wave motion - chapter 8 - grandinetti · wave motion chapter 8 p. j. grandinetti chem. 4300 sept...

Wave MotionChapter 8

P. J. Grandinetti

Chem. 4300

Sept. 20, 2017

P. J. Grandinetti (Chem. 4300) Wave Motion Sept. 20, 2017 1 / 108

Wave Motion

DefinitionA wave is a self sustaining disturbance in a continuous medium and canmove through space without transporting the medium.

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

Wave leads to local displacements of medium away from itsequilibrium position.

transverse wave: wave displacement is perpendicular to direction ofwave propagation

longitudinal wave: wave displacement is parallel to direction of wavepropagation


Web Video Demos

Web Links: Crash Course PhysicsWavesSoundMusicLightLight 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

What are the equations describing wave motion?

Assume that we can describe a particular wave shape by amathematical function, f (x, t), called a wave function

Wave function, f (x, t), describes spatial and time dependence ofmedium displacement.

We find that a complete set of wave functions describing all possiblewave behaviors are the set of solutions to a single partial differentialequation (PDE) called the wave equation for that medium.


Wave EquationsWe divide wave equations into two broad classes depending on whetherthey are linear or non-linear partial differential equations.

DefinitionOrder of differential equation is order of highest derivative in equation.

Definition

Ordinary differential equation, F(

t, f ,dfdx, ...,

dnfdxn

)= 0,

is linear if F is linear function of variables f , dy∕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.


Order and Linearity of Differential EquationsExampleDetermine order of these differential equations and whether or not it’s alinear differential equation.

t2 d2ydt2 + t

dydt

+ 2y = sin t 2nd order, linear

dydt

+ ty2 = 0 1st order, non-linear

(1 + y2)d2ydt2 + t

dydt

+ y = et 2nd order, non-linear

d3ydt3 + t

dydt

+ (cos2 t)y = t3 3rd order, linear

d4ydt4 +

d3ydt3 +

d2ydt2 +

dydt

+ y = 1 4th order, linear

d2ydt2 + sin(t + y) = sin t 2nd order, non-linear


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

DefinitionSuperposition Principle: Any solution to a linear PDE can be addedtogether with other solutions to form further solutions.

∑n

an 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—superpositionprinciple does not generally apply.


The Classical Wave equationIn 1746 the French mathematician Jean le Rond d’Alembert discovered anequation that describes all linear 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 ondetails of medium in which wavetravels


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 =√𝜇∕T, 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 and𝜌 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 functionarguments have fixed relationship u = x − vpt then wave retains its shapeas it moves to right through space with time.

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

sign of vp to get left traveling wave.


The Classical Wave equationBy re-expressing wave equation in terms of two variables u = x − vpt andv = x + vpt d’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(x − vpt) represents wave moving left to right, i.e., in positive x direction,

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


Traveling Sinusoidal (Harmonic) WavesSinusoidal (harmonic) waves are useful in study of waves for many reasons.Main reason is that superposition principle allows us to decomposearbitrary wave shapes into linear combinations of sinusoidals.A sinusoidal wave is

f (x, t) = A cos[k(x − vpt

)+ 𝛿

]A is wave amplitudevp is wave velocity𝛿 is phase constantk is wave number

k = 2𝜋𝜆

𝜆 is wavelength—spatial periodicity of wave


Traveling Sinusoidal (Harmonic) WavesSinusoidal wave moving forward as a function of time.

wave phase velocity, vp, is speedthat dot moves.period of traveling sinusoidalwave—duration of time of onecycle—is T = 2𝜋∕(kvp)frequency is inverse of period,𝜈 = 1∕T = kvp∕(2𝜋)angular frequency is𝜔 = 2𝜋𝜈 = 2𝜋∕T = kvp

It is common to write wavefunctions in terms of 𝜔 insteadof 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,

becomes

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

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

)+ 𝛿

], left traveling wave,

becomes

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


Traveling Sinusoidal Waves

It is also common to use complex notation to describe wave functions. Incomplex notation right and left traveling sinusoidal wave function arewritten

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

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

respectively, where symbol ℜ means take real part of a complex number.

ℑ means take imaginary part of a complex number.


Complex Variables Review


Complex Variables ReviewComplex variables are a mathematical tool that simplifies equationsdescribing oscillations.Consider the 2D motion of this vector.

