Classical Wave Motion P. J. Grandinetti · Wave Motion Deﬁnition A wave is a self sustaining disturbance in a continuous medium and can move through space without transporting the

Chapter 08Classical Wave Motion

P. J. Grandinetti

Chem. 4300

P. J. Grandinetti Chapter 08: Classical Wave Motion

Wave Motion

DefinitionA wave is a self sustaining disturbance in a continuous medium and can move through spacewithout transporting the medium.

Even though wave does not transport medium it does transport energy and momentum.

Wave leads to local displacements of medium away from its equilibrium position.

transverse wave: wave displacement is perpendicular to direction of wave propagation

longitudinal wave: wave displacement is parallel to direction of wave propagation


Earthquake Body WavesLongitudinal Transverse

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Seismic waves are called P-waves (primarywaves) and S-waves (secondary waves),respectively, because P-waves travel faster andarrive before S-waves.

Explains why earthquakes often begin withquick upward-downward shock followed byarrival of waves producing side-to-side motion.

Time difference between arrival of P andS-waves used to determine distance tohypocenter where seismic waves are generated.

epicenter is point on earth’s surface directlyabove hypocenter.


What are the equations describing wave motion?Assume that we can describe a particular wave shape by a mathematical function:

f (r⃗, t) ≡ a wave function

Wave function, f (r⃗, t), describes spatial and time dependence of medium displacement.

All possible wave behaviors in a given medium are described by set of solutions to a singlepartial differential equation (PDE) called the wave equation for that medium


Wave Equations

Divide wave equations into 2 broad classes: linear or non-linear partial differential equations.

Linear Partial Differential Equations

Ordinary differential equation, F(

t, f ,dfdx, ...,

dnfdxn

)= 0

is linear if F is linear function of variables f , df∕dx, ..., dnf∕dxn.General linear ordinary differential equation of order n has form

a0(t)dnfdxn + a1(t)

dn−1fdxn−1 +⋯ + an(t)f = g(t)

Similar definition applies to partial differential equations.

What’s so special about Linear Partial Differential Equations?


Superposition PrincipleWhat’s so special about Linear Partial Differential Equations?

DefinitionSuperposition Principle: Any solution to a linear PDE can be added together with othersolutions to form further solutions.∑

nan fn(x, t) = g(x, t) If all fn(x, t) are solutions of a linear PDE, then g(x, t) is

also a solution of same PDE.

All waves modeled by linear PDEs obey the superposition principle.

For non-linear waves—waves modeled by non-linear PDEs—superposition principle does notgenerally apply.


The Classical Wave equationIn 1746 French mathematician Jean le Rond d’Alembert discovered equation that describes alllinear waves in one-dimension

Jean le Rond d’Alembert1717 - 1783

𝜕2f (x, t)𝜕t2 = v2

p𝜕2f (x, t)𝜕x2

f (x, t) is the wave functionvp is wave speed—depends on details ofmedium in which wave travels


The Classical Wave equation

𝜕2f (x, t)𝜕t2 = v2

p𝜕2f (x, t)𝜕x2

f (x, t) is the wave functionvp is the wave speed—depends on medium in which wave travels.

Ideal string wave: vp =√

T∕𝜇, where 𝜇 is mass density of string and T is string tension.

Sound waves in gases: vp =√𝛾RT∕M, where 𝛾 is ratio of heat capacities, 𝛾 = Cp∕Cv,

and M is molar mass of gas.

Sound waves in liquids: vp =√

K∕𝜌, where K is bulk modulus and 𝜌 is liquid density.

Light: vp = 1∕√𝜇𝜖, where 𝜇 is magnetic permeability and 𝜖 is electric permittivity of

space.


The Classical Wave equationTwo arguments to wave function: position, x, and time, t.If wave function arguments have fixed relationship u = x − vptthen wave retains its shape as it moves from left to right through space with time.

As wave equation involves v2p we get another class of solutions by changing sign of vp to

get a right to left traveling wave.


The Classical Wave equationBy re-expressing wave equation in terms of two variables u = x − vpt and v = x + vptd’Alembert showed that wave equation becomes

𝜕2f𝜕u𝜕v

= 0

General solution to this equation is

f (u, v) = g(u) + h(v)

orf (x, t) = g(x − vpt) + h(x + vpt)

g(u) = g(x − vpt) represents wave moving left to right, i.e., in positive x direction,

h(v) = h(x + vpt) represents wave moving right to left, i.e., in negative x direction.


Traveling Sinusoidal (Harmonic) Waves

λ

A

A

A

Sinusoidal (harmonic) waves are useful in study ofwaves for many reasons.

Main reason is that superposition principle allowsus to decompose arbitrary wave shapes into linearcombinations of sinusoidals.

A sinusoidal wave is f (x, t) = A cos[k(x − vpt

)+ 𝛿

]▶ A is wave amplitude▶ vp is wave velocity▶ 𝛿 is phase constant▶ k is wave number

k = 2𝜋𝜆

▶ 𝜆 is wavelength—spatial periodicity of wave


Traveling Sinusoidal (Harmonic) Waves

λ

A

A

A

wave phase velocity, vp, is speed that dotmoves.period of traveling sinusoidal wave—durationof time of one cycle—is T = 2𝜋∕(kvp)frequency is inverse of period,𝜈 = 1∕T = kvp∕(2𝜋)angular frequency is 𝜔 = 2𝜋𝜈 = 2𝜋∕T = kvp

It is common to write wave functions in termsof 𝜔 instead of vp


Traveling Sinusoidal WavesAngular frequency is 𝜔 = kvp. Rewrite wave functions in terms of 𝜔.

fR(x, t) = A cos[k(x − vpt

)+ 𝛿

], right traveling wave

becomesfR(x, t) = A cos(kx − 𝜔t + 𝛿), right traveling wave

fL(x, t) = A cos[k(x + vpt

)+ 𝛿

], left traveling wave

becomesfL(x, t) = A cos(kx + 𝜔t + 𝛿), left traveling wave


Traveling Sinusoidal Waves - Complex Number Notation

It is also common to use complex notation to describe wave functions.In complex notation right and left traveling sinusoidal wave function are written

fR(x, t) = ℜ{Aei(kx−𝜔t+𝛿)} = A cos(kx − 𝜔t + 𝛿) right traveling wave

andfL(x, t) = ℜ{Aei(kx+𝜔t+𝛿)} = A cos(kx + 𝜔t + 𝛿) left traveling wave

▶ symbol ℜ means take real part of a complex number.

▶ symbol ℑ means take imaginary part of a complex number.


Standing Waves


Standing Waves

Familiar standing waves are those within musical instruments which generate instrumentsound, such as standing wave of violin string, standing wave of drum head, or standingwave inside wind instrument.

Think of standing waves as superposition of two traveling waves of same frequency andamplitude traveling in opposite directions.

Begin by considering sinusoidal standing waves on a string, where all parts of the waveoscillate in harmonic motion at the same frequency, 𝜔, and with the same phase, 𝛿.Sinusoidal standing wave function has same time dependence at all positions, x, in thestanding wave.


Sinusoidal Standing Wave Function Solutions for a Vibrating StringSeparation of VariablesTo find standing wave functions start with wave equation:

𝜕2f (x, t)𝜕t2 = v2

p𝜕2f (x, t)𝜕x2

Solve PDE using separation of variables—factor wave function into 2 parts:

f (x, t) = X(x)T(t)

X(x) is called stationary state wave function.Substitute X(x)T(t) into classical wave equation

X(x)𝜕2T(t)𝜕t2 = v2

pT(t)𝜕2X(x)𝜕x2

and rearrange1

T(t)𝜕2T(t)𝜕t2 =

v2p

X(x)𝜕2X(x)𝜕x2


Sinusoidal Standing Wave Function Solutions for a Vibrating String

1T(t)

𝜕2T(t)𝜕t2 =

v2p

X(x)𝜕2X(x)𝜕x2

Left hand side depends only on t.

