
= v
y
@A
y
@x
@A
x
@y
v
z
@A
x
@z
@A
z
@x
= v
y
@A
y
@x
+v
z
@A
z
@x
+v
x
@A
x
@x
v
y
@A
x
@y
v
z
@A
x
@z
v
x
@A
x
@x
=
@
@x
(v A)
dA
x
dt
+
@A
x
@t
;(18)
since
@
@x
(v A) =
@
@x
(v
x
A
x
+v
y
A
y
+v
z
A
z
) = v
x
@A
x
@x
+v
y
@A
y
@x
+v
z
@A
z
@x
(19)
and
dA
x
dt
=
@A
x
@t
+v
x
@A
x
@x
+v
y
@A
x
@y
+v
z
@A
x
@z
:(20)
Substituting Eqs.(17) and (18) into Eq.(16),we nally obtain
F
x
= q
h
@'
@x
@A
x
@t
+
@
@x
(v A)
dA
x
dt
+
@A
x
@t
i
= q
h
@
@x
'v A
dA
x
dt
i
=
q
h
@
@x
'v A
+
d
dt
@
@v
x
'v A
i
;(21)
which can be written as
F
x
=
@U
@x
+
d
dt
@U
@v
x
;(22)
where we introduce the function
U = q'qv A= q'qA v:(23)
Note that the termin the righthand side of this equation coincides with the potential
V dened by Eq.(2.119) fromour Lecture notes in a particular case of a single particle
of charge q.
Comparing the expression for the Lorentz force in the form(22) with the denition
(2) of the generalized force in terms of the velocitydependent potential,we see that
in our case this potential is dened just by the equation (23).This observation
immediately yields the Lagrangian in the desired form given by Eq.(6),
L = T U =
1
2
mv
2
q'+qA v:(24)
Problem 15.Calculate the conjugate momentum p and the energy function h
for a particle of the mass m and charge q in an electromagnetic eld given by the
scalar and vector potentials'and A.
Solution:
The xcomponent of the conjugate momentum is
p
x
=
@L
@ _x
=
@
@v
x
(
1
2
m[v
2
x
+v
2
y
+v
2
z
] q'+q[A
x
v
x
+A
y
v
y
+A
z
v
z
]
)
= mv
x
+qA
x
:(25)
The same for y and zcomponents
p
y
= mv
y
+qA
y
(26)
and
p
z
= mv
z
+qA
z
:(27)
The second terms in these expressions play the role of a potential momentum.
The energy function is
h =
P
j
_q
j
@L
@ _q
j
L =
P
j
_q
j
p
j
L = v
x
p
x
+v
y
p
y
+v
z
p
z
mv
2
q'+qA v
= v
x
mv
x
+qA
x
+v
y
mv
y
+qA
y
+v
z
mv
z
+qA
z
1
2
mv
2
q'+qA v
= mv
2
+qA v
1
2
mv
2
q'+qA v
=
1
2
mv
2
+q'= T +q':(28)
If A and'are independent of t,then L does not depend on t explicitly and the
energy function is constant,that is
T +q'= const:(29)
Since the second term in this equation is nothing but the potential energy of a charge
particle,we see that in this case the total energy of the system is conserved as it
might be.
Problem16.A particle of the mass mand charge q moves in a constant magnetic
eld
B = (0;0;B):(30)
Show that the orbit of a particle is a helix.
Solution:
Let specify the scalar and vector potentials for the case when the electric eld is
absent,
E = 0;(31)
and magnetic eld has only nonzero component B
z
= B.Keeping in mind that this
component is expressed in terms of the vector potential A as
B
z
=
@A
y
@x
@A
x
@y
;(32)
we can choose the potentials in the form
A= (0;Bx;0);'= 0:(33)
With this choice,the Lagrangian is
L =
m
2
( _x
2
+ _y
2
+ _z
2
) +qBx_y:(34)
The corresponding Lagrange's equation are written as
8
>
<
>
:
d
dt
(m_x) qB_y = 0
d
dt
(m_y +qBx) = 0
d
dt
(m_z) = 0:
(35)
From the second and third equations it follows
_y = C !x (36)
and
z = z
0
+Dt;(37)
where C;D;z
0
are constants and the substitution
!=
qB
m
(38)
has been made.With the value of the _y given by (36),the rst equation from (35)
takes the form
x !(C !x) = 0;(39)
or
x +!
2
(x x
0
) = 0;(40)
where
x
0
=
C
!