How would you describe this mathematically?You probably would suggest: x(t) = r cos𝜔t and y(t) = r sin𝜔t


Complex Variables ReviewWith complex notation we combine two equations into one

Start with x(t) = r cos𝜔t and y(t) = r sin𝜔t

First we define the square root of −1 as

if i =√−1 then i2 = −1

Second we define complex variable z as

z = x + iy

x is the real part and y is the imaginary part of z.

Two circular motion equations become one circular motion equation

z(t) = r cos𝜔t + ir sin𝜔t


Complex Variables ReviewEuler’s formulaIn 1748 Euler showed that

ei𝜃 = cos 𝜃 + i sin 𝜃 Euler’s formula

With Euler’s formula

z(t) = r cos𝜔t + ir sin𝜔t becomes z(t) = rei𝜔t

Any complex number can be written in the form

z = x + iy = |z|ei𝜃

where |z| is the magnitude of the complex number

|z| = √x2 + y2

and 𝜃 is the argument of the complex number

tan 𝜃 = y∕x


Complex Variables ReviewComplex Conjugate

Complex conjugate of complex number is obtained by changing sign ofimaginary part

if z = x + iy then z∗ = x − iy

z∗ is the complex conjugate of z.

Complex conjugate can also be formed by changing the sign of 𝜃

if z = |z|ei𝜃 then z∗ = |z|e−i𝜃

Couple useful identities

zz∗ = (x + iy)(x − iy) = x2 + iyx − ixy + y2 = x2 + y2 = |z|2and

(z1z2z3 ⋯)∗ = z∗1z∗2z∗3 ⋯


Standing Waves

Familiar standing waves are those within musical instruments whichgenerate sound of the instrument, such as standing wave of a violinstring, standing wave of a drum head, or standing wave inside a windinstrument.

In standing wave all parts of wave oscillate in harmonic motion atsame frequency, 𝜔, and with same phase, 𝛿.

Wave function should have same time dependence at all positions, x,in standing wave.

Think of standing waves as superposition of two traveling waves ofsame frequency and amplitude traveling in opposite directions.


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


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


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 mustbe always be equal to 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


Standing Wave Function SolutionsTime dependent part of wave function

d2T(t)dt2 + 𝜔2 T(t) = 0

For T(t) we proposeT(t) = A cos ct + B sin ct

Substitute into ODE for T(t)

(c2 − 𝜔2) cos ct + (c2 − 𝜔2) sin kt = 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 sin(𝜃 + 𝛿) where

C =√

A2 + B2 and tan 𝛿 = BA

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


Standing Wave Function SolutionsSpatial dependent part of wave function

d2X(x)dx2 + 𝜔2

v2p

X(x) = 0

For X(x) we propose

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

Substitute into ODE for X(x)(k2 − 𝜔2

v2p

)cos kx +

(k2 − 𝜔2

v2p

)sin kx = 0

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

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


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 givesX(x = 0) = D = 0

2nd boundary condition givesX(x = L) = F sin kL = 0

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

Xn(x) = F sin knx, kn = n𝜋L

From earlier fixed relationship 𝜔 = vpk, we have

𝜔n = vpkn =n𝜋vp

Lor 𝜈n =

nvp

2LP. J. Grandinetti (Chem. 4300) Wave Motion Sept. 20, 2017 27 / 108

Standing Wave—Harmonic Functions (Normal Modes)

TakingXn(x) = F sin knx

together with

Tn(t) = C 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 (normalmodes) for vibrating string.


Fourier Series Analysis of Arbitrary Wave ShapeArbitrary wave shape with these boundary conditions can be expressed aslinear combination of these harmonic functions,

f (x, t) =∞∑

n=1an sin knx cos(𝜔nt + 𝛿n)

an is amplitude of each normal mode contribution to arbitrary wave shape.

Joseph Fourier1768-1830

DefinitionAt any instant in time we can decompose any wavefunction, f (x, t), in terms of the coefficients, anusing a Fourier series analysis

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

2


Standing Wave Function SolutionsFourier Series Analysis

Wave function of plucked string

f (x, 0) = 2dL

x, 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 App Demos

String WavesBar WavesFourier Series


http://www.grandinetti.org/web-app-string-wave

http://www.falstad.com/barwaves

http://www.falstad.com/fourier

Traveling wave packet


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

Unlike standing wave, traveling wave has continuous range ofwavelengths possible.At given instant decompose it into a sum of sinusoidal (singlewavelength) waves.Discrete Fourier series needs to become continuous integral...