Right hand side depends only on x.

To be true for all x and t both sides must equal same constant value


Sinusoidal Standing Wave Function Solutions for a Vibrating String

1T(t)

𝜕2T(t)𝜕t2 =

v2p

X(x)𝜕2X(x)𝜕x2 = −𝜔2

Left hand side depends only on t.item Right hand side depends only on x.For equality to remain true for all x and t two sides of equation must be always be equalto same constant value.define −𝜔2 as the separation constantTurns 1 PDE into 2 ODEs:

d2T(t)dt2 + 𝜔2 T(t) = 0 and d2X(x)

dx2 + 𝜔2

v2p

X(x) = 0

PDE ≡ partial differential equationODE ≡ ordinary differential equation


Sinusoidal Standing Wave Function Solutions for a Vibrating StringTime dependent part of wave function

d2T(t)dt2

+ 𝜔2 T(t) = 0

For T(t) we propose sinusoidal solutionT(t) = A cos ct + B sin ct

Substitute into ODE for T(t)

A(c2 − 𝜔2) cos ct + B(c2 − 𝜔2) sin ct = 0

Set c = 𝜔 and obtain valid solutions for T(t):T(t) = A cos𝜔t + B sin𝜔t

Use trigonometric identity A sin 𝜃 + B cos 𝜃 = C cos(𝜃 + 𝛿) where

C =√

A2 + B2 and tan 𝛿 = −AB

and rearrange T(t) toT(t) = C cos(𝜔t + 𝛿)


Sinusoidal Standing Wave Function Solutions for a Vibrating StringSpatial dependent part of wave function

d2X(x)dx2 + 𝜔2

v2p

X(x) = 0

For X(x) we propose sinusoidal solution

X(x) = D cos kx + F sin kx

Substitute into ODE for X(x)

D

(k2 − 𝜔2

v2p

)cos kx + F

(k2 − 𝜔2

v2p

)sin kx = 0

Set k = 𝜔∕vp and obtain valid solutions for X(x):

X(x) = D cos kx + F sin kx where (k = 𝜔∕vp)


Sinusoidal Standing Wave Function SolutionsSpatial dependent part of wave function - Boundary ConditionsNext impose boundary conditions for string fixed at x = 0 and x = L.For X(x)

X(x) = D cos kx + F sin kx, where (k = 𝜔∕vp)1st boundary condition gives

X(x = 0) = D = 02nd boundary condition gives

X(x = L) = F sin kL = 0Requires that kL = n𝜋, where n = 1, 2, 3,….Solutions for X(x) become

Xn(x) = Fn sin knx, kn = n𝜋L

From earlier fixed relationship 𝜔 = vpk, we have

𝜔n = vpkn =n𝜋vp

Lor 𝜈n =

nvp

2LP. J. Grandinetti Chapter 08: Classical Wave Motion

Sinusoidal Standing Wave—Harmonic Functions (Normal Modes)

TakingXn(x) = Fn sin knx

together with

Tn(t) = Cn cos(𝜔nt + 𝛿)

we have

fn(x, t) = Xn(x)Tn(t) = an sin knx cos(𝜔nt + 𝛿n)

Amplitude redefined with symbol an.fn(x, t) are harmonic functions (normal modes) forvibrating string.


Fourier Series Analysis of Arbitrary Wave Shape


Fourier Series Analysis of Arbitrary Wave ShapeArbitrary wave with boundary conditions expressed as linear combination of harmonics,

f (x, t) =∞∑

n=1an sin knx cos(𝜔nt + 𝛿n)

an is amplitude of each normal mode contribution to arbitrary wave shape.Assume at t = 0 that string is at maximum displacement and all parts of string have zerodisplacement velocity. This leads to(

𝜕f (x, t)𝜕t

)t=0

= −∞∑

n=1an𝜔n sin knx sin(𝛿n) = 0

which can be satisfied by setting 𝛿n = 0. This leaves us with

f (x, t) =∞∑

n=1an sin knx cos𝜔nt


Fourier Series Analysis of Arbitrary Wave ShapeArbitrary wave with boundary conditions expressed as linear combination of harmonics,

f (x, t) =∞∑

n=1an sin knx cos𝜔nt

Joseph Fourier1768-1830

DefinitionAt any instant in time we can decompose any wave function,f (x, t), in terms of the coefficients, an using a Fourier seriesanalysis

an = 2L ∫

L

0f (x, t) sin

(n𝜋L

x)

dx, n = 1, 2,…


Fourier Series AnalysisImagine harp or guitar string of length L is plucked at the center of string.

Calculate first 3 normal mode amplitudes at t = 0 when string is released.

f (x, 0) = dL∕2

x, 0 ≤ x ≤ L∕2, left of pluck

f (x, 0) = 2d − dL∕2

x, L∕2 ≤ x ≤ L, right of pluck


Standing Wave Function SolutionsTo use Fourier series analysis we 1st define wave function f (x, 0) in 2 parts:

f (x, 0) = 2dL

x, 0 ≤ x ≤ L∕2, left of pluck

f (x, 0) = 2d − 2dL

x, L∕2 ≤ x ≤ L, right of pluck

Break integral into 2 parts

an = 2L ∫

L

0f (x, 0) sin

(n𝜋L

x)

dx,

= 2L

[∫

L∕2

0

2dL

x sin(n𝜋

Lx)

dx + ∫L

L∕2

(2d − 2d

Lx)sin

(n𝜋L

x)

dx

]Evaluating integral gives

an = 8dn2𝜋2 sin n𝜋

2P. J. Grandinetti Chapter 08: Classical Wave Motion

Standing Wave Function SolutionsFourier Series Analysis - Summary

Wave function of plucked stringf (x, 0) = 2d

Lx, 0 ≤ x ≤ L∕2

f (x, 0) = 2d − 2dL

x, L∕2 ≤ x ≤ L

Fourier Series Analysis plucked string wave function

1 2

2

-20

3 4

4

5 6

68 an = 8d

n2𝜋2 sin n𝜋2

f (x, t) =∞∑

n=1an sin knx cos(𝜔nt + 𝛿n)


Web Apps by Paul Falstad

String WavesBar WavesFourier Series


http://www.falstad.com/loadedstring/

http://www.falstad.com/barwaves

http://www.falstad.com/fourier/Fourier.html

Traveling wave packet


Traveling wave packetConsider wave packet traveling through space as function of time.

Unlike standing wave, traveling wave has continuous range of wavelengths possible.

At given instant decompose it into a sum of sinusoidal (single wavelength) waves.

Discrete Fourier series needs to become continuous integral...


Fourier transform decomposes wave packet into component sinusoidsInstead of Fourier series we use Fourier integral transforms:

a(k, t) = 1√2𝜋 ∫

∞

−∞f (x, t) e−ikxdx, and f (x, t) = 1√

2𝜋 ∫∞

−∞a(k, t) eikxdk.

Fourier transform relates 2 equally valid wave descriptions: f (x, t) and a(k, t).

With f (x, t) we speak of wave packet in terms of a position basis,

With a(k, t) we speak of wave packet in terms of a wave number basis.

Note: integral of wave numbers, k, extends over negative values.

A negative k value is not associated with a negative wavelengthbut rather a wave traveling in the opposite direction.


Wave Packets and the Fourier TransformSum of sinusoids leads to localization of wave into packet centered at given position.

++++++++++++++++++

=

Inverse process:Given f (x, 0) find a(k, 0) using Fourier transform.


