Charged particle in a electromagnetic field 2

Charged particle in a electromagnetic field 2

Seminar 4: CHARGED PARTICLE IN ELECTROMAGNETIC FIELD Introduction

Let take Lagrange’s equations in the form that follows from D’Alembert’s principle,

d dt

∂T

∂ ˙qj

!

− ∂T

and suppose that the generalized force is derivable not from the potential V (qj) but from a more general function U (qj, ˙qj), by the prescription

Qj = −∂U

∂qj

+ d dt

∂U

∂ ˙qj

!

Substituting this expression for the force into Eqs (1) yields

d dt

∂T

∂ ˙qj

!

− ∂T

∂qj = −

∂U

∂qj +

d dt

∂U

∂ ˙qj

!

or

d dt

∂(T − U )

∂ ˙qj

!

− ∂(T − U )

∂qj

It follows that these equations can be written in the form of Lagrange’s equations,

d dt

∂L

∂ ˙qj

!

− ∂L

∂qj

if we use as the Lagrangian the function

The function U is called usually the velocity-dependent potential (sometimes the term generalized 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 field As you know, a charge

q moving with the velocity v in an electromagnetic field, containing both an electric,

E, and magnetic, B, fields, experiences a force (Note: 1/c appears in Gauss system

of units, in SI system it will be absent)

F = qhE + 1

c(v × B)

i

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 ϕ(r, t) and B(r, t) by

E = −∇ϕ − 1

c

∂A

and

Here, ∇ is the differential operator defined in Cartesian coordinates by

∂x,

∂

∂y,

∂

∂z

!

so that

∇ϕ = ∂ϕ

∂xx +ˆ

∂ϕ

∂yy +ˆ

∂ϕ

and

∇ × A =

ˆ

x yˆ ˆ

∂

∂x

∂

∂y

∂

∂z

Ax Ay Az

where ˆx, ˆy and ˆz are the unit vectors along x-, y- and z axes, respectively

Notice that the electromagnetic field defined by Eqs (8-9) don’t change when the potentials are transformed according to:

ϕ0 = ϕ − ∂ψ

∂t, A

0

where ψ 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 field given by the scalar and vector potentials ϕ and A is

L = 1

2mv

2− qϕ + q

Solution:

Taking the vectors of the electric and magnetic fields E and B represented in terms

of the scalar and vector potentials in accordance with Eqs (8) and (9), we have for the Lorentz force

F = q

"

−∇ϕ − 1

c

∂A

∂t +

1 c



Let calculate, say, the x-component of this vector force,

Fx = q

"

−(∇ϕ)x−1

c

∂Ax

∂t +

1 c



v × (∇ × A)

x

#

Using the definitions (11) and (12), we obtain

(∇ϕ)x = ∂ϕ

and



v × (∇ × A)

x = vy∂Ay

∂x − ∂A x

∂y



− vz∂A x

∂z −∂A z

∂x



= vy∂Ay

∂x + vz∂A z

∂x + vx∂A x

∂x

−vy∂A x

∂y − vz∂A x

∂z − vx∂A x

∂x

= ∂x∂ (v · A) −dAx

dt + ∂Ax

since

∂

∂x(v · A) =

∂

∂x(vxAx+ vyAy+ vzAz) = vx

∂Ax

∂x + vy

∂Ay

∂x + vz

∂Az

and

dAx

dt =

∂Ax

∂t + vx

∂Ax

∂x + vy

∂Ax

∂y + vz

∂Ax

Substituting Eqs (17) and (18) into Eq (16), we finally obtain

Fx = qh−∂ϕ∂x − 1

c

∂A x

∂t + 1c∂x∂ (v · A) −1cdAx

dt + 1c∂Ax

∂t

i

= qh− ∂

∂x



ϕ − 1cv · A−1

c

dA x

dt

i

= qh− ∂

∂x



ϕ − 1cv · A+dtd ∂v∂

x



ϕ − 1cv · Ai, (21) which can be written as

Fx= −∂U

∂x +

d dt

∂U

where we introduce the function

U = qϕ − q

cv · A = qϕ −

q

Note that the term in the right-hand side of this equation coincides with the potential

V defined by Eq (2.119) from our 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 definition (2) of the generalized force in terms of the velocity-dependent potential, we see that

in our case this potential is defined just by the equation (23) This observation immediately yields the Lagrangian in the desired form given by Eq (6),

L = T − U = 1

2mv

2− qϕ + q

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 field given by the scalar and vector potentials ϕ and A

Solution:

The x-component of the conjugate momentum is

px = ∂L

∂ ˙x =

∂

∂vx

(

1

2m[v

2

x+ vy2+ v2z] − qϕ +q

c[Axvx+ Ayvy+ Azvz]

)