Fourier transform decomposes wave packet intocomponent sinusoids

Instead 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.

Note: integral of wave numbers, k, extends over negative values. Negativek values are not associated with negative wavelength but rather wavetraveling in opposite direction.

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

With f (x, t) we speak of wave packet in terms of position and time bases,

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


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

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

=

Inverse process: Given f (x, 0) find a(k, 0) using Fourier transform.P. J. Grandinetti (Chem. 4300) Wave Motion Sept. 20, 2017 37 / 108

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

Using Euler’s relation we 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

⎤⎥⎥⎥⎥⎥⎦P. J. Grandinetti (Chem. 4300) Wave Motion Sept. 20, 2017 38 / 108

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 — odd function of x — is zero

a(k, 0) = A√2𝜋 ∫

∞

−∞exp

{−1

2x2

𝜎2x

}cos kx dx

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

√𝜋∕b e−k2∕(4b). Setting b = 1∕(2𝜎2

x )

a(k, 0) = A√2𝜋

[√2𝜋𝜎2

x e−𝜎2x k2∕2

]= A𝜎k

exp

{−1

2k2

𝜎2k

}where we identify 𝜎k = 1∕𝜎x


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

Fourier transform of

f (x, 0) = A exp{−1

2x2

𝜎2x

}is a(k, 0) = A

𝜎kexp

{−1

2k2

𝜎2k

}

Fourier transform of a Gaussian function is another Gaussian function.

Gaussian is the only function with this behavior under a Fourier transform.

Wave packet with a Gaussian distribution in wave numbers produces aGaussian shaped wave packet in space.



A wave packet moving forward in time is

f (x, t) = 1√2𝜋 ∫

∞

−∞a(k, 0)ei(kx−𝜔t)dk

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 wavenumber basis as

a(k, t) = 1√2𝜋 ∫

∞

−∞f (x, 0)e−i(kx−𝜔t)dk.

Both views are equally valid, although your first preference is more likelyto imagine the wave packet in position basis.



Wave packet shape will not change with time if wave functionargument is x − vpt or x + vpt (Recall k

(x ± vpt

)= kx ± 𝜔t)

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

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

In real media wave packet may not move in space at phase velocity.

In real media 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 itsdistribution of frequencies and back with analogous transforms,

F(𝜔) = 1√2𝜋 ∫

∞

−∞f (t)e−i𝜔tdt and f (t) = 1√

2𝜋 ∫∞

−∞F(𝜔)ei𝜔td𝜔

We can even define the 2D Fourier transforms between wave function inposition and time bases to wave function in wave number and frequencybases.

f (x, t) = 1√2𝜋 ∫

∞

−∞ ∫∞

−∞a(k, 𝜔)eikxei𝜔td𝜔 dk.

and

a(k, 𝜔) = 1√2𝜋 ∫

∞

−∞ ∫∞

−∞f (x, t)e−ikxe−i𝜔tdt dx,


Uncertainty principle


What is the location of a wave packet in x?How can you define location of wave packet, f (x, t), at a given instant, t?Take p(x)dx as probability that wave packet has non-zero amplitude in dxcentered at x and calculate mean position

x = ∫∞

−∞x p(x, t)dx

But how do you get p(x, t) for wave function f (x, t)?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)|2Then 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%.P. J. Grandinetti (Chem. 4300) Wave Motion Sept. 20, 2017 45 / 108

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

Similarly, define wave packet width—range of x over which it extends—asstandard 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?Calculate 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

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

√𝜋∕a. So we set a = 1∕𝜎2

x and obtain

N = A2√𝜋𝜎2

x

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


What is the width of a Gaussian wave packet in x?2nd calculate x

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.3rd calculate x2

x2 = 1N ∫

∞

−∞x2 |f (x, 0)|2dx = 1√

𝜋𝜎2x∫

∞

−∞x2 ||||e− 1

2 x2∕𝜎2x||||2dx = 𝜎2

x∕2

Finally Plugging 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?

We can also calculate Δk, the width of wave packet in wavenumbers.