What is the Fourier transform of a Gaussian wave packet?Take Gaussian shaped wave packet

f (x, t) = A exp

{−1

2(x − vpt)2

𝜎2x

}Taking Fourier transform of f (x, 0):

a(k, 0) = 1√2𝜋 ∫

∞

−∞f (x, 0)e−ikxdx = 1√

2𝜋 ∫∞

−∞A exp

{−1

2x2

𝜎2x

}e−ikxdx

Use Euler’s relation, e−ikx = cos kx − i sin kx, to break integral into 2 parts,

a(k, 0) = A√2𝜋

[∫

∞

−∞exp

{−1

2x2

𝜎2x

}cos kx dx−i∫

∞

−∞exp

{−1

2x2

𝜎2x

}sin kx dx

]Use Euler’s relation, e−ikx = cos kx − i sin kx, to break integral into 2 parts,

a(k, 0) = A√2𝜋

⎡⎢⎢⎢⎢⎢⎣∫

∞

−∞exp

{−1

2x2

𝜎2x

}cos kx

⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟Even function

dx−i∫∞

−∞exp

{−1

2x2

𝜎2x

}sin kx

⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟Odd function

dx

⎤⎥⎥⎥⎥⎥⎦Recall f (x) = f (−x) is an even function, and f (x) = −f (−x) is an odd function.


What is the Fourier transform of a Gaussian wave packet?

a(k, 0) = A√2𝜋

⎡⎢⎢⎢⎢⎢⎣∫

∞

−∞exp

{−1

2x2

𝜎2x

}cos kx

⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟Even function

dx − i∫∞

−∞ ��:0exp

{−1

2x2

𝜎2x

}sin kx

⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟Odd function

dx

⎤⎥⎥⎥⎥⎥⎦2nd integrand is odd function of x. Its integral is zero, leaving us with

a(k, 0) = A√2𝜋 ∫

∞

−∞exp

{−1

2x2

𝜎2x

}cos kx dx

Integral table: ∫ ∞−∞ e−bx2 cos kx dx =

√𝜋∕b e−k2∕(4b). Identifying b = 1∕(2𝜎2

x )

a(k, 0) = A√2𝜋

[√2𝜋𝜎2

x e−𝜎2x k2∕2

]= A𝜎k

exp

{−1

2k2

𝜎2k

}where we identify 𝜎x = 1∕𝜎k


What is the Fourier transform of a Gaussian wave packet?Executive Summary

Fourier transform of a Gaussian shaped wave packet in position space is a Gaussianshaped wave packet in wavenumber space.

f (x, 0) = A exp{−1

2x2

𝜎2x

}FT⟷ a(k, 0) = A

𝜎kexp

{−1

2k2

𝜎2k

}where 𝜎k = 1∕𝜎x

Gaussian is the only function with this behavior under a Fourier transform.P. J. Grandinetti Chapter 08: Classical Wave Motion

Traveling wave packetWave packet moving forward in time is

f (x, t) = 1√2𝜋 ∫

∞

−∞a(k, 0)ei(kx−𝜔t)dk

This is our recipe for adding together many traveling sinusoidal waves with differentwavelengths and amplitude, a(k), to form the localized wave packet.

Conversely, we can view wave packet moving forward in time in wave number basis as

a(k, t) = 1√2𝜋 ∫

∞

−∞f (x, 0)e−i(kx−𝜔t)dx

Both views are equally valid, although your first preference might be to consider the wavepacket in position basis.


Traveling wave packet

Wave packet shape will not change with time if wave function argument is x − vpt orx + vpt (Recall k

(x ± vpt

)= kx ± 𝜔t)

Therefore, wave packet shape will remain unchanged with time if phase velocity,𝜔∕k = vp, is identical for all wave numbers.

In real media, ...▶ we find that 𝜔 ≈ kvp is an approximation.

▶ wave packet may not move through space at phase velocity, vp.

▶ wave packet may change shape as it moves in space.

More on the breakdown of this approximation later ...


The Fourier transform – one last thingTime and Frequency BasesFourier transform can also decompose time dependent function into its distribution offrequencies and back with analogous transforms,

F(𝜔) = 1√2𝜋 ∫

∞

−∞f (t)e−i𝜔tdt and f (t) = 1√

2𝜋 ∫∞

−∞F(𝜔)ei𝜔td𝜔

We can even use 2D Fourier transforms to relate wave function in position and time bases towave function in wave number and frequency bases.

a(k, 𝜔) = 1√2𝜋 ∫

∞

−∞ ∫∞

−∞f (x, t)e−ikxe−i𝜔tdt dx,

and

f (x, t) = 1√2𝜋 ∫

∞

−∞ ∫∞

−∞a(k, 𝜔)eikxei𝜔td𝜔 dk.


Uncertainty principle

ΔxΔk ≥ 1∕2

Δx is uncertainty in position of wave packet.Δk is uncertainty in wavenumber of wave packet.


What is the location of a wave packet in x?Define probability that wave packet has non-zero amplitude in dx centered at x as p(x)dx.Probabilities are always positive but f (x, t) can be positive or negativeUse square of f (x, t), also known as wave function intensity,

I(x, t) = f (x, t) f ∗(x, t) = |f (x, t)|2and define probability as

p(x, t) =|f (x, t)|2

∫∞

−∞|f (x, t)|2dx

= 1N|f (x, t)|2, where N = ∫

∞

−∞|f (x, t)|2dx

Scale by N so probability of finding wave packet over all x values is 100%.To find location, calculate mean position

x = ∫∞

−∞x p(x, t)dx


What is the location and width of a wave packet in x?

Define wave packet mean position as

x = ∫∞

−∞x p(x, t)dx = 1

N ∫∞

−∞x |f (x, t)|2dx

Define wave packet width—range of x over which it extends—as standard deviation

Δx =√

x2 − (x)2

where

x2 = ∫∞

−∞x2 p(x, t)dx = 1

N ∫∞

−∞x2 |f (x, t)|2dx


What is the width of a Gaussian wave packet in x?

ProblemCalculate the width at t = 0 of a Gaussian shaped wave packet given by

f (x, t) = A exp

{−1

2(x − vpt)2

𝜎2x

}

1st: find normalization constant at t = 0:

N = ∫∞

−∞|f (x, 0)|2dx = A2 ∫

∞

−∞

||||e− 12 x2∕𝜎2

x||||2 dx = A2 ∫

∞

−∞e−x2∕𝜎2

x dx

Integral tables tell us ∫ ∞−∞ e−ax2dx =

√𝜋∕a. So we set a = 1∕𝜎2

x and obtain

N = A2√𝜋𝜎2

x


What is the width of a Gaussian wave packet in x?ProblemCalculate the width at t = 0 of a Gaussian shaped wave packet given by

f (x, t) = A exp

{−1

2(x − vpt)2

𝜎2x

}

2nd: calculate x

x = 1N ∫

∞

−∞x |f (x, 0)|2dx = A2

A2√𝜋𝜎2

x∫

∞

−∞x||||e− 1

2 x2∕𝜎2x||||2

⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟Odd function

dx = 0

x = 1N ∫

∞

−∞x |f (x, 0)|2dx = A2

A2√𝜋𝜎2

x∫

∞

−∞ ��*0

x||||e− 1

2 x2∕𝜎2x||||2

⏟⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏟Odd function

dx = 0

Integrand is odd function of x and so integral goes to zero.P. J. Grandinetti Chapter 08: Classical Wave Motion

What is the width of a Gaussian wave packet in x?ProblemCalculate the width at t = 0 of a Gaussian shaped wave packet given by

f (x, t) = A exp

{−1

2(x − vpt)2

𝜎2x

}3rd: Calculate x2

x2 = 1N ∫

∞

−∞x2 |f (x, 0)|2dx = 1√

𝜋𝜎2x∫

∞

−∞x2 ||||e− 1

2 x2∕𝜎2x||||2dx = 𝜎2

x∕2

4th: plug x = 0 and x2 = 𝜎2x∕2 into Δx expression

Δx =√

x2 − (x)2 =√𝜎2

x∕2 − (0)2 = 𝜎x∕√

2

Δx, is standard deviation of wave packet intensity, |f (x, 0)|2,in terms of 𝜎x, the standard deviation of wave packet f (x, 0).


