Seminar 4: CHARGED PARTICLE IN ELECTROMAGNETIC FIELD
Introduction
Let take Lagrange’s equations in the form that follows from D’Alembert’s principle,
d ∂T
dt ∂ q̇j
!
−
∂T
= Qj ,
∂qj
(1)
and suppose that the generalized force is derivable not from the potential V (qj ) but
from a more general function U (qj , q̇j ), by the prescription
!
∂U
d ∂U
Qj = −
+
.
∂qj dt ∂ q̇j
(2)
Substituting this expression for the force into Eqs. (1) yields
d ∂T
dt ∂ q̇j
!
!
∂T
∂U
d ∂U
−
=−
+
,
∂qj
∂qj dt ∂ q̇j
or
(3)
!
d ∂(T − U )
∂(T − U )
= 0.
−
dt
∂ q̇j
∂qj
(4)
It follows that these equations can be written in the form of Lagrange’s equations,
d ∂L
dt ∂ q̇j
!
−
∂L
= 0,
∂qj
(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
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)
h
i
F = q E + (v × B) ,
(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 ϕ(r, t) and B(r, t) by
E = −∇ϕ −
∂A
∂t
(8)
and
B = ∇ × A.
(9)
Here, ∇ is the differential operator defined in Cartesian coordinates by
!
∂ ∂ ∂
, ,
,
∂x ∂y ∂z
(10)
∂ϕ
∂ϕ
∂ϕ
x̂ +
ŷ +
ẑ
∂x
∂y
∂z
(11)
∇≡
so that
∇ϕ =
and
x̂
∂
∂x
ŷ
ẑ ∂
∂ ,
∇×A=
∂y
∂z Ax Ay Az
where x̂, ŷ and ẑ are the unit vectors along x-, y- and z axes, respectively.
(12)
Notice that the electromagnetic field defined by Eqs. (8-9) don’t change when the
potentials are transformed according to:
ϕ0 = ϕ −
∂ψ
,
∂t
A0 = A + ∇ψ,
(13)
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
1
L = mv 2 − qϕ + qA · v.
2
(14)
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
"
#
∂A F = q −∇ϕ −
+ v × (∇ × A) .
∂t
(15)
Let calculate, say, the x-component of this vector force,
#
"
∂Ax + v × (∇ × A) .
Fx = q −(∇ϕ)x −
x
∂t
(16)
Using the definitions (11) and (12), we obtain
(∇ϕ)x =
∂ϕ
∂x
(17)
and
v × (∇ × A)
x
= vy
∂Ay
∂x
−
∂Ax
∂y
− vz
∂Ax
∂z
−
∂Az
∂x
y
z
x
= vy ∂A
+ vz ∂A
+ vx ∂A
∂x
∂x
∂x
x
x
x
−vy ∂A
− vz ∂A
− vx ∂A
∂y
∂z
∂x
=
∂
(v
∂x
· A) −
dAx
dt
+
∂Ax
,
∂t
(18)
since
∂
∂
∂Ax
∂Ay
∂Az
(v · A) =
(vx Ax + vy Ay + vz Az ) = vx
+ vy
+ vz
∂x
∂x
∂x
∂x
∂x
(19)
and
dAx
∂Ax
∂Ax
∂Ax
∂Ax
=
+ vx
+ vy
+ vz
.
dt
∂t
∂x
∂y
∂z
Substituting Eqs. (17) and (18) into Eq. (16), we finally obtain
h
−
Fx = q − ∂ϕ
∂x
∂Ax
∂t
+
h
∂
(v
∂x
· A) −
dAx
dt
+
∂
q − ∂x
ϕ−v·A +
∂Ax
∂t
i
d ∂
dt ∂vx
h
(20)
∂
= q − ∂x
ϕ−v·A −
i
ϕ−v·A ,
dAx
dt
i
=
(21)
which can be written as
Fx = −
d ∂U
∂U
,
+
∂x
dt ∂vx
(22)
where we introduce the function
U = qϕ − qv · A = qϕ − qA · v.
(23)
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),
1
L = T − U = mv 2 − qϕ + qA · v.
2
(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 field given by the
scalar and vector potentials ϕ and A.
Solution:
The x-component of the conjugate momentum is
)
(
∂L
∂ 1
px =
=
m[vx2 + vy2 + vz2 ] − qϕ + q[Ax vx + Ay vy + Az vz ] = mvx + qAx . (25)
∂ ẋ
∂vx 2
The same for y- and z-components
py = mvy + qAy
(26)
pz = mvz + qAz .
(27)
and
The second terms in these expressions play the role of a potential momentum.
The energy function is
h=
∂L
j q̇j ∂ q̇j − L =
P
− qϕ + qA · v
2
j q̇j pj − L = vx px + vy py + vz pz − mv − qϕ + qA · v
P
= vx mvx + qAx + vy mvy + qAy + vz mvz + qAz −
= mv 2 + qA · v −
1
mv 2
2
1
mv 2
2
− qϕ + qA · v = 21 mv 2 + 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
T + qϕ = 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.
Problem 16. A particle of the mass m and charge q moves in a constant magnetic
field
B = (0, 0, B).
Show that the orbit of a particle is a helix.
(30)
Solution:
Let specify the scalar and vector potentials for the case when the electric field is
absent,
E = 0,
(31)
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 ∂Ax
−
,
∂x
∂y
(32)
we can choose the potentials in the form
A = (0, Bx, 0),
ϕ = 0.
(33)
With this choice, the Lagrangian is
L=
m 2
(ẋ + ẏ 2 + ż 2 ) + qBxẏ.
2
(34)
The corresponding Lagrange’s equation are written as
 d