We use same approach with

Δk =√

k2 − (k)2

wherek = ∫

∞

−∞k p(k) dk and k2 = ∫

∞

−∞k2 p(k) dk


What is the width of a Gaussian wave packet in k?For Gaussian wave packet we obtain p(k) from |a(k, 0)|2

a(k, 0) = A𝜎k

exp

{−1

2k2

𝜎2k

}1st normalize wave function at t = 0

N = ∫∞

−∞|a(k, 0)|2dk = A2

𝜎2k∫

∞

−∞

||||e− 12 k2∕𝜎2

k||||2 dk =

A2√𝜋

𝜎k

Calculating k gives

k = 1N ∫

∞

−∞k |a(k, 0)|2 dk = 0

since the integrand is an odd function.Calculating k2 gives

k2 = 1N ∫

∞

−∞k2 |a(k, 0)|2 dk = 𝜎2

k∕2


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

Plugging 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.

But remember that 𝜎x = 1∕𝜎k, so product becomes

ΔxΔk = 1∕2

This product is smallest for Gaussian wave packet. For any other wavepacket shape it will always be larger, that is,

ΔxΔk ≥ 1∕2

This is the uncertainty principle between wave packet position and wavenumber.The more localized a wave packet is in space, that is, with smaller Δx, thegreater the uncertainty in wave lengths or wave numbers of which thatwave packet is composed.The more localized wave packet is in wave number (inverse) space (withsmaller Δk) the greater the uncertainty in position of wave packet.


The uncertainty principle

A similar relationship exists between time and frequency

ΔxΔk ≥ 1∕2

to go with the relationship exists between position and wavenumber

Δ𝜔Δt ≥ 1∕2

These 2 uncertainty principles appear in study of the classical waveequation ...

... and both also lie at heart of uncertainty principle that we encounter inquantum mechanics.


Dispersive Waves


Dispersive Waves

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

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


Dispersive Waves

Relationship between 𝜔 and k is called dispersion relation.

In plot of 𝜔 versus k, dispersive medium has non-linear relationship.Curvature in plot of 𝜔 versus k leads to dispersion.

non-dispersivemedium

normaldispersivemedium

anomalousdispersivemedium

00

k

ω

No dispersion occurs when𝜔∕k = vp.“Normal” dispersion is when vpdecreases with increasing k.“Anomalous” dispersion is whenvp increases with increasing k


Dispersive Wave ExamplesString waves in stiff string have anomalous dispersion with

𝜔2 =√

T𝜇

k2 + 𝛼k4

𝛼 is string stiffness.Capillary waves traveling along phase boundary of 2 fluids aredispersive

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) is refractive index of medium. This dispersion causes lightdispersing into rainbow through a prism.

Why does laser light travel long distances without significant dispersion?P. J. Grandinetti (Chem. 4300) Wave Motion Sept. 20, 2017 57 / 108

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

ideal media

Monochromatic wave with k0 = 2𝜋∕𝜆0 travels at slower speed of 𝜔0∕k0 inthis medium compared to that predicted by initial slope at k = 0.Monochromatic waves never disperse, even in highly dispersive medium,since they are 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 phasevelocity represented by range 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 series

𝜔(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 +⋯

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 movesthrough space. It is defined 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: Phase and Group velocity


https://www.youtube.com/watch?v=tlM9vq-bepA

Dispersive WavesTruncating dispersion relation series expansion at group velocity gives wavepacket that travels without changing shape, that is, without dispersing.To describe dispersion we add 𝛼 term to series

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)4

Δx0 is the initial width of the wave packet.P. J. Grandinetti (Chem. 4300) Wave Motion Sept. 20, 2017 64 / 108

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-dimensional 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

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


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 tov2

p

X(x)𝜕2X(x)𝜕x2 +

v2p

Y(y)𝜕2Y(y)𝜕y2 = 1

T(t)𝜕2T(t)𝜕t2 = −𝜔2

Right hand side is function of t only, whereas left hand side depends onlyon x and y. For equality to remain true for all values of x, y and t bothsides 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

Thus we converted PDE into three uncoupled ODEs,

d2T(t)dt2 + 𝜔2T(t) = 0 d2X(x)

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

dy2 + k2yY(y) = 0

which obey the constraint

𝜔2∕v2p = k2

y + k2x

Sinusoids will be solutions to all 3 ODEs, and traveling 2D sinusoidal wavecan be written as simple product of sinusoidal functions, e.g.,