What is the width of a wave packet in k?

ProblemCalculate the width in wavenumbers, Δk, at t = 0 of a Gaussian shaped wave packet given by

f (x, 0) = A exp{−1

2x2

𝜎2x

}↔ a(k, 0) = A

𝜎kexp

{−1

2k2

𝜎2k

}where 𝜎k = 1∕𝜎x

Use same approach with

Δk =√

k2 − (k)2

wherek = ∫

∞

−∞k p(k) dk and k2 = ∫

∞

−∞k2 p(k) dk

1st: Normalize wave function at t = 0

N = ∫∞

−∞|a(k, 0)|2dk = A2

𝜎2k∫

∞

−∞

||||e− 12 k2∕𝜎2

k||||2 dk =

A2√𝜋

𝜎k


What is the width of a Gaussian wave packet in k?2nd: Calculate k

k = 1N ∫

∞

−∞k |a(k, 0)|2 dk = 0

since the integrand is an odd function.

3rd: Calculate k2

k2 = 1N ∫

∞

−∞k2 |a(k, 0)|2 dk = 𝜎2

k∕2

4th: Plug k = 0 and k2 = 𝜎2k∕2 into Δk expression

Δk =√

k2 − (k)2 =√𝜎2

k∕2 − (0)2 = 𝜎k∕√

2

As before we find Δk, the standard deviation of |a(k, 0)|2,in terms of 𝜎k, standard deviation of wave packet amplitude, a(k, 0).


The uncertainty principleFor Gaussian wave packet we found Δx = 𝜎x∕

√2 and Δk = 𝜎k∕

√2.

Product isΔxΔk = 𝜎x𝜎k∕2.

Recall 𝜎x = 1∕𝜎k, so product becomesΔxΔk = 1∕2

Product is smallest for Gaussian wave packet. For any other wave packet shape it will always belarger, that is,

ΔxΔk ≥ 1∕2

This is the uncertainty principle between wave packet position and wave number.▶ The more localized a wave packet is in space (i.e., smaller Δx) the greater the uncertainty in

wave number (i.e., larger Δk) of the wave packet.▶ The more localized wave packet is in wave number (i.e., smaller Δk) the greater the

uncertainty in position (i.e., larger Δx) of the wave packet.


Uncertainty principles

Uncertainty Principle for Position and Wavenumber

ΔxΔk ≥ 1∕2

You can similarly derive the

Uncertainty Principle for Time and Frequency

Δ𝜔Δt ≥ 1∕2

These 2 uncertainty principles appear in study of the classical wave equation andboth lie at heart of uncertainty principles we will encounter in quantum mechanics.


Dispersive Waves


Dispersive Waves

In some media simple relationship 𝜔∕k = vp may not hold identically for all wavenumbers(k = 2𝜋∕𝜆) present in wave packet.

We might see wave packet broaden or disperse as it moves through space.


Dispersive Waves

𝜔(k) has non-linear dependence on k in dispersive medium.

non-dispersivemedium

normaldispersivemedium

anomalousdispersivemedium

00

k

ωCurvature in plot of 𝜔 versus k leads to wave packetdispersion.

Relationship between 𝜔 and k is called dispersionrelation.

No dispersion occurs when 𝜔 is linearly dependent on k,i.e., 𝜔 = vpk.

“Normal” dispersion is when vp decreases with increasingk.

“Anomalous” dispersion is when vp increases withincreasing k


Dispersive Wave ExamplesString waves on stiff string have anomalous dispersion with

𝜔2 =√

T𝜇

k2 + 𝛼k4, 𝛼 → string stiffness

Capillary waves traveling along phase boundary of 2 fluids are dispersive

𝜔2 =𝛾

𝜌 + 𝜌′|k|3

𝛾 is surface tension, 𝜌 and 𝜌′ are heavier and lighter fluid densities, respectively.Light traveling in a dispersive medium where

𝜔 =c0kn(k)

, n(k) → refractive index of medium

This causes light to disperse into rainbow through a prism.Why does laser light travel long distances without significant dispersion?


Dispersive WavesNormal dispersion: phase velocity, vp = 𝜔∕k, decreases with increasing wave number.

ideal media

Monochromatic wave with k0 = 2𝜋∕𝜆0 travels at slower speed of 𝜔0∕k0 in this medium comparedto that predicted by initial slope at k = 0.Monochromatic waves (e.g., lasers) never disperse, even in highly dispersive medium, since theyare composed of a single wavelength.


Dispersive WavesWhat about wave packet that contains range of wavelengths?

Since vp = 𝜔∕k, each wavelength in shaded region has a different phase velocity represented byrange of slopes shown.


Dispersive WavesPerform Taylor series of dispersion relationship, 𝜔(k), about k0

𝜔(k) ≈ 𝜔(k0) +(𝜕𝜔(k0)𝜕k

)(k − k0) +

12

(𝜕2𝜔(k0)𝜕k2

)(k − k0)2 +⋯

Defining 𝜔0 = 𝜔(k0), and

vg =𝜕𝜔(k0)𝜕k

and 𝛼 =𝜕2𝜔(k0)𝜕k2

Rewrite series𝜔(k) ≈ 𝜔0 + vg(k − k0) +

12𝛼(k − k0)2 +⋯

Series for wave k distribution confined to small range centered on k0.


Dispersive Waves — Group Velocity

𝜔(k) ≈ 𝜔0 + vg(k − k0) +12𝛼(k − k0)2 +⋯

Substitute 1st two terms of this series into Fourier expansion of wave

f (x, t) = 1√2𝜋 ∫

∞

−∞a(k, 0)ei(kx−𝜔t)dk ≈ 1√

2𝜋 ∫∞

−∞a(k)ei[kx−(𝜔0+vg(k−k0))t]dk

Setting Δk = k − k0 and rearranging gives

f (x, t) ≈ 1√2𝜋

ei(k0x−𝜔0t)

⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟Sinusoid wave

×∫∞

−∞a(Δk)eiΔk(x−vgt)dΔk

⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟Wave shape

In this form wave packet is sinusoidal wave multiplied by shape function.Sinusoid is moving at a speed of vp = 𝜔0∕k0 — Phase VelocityShape function is moving at a speed of vg — Group Velocity


Dispersive Waves—Group velocityDefinitionGroup velocity is the speed that mean position of wave packet moves through space. It isdefined as slope of dispersion curve at k0.

