Seminar 4: CHARGED PARTICLE IN ELECTROMAGNETIC FIELD

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Nov 16, 2013 (3 years and 6 months ago)

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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 velocity-dependent 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 = rA:(9)
Here,r is the dierential operator dened in Cartesian coordinates by
r 

@
@x
;
@
@y
;
@
@z
!
;(10)
so that
r'=
@'
@x
^x +
@'
@y
^y +
@'
@z
^z (11)
and
rA=





^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 dened by Eqs.(8-9) 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 (rA)

#
:(15)
Let calculate,say,the x-component of this vector force,
F
x
= q
"
(r')
x

@A
x
@t
+

v (rA)

x
#
:(16)
Using the denitions (11) and (12),we obtain
(r')
x
=
@'
@x
(17)
and

v (rA)

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 right-hand side of this equation coincides with the potential
V dened 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 denition
(2) of the generalized force in terms of the velocity-dependent potential,we see that
in our case this potential is dened 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 x-component 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 z-components
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 non-zero 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 z-component,this denes a helix as a trajectory of a
particle.
Problem 17.Find the eigenfrequencies for an isotropic three-dimensional 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 x-axes,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 conguration 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 z-direction 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 dierential 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)