𝜓(x, y, t) = ℜ{Aeikxxeikyye−i𝜔t}


2D Waves

𝜓(x, y, t) = ℜ{Aeikxxeikyye−i𝜔t}

Solution can also be written

𝜓(r⃗, t) = Aei(k⃗⋅r⃗−𝜔t)

k⃗ is called the wave vector and defines direction of wave propagation,

k⃗ = kxe⃗x + kye⃗y

Wave vector is related to wavelength by

|k⃗| = 2𝜋∕𝜆, where |k⃗| = √k2

x + k2y

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


2D Waves

At a given instant in time, phase of plane wave is constant along a linedefined by

𝜙 = kxx + kyy = k⃗ ⋅ r⃗

x

y

Lines

of co

nstan

t pha

se


Rectangular membrane


2D standing waves–Rectangular membraneStretch rubber sheet over opening of rectangular box and glue sheet around sides of boxStanding waves of rectangular membrane attached to fixed supports alonglines 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) = 0

𝜓(x, y, t) = A sin kxx sin ky cos(𝜔t + 𝛿),𝜓(x, y, t) must go to zero along edges: 𝜓(x = Lx, y, t) = 𝜓(x, y = Ly, t) = 0

kxLx = n𝜋 where n = 1, 2,… and kyLy = m𝜋 where m = 1, 2,…

Vibration frequencies constrained to discrete values

𝜔n,m = vp

√(n𝜋Lx

)2+(

m𝜋Ly

)2

Take special case of square box, Lx = Ly = L

𝜔n,m =𝜋vp

L

√n2 + m2

Some frequencies are degenerate (have identical values).P. J. Grandinetti (Chem. 4300) Wave Motion Sept. 20, 2017 74 / 108

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

𝜓n,m(x, y, t) = am,n sin kxx sin kyy cos(𝜔m,nt + 𝛿m,n)

As before any rectangular membrane wave can be expressed assuperposition of normal mode vibrations

𝜓(x, y, t) =∞∑

n=1

∞∑m=1

an,m sin n𝜋xLx

sinm𝜋yLy

cos(𝜔n,mt + 𝛿n,m)

Given 𝜓(x, y, t), a 2D Fourier series analysis gives an,m,

an,m = 4LxLy ∫

Lx

0 ∫Ly

0𝜓(x, y, t) sin

(n𝜋Lx

x)sin

(m𝜋Ly

x)

dxdy, n,m = 1, 2,…


2D standing waves–Rectangular membraneNormal modes of a rectangular membrane.

The nodal lines for the first few values of m and n..m = 1

1

2

2

3

3n


Web App Demos

Rectangular Membrane Waves


http://www.grandinetti.org/rectangular-membrane-wave

Circular membrane


2D standing waves–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 fixedsupports.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–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

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).


2D standing waves–Circular membraneODE for angular part, Φ(𝜙),

d2Φ(𝜙)d𝜙2 + m2Φ(𝜙) = 0

has solutionΦ(𝜙) = Ae±im𝜙

We require wave functions to be single valued, therefore

Φ(𝜙 + 2𝜋) = Φ(𝜙)

which meansAe±im(𝜙+2𝜋) = Ae±im𝜙 or e±im2𝜋 = 1

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


2D standing waves–Circular membrane

ODE 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 ODE reduces to

d2R(r)dr2 +

(𝜔vp

)2R(r) = 0

This has form of 1D wave equation. In other words, if membrane extendedradially to infinity then at large r circular waves appear as plane wavesradiating outward.

But we need exact solution for arbitrary value of r.


2D standing waves–Circular membraneHow do we find exact 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).P. J. Grandinetti (Chem. 4300) Wave Motion Sept. 20, 2017 83 / 108

Bessel functions of 1st kind

Look like oscillating sinusoid that decays as 1∕√

u.

-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 Roots are indicated as u(n)m .

For solutions with negativem use identityJ−m(u) = (−1)mJm(u)Boundary condition,𝜓(r = a, 𝜙, t) = 0, requiresu = ka be a roots of Jm(u),

Jm(kna) = 0 or kna = u(n)m

With this constraint 𝜔becomes𝜔n = vpkn = 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 atlarge u.


2D normal modes of a circular membrane𝜓m,n(r, 𝜙, t) = AJm(knr) cosm𝜙 cos(𝜔nt+𝛿n)

m = 0

1

1

2

3

2 3

n

Each normal mode hasn − 1 radial nodes and mangular nodes.