Phase Velocity: vp = 𝜔0∕k0 Group Velocity: vg =𝜕𝜔(k0)𝜕k


Dispersive Waves—Phase and Group velocity

Web Link: https://www.youtube.com/user/meyavuz/playlistsGo to Group and phase velocity playlist

Case 1: Group Velocity larger than Phase VelocityCase 2: Zero Group VelocityCase 3: Negative Group VelocityCase 4: Zero Phase VelocityCase 5: Positive Phase and Negative Group Velocity


https://www.youtube.com/user/meyavuz/playlists

https://www.youtube.com/user/meyavuz/playlists

https://www.youtube.com/watch?v=tlM9vq-bepA&list=PLrBvxu47JdxCqx4AvPZ3dXMkWfbRQPDnL&index=1

https://www.youtube.com/watch?v=v9DPzMoWpc0&list=PLrBvxu47JdxCqx4AvPZ3dXMkWfbRQPDnL&index=2

https://www.youtube.com/watch?v=ePJdV75fT5o&list=PLrBvxu47JdxCqx4AvPZ3dXMkWfbRQPDnL&index=3

https://www.youtube.com/watch?v=iVJNcANWmI0&list=PLrBvxu47JdxCqx4AvPZ3dXMkWfbRQPDnL&index=4

https://www.youtube.com/watch?v=MMV2Zt7x430&list=PLrBvxu47JdxCqx4AvPZ3dXMkWfbRQPDnL&index=5

Dispersive WavesTruncating dispersion relation series expansion at group velocity term gives wave packetthat travels without changing shape, that is, without dispersing.To describe dispersion we add 𝛼 term of series expansion of 𝜔(k) and obtain

f (x, t) ≈ 1√2𝜋

ei(k0x−𝜔0t) ∫∞

−∞a(Δk)eiΔk(x−vgt)−i𝛼Δk2t∕2dΔk

For Gaussian wave packet (after some math) we find wave intensity

|f (x, t)|2 = A2[1 + 𝛼2t2∕(Δx0)4

]1∕2exp

[− 1

2(x − vgt)2

(Δx0)2[1 + 𝛼2t2∕(Δx0)4]⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

(Δx)2

]

Gaussian wave packet standard deviation increases with t.

|Δx| = Δx0

√1 + 𝛼2t2

(Δx0)4where Δx0 is initial width of wave packet.


Dispersive WavesGaussian wave packet standard deviation increases with t.

|Δx| = Δx0

√1 + 𝛼2t2

(Δx0)4

Δx0 is the initial width of the wave packet.


Two- and Three-dimensional Waves

Examples of 2- and 3-dimensional waves include water waves, seismic waves, membrane waves(e.g., a drum head, soap film, etc.), and electromagnetic waves.


Two- and Three-dimensional Waves

After d’Alembert, Euler generalized wave equation to 2 and 3 dimensions,

𝜕2𝜓(r⃗, t)𝜕t2 = v2

p∇2𝜓(r⃗, t)

∇2 is the Laplacian operator.

In 2 and 3 dimensions, it is given by

∇2 = 𝜕2

𝜕x2 + 𝜕2

𝜕y2 , and ∇2 = 𝜕2

𝜕x2 + 𝜕2

𝜕y2 + 𝜕2

𝜕z2


2D WavesLet’s start with solutions of the 2D wave wave equation,

𝜕2𝜓(x, y, t)𝜕x2 +

𝜕2𝜓(x, y, t)𝜕y2 = 1

v2p

𝜕2𝜓(x, y, t)𝜕t2

To solve 2D wave equation we again employ separation of variables,

𝜓(x, y, t) = X(x)Y(y)T(t)

Plugging expression above into wave equation gives

Y(y)T(t)𝜕2X(x)𝜕x2 + X(x)T(t)

𝜕2Y(y)𝜕y2 = 1

v2p

X(x)Y(y)𝜕2T(t)𝜕t2

Dividing both sides by 𝜓(x, y, t) leads to

v2p

X(x)𝜕2X(x)𝜕x2 +

v2p

Y(y)𝜕2Y(y)𝜕y2 = 1

T(t)𝜕2T(t)𝜕t2

= −𝜔2

Right side is function of t only, whereas left side depends only on x and y. For equality to remain truefor all values of x, y and t both sides of equation must be equal to separation constant : −𝜔2.


2D WavesThis allows us to turn PDE into ODE for T(t)

1T(t)

d2T(t)dt2 = −𝜔2

We are left with PDE for X(x) and Y(y):

v2p

X(x)𝜕2X(x)𝜕x2 +

v2p

Y(y)𝜕2Y(y)𝜕y2 = −𝜔2

2nd equation can be further rearranged to

1X(x)

𝜕2X(x)𝜕x2 = − 1

Y(y)𝜕2Y(y)𝜕y2 − 𝜔2

v2p= −k2

x

where we now introduce second separation constant: −k2x .


2D Waves

Further defining k2y = 𝜔2∕v2

p − k2x we converted PDE into 3 uncoupled ODEs,

d2T(t)dt2 + 𝜔2T(t) = 0, d2X(x)

dx2 + k2xX(x) = 0, and d2Y(y)

dy2 + k2yY(y) = 0

which must obey constraint𝜔2∕v2

p = k2y + k2

x

Sinusoids will be solutions to all 3 ODEs, and traveling 2D sinusoidal wave can be written assimple product of sinusoidal functions, e.g.,

𝜓(x, y, t) = ℜ{Aeikxxeikyye−i𝜔t}


2D Waves

𝜓(x, y, t) = ℜ{Aeikxxeikyye−i𝜔t} = ℜ{Aei(kxx+kyy)e−i𝜔t}

Define wave vector, k⃗,k⃗ = kxe⃗x + kye⃗y

then k⃗ ⋅ r⃗ = kxx + kyy, and solution can be written

𝜓(r⃗, t) = ℜ{Aei(k⃗⋅r⃗−𝜔t)}

k⃗ defines direction of wave propagationWave vector is related to wavelength by

|k⃗| = 2𝜋∕𝜆, where |k⃗| = √k2

x + k2y


2D WavesAt a given instant in time, phase of plane wave is constant along a line defined by

𝜙 = kxx + kyy = k⃗ ⋅ r⃗

x

y

Lines

of co

nstan

t pha

se

This traveling wave should be called a 2D “line wave”, but in anticipation of 3D case it iscalled a plane wave.


Rectangular membrane


2D standing waves–Rectangular membraneStretch rubber sheet over opening of rectangular box and glue sheet around sides of box

Standing waves of rectangular membrane attached to fixed supports along lines x = 0, y = 0,x = Lx, and y = Ly.𝜓(x, y, t) must go to zero along edges: 𝜓(x = 0, y, t) = 𝜓(x, y = 0, t) = 0Solution has general form: 𝜓(x, y, t) = A sin(kxx) sin(kyy) cos(𝜔t + 𝛿),𝜓(x, y, t) must go to zero along edges: 𝜓(x = Lx, y, t) = 𝜓(x, y = Ly, t) = 0, so

kxLx = nx𝜋 where nx = 1, 2,… and kyLy = ny𝜋 where ny = 1, 2,…

Vibration frequencies constrained to discrete values

𝜔nx,ny= vp

√k2

x + k2y = vp

√(nx𝜋Lx

)2+(ny𝜋

Ly

)2

In special case of square box, Lx = Ly = L we have 𝜔nx,ny= 𝜋vp

L

√n2

x + n2y

Some frequencies are degenerate (have identical values).P. J. Grandinetti Chapter 08: Classical Wave Motion

2D standing waves–Rectangular membraneStretch rubber sheet over opening of rectangular box and glue sheet around sides of box

Normal modes for a rectangular membrane are

𝜓nx,ny(x, y, t) = anx,ny

sin(kxx) sin(kyy) cos(𝜔nx,nyt + 𝛿nx,ny

)

Any rectangular membrane wave can be expressed as superposition of normal mode vibrations

𝜓(x, y, t) =∞∑

nx=1

∞∑ny=1

anx,nysin

(nx𝜋Lx

x)sin

(ny𝜋Ly

y)cos(𝜔nx,ny

t + 𝛿nx,ny)

Given 𝜓(x, y, t), a 2D Fourier series analysis gives anx,ny(t),

anx,ny(t) = 4

LxLy ∫Lx

0 ∫Ly

0𝜓(x, y, t) sin

(nx𝜋Lx

x)sin

(ny𝜋Ly

y)

dx dy, nx, ny = 1, 2,…


2D standing waves–Rectangular membraneNormal modes of a rectangular membrane.The nodal lines for the first few values of nx and ny..

