Lecture 4

Chapter 2 Wave Motion

 Plane waves  3D Differential wave equation  Spherical waves  Cylindrical waves 3-D waves: plane waves (simplest 3-D waves) All the surfaces of constant phase of disturbance form parallel planes that are perpendicular to the propagation direction 3-D waves: plane waves (simplest 3-D waves) All the surfaces of constant phase of disturbance form parallel planes that are perpendicular to the propagation direction

An equation of plane that is ˆ ˆ ˆ perpendicular to k  kx i  k y j  kzk Unit vectors   k  r  const  a All possible coordinates of vector r are on a plane  k Can construct a set of planes over which  varies in space harmonically:     r  Asink  r     or  r  Acosk  r     or  r  Aeikr Plane waves     r  sink  r  The spatially repetitive nature can be expressed as:     k     r  r    k  In exponential form:          r  Aeikr  Aeikr  k / k  Aeikr eik For that to be true: ei2  1 k  2 2 k   Vector k is called propagation vector Plane waves: equation

    r  Aeik r This is snap-shot in time, no time dependence

To make it move need to add time dependence the same way as for one-dimensional wave:     r,t Aeik r t  Plane wave equation Plane wave: propagation velocity

Can simplify to 1-D case assuming that wave propagates along x:   ˆ   r || i  ikxt   r,t  Aeikr t   r,t Ae

We have shown that for 1-D wave phase velocity is:

 That is true for any direction of k v   + propagate with k k - propagate opposite to k

More general case: see page 26 Example: two plane waves

Same wavelength: k1= k2=k=2/, Write equations for both waves. Solution:     Aeikr t Same speed v: 1=2==kv Dot product:   k  r  kx x  k y y  kz z      Wave 1: k1  r  k1z Wave 2: k2  r  k2 sin  y  k2 cos  z direction   ik1zt ik2 y sin z cos t 1  A1e  2  A2e

1  A1 coskz  t  2  A2 cosky sin  z cos  t

Note: in overlapping region  = 1 + 2 Plane waves: Cartesian coordinates

    ik r t   r,t  Ae k  r  kx x  k y y  kz z

Wave eq-ns in Cartesian coordinates:

 x, y, z,t  Aeikx xk y ykz zt      x, y, z,t  Aeik  x y z t , ,  - direction cosines of k  2 2 2 k  kx  k y  kz   2  2   2  1 Importance of plane waves: • easy to generate using any harmonic generator • any 3D wave can be expressed as superposition of plane waves Three dimensional differential wave equation    Taking second derivatives for  x, y, z,t Aeik  x y z t can derive the following:   2 2   2 2    k    2 x2 2  t + 2 2 1     2 2       k  2 t2 y2 +  2    2 2 combine and use:  v   k  k z2   2 2 2        k 2  x2 y2 z2 3-D differential wave equation 2 2 2 1 2    x2 y2 z2 v2 t2   Three dimensional differential wave equation 2 2 2 1 2 Alternative expression    x2 y2 z2 v2 t2 Use Laplacian operator:  2 2 2 2 2 1   2          2 2 x2 y2 z2 v t

    x, y, z,t Aeik  x y z t Using =kv, we can rewrite    x, y, z,t Aeik  x yzvt  function of x  y  z  vt It can be shown, that:  x, y, z,t  f x  y  z  vt f, g are plane-wave  solutions of the diff. eq- n, provided that are twice differentiable.  x, y, z,t  g x  y  z  vt Not necessarily  harmonic! In more general form, the combination is also a solution:      x, y, z,t  C1 f r  k / k  vt C2 gr  k / k  vt Example

Given expression:   x , t    ax  bt  c  2 , where a>0, b>0 Does it correspond to a traveling wave? What is its speed? Solution: 1. Function must be twice differentiable    2ax  bt  c a  2ax  bt  c b x t 2 2   2    2 2  2a 2  2b x t  2. Speed: 2 2 2 1 2    x2 y2 z2 v2 t2 1 b 2a2  2b2 v  v2 a Direction: negative x direction Example

Given expression   x , t   ax  2  bt ,where a>0, b>0: Does it correspond to a traveling wave? What is its speed? Solution: 1. Function must be twice differentiable    2x3  b x t 2 2    6ax4  0 2  t2 x  2. Wave equation: 2 2 2 1 2    x2 y2 z2 v2 t2

6ax4  0 Is not solution of wave equation! This is not a wave traveling at constant speed! Spherical waves