Arbitrary circular membrane wave is superposition of normal modes

𝜓(r, 𝜙, t) =∞∑

n=1

∞∑m=−∞

am,nJm(knr) cosm𝜙 cos(𝜔nt + 𝛿n)


Web App Demos

Circular Membrane Waves


http://www.grandinetti.org/web-app-circular-membrane-wave

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

using 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


describes a three dimensional plane wave. Plane waves are effectively onedimensional but travel in three dimensions.


Web App Demos

3D Box Waves


http://www.grandinetti.org/web-app-3d-box-wave

3D Waves in spherical coordinates


3D Waves in spherical coordinatesNext let’s examine the 3D wave equation in spherical coordinates

∇2𝜓(r, 𝜃, 𝜙, t) = 1v2

p

𝜕2𝜓(r, 𝜃, 𝜙, t)𝜕2t

where ∇2 becomes

∇2 = 𝜕2

𝜕r2 + 2r𝜕𝜕r

+ 1r2

[1

sin 𝜃𝜕𝜕𝜃

(sin 𝜃 𝜕

𝜕𝜃

)+ 1

sin2 𝜃

(𝜕2

𝜕𝜙2

)]Using separation of variables we can write the wave function in the form

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,1

R(r)Θ(𝜃)Φ(𝜙)∇2

(R(r)Θ(𝜃)Φ(𝜙)

)= 1

v2p

1T(t)

𝜕2T(t)𝜕2t

= −𝜔2

and obtain an ODE for T(t),d2T(t)

dt2 + 𝜔2T(t) = 0



Leaves us with PDE involving 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

and 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)

We transform the differential for R(r) 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)where 𝜆 = 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



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) = Ae±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.1st defining v = cos 𝜃, where

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 (Chem. 4300) Wave Motion Sept. 20, 2017 97 / 108

3D Waves in spherical coordinatesTaken together we have

ddv

[(1 − v2)dΘ(v)

dv

]+[𝓁(𝓁 + 1) − m2

1 − v2

]Θ(v) = 0

This is the associated Legendre differential equation whose solutions arethe associated Legendre 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 harmonicfunctions 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.Highlighted are the polynomials associated with the sectoral harmonics.

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


3D Waves in spherical coordinatesSpherical harmonics with m = 0...

are proportional to just Legendre polynomials.

do not depend upon longitude and are called zonal harmonics becausetheir roots divide the sphere into zones.

when viewed down the y axis the nodal lines of the normal modesappear as shown below.

z

x



Spherical harmonics with |m| = 𝓁...

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

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

x

y


3D Waves in spherical coordinatesSpherical 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 is2|m|.nodal lines of the normal modes appear as shown below for the caseof 𝓁 = 7 and m = 4.

z

x

x

y



The spherical harmonics are a complete orthonormal basis for expandingthe angular dependence of a function, and satisfy an orthogonality relation:

∫2𝜋

0 ∫𝜋

0Y∗𝓁1,m1

(𝜃, 𝜙)Y𝓁2,m2(𝜃, 𝜙) = 𝛿𝓁1,𝓁2

𝛿m1,m2

Any arbitrary real function f (𝜃, 𝜙) can be expanded in terms of complexspherical 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 1600s Christiaan Huygens showed that light behaves like a wave. Heshowed that light undergoes diffraction and interference.In diffraction plane waves spread out (diffract) as they pass through slit.With slit width larger than wavelength wave undergoes less diffraction.

planewaves

wavecrest

wavetrough

planewaves

wavecrest

wavetrough


Diffraction and InterferenceWith interference we observe constructive effects where two wave crestsmeet or two wave troughs meet and destructive effects where wave crestand trough meet.

planewaves

wavecrest

wavetrough

point ofconstructiveinterference

point ofdestructiveinterference

Huygens discovery that lightdiffracts and interferes wassurprising because at the timelight was thought to becomposed of particles.Only waves known at the timewere mechanical waves whichrequire a physical medium.After Huygens the search beganto discover the “mechanicalmedium” for light or what wascalled the luminiferous ether.


Web App Demos

2D Wave Interference


http://www.grandinetti.org/2d-wave-interference

wave motion - chapter 8 - grandinetti · wave motion chapter 8 p. j. grandinetti chem. 4300 sept...

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