nx = 1

1

2

2

3

3ny


Web App Demos

Rectangular Membrane WavesWeb Video: Square Chladni Plate


http://www.falstad.com/membrane/fullscreen.html

https://www.youtube.com/watch?v=wvJAgrUBF4w

Gradient and Laplacian Operators in Polar (2D)/Spherical (3D) Coordinatesz

y

x

Gradient Operator in Cartesian and Spherical Coordinates

∇⃗ = e⃗x𝜕𝜕x

+ e⃗y𝜕𝜕y

+ e⃗z𝜕𝜕z

= e⃗r𝜕𝜕r

+ e⃗𝜃1r𝜕𝜕𝜃

+ e⃗𝜑1

r sin 𝜃𝜕𝜕𝜑

Laplacian Operator in Cartesian and Spherical Coordinates

∇2 = 𝜕2

𝜕x2 + 𝜕2

𝜕y2 + 𝜕2

𝜕z2 = 𝜕2

𝜕r2 + 2r𝜕𝜕r

+ 1r2

[1

sin 𝜃𝜕𝜕𝜃

(sin 𝜃 𝜕

𝜕𝜃

)+ 1

sin2 𝜃

(𝜕2

𝜕𝜙2

)]P. J. Grandinetti Chapter 08: Classical Wave Motion

Circular membrane


2D standing waves–Normal modes of a Circular membraneStretch rubber sheet over opening of large diameter pipe and glue sheet around sides.

Standing waves of circular membrane of radius a attached to fixed supports.Wave function should go to zero along circular edges of membrane.In polar coordinates: 𝜓(r = a, 𝜙, t) = 0.Wave equation in terms of polar coordinates,

𝜕2𝜓(r, 𝜙, t)𝜕r2 + 1

r𝜕𝜓(r, 𝜙, t)

𝜕r+ 1

r2𝜕2𝜓(r, 𝜙, t)

𝜕𝜙2 = 1v2

p

𝜕2𝜓(r, 𝜙, t)𝜕2t

Using separation of variables we write

𝜓(r, 𝜙, t) = R(r)Φ(𝜙)T(t)

Plug 𝜓(r, 𝜙, t) into wave equations and identify first separation constant,

v2p

R(r)

(𝜕2R(r)𝜕r2 + 1

r𝜕R(r)𝜕r

)+

v2p

r2Φ(𝜙)𝜕2Φ(𝜙)𝜕𝜙2 = 1

T(t)𝜕2T(t)𝜕t2 = −𝜔2


2D standing waves–Normal modes of a Circular membraneWe have ODE for time dependent part, T(t),

d2T(t)dt2 + 𝜔2T(t) = 0

We’ve already worked out T(t) solutions for this ODE.We are left with a PDE involving R(r) and Φ(𝜙).

v2p

R(r)

(𝜕2R(r)𝜕r2 + 1

r𝜕R(r)𝜕r

)+

v2p

r2Φ(𝜙)𝜕2Φ(𝜙)𝜕𝜙2 = −𝜔2

Rearranging using 1st separation constant we identify 2nd separation constant,

r2

R(r)

(𝜕2R(r)𝜕r2 + 1

r𝜕R(r)𝜕r

)+(

r𝜔vp

)2= − 1

Φ(𝜙)𝜕2Φ(𝜙)𝜕𝜙2 = m2

and obtain ODEs for R(r), and Φ(𝜙).P. J. Grandinetti Chapter 08: Classical Wave Motion

2D standing waves–Normal modes of a Circular membraneODE for angular part, Φ(𝜙),

d2Φ(𝜙)d𝜙2 + m2Φ(𝜙) = 0

has solutionΦ(𝜙) = Aeim𝜙 + Be−im𝜙

We require wave functions to be single valued, therefore

Φ(𝜙 + 2𝜋) = Φ(𝜙)

which means

Aeim(𝜙+2𝜋) + Be−im(𝜙+2𝜋) = Aeim𝜙 + Be−im𝜙, or e±im2𝜋 = 1

leading to constraints that m to m = 0,±1,±2,…


2D standing waves–Normal modes of a Circular membraneODE for radial part, R(r), is harder to solve.

d2R(r)dr2 + 1

rdR(r)

dr+

[(𝜔vp

)2− m2

r2

]R(r) = 0

At large r values we can neglect 1∕r and 1∕r2 terms and ODEreduces to

d2R(r)dr2 +

(𝜔vp

)2R(r) = 0

This has form of 1D wave equation. In other words, ifmembrane extended radially to infinity then at large r circularwaves appear as plane waves radiating outward.

But we need exact solution for arbitrary value of r.


2D standing waves–Normal modes of a Circular membraneWhat are solutions of ODE for R(r)

d2R(r)dr2 + 1

rdR(r)

dr+

[(𝜔vp

)2− m2

r2

]R(r) = 0

1st we define u = kr = 𝜔vp

r gives du = kdr.

With R(r)dr = R(u)du we have R(r) = kR(u) and rewrite the radial ODE as

u2 d2R(u)du2 + u

dR(u)du

+(u2 − m2)R(u) = 0

This ODE is known as Bessel’s equation.Solutions to Bessel’s equation are the Bessel functions.It has 2 linearly independent solutions:

▶ Jm(u), Bessel functions of 1st kind▶ Ym(u), Bessel functions of 2nd kind.▶ For standing waves of circular membrane we only need Jm(u).


Bessel functions of 1st kind

-0.2

0.20.40.6

-0.2

0.20.4

-0.2

0.20.0

0.0

0.0

0.0

0.4

5 10 15 20 25

-0.4-0.2

0.20.40.60.81.0

Look like oscillating sinusoid thatdecays as 1∕

√u.

Roots are indicated as u(n)m .For solutions with negative m useidentity J−m(u) = (−1)mJm(u)Boundary condition for circularmembrane of radius a:𝜓(r = a, 𝜙, t) = 0, requires u = ka bea roots of Jm(u),

Jm(km,na) = 0 where km,na = u(n)m

With this constraint 𝜔 becomes𝜔m,n = vpkm,n = vpu(n)m ∕a


Roots, u(n)m , of Jm(u(n)

m ) = 0, for Bessel functions of 1st kind

-0.2

0.20.40.6

-0.2

0.20.4

-0.2

0.20.0

0.0

0.0

0.0

0.4

5 10 15 20 25

-0.4-0.2

0.20.40.60.81.0

nu(n)m 1 2 3 4 5 6 7u(n)0 2.4048 5.5201 8.6537 11.7915 14.9309 18.0711 21.2116u(n)1 3.8317 7.0156 10.1735 13.3237 16.4706 19.6159 22.7601u(n)2 5.1356 8.4172 11.6198 14.7960 17.9598 21.117 24.2701u(n)3 6.3802 9.7610 13.0152 16.2235 19.4094 22.5827 25.7482u(n)4 7.5883 11.0647 14.3725 17.6160 20.8269 24.019 27.1991

Not spaced periodic in u in general but become periodic asymptotically at large u.P. J. Grandinetti Chapter 08: Classical Wave Motion

2D normal modes of a circular membrane

Circular membrane normal modes are:

𝜓m,n(r, 𝜙, t) = am,nJm(km,nr) cos(m𝜙) cos(𝜔m,nt + 𝛿m,n)+ bm,nJm(km,nr) sin(m𝜙) cos(𝜔m,nt + 𝛿m,n)

Each normal mode has

n − 1 radial nodes

m angular nodes.

m = 0

1

1

2

3

2 3

n

Arbitrary wave is superposition of normal modes

𝜓(r, 𝜙, t) =∞∑

n=1

∞∑m=−∞

𝜓m,n(r, 𝜙, t) =∞∑

n=1

∞∑m=−∞

Jm(km,nr)(am,n cos(m𝜙) + bm,n sin(m𝜙)