 dt (mẋ) − qB ẏ


d
(mẏ + qBx)
dt
d
(mż) = 0.
dt
=0
=0
(35)
From the second and third equations it follows
ẏ = C − ωx
(36)
z = z0 + Dt,
(37)
and
where C, D, z0 are constants and the substitution
ω=
qB
m
(38)
has been made. With the value of the ẏ given by (36), the first equation from (35)
takes the form
ẍ − ω(C − ωx) = 0,
(39)
ẍ + ω 2 (x − x0 ) = 0,
(40)
or
where
x0 =
C
.
ω
(41)
The general solution of Eq. (40) can be written as
x − x0 = a cos(ωt + δ).
(42)
ẏ = C − ωx = ω(x0 − x) = −aω cos(ωt + δ).
(43)
Then
Integrating this equation yields
y = −a sin(ωt + δ) + y0 .
(44)
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 harmonic 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
1
1
V = kr2 + qϕ − qA · v ≡ mω02 r2 + qϕ − qA · v,
2
2
(46)
where we introduced the natural angular frequency of an isotropic oscillator
s
ω0 =
k
,
m
(47)
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,
1
1
A = − By, Bx, 0 .
2
2
(48)
Indeed, with this choice we have
E = −∇ϕ = x̂E
(49)
and
∂ h1 i
∂ h 1 i 1
1
∂Ay ∂Ax
−
=
Bx −
(50)
− By = B + B = B,
∂x
∂y
∂x 2
∂y 2
2
2
as it should be. It is instructive to notice that in previous problem to obtain the
Bz =
same result for B we used another choice of A given by Eq. (33).
Now we are able to write the total Lagrangian as
1
1
1
L = m(ẋ2 + ẏ 2 + ż 2 ) − mω02 (x2 + y 2 + z 2 ) + qEx + qB(−ẋy + xẏ).
2
2
2
(51)
This Lagrangian create Lagrange’s equations

qB
2

 ẍ + ω0 x − m ẏ
− qE
=0
m
qB
2
ÿ + ω0 y + m ẋ = 0


z̈ + ω02 z = 0.
(52)
The general solution of the last equation is
z = z0 cos(ω0 t + δ),
(53)
that is the oscillation in z-direction takes place with the natural angular frequency
ω0 . By the change of variables
x = x0 +
qE
,
mω02
(54)
the first two equations can be represented in a more symmetric form
0
ẍ
+ ω02 x0 − ωL ẏ = 0
ÿ + ω02 y + ωL ẋ0 = 0,
(55)
where
qB
.
(56)
m
This is the system of the coupled linear differential equations of the second order,
ωL =
and hence we can try a solution of type
x0 = Aeiωt ,
y = Beiωt .
(57)
Then the system (55) changes to the system of the algebraic equations which is
written in matrix form as
2
ω
− ω2
iωL ω
0
−iωL ω
ω02 − ω 2
A
B
= 0.
(58)
So the secular equation is
2
2
ω0 − ω
iωL ω
−iωL ω = (ω02 − ω 2 )2 − (ωL ω)2 = 0.
ω02 − ω 2 (59)
This equation is equivalent two the pair of equations,
2
ω
+ ωL ω − ω02 = 0;
ω − ωL ω − ω02 = 0,
(60)
2
which has two positive roots
ω+ = 21 [ωL +
q
ω− = 12 [−ωL +
ωL2 + 4ω02 ],
q
ωL2 + 4ω02 ].
(61)
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
qE
mω02
along its direction. For a weak
magnetic field,
ωL << ω0 ,
(62)
the frequencies ω± are approximated to the form
ω+ = ω0 +
ωL
,
2
ω− = ω0 −
ωL
,
2
(63)
while in an opposite case of a strong magnetic field,
ωL >> ω0 ,
(64)
we have
ω+ ≈
1
2
h
ωL + 1 +
ω− ≈
1
2
h
2ω02
2
ωL
−ωL 1 +
i
2ω02
2
ωL
= ωL +
i
=
ω02
.
ωL
ω02
,
ωL
(65)