= mvx+q

cAx (25) The same for y- and z-components

py = mvy +q

and

pz = mvz+ q

The second terms in these expressions play the role of a potential momentum The energy function is

h =P

jq˙j∂ ˙∂Lq

j − L =P

jq˙jpj − L = vxpx+ vypy+ vzpz −1

2mv2− qϕ +qcA · v

= vxmvx+ qcAx+ vymvy +qcAy+ vzmvz +qcAz−1

2mv2 − qϕ +q

cA · v

= mv2+qcA · v −12mv2− qϕ + q

cA · v= 12mv2+ qϕ = T + qϕ (28)

If A and ϕ are independent of t, then L does not depend on t explicitly and the energy function is constant, that is

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

Problem 16 A particle of the mass m and charge q moves in a constant magnetic field

Show that the orbit of a particle is a helix

Let specify the scalar and vector potentials for the case when the electric field is absent,

and magnetic field has only non-zero component Bz = B Keeping in mind that this component is expressed in terms of the vector potential A as

Bz = ∂Ay

∂x −∂Ax

we can choose the potentials in the form

With this choice, the Lagrangian is

L = m

2( ˙x

2+ ˙y2+ ˙z2) + q

The corresponding Lagrange’s equation are written as







d

dt(m ˙x) −qcB ˙y = 0

d

dt(m ˙y +qcBx) = 0

d

dt(m ˙z) = 0

(35)

From the second and third equations it follows

˙

and

where C, D, z0 are constants and the substitution

ω = qB

has been made With the value of the ˙y given by (36), the first equation from (35) takes the form

¨

or

¨

where

x0 = C

The general solution of Eq (40) can be written as

Then

˙

y = C − ωx = ω(x0− x) = −aω cos(ωt + δ) (43) Integrating this equation yields

Finally, combining Eqs (42) and (44) we obtain

(x − x0)2+ (y − y0)2 = a2 (45) Together with Eq (37) for z-component, this defines 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, fields 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

2kr

2+ qϕ −q

cA · v ≡

1

2mω

2

0r2+ qϕ − q

where we introduced the natural angular frequency of an isotropic oscillator

ω0 =

s

k

and r = (x, y, z) is the displacement of the particle from the origin We can easily check also that the configuration of electric and magnetic field given in the problem can be realized by the choice of the scalar and vector potentials in the form

ϕ = −Ex, A =−1

2By,

1

2Bx, 0



Indeed, with this choice we have

Bz = ∂Ay

∂x − ∂Ax

∂y =

∂

∂x

h1

2Bx

i

− ∂

∂y

h

−1

2By

i

= 1

2B +

1

2B = 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

2m( ˙x

2

+ ˙y2+ ˙z2) −1

2mω

2

0(x2+ y2+ z2) + qEx + 1

2cqB(− ˙xy + x ˙y). (51) This Lagrangian create Lagrange’s equations







¨

x + ω2

0x − qBmcy −˙ qEm = 0

¨

y + ω02y + qBmc˙x = 0

¨

z + ω02z = 0

(52)

The general solution of the last equation is

that is the oscillation in z-direction takes place with the natural angular frequency

ω0 By the change of variables

x = x0+ qE

mω2 0

the first two equations can be represented in a more symmetric form



¨

x0+ ω2

0x0− ωLy = 0˙

¨

y + ω2

where

ωL= qB

This is the system of the coupled linear differential equations of the second order, and hence we can try a solution of type

Then the system (55) changes to the system of the algebraic equations which is written in matrix form as



ω02− ω2 −iωLω

iωLω ω20− ω2

 

A B



So the secular equation is

ω2

0 − ω2 −iωLω

iωLω ω2

0 − ω2

= (ω02− ω2)2− (ωLω)2 = 0 (59)

This equation is equivalent two the pair of equations,



ω2+ ωLω − ω02 = 0;

ω2− ωLω − ω2

which has two positive roots

ω+= 12[ωL+qω2

L+ 4ω2

0],

ω− = 12[−ωL+qω2

L+ 4ω2

Hence the oscillator in a combined electric and magnetic field exhibits the three eigenfrequencies ω0, ω+ and ω− Note that first mode does not depend on the applied fields at all, and the last two modes are caused by the magnetic field alone, whereas the electric field only causes a displacement mωqE2 along its direction For a weak magnetic field,

the frequencies ω± are approximated to the form

ω+ = ω0+ ωL

2 ,

ω− = ω0 −ω L

while in an opposite case of a strong magnetic field,

we have

ω+ ≈ 1 2

h

ωL+1 + 2ωω22

L

i

= ωL+ωω2

L,

ω−≈ 1 2

h

−ωL1 + 2ωω22

L

i

= ωω2

Show that the Lagrangian of a particle with the charge q moving with the velocity

v in an electromagnetic field given by the scalar and... 4

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 field. .. three-dimensional har-monic oscillator realized as a particle of charge q placed in a uniform magnetic, B, and electric, E, fields which are mutually perpendicular and take their directions along z- and x-axes,