)cos(𝜔m,nt + 𝛿m,n)


Demos

Web App: Circular Membrane WavesWeb Video: Circular Centered Chladni Plate


http://www.falstad.com/circosc/fullscreen.html

https://www.youtube.com/watch?v=CGiiSlMFFlI

Three-dimensional Waves


3D WavesShifting to wave solutions in 3D where the wave equation is

𝜕2𝜓(x, y, z, t)𝜕x2 +

𝜕2𝜓(x, y, z, t)𝜕y2 +

𝜕2𝜓(x, y, z, t)𝜕z2 = 1

v2p

𝜕2𝜓(x, y, z, t)𝜕t2

use separation of variables,𝜓(x, y, z, t) = X(x)Y(y)Z(z)T(t)

we find harmonic wave solution that satisfies 3D wave equation


k⃗ is wave vector that defines direction of wave propagation,k⃗ = kxe⃗x + kye⃗y + kze⃗z

Wave vector is related to wavelength by |k⃗| = 2𝜋∕𝜆, where

|k⃗| = √k2

x + k2y + k2

z


3D Waves

Three dimensional plane wave described by


Plane waves are effectively one dimensional buttravel in three dimensions.


Web Apps by Paul Falstad

3D Box Waves


http://www.falstad.com/modebox/fullscreen.html

3D Waves in spherical coordinates


3D Waves in spherical coordinatesNext examine 3D wave equation in spherical coordinates

∇2𝜓(r, 𝜃, 𝜙, t) = 1v2

p

𝜕2𝜓(r, 𝜃, 𝜙, t)𝜕2t

where∇2 = 𝜕2

𝜕r2 + 2r𝜕𝜕r

+ 1r2

[1

sin 𝜃𝜕𝜕𝜃

(sin 𝜃 𝜕

𝜕𝜃

)+ 1

sin2 𝜃

(𝜕2

𝜕𝜙2

)]Separation of variables suggests wave function

𝜓(r, 𝜃, 𝜙, t) = R(r) Θ(𝜃) Φ(𝜙)T(t)

Plug 𝜓(r, 𝜃, 𝜙, t) into wave equation and identify 1st separation constant,

1R(r)Θ(𝜃)Φ(𝜙)

∇2(

R(r)Θ(𝜃)Φ(𝜙))

= 1v2

p

1T(t)

𝜕2T(t)𝜕2t

= −𝜔2

Obtain an ODE for T(t) d2T(t)dt2

+ 𝜔2T(t) = 0


3D Waves in spherical coordinatesRemaining PDE involves R(r), Θ(𝜃), Φ(𝜙),

∇2(

R(r)Θ(𝜃)Φ(𝜙))+ 𝜔2R(r)Θ(𝜃)Φ(𝜙) = 0

Using the 1st separation constant we identify a 2nd separation constant,

r2

R(r)𝜕2R(r)𝜕r2 + 2r

R(r)𝜕R(r)𝜕r

+ 𝜔2r2

v2p

= −[

1Θ(𝜃) sin 𝜃

𝜕𝜕𝜃

(sin 𝜃 𝜕Θ(𝜃)

𝜕𝜃

)+ 1

Φ(𝜙) sin2 𝜃𝜕2Φ(𝜙)𝜕𝜙2

]= 𝜆

Obtain ODE for radial part, R(r),

r2 d2R(r)dr2 + 2r

dR(r)dr

+

(r2𝜔2

v2p

− 𝜆

)R(r) = 0


3D Waves in spherical coordinatesWith change in variables

u = (𝜔∕vp)r with du = (𝜔∕vp)dr

andR(r)dr = R(u)du with R(r) = (𝜔∕vp)R(u)

differential for R(r) transforms into

u2 d2R(u)du2 + 2u

dR(u)du

+(u2 − 𝜆

)R(u) = 0

Recognize as spherical Bessel differential equation.Solutions to this ODE are the spherical Bessel functions: j𝓁(u) and y𝓁(u) where𝜆 = 𝓁(𝓁 + 1) and are given by

j𝓁(u) = (−u)𝓁(1

uddu

)𝓁 sin uu

y𝓁(u) = −(−u)𝓁(1

uddu

)𝓁 cos uu


Spherical Bessel Functions

0.25

0.00

−0.25

−0.50

−0.75

−1.00

−1.25

−1.50

0 5 10 15 20

y (x)0y (x)1y (x)2

x

1.00

0.80

0.60

0.40

0.20

0.00

−0.20

−0.40

0 5 10 15 20

j (x)0j (x)1j (x)2

x



Using 2nd separation constant we identify 3rd separation constant,

sin 𝜃Θ(𝜃)

𝜕𝜕𝜃

(sin 𝜃 𝜕Θ(𝜃)

𝜕𝜃

)+ 𝓁(𝓁 + 1) sin2 𝜃 = − 1

Φ(𝜙)𝜕2Φ(𝜙)𝜕𝜙2 = m2

Obtain an ODE for Φ(𝜙)d2Φ(𝜙)

d𝜙2 + m2Φ(𝜙) = 0

As before we have solution Φ(𝜙) = Aeim𝜙 + Be−im𝜙.

Constraint of single valued wave function restricts m to integer values: m = 0,±1,±2,….


3D Waves in spherical coordinatesFinally, we have ODE for Θ(𝜃),

1sin 𝜃

dd𝜃

(sin 𝜃 d

d𝜃Θ(𝜃)

)+(𝓁(𝓁 + 1) − m2

sin2 𝜃

)Θ(𝜃) = 0

To find these solutions we make change in variables from 𝜃 to v = cos 𝜃.Differentiation with respect to v is

ddv

= d𝜃dv

dd𝜃

=( dv

d𝜃

)−1 dd𝜃

= − 1sin 𝜃

dd𝜃

(1 − v2) ddv

= −1 − cos2 𝜃sin 𝜃

dd𝜃

= −sin2 𝜃sin 𝜃

dd𝜃

= − sin 𝜃 dd𝜃

andddv

(1 − v2) ddv

= − 1sin 𝜃

dd𝜃

[− sin 𝜃 d

d𝜃

]= 1

sin 𝜃d

d𝜃

(sin 𝜃 d

d𝜃

)P. J. Grandinetti Chapter 08: Classical Wave Motion

3D Waves in spherical coordinatesODE for Θ(𝜃) with change of variable v = cos 𝜃 becomes

ddv

[(1 − v2) d

dvΘ(v)

]+[𝓁(𝓁 + 1) − m2

1 − v2

]Θ(v) = 0

This is the associated Legendre differential equation whose solutions are the associatedLegendre polynomials,

Θ(𝜃) = Pm𝓁 (cos 𝜃)

These polynomials vanish for |m| > 𝓁, so only m values ofm = −𝓁,−𝓁 + 1,… , 0,… ,𝓁 − 1,𝓁 are allowed.Together with Φ(𝜙) we have

Θ(𝜃)Φ(𝜙) = Pm𝓁 (cos 𝜃)e

im𝜙

This combination is proportional to well known spherical harmonic functions according to

Y𝓁,m(𝜃, 𝜙) = (−1)m√

(2𝓁 + 1)4𝜋

√(𝓁 − m)!(𝓁 + m)!

Pm𝓁 (cos 𝜃)e

im𝜙


Associated Legendre Polynomials and their roots.Sectoral harmonics are highlighted.