Spherical waves originate from a point 2-D concentric water waves source and propagate at constant speed in all directions: waveforms are concentric spheres. Isotropic source - generates waves in all directions. spherical wave Symmetry: introduce spherical coordinates x  r sin cos y  r sin sin  z  r cos Symmetry: the phase of wave should only depend on r, not on angles:   r  r,  ,  r  1 2 2  Spherical waves v2 t2

Since  depends only on r:

2 1   2     2 r   r r  r  1 2 evaluates to the same 2  r r r2 Wave equation: x  r sin cos  y  r sin sin 1 2 1 2  r  ×r z  r cos r r2 v2 t 2  2 1   2     2 r  2 1 2 r r  r  r  r r2 2 t 2 1     v  2  sin  r sin      1 2  r2 sin2   2 Spherical waves  2 1 2 r  r r2 v2 t 2 This is just 1-D wave equation In analogy, the solution is:

r r,t  f r  vt  f r  vt - propagates outwards (diverging)  r,t   r + propagates inward (converging) Note: solution blows up at r=0

In general, superposition works too: f r  vt gr  vt  r,t  C  C 1 r 2 r Harmonic Spherical waves

f r  vt  r,t  r In analogy with 1D wave: Harmonic spherical wave A  r,t  coskr  vt r

A ik rvt A  r,t  e - source strength r Single propagating Constant phase at any given time: kr=const pulse Amplitude decreases with r A Spherical harmonic waves A  r,t  coskr  vt r

Decreasing amplitude makes sense: Waves can transport energy (even though matter does not move)

The area over which the energy is distributed as wave moves outwards increases

Amplitude of the wave must drop! Note: spherical waves far from source approach plane waves: Cylindrical waves Wavefronts form concentric cylinders of infinite length Symmetry: work in cylindrical coordinates  x  r cos  r  r,  , z  r  y  r sin z  z 1     1 2 r   r r  r  v2 t 2 It similar to Bessel’s eq-n. At larger r the solution can be approximated: Harmonic cylindrical wave A  r,t  coskr  vt 2 1     r   r  r r  r  A 2 2 ik rvt 1    r,t  e   r r2  2 z2 Cylindrical waves

Harmonic cylindrical wave A  r,t  coskr  vt r A  r,t  eik rvt r

Can create a long wave source by cutting a slit and directing plane waves at it: emerging waves would be cylindrical. Lecture 5 Chapter 3

Electromagnetic theory, Photons. and Light

 Basic laws of electromagnetic theory  Maxwell’s equations  Electromagnetic waves  Polarization of EM waves  Energy and momentum Basic laws of electromagnetic theory Electric field Coulomb force law:  1 Q1Q2 F  F  2 40 r electric permittivity of free space

Q1 Q2 Black box F F

  F  EQ1 Interaction occurs via electric field  Electric field can exist even when charge 1 Q2 E  2 rˆ disappears (annihilation in black box) 40 r Basic laws of electromagnetic theory

Magnetic field Moving charges create magnetic field

permeability of free space  The Biot-Savart law for   qv  rˆ B  0 moving charge 4 r2

Magnetic field interacts    with moving charges: Fmagnetic  qv  B

Charges interact with both fields:     F  qE  qv  B (Lorentz force) Basic laws of electromagnetic theory Gauss’s Law: electric Karl Friedrich Gauss (1777-1855) Electric field flux from an enclosed volume is proportional to the amount of charge inside 1  E  q 0   If there are no charges (no 1 sources of E field), the flux E  dS  q   S  is zero: 0 E  dS  0 S

More general form:   1 E  dS  dV S  V 0 Charge density Basic laws of electromagnetic theory Gauss’s Law: magnetic Magnetic field flux from an enclosed volume is zero (no magnetic monopoles)

 M  0   B  dS  0 S Basic laws of electromagnetic theory Faraday’s Induction Law 1822: Michael Faraday Changing magnetic field can result in variable electric field d emf   M dt normal to area   d   Formal E  dl    B  dS area version C dt A  dS  nˆdA

   Changing current in the d  B  dS  B  nˆdA solenoid produces changing M magnetic field B. Changing d M  BdAcos magnetic field flux creates angle between B and electric field in the outer wire. normal to the area dA Basic laws of electromagnetic theory Ampère’s Circuital Law

1826: (Memoir on the Mathematical Theory of Electrodynamic Phenomena, Uniquely Deduced from Experience) A wire with current creates magnetic field around it

