Chapter 9

Particle in Constant Magnetic Field

As a simple example, we consider the motion of a charged particle in a magnetic field. Our

relativistic eqn. motion is

mduµ

dτ=e

cuνF

µν (9.1)

and so in component form

µ = i,dpi

dτ=e

cujF

ij; F ij = εijkBk, pi = mui = m

dxi

dτ(9.2)

or

dpi

dτ=e

c(~u× ~B)i (9.3)

We can rewrite Eq.9.3 in terms of d/dt since

dpi

dτ=dt

dτ

dpi

dt, ui =

dxi

dτ=dt

dτ

dxi

dt(9.4)

so

d~p

dt=e

c(~v × ~B) (9.5)

70

Note however that

pi = mdxi

dτ= m

dt

dτ

dxi

dt= mγvi; γ =

1√1− v2/c2

(9.6)

So Eq.9.5 is indeed different from the non.rel formula. For the time component:

µ = 0,dp0

dτ=e

cuνF

0ν ;e

cuiF

0i = 0; (9.7)

Hence

p0 =mc√

1− v2/c2= const (9.8)

This means that γ=const and Eq.9.5 reduces to

d~v

dt=

e

γmc~v × ~B =

ec

E~v × ~B; E = mc2 (9.9)

This is now identitical to the non-rel. case with m → E/c2. Consider the magnetic field

along z-axis

~B = (0, 0, B) (9.10)

Then

dvxdt

= ωBvy; ωB =ecB

E(9.11)

dvydt

= −ωBvx; (9.12)

dvzdt

= 0 (9.13)

Thus particle is a free particle in the z-direction:

vz = const; z(t) = z0 + vz0t (9.14)

and moves in a circle in x-y plane i.e., we can integrate Eq.9.11 and Eq.9.12 to get

vx = v⊥sin(ωBt+ φ) (9.15)

71

vy = v⊥cos(ωBt+ φ) (9.16)

with

v2x(t) + v2

y(t) = v2⊥ = const (9.17)

Integrating once more gives

x(t) = − v⊥ωB

cos(ωBt+ φ); y(t) =v⊥ωB

sin(ωBt+ φ) (9.18)

and so the radius of circle is obtained from

ρ2 = x2(t) + y2(t) =v2⊥ω2B

(9.19)

or

ρ =v⊥ωB

(9.20)

Inserting in ωB gives

ρ =mγv⊥c

eB=p⊥c

eB; p⊥ = mom. in x− y plane (9.21)

This is the same as in non-rel. case except that now

p⊥ =mv⊥√

1− v2/c2; v2 = v2

⊥ + v2z0

(9.22)

Thus in general the particle moves in a helix gives determination of x-y component of mo-

mentum.

72