Pm𝓁 (cos 𝜃) Polynomial Roots

P00(cos 𝜃) 1

P01(cos 𝜃) cos 𝜃 90◦

P11(cos 𝜃) − sin 𝜃 0◦ 180◦

P02(cos 𝜃)

12(3 cos2 𝜃 − 1) 54.73561◦ 125.2644◦

P12(cos 𝜃) −3 cos 𝜃 sin 𝜃 0◦ 90◦ 180◦

P22(cos 𝜃) 3 sin2 𝜃 0◦ 180◦

P03(cos 𝜃)

12(5 cos2 𝜃 − 3) cos 𝜃 39.23152◦ 90◦ 140.7685◦

P13(cos 𝜃) − 3

2(5 cos2 𝜃 − 1) sin 𝜃 0◦ 63.43495◦ 116.56505◦ 180◦

P23(cos 𝜃) 15 cos 𝜃 sin2 𝜃 0◦ 90◦ 180◦

P33(cos 𝜃) −15 sin3 𝜃 0◦ 180◦

P04(cos 𝜃)

18(35 cos4 𝜃 − 30 cos2 𝜃 + 3) 30.55559◦ 70.124281◦ 109.87572◦ 149.44441◦

P14(cos 𝜃) − 5

2(7 cos2 𝜃 − 3) cos 𝜃 sin 𝜃 0◦ 49.10660◦ 90◦ 130.89340◦ 180◦

P24(cos 𝜃)

152(7 cos2 𝜃 − 1) sin2 𝜃 0◦ 67.79235◦ 112.20765◦ 180◦

P34(cos 𝜃) −105 cos 𝜃 sin3 𝜃 0◦ 90◦ 180◦

P44(cos 𝜃) 105 sin4 𝜃 0◦ 180◦

P05(cos 𝜃)

18

(63 cos4 𝜃 − 70 cos2 𝜃 + 15

)cos 𝜃 25.01732◦ 57.42052◦ 90◦ 122.57948◦ 154.98268◦

P15(cos 𝜃) − 15

8

(21 cos4 𝜃 − 14 cos2 𝜃 + 1

)sin 𝜃 0◦ 40.08812◦ 73.42728◦ 106.57272◦ 139.91188◦ 180◦

P25(cos 𝜃)

1052

(3 cos2 𝜃 − 1

)cos 𝜃 sin2 𝜃 0◦ 54.73561◦ 90◦ 125.2644◦ 180◦

P35(cos 𝜃) − 105

2

(9 cos2 𝜃 − 1

)sin3 𝜃 0◦ 70.52878◦ 109.47122◦ 180◦

P45(cos 𝜃) 945 cos 𝜃 sin4 𝜃 0◦ 90◦ 180◦

P55(cos 𝜃) −945 sin5 𝜃 0◦ 180◦


Laplace’s spherical harmonics

Y𝓁,m(𝜃, 𝜙) = (−1)m√

(2𝓁 + 1)4𝜋

√(𝓁 − m)!(𝓁 + m)!

Pm𝓁 (cos 𝜃)e

im𝜙

First introduced by Pierre Simon de Laplace in 1782.Spherical harmonics for 𝓁 = 0 to 𝓁 = 2.

Y0,0(𝜃, 𝜙) = 1√4𝜋

Y1,0(𝜃, 𝜙) =√

34𝜋

cos 𝜃

Y1,±1(𝜃, 𝜙) = ∓√

34𝜋

√12

⋅ sin 𝜃 e±i𝜙

Y2,0(𝜃, 𝜙) =√

54𝜋

⋅12(3 cos2 𝜃 − 1)

Y2,±1(𝜃, 𝜙) = ∓√

54𝜋

√16

⋅ 3 cos 𝜃 sin 𝜃 e±i𝜙

Y2,±2(𝜃, 𝜙) =√

54𝜋

√124

⋅ 3 sin2 𝜃 e±i2𝜙

Spherical harmonics for 𝓁 = 3.

Y3,0(𝜃, 𝜙) =√

74𝜋

⋅12(5 cos2 𝜃 − 3) cos 𝜃

Y3,±1(𝜃, 𝜙) = ∓√

74𝜋

√1

12⋅

32(5 cos2 𝜃 − 1) sin 𝜃 e±i𝜙

Y3,±2(𝜃, 𝜙) =√

74𝜋

√1

120⋅ 15 cos 𝜃 sin2 𝜃 e±i2𝜙

Y3,±3(𝜃, 𝜙) = ∓√

74𝜋

√1

720⋅ 15 sin3 𝜃 e±i3𝜙



Spherical harmonics with m = 0...

are proportional to just Legendre polynomials.

do not depend upon longitude and are called zonal harmonics because their roots dividethe sphere into zones.

when viewed down the y axis the nodal lines of the normal modes appear as shown below.z

x



Spherical harmonics with |m| = 𝓁...

have no zero crossings in latitude, and functions are referred to as sectoral harmonics.

when viewed down z axis nodal lines of normal modes appear as shown below.

x

y



Spherical harmonics with other values of 𝓁 and m...

divide spherical surface into tesserae and are called tesseral harmonics.

number of zonal nodes is 𝓁 − |m| while number of sectoral nodes is 2|m|.nodal lines of the normal modes appear as shown below for the case of 𝓁 = 7 and m = 4.

z

x

x

y


3D Waves in spherical coordinatesThe spherical harmonics are a complete orthonormal basis for expanding the angulardependence of a function, and satisfy an orthogonality relation:

∫2𝜋

0 ∫𝜋

0Y∗𝓁1,m1

(𝜃, 𝜙)Y𝓁2,m2(𝜃, 𝜙) sin 𝜃 d𝜃 d𝜙 = 𝛿𝓁1,𝓁2

𝛿m1,m2

where 𝛿m,n is the Kronecker delta function defined as

𝛿mn =

{0 if m ≠ n,1 if m = n.

Any arbitrary real function f (𝜃, 𝜙) can be expanded in terms of complex spherical harmonics by

f (𝜃, 𝜙) =∞∑𝓁=0

𝓁∑m=0

a𝓁,mY𝓁,m(𝜃, 𝜙)

wherea𝓁,m = ∫

𝜋

0 ∫𝜋

−𝜋f (𝜃, 𝜙)Y∗

𝓁,m(𝜃, 𝜙) sin 𝜃 d𝜃 d𝜙


Diffraction and Interference


Diffraction and InterferenceIn diffraction plane waves spread out (diffract) as they pass through slit. With slit width largerthan wavelength wave undergoes less diffraction.

planewaves

wavecrest

wavetrough

planewaves

wavecrest

wavetrough


Diffraction and InterferenceWith interference we observe constructive effects where two wave crests meet or two wavetroughs meet and destructive effects where wave crest and trough meet.

planewaves

wavecrest

wavetrough

point ofconstructiveinterference

point ofdestructiveinterference

In 1600s Christiaan Huygens showed that lightbehaves like a wave. He showed that lightundergoes diffraction and interference.Huygens discovery that light diffracts and interfereswas surprising because at the time light wasthought to be composed of particles.Only waves known at the time were mechanicalwaves which require a physical medium.After Huygens the search began to discover the“mechanical medium” for light or what was calledthe luminiferous ether. To be continued...


Web App Demos

2D Wave Interference


http://www.falstad.com/wave2d/fullscreen.html

Web Videos

Web Links: Crash Course Physics ReviewWavesSoundMusicLightLight Interference


https://www.youtube.com/watch?v=TfYCnOvNnFU

https://www.youtube.com/watch?v=qV4lR9EWGlY

https://www.youtube.com/watch?v=XDsk6tZX55g

https://www.youtube.com/watch?v=IRBfpBPELmE

https://www.youtube.com/watch?v=-ob7foUzXaY

Documents

Classical Wave Motion P. J. Grandinetti · Wave Motion Deﬁnition A wave is a self sustaining disturbance in a continuous medium and can move through space without transporting the