All the currents in the universe contribute to B but only ones inside the path result in nonzero path integral Ampere’s law   B  dl  0 Iinside_ path C      Incomplete! B  dl  0 J  dS C A Current density Basic laws of electromagnetic theory Ampère’s-Maxwell’s Law Maxwell considered all known laws and noticed asymmetry:   Gauss’s B  dS  0 S   1 Gauss’s E  dS  q S  0   d   Changing magnetic field Faraday’s E  dl    B  dS C dt A leads to changing electric     field Ampère’s B  dl  0 J  dS C A No similar term here

Hypothesis: changing electric field leads to variable magnetic field Basic laws of electromagnetic theory Ampère’s law Ampère’s-Maxwell’s Law     B  dl  0 J  dS C A

The B will depend on area:     B  dl  0 J  dS  0i C A 1     B  dl  0 J  dS  0 C A 2 Workaround: Include term that takes into account changing electric field flux in

area A2:       E   B  dl  0  J  0   dS Ampère’s-Maxwell’s Law: C A    t  displacement current density Maxwell equations   Gauss’s B  dS  0 S   1 In vacuum Gauss’s E  dS  q S  (free space) 0   d   Faraday’s E  dl    B  dS C dt A   Ampère-     E   B  dl  0  J  0   dS C A   Maxwell’s  t  +     Lorentz force: F  qE  qv  B fields are defined through interaction with charges

Inside the media electric and magnetic fields are scaled. To account

for that the free space permittivity 0 and 0 are replaced by  and :

  KE0   KM 0

dielectric constant, KE>1 relative permeability Maxwell equations   Gauss’s B  dS  0 S   1 In matter Gauss’s E  dS  q S     d   Faraday’s E  dl    B  dS C dt A       E   Ampère- B  dl   J     dS C A   Maxwell’s  t  +     Lorentz force: F  qE  qv  B fields are defined through interaction with charges Maxwell equations: free space, no charges

Current J and charge  are zero Integral form of Maxwell equations in free space:   no magnetic ‘charges’ B  dS  0 S   no electric charges E  dS  0 S     dB  changing magnetic field E  dl    dS C A  creates changing electric  dt  field     changing electric field  E B  dl  00  dS creates changing magnetic C A t field There is remarkable symmetry between electric and magnetic fields! Maxwell equations: differential form (free space)     Notation:   iˆ  ˆj  kˆ   x y z   E  0   2 2 2   2    Laplacian:          B  0 x2 y 2 z2    B    E E E   E     E  div(E)  x  y  z  0 t x y z      E E E  Ez y ˆ  Ex Ez  ˆ  y Ex  ˆ   B      E    i     j    k 0 0 t  y z   z x   x y     E E B   E  curl(E) z  y   x y z t E E B x  z   y z x t E E B y  x   z x y t Electromagnetic waves

(free space) Changing E field creates B field  Changing B field creates E field   B   E   t    E Is it possible to create self-sustaining   B    EM field? 0 0 t Can manipulate mathematically into:    2E  2B 2E    2B    0 0 t 2 0 0 t 2       iˆ  ˆj  kˆ Electromagnetic waves x x x     2 2 2 2 2 2         E   B 2 2 2 2E    2B    x y z 0 0 t 2 0 0 t 2 2E 2E 2E  2E 2B 2B 2B  2B x  x  x  x x  x  x  x x2 y 2 z2 0 0 t 2 x2 y 2 z2 0 0 t 2 2E 2E 2E  2E 2B 2B 2B  2B y  y  y  y y  y  y  y x2 y 2 z2 0 0 t 2 x2 y 2 z2 0 0 t 2 2E 2E 2E 2E  2B 2B 2B 2B z  z  z    z z  z  z    z x2 y 2 z2 0 0 t 2 x2 y 2  z2 0 0 t 2  2 2 2 1 2 Resembles wave equation: 2     x2 y 2 z2 v2 t 2

Each component of the EM field obeys the scalar wave equation, provided that 1  v  00 Light - electromagnetic wave? 1  Maxwell in ~1865 found that EM wave must move at speed v  00

At that time permittivity 0 and permeability 0 were known from electric/magnetic force measurements and Maxwell calculated 1   310,740 km/s v  00 Speed of light was also measured by Fizeau in 1949: 315,300 km/s Maxwell wrote: This velocity is so nearly that of light, that it seems we have strong reason to conclude that light itself (including radiant heat, and other radiations if any) is an electromagnetic disturbance in the form of waves propagated through the electromagnetic field according to electromagnetic laws. celer (lat. - fast) Exact value of speed of light: c = 2.997 924 58 × 108 m/s Electromagnetic wave

