Classical Mechanics and Electromagnetism -  MKS vs CGS -  Maxwell equations -  Electromagnetic waves -  special relativity -  coordinate system -  harmonic oscillators

MKS CGS

Speed of the light

Electric field, potential

Charge density, current

Magnetic induction

Vacuum impedance

€

1/ ε0µ0

€

E / 4πε 0

€

φ / 4πε 0

€

E

€

φ

€

ρ

€

j

€

q

€

q 4πε 0

€

ρ 4πε 0

€

j 4πε 0

€

B µ0 /4π

€

4π /c

€

µ0 /ε 0 ≈ 377Ω

€

c

€

B

Maxwell’s equation

  James Clerk Maxwell, “On Physical Lines of Force”, 1861-1862

Differential form Integral form

Gauss’s Law

Farady’s Law

Ampere’s Law

€

∇ • E = ρ /ε 0

€

E • dA∫∫ =Q /ε0

€

∇ • B = 0

€

B • dA∫∫ = 0

€

∇ × E = −∂B∂t

€

∇ × B = µ0

J + µ0ε 0

∂ E ∂t

€

E •dl∫ = −∂φB∂t

€

B •dl∫ = −∂φE∂t

€

∇ • J + ∂ρ

∂t= 0Continuity:

Electric field of a beam of charged particles

  Uniformly distributed beam of charged particles with charge intensity

€

∇ • E = ρ /ε 0

€

Er =

ρ2ε 0

r for r < R

ρ2ε 0

R2

rfor r > R

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

€

ρ

€

E

€

R

€

ρ

Magnetic field of a uniform current

  Uniformly distributed beam of charged particles with current intensity of

€

∇ × B = µ

J

€

Bφ =

J2µ

r for r < R

J2µ

R2

rfor r > R

⎧

⎨ ⎪

⎩ ⎪

€

J

€

B

€

R

€

J

Electromagnetic wave propagation

  propagation of an electromagnetic wave in vacuum

€

(∇2 − µε∂ 2

∂t 2) E = 0

(∇2 − µε∂ 2

∂t 2) B = 0

⎧

⎨ ⎪

⎩ ⎪

€

E ( r ,t) =

E 0e

iωt− i k ⋅ r

B ( r ,t) =

B 0e

iωt− i k ⋅ r

⎧ ⎨ ⎩

€

ω

€

k = k =

2πλ

: angular frequency,

: wave number, is the wavelength

€

λ

€

k =ωc

: dispersion relation

Traveling wave in vacuum

  transverse wave:

€

n

€

E

€

B

€

E ⋅ n = 0

€

B ⋅ n = 0 Propagation direction

€

B 0 = µε

n × E 0

Special Relativity

  The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of the two systems in uniform translational motion relative to each other.

  Invariant of the speed of the light, i.e. the speed of the light in vacuum is the same for all reference systems

  Nothing can be faster than the speed of the light

Lorentz Transformation

  two space-time reference systems, (t, x, y, z) and (t’, x’, y’, z’) moving w.r.t. at a velocity of v along the axis of x

€

ʹ′ t = γ t − vc 2

x⎛

⎝ ⎜

⎞

⎠ ⎟

€

ʹ′ x = γ x − vt( )

€

ʹ′ y = y

€

ʹ′ z = z

€

γ =1

1− v2

c 2

x

y

z

X’

Y’

Z’

: Lorentz factor

Special Relativity

  Time dilution: a clock in the moving reference frame(primed frame) runs slower than in the rest frame of the clock

  Length contraction: the length of a project in the direction of primed frame relative to the rest frame is shorter in the moving frame €

Δt '= γ Δt − vc 2Δx

⎛

⎝ ⎜

⎞

⎠ ⎟

€

Δx = γ Δx'+vΔt'( )

€

Δt '= γΔt

€

Δx '= Δxγ

€

Δx = 0

€

Δt '= 0

Space-time Four vector

  definition

€

S = (ct,x,y,z)

€

| S |= (ct)2 − x 2 − y 2 − z2

  Length of four vector: Lorentz invariant

  Lorentz invariant: length of four vector

€

S ' 2 = (ct')2 − x '2 −y '2 −z'2 = c 2t 2 − x 2 − y 2 − z2 = S2

Four momentum

  Conservation of Four momentum:   Energy conservation   Momentum conservation

€

v

€

P = γmc, p ( )

Fixed target experiment

  In the lab frame

  In the center of mass frame:

€

v 1

€

v 2 = 0

€

P 1 = E1 /c,

p 1( )

€

Ecm /2c, p ( )

€

P 1 + P 2

2LF =

P 1 + P 2

2CM

€

P 2 = m0c, 0( )

€

Ecm /2c, − p ( )

€

E 2cm = (E1 +m0c

2) − c 2p12

Lorentz transformation of E&M fields

  Prime reference system boost w.r.t. the unprime system along x direction

€

ʹ′ E y = γ(Ey − βBz )

x

y

z

X’

Y’

Z’

€

ʹ′ E x = Ex

€

ʹ′ E z = γ(Ez + βBy )

€

ʹ′ B y = γ(By + βEz )

€

ʹ′ B x = Bx

€

ʹ′ B z = γ(Bz − βEy )

€

v

X’

Y’

Z’

Electromagnetic field of a moving charge

  Ratio of transverse electric field and longitudinal electric field is ~vt/b,

  Generate transverse magnetic field   Electric field lines becomes whiskbroom shape instead of

isotropically distributed

x

y

z

€

v Q

Observation point

b

Lorentz force

  A moving charged particle with velocity V in a electric-magnetic field

€

F = q

E + v ×

B ( )

  Home work: Design a Wien filter, a device of static electric and magnetic field arranged in a way that only allow charged particle q with a specific velocity v to pass. In other words, select electric field and magnetic field which is transparent to a 35keV electron beam. Please also describe the trajectory of electron in side the wien filter.