:(41)
The general solution of Eq.(40) can be written as
x x
0
= acos(!t +):(42)
Then
_y = C !x =!(x
0
x) = a!cos(!t +):(43)
Integrating this equation yields
y = asin(!t +) +y
0
:(44)
Finally,combining Eqs.(42) and (44) we obtain
(x x
0
)
2
+(y y
0
)
2
= a
2
:(45)
Together with Eq.(37) for zcomponent,this denes a helix as a trajectory of a
particle.
Problem 17.Find the eigenfrequencies for an isotropic threedimensional har
monic oscillator realized as a particle of charge q placed in a uniform magnetic,B,
and electric,E,elds which are mutually perpendicular and take their directions
along z and xaxes,respectively.
Solution:
As the particle is an isotropic harmonic oscillator and has the charge q,its potential
energy may be written as
V =
1
2
kr
2
+q'qA v
1
2
m!
2
0
r
2
+q'qA v;(46)
where we introduced the natural angular frequency of an isotropic oscillator
!
0
=
s
k
m
;(47)
and r = (x;y;z) is the displacement of the particle from the origin.We can easily
check also that the conguration of electric and magnetic eld given in the problem
can be realized by the choice of the scalar and vector potentials in the form
'= Ex;A=
1
2
By;
1
2
Bx;0
:(48)
Indeed,with this choice we have
E = r'= ^xE (49)
and
B
z
=
@A
y
@x
@A
x
@y
=
@
@x
h
1
2
Bx
i
@
@y
h
1
2
By
i
=
1
2
B +
1
2
B = B;(50)
as it should be.It is instructive to notice that in previous problem to obtain the
same result for B we used another choice of A given by Eq.(33).
Now we are able to write the total Lagrangian as
L =
1
2
m( _x
2
+ _y
2
+ _z
2
)
1
2
m!
2
0
(x
2
+y
2
+z
2
) +qEx +
1
2
qB(_xy +x_y):(51)
This Lagrangian create Lagrange's equations
8
>
<
>
:
x +!
2
0
x
qB
m
_y
qE
m
= 0
y +!
2
0
y +
qB
m
_x = 0
z +!
2
0
z = 0:
(52)
The general solution of the last equation is
z = z
0
cos(!
0
t +);(53)
that is the oscillation in zdirection takes place with the natural angular frequency
!
0
.By the change of variables
x = x
0
+
qE
m!
2
0
;(54)
the rst two equations can be represented in a more symmetric form
x
0
+!
2
0
x
0
!
L
_y = 0
y +!
2
0
y +!
L
_x
0
= 0;
(55)
where
!
L
=
qB
m
:(56)
This is the system of the coupled linear dierential equations of the second order,
and hence we can try a solution of type
x
0
= Ae
i!t
;y = Be
i!t
:(57)
Then the system (55) changes to the system of the algebraic equations which is
written in matrix form as
!
2
0
!
2
i!
L
!
i!
L
!!
2
0
!
2
A
B
= 0:(58)
So the secular equation is
!
2
0
!
2
i!
L
!
i!
L
!!
2
0
!
2
= (!
2
0
!
2
)
2
(!
L
!)
2
= 0:(59)
This equation is equivalent two the pair of equations,
!
2
+!
L
!!
2
0
= 0;
!
2
!
L
!!
2
0
= 0;
(60)
which has two positive roots
!
+
=
1
2
[!
L
+
q
!
2
L
+4!
2
0
];
!
=
1
2
[!
L
+
q
!
2
L
+4!
2
0
]:(61)
Hence the oscillator in a combined electric and magnetic eld exhibits the three
eigenfrequencies!
0
;!
+
and!
.Note that rst mode does not depend on the applied
elds at all,and the last two modes are caused by the magnetic eld alone,whereas
the electric eld only causes a displacement
qE
m!
2
0
along its direction.For a weak
magnetic eld,
!
L
<<!
0
;(62)
the frequencies!
are approximated to the form
!
+
=!
0
+
!
L
2
;
!
=!
0
!
L
2
;(63)
while in an opposite case of a strong magnetic eld,
!
L
>>!
0
;(64)
we have
!
+
1
2
h
!
L
+
1 +
2!
2
0
!
2
L
i
=!
L
+
!
2
0
!
L
;
!
1
2
h
!
L
1 +
2!
2
0
!
2
L
i
=
!
2
0
!
L
:(65)
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