Assume: reference frame is chosen so that E=(Ex,0,0) longitudinal wave, propagates along x E E E   x  y  z  0 x y z Ex   E  0  0 Ex does not vary with x   x   B  0    B This cannot be a wave!   E   t Conclusion: it must be transverse wave,    i.e. E =0. Similarly B =0.  E x x   B  00 t Since E is perpendicular to x, we must specify its direction as a function of time Direction of vector E in EM wave is called polarization Simple case: polarization is fixed, i.e. direction of E does not change Polarized electromagnetic wave We are free to chose y-axis so that E field propagating along x is

polarized along y: (0, Ey ,0).   E   E  0 E Ey B y Bz   z    x     B  0 y z t x t E E B  x  z   y   B Also: Bx=By=const (=0)   E   z x t t Ey E B   x   z   E x y t   B    0 0 t E-field of wave has only y component B-field of wave has only z component (for polarized wave propagating along x) In free space, the plane EM wave is transverse Harmonic polarized electromagnetic wave

Harmonic functions are solution for wave equation:

E y x,t  E0 y cost  x / c  

polarized along y axis propagates along x axis Find B:

E B E y y z B   dt   z  x t x  1 B x,t  E cost  x / c   z c 0 y

E y  cBz This is true for any wave: - amplitude ratio is c - E and B are in-phase Harmonic polarized electromagnetic wave Electromagnetic waves

* direction of propagation is in the direction of cross-product:   E  B

* EM field does not ‘move’ in space, only disturbance does. Changing E field creates changing B field and vice versa Energy of EM wave

It was shown (in Phys 272) that field energy densities are:

0 2 1 2 uE  E uB  B 2 20

-1/2 Since E=cB and c=(00) :

uE  uB - the energy in EM wave is shared equally between electric and magnetic fields  2 1 2 2 Total energy: u  uE  uB  0E  B (W/m ) 0 The Poynting vector

EM field contains energy that propagates through space at speed c Energy transported through area A in time t: uAct  Energy S transported by a wave through unit area in unit time: E   c2 uAct 2 1 1 S   uc  c 0E  c 0EcB  0EB  EB At 0 0 0

The Poynting vector:  1   power flow per unit area for a S  E  B wave, direction of propagation John Henry Poynting  is direction of S. (1852-1914) 0 (units: W/m2) The Poynting vector: polarized harmonic wave  1   Polarized EM wave: S  E  B      0 E  E0 cosk  r  t     B  B0 cosk  r  t

Poynting vector:     1 2  S  E0  B0 cos k  r t 0 This is instantaneous value: S is oscillating Light field oscillates at ~10 15 Hz - most detectors will see average value of S. Irradiance      Average value for periodic function: 1 2  S  E0  B0 cos k  r  t need to average one period only. 0

It can be shown that average of cos2 is: cos2 t  1 2 T 1 c S  E B  0 E 2 And average power flow per unit time: T 0 0 0 20 2 Irradiance: Alternative eq-ns: For linear isotropic c c dielectric: I  S  0 E 2 I  c E 2  B2 2 T 0 0 T T I  v E 2 0 T Irradiance is proportional to the square of the amplitude of the E field

Usually mostly E-field component interacts with matter, and we will refer to E as optical field and use energy eq-ns with E Optical power  radiant flux  total power falling on some area (Watts) Spherical wave: inverse square law Spherical waves are produced by point sources. As you move away from the source light intensity drops Spherical wave eq-n: A  r,t  coskr  vt r    E    B   E  0 cosk  r  t B  0 cosk  r  t r r      1  E0 B0  2  S    cos k  r  t 0  r r  c 1 I  S  0 E 2 T 2 r2 0 Inverse square law: the irradiance from a point source drops as 1/r2 Classical EM waves versus photons The energy of a single light photon is E=h

The Planck’s constant h = 6.626×10-34 Js  c E  h  h  4 1019 J Visible light wavelength is  ~ 0.5 m 1  Example: laser pointer output power is ~ 1 mW number of photons emitted every second: 3 P 10 J/s 15  19  2.510 photons/s E1 4 10 J/photon Conclusion: in many every day situations the quantum nature of light is not pronounced and light could be treated as a classical EM wave