Harmonic oscillator

  Equation of motion   In absence of any frictions or other external force

€

d2xdt 2

+ω 2x = 0

x

€

x(t) = x0 cos(ωt + χ)

k

€

dx(t)dt

= −x0ω sin(ωt + χ)

Driven Harmonic oscillator

  In the presence of an external driving force

€

d2xdt 2

+ω 2x = f (t)

€

x(t) = x0 cos(ωt + χ) +1ω

sinω (t − t ') f (t ')dt '0

t

∫

Driven Harmonic oscillator

  Equation of motion

€

d2x(t)dt 2

+ω 2x(t) = f (t)

€

f (t) = Cmeiωmt

€

x(t) = Aeiωt + Ameiωmt

€

d2x(t)dt 2

+ω 2x(t) = Cmeiωmt

€

= Cmeiωmt

m= 0∑

  for

  Assume solution is like

€

Am =Cm

ω 2 −ωm2

Resonance Response

€

x(t) = Aeiωt +Cm

ω 2 −ωm2

  Response of the harmonic oscillator to a periodic force is

€

ωm

€

ωm

€

ω

Coupled Harmonic Oscillator

  Equation of motion

€

x' '+ωx2x = q2y

€

y' '+ωy2y = q2x

  Assume solutions are:

€

x = Aeiωt

€

y = Beiωt

€

−ω 2A +ωx2A = q2B

€

−ω 2B +ωy2B = q2A

€

(ωx2 −ω 2)(ωy

2 −ω 2) = q4

€

ω 2 =ωx2 +ωy

2 ± (ωx2 −ωy

2)2 + 4q4

2

Coupled Harmonic Oscillator

  The two frequencies of the harmonic oscillator are functions of the two unperturbed frequencies

  When the unperturbed frequencies are the same, a minimum frequency difference

€

ω 2 =ωx2 +ωy

2 ± (ωx2 −ωy

2)2 + 4q4

2

€

ωx

€

ωy

€

ω1

€

ω2

€

Δω ≈q2

ω

Nonlinear Harmonic oscillator

  Harmonic oscillator with high order terms

€

d2xdt 2

+ω 2x = aixi

i=2

n

∑  Pendulum

€

d2θdt 2

+ω 2 sinθ = 0

€

ω =gl

for

€

θ << l

€

d2θdt 2

+ω 2θ ≈ω 2 θ3

6

Exact Solution of Pendulum Equation

  Unlike linear linear oscillator, pendulum has oscillation period as function of its amplitude.

  Stable condition:

  Separatrix:

€

d2θdt 2

+ω 2 sinθ = 0

for C is a constant

€

12ddt

dθdt

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

−ω 2 ddtcosθ = 0

€

12ω 2

dθdt

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

= cosθ +C

€

−1 < C <1

€

C = 0

€

dθdt

= 0

Phase Space

  Pendulum

€

1ω

˙ θ

€

θ /2π

  Linear harmonic

€

xx0

⎛

⎝ ⎜

⎞

⎠ ⎟

2

+˙ x

x0ω

⎛

⎝ ⎜

⎞

⎠ ⎟

2

=1€

˙ x

€

x

Separatrix, boundary between stable and unstable region on which motion is frozen

Nonlinear Oscillator Frequency

  Pendulum period expression

€

12Tω 0 =

12

dθcosθ − cosθ0( )−θ 0

θ 0

∫

€

TT0

=2πK sin2 θ0

2⎛

⎝ ⎜

⎞

⎠ ⎟

⎛

⎝ ⎜

⎞

⎠ ⎟

€

12ω 2

dθdt

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

= cosθ +C

€

ω 0dt =12

dθ| cosθ − cosθ0 |

Nonlinear Resonance

  Unlike linear harmonic resonance, the frequency of a nonlinear oscillator is a function of amplitude   Small angle pendulum, i.e. small x

  Nonlinear resonance amplitude doesn’t grow unlimited because of the detuning, i.e. frequency moves away from resonant condition as amplitude grows

€

(2 /π)K(x) ≈1+ x /4

€

ω ≈ω 0 1−θ02

16⎛

⎝ ⎜

⎞

⎠ ⎟

Home works

  The Stanford LINAC accelerates electrons to 50 GeV in a distance of 2 miles at a constant rate of energy gain. For an observer who rides precariously on an electron, how long does this journey last?

  A positron beam accelerated to 50 GeV in the linac hits a fixed hydrogen target. What is the available energy from a collision with a target electron assumed to be at rest? Compare this with that obtained in a linear collider where electrons and positrons from two linacs collide head on with the same energy.