The electromagnetic waves

The transverse and longitudinal waves.

The electromagnetic waves are transverse.

The acoustic waves in air are longitudinal.Also acoustic waves inside liquid are longitudinal.

But, the waves on the surface of water are transverse!

A transverse wave travelling to the right along the string.

Caution: distinguish between the propagation of the wave along the string, and the motion of a particle of the string!

v fTλ λ= =

In many cases the speed of waves does not depend on the wave-length. In the case of light in the vacuum (only in the vacuum!), this constant speed is denoted by c.

A transverse wave travelling to the right along the string.

2π 2π λω = k = v = = ω/k ω = vkT λ T

( 0; ) cos2 ( / ) cos( ; 0) cos2 ( / ) cos( ; ) cos2 ( / / )

y x t A t T A ty x t A x A kxy x t A x t T

π ωπ λ

π λ

= = == = == −

2( ; ) cos ( / )

2cos ( / ) cos ( )

cos ( ) cos( )

y x t A x v tT

A x v t A x vt

A k x vt A kx t

π

πωλω

= − =

= − = −

= − = −

A transverse wave travelling to the right/left along the string.

2π 2π λω = k = v = = ω/k ω = vkT λ T

( ; ) cos( )

( ; ) cos( )

y x t A kx tdenotes

y x tphase

A kx t

ϕωϕ

ω ϕ= +

= − +

+

⇒λv = = ω/kT

phase velocity of the wave

Wave equation

2π 2π λω = k = v = = ω/k ω = vkT λ T

( ; ) cos( )

( ; ) cos( )

right

left

y x t A kx t

denote phasesy x t A kx t

ϕω

ϕω ϕ= +

= − +

+

2 2

2 2 2

( , ) 1 ( , )y t x y t xx v t

∂ ∂=

∂ ∂Caution: partial derivatives

Wave equation

2π 2π λω = k = v = = ω/k ω = vkT λ T

/( ; ) cos( )right lefty x t A kx tω ϕ= +m

2 22

2 2

( , ) ( , )y t x y t xvx t

∂ ∂=

∂ ∂RHS is the transverse accelerationtherefore v2 is proportional to a force and inversely proportional to a mass

2

2

/[ ] [ / ] [( / ) / ]v F

k m F x xμ

ω μ

=

= =

Wave equation

2π 2π λω = k = v = = ω/k ω = vkT λ T

/( ; ) cos( )right lefty x t A kx tω ϕ= +m

2 22

2 2

( , ) ( , )y t x y t xvx t

∂ ∂=

∂ ∂

2 /v F μ=

2 2 2 2 2

( ; )

sin( ) sin( )1sin ( )2

y yy yP x t F v Fx t

F Ak kx t A kx t

F A kx t F A

ω ϕ ω ω ϕ

μ ω ω ϕ μ ω

∂ ∂= = −

∂ ∂⋅ − − ⋅ − − =

⋅ − − = ⋅

Power oscillates between and zero2 2F Aμ ω ⋅

The inverse square law

2/(4 )Intensity P rπ=

Principle of the superposition for the linear wave equation:

2 22

2 2

( , ) ( , )y t x y t xvx t

∂ ∂=

∂ ∂sum of solutions is also a solution

Principle of the superposition for the linear wave equation: sum of solutions is also a solution

Specific solution is determined by the boundary or initial conditions

Reflection at the boundaryReflection at the boundary

( ; ) ( ; ) ( ; )

cos( )cos( )

left righty x t y x t y x t

A kx tA kx t

ωω

= +

= − + ++ −

⇐

at x=0, at any time y(0,t)=0

Principle of the superposition for the linear wave equation: sum of solutions is also a solution.interference, standing waves, etc

Specific solution is determined by the boundary or initial conditions

( ; ) ( ; ) ( ; )

cos( ) cos( )2 sin sin

0,1,2,3...

0, ...

left righty x t y x t y x t

A kx t A kx tA kx t

kL n n

L

ω ωωπ

= +

= − + + −= ⋅

= =

=λ 3λ,λ,2 2

∫Φ

−=⋅dt

drdE Brr

∫ =⋅S

enclosedQSdE0ε

rr

∫S

0B d S =ur ur

0 0( )EcdBdr Idt

μ ε Φ= +ur r∫

Maxwell’s equations: two Gauss’s laws + Faraday’s and Ampere’s laws

2 22 2

2 20 0

( , ) ( , ) 1y t x y t xv cx t ε μ

∂ ∂= =

∂ ∂

!!! !!!( , ) ( , )y zE t x B t xx t

∂ ∂= −

∂ ∂

0 0

( , )( , ) yz E t xB t xx t

ε μ∂∂

− =∂ ∂

2 2

0 02 2

( , ) ( , )y yE t x E t xx t

ε μ∂ ∂

=∂ ∂

The electromagnetic waves equation;attention: waves are transverse

∫Φ

−=⋅dt

drdE Brr

0 0EdBdr

dtμε Φ=

ur r

∫

The speed of the electromagnetic waves in vacuum

2

0 0

8 212 2 2 7 2

1

1 (3 10 / )(8.85 10 / ) (4 10 / )

c

m sC Nm N A

ε μ

π− −

= =

= ×× ⋅ ×

0 0

2

( , )( , )

