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 = 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.(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 (rA)

#

:(15)

Let calculate,say,the x-component 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 right-hand 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 velocity-dependent 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 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 denes 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 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 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 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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