1 | | | |

yz E t xB t xx t

Bk E E c Bc

ε μ

ω

∂∂− =

∂ ∂

= ⇒ =

The speed and direction of propagation of the electromagnetic waves

83 10 /c m s

E B direction of propagationy z x

≈ ×

× ⇒× =

ur ur

In this example the wave is linearly polarized:E (and B) oscillates strictly along one direction

The electromagnetic spectrum

8

9

3 10 / 310 10

c m s cmf Hz

λ ⋅= = =⋅

Propagation of the electromagnetic waves

83 10 /c m s

E B direction of propagation

≈ ×

× ⇒× =y z x

ur ur

( , ) ( , )

( ; ) cos( )( ; ) cos( )

( ; ) cos( )

y z

y

z

E t x B t xx t

y x t A kx tE x t E kx t

B x t B kx t

ω ϕω

ω

∂ ∂= −

∂ ∂= − += −

= −

The electromagnetic wave in the medium

2 2

0 0

1 1

/

/magn

c v

n c v KK K

v c n

ε μ εμ= =

= = ≈

=

index of refraction

K depends very strongly on frequency:index of refraction of water in visible light is 1.8,while in a steady electric field (i.e., when f=0) K=80 !

Kmagn is usually close to 1, but in ferrites Kmagn=10000

analog of the kinetic energy

20

2

02

0

( , )2

( , )2

( , )

E x tu

B x tu

u E x t

ε

μ

ε

= ⇒

= ⇒

=

E

B

analog of the potential energy

What oscillates in the electromagnetic waves:two kind of energies in electromagnetism

Energy pulsates: 2 2

0 0

22 2 0

0

( , ) ( , )

cos ( , )2

u E x t u E x t

EE x t

ε ε

εε

= ⇒ = =

=

another ½ originating from pulsation

Energy transferred per unit time per unit cross-sectional area;Poynting vector

20

20

0

0

[ ( , )]

( , )

( , ) ( , )( , )

dU udV E x t cdt AdUS c E x tA dt

E x t B x tcu x t

E BS P S d A

ε

ε

μ

μ

= = ⋅ ⋅

= = =⋅

=

×= = ⋅ur ur

ur ur ur∫

energy-”train” delivery

2

0 0

0

20

( ; ) cos( )

( ; ) cos( )

cos ( )

2

energy train: [ ( , )]

y

z

E x t E kx t

B x t B kx t

E B EBS kx t l

Power dUIA Adt

EBS c u

dU udV E x t cdt A

ω

ω

ωμ μ

μ

ε

= −

= −

×= = −

≡ ≡ ≡

= =

= = ⋅ ⋅

ur urur r

Power transferred per unit cross-sectional area; Poynting vector; intensity of light

Power transferred per unit cross-sectional area; Poynting vector; momentum; pressure

2( ) / 2for particles: /

( )for light: ( )

d p dp mv p m vdp dp

d pc p cpdp

ε

ε ε

= = =

= ⇒ =

Velocity of propagation is determined by the relation between the energy and momentum:

02momentum "train" delivery

/( ) //( ) / pressure /rad

EBI S c u

dp dt A dp dU cdp dt A I c P I cα

μ≡ = =

⋅ =⋅ = ⇒ =

Black body (everything is absorbed): Prad=I/c

Mirror (everything is reflected): Prad=2I/c

02/( ) / pressure /

/rad

rad

EBI S c u

dp dt A I c P I cP I c

αμ

α

≡ = =

⋅ = ⇒ ==

Black body (everything is absorbed): Prad=I/c

Mirror (everything is reflected): Prad=2I/c

Power transferred per unit cross-sectional area; Poynting vector; momentum; pressure

Example 32.5

3 2 2 3

3 26 2 5

8 2

(1.4 10 / )(4 ) 5.6 10 5.6

1.4 10 / (4.7 10 )(4 ) 1.9 103 10 /

power

rad

P IA W m m W kW

W m NF P A A m Nm s m

− −

= = × = × =

×= = = × = ×

×

Reflection in a perfect conductor (mirror)

( ; ) cos( )

( ; ) cos( )( ; ) cos( )

( ; ) cos( )

( ; ) ( ; ) ( ; ) 2 sin sin

( ; ) ( ; ) ( ; ) 2 cos cos

y in

z in

y ref

z ref

y y in y ref

z z in z ref

E x t E kx t

B x t B kx tE x t E kx t

B x t B kx t

E x t E x t E x t E kx t

B x t B x t B x t B kx t

ω

ωω

ω

ω

ω

= + +

= − += − −

= −

= + = − ⋅

= + = − ⋅

−

To what moment this figure does correspond? What will happen with E and B in the next moment?

Principle of the superposition for the linear wave equation: sum of solutions is also a solution

Specific solution is determined by the boundary or initial conditions

Electric field reflection at mirror

( ; ) ( ; ) ( ; )

cos( )cos( )

left righty x t y x t y x t

A kx tA kx t

ωω

= +

= − + ++ −

⇐

at x=0, at any time y(0,t)=0

Magnetic field reflection at mirror