# Lecture 5 Energy and Momentum

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Wave Phenomena Physics 15c

Lecture 5 Energy and Momentum

What We Did Last Time Studied (finally!) continuous waves Started from mass-spring transmission line Found the wave equation and its normal modes

Solutions represent waves traveling at constant velocity

Used Fourier series to show the waveform can be arbitrary

ρl

∂2

∂t 2 ξ(x,t) = K∂2

∂x2 ξ(x,t) ξ(x,t) = Aexp(i(ω t ± kx))

cw =

Kρl

Recap: How We Did It N coupled pendulums turns into continuous transmission line as N infinity N equations of motion turn into a wave equation

Linear algebra assures that the coupled pendulums have N normal mode solutions This turns into a factorized solution

Given this tip, we can solve the wave equation to get

Fourier series allows us to break arbitrary function into sin and cos, for which we have wave solutions Since all solutions travel at the same speed, and because of

linearity, we conclude that arbitrary wave form can be created, and transmitted at the same speed

ξ(x,t) = a(x)eiω t

ξ(x,t) = Aexp(i(ω t ± kx))

Arbitrary Traveling Waves In fact, the wave equation is satisfied by any function of the form ξ(x, t) = f(x ± cwt)

LHS = ρl

∂2

∂t 2 f (x ± cwt)

= ρlcw2 ′′f (x ± cwt)

= K ′′f (x ± cwt)

ρl

∂2

∂t 2 ξ(x,t) = K∂2

∂x2 ξ(x,t)

RHS = K∂2

∂x2 ξ(x ± cwt)

= K ′′f (x ± cwt)

cw =

Kρl

This seems just too simple after all that work!

Subtleties The conclusion appears stupidly simple We can generate any wave, and it will travel Examples: sound, radio waves

But there are non-trivial assumptions Linearity Constant velocity for all normal modes, i.e. for all ω

Bad examples: water waves, light passing glass Those are called dispersive media

Will come back to this soon

Goals for Today Energy and momentum carried by the waves Calculate the energy and momentum densities Find how fast they are transmitted There is a general relationship between the two

Calculate the power and the force needed to create the waves Must match the energy and momentum transfer

Energy Density

Consider forward-going normal-mode waves:

Energy is in two forms Kinetic energy of the masses Potential energy of the springs

ξ(x − Δx,t) ξ(x,t) ξ(x + Δx,t)

ξ(x,t) = ξ0 cos(kx −ω t)

Kinetic Energy Velocity of the mass at x is

Kinetic energy is

There is a mass at every Δx The energy density (in Joules/meter) is

EK =

12

mv 2 =12

m∂ξ(x,t)

∂t⎛⎝⎜

⎞⎠⎟

2

dEK

dx= lim

Δx→0

EK

Δx= lim

Δx→0

12

mΔx

∂ξ(x,t)∂t

⎛⎝⎜

⎞⎠⎟

2

=12ρl

∂ξ(x,t)∂t

⎛⎝⎜

⎞⎠⎟

2

v(x,t) = ∂ξ(x,t)

∂t

Spring Energy The spring between x and x + Δx has an energy of

Taylor expansion

There is a spring at every Δx The energy density (in Joules/meter) is

ES =

12

kS ξ(x + Δx,t) − ξ(x,t)( )2

dES

dx= lim

Δx→0

ES

Δx= lim

Δx→0

12

kSΔx∂ξ(x,t)∂x

⎛⎝⎜

⎞⎠⎟

2

=12

K∂ξ(x,t)∂x

⎛⎝⎜

⎞⎠⎟

2

ES =

12

kS

∂ξ(x,t)∂x

Δx⎛⎝⎜

⎞⎠⎟

2

Kinetic + Spring Energy Recall the general solution:

Since for any waveform

We have not used any specific shape of ξ(x, t)

The total energy density (in J/m) is

dEK

dx=

12ρl

∂ξ(x,t)∂t

⎛⎝⎜

⎞⎠⎟

2

=12ρ

lcw

2 ′f (x ± cwt)( )2

dES

dx=

12

K∂ξ(x,t)∂x

⎛⎝⎜

⎞⎠⎟

2

=12

K ′f (x ± cwt)( )2

ξ(x,t) = f (x ± cwt)

cw =

Kρl

dEK

dx=

dES

dx

Kinetic

Spring

dEdx

= ρl

∂ξ(x,t)∂t

⎛⎝⎜

⎞⎠⎟

2

= K∂ξ(x,t)∂x

⎛⎝⎜

⎞⎠⎟

2

Energy Density in Normal Mode Now consider the normal mode ξ0cos(kx – ωt)

Energy density comes in packets The packets travel with the

waves, with the velocity cw

x

t

Wave Energy density

dEdx

= ρl

∂ξ0 cos(kx −ω t)∂t

⎛

⎝⎜⎞

⎠⎟

2

= ρlω2ξ0

2 sin2(kx −ω t)

Energy Transfer Rate The energy density averages to

This travels at velocity cw The average rate of energy transfer (in Joules/second = Watts) is

This much of power is needed to create this wave Let’s check

dEdx

= ρlω2ξ0

2 sin2(kx −ω t)

dEdx

=12ρlω

2ξ02 Average energy density

of a normal-mode wave

dEdt

=12

cwρlω2ξ0

2 =12

Kρlω2ξ0

2

cw =

Kρl

Creating Waves

Motor creates waves by moving as ξ0cosωt The power is given by (force) x (velocity) Velocity = −ξ0ω sinωt What is the force?

Creating Waves

Instead of the motor, imagine there were more masses and springs, doing the wave together The spring between x = −Δx and x = 0 is pushing the first “real” mass by

This is the force you need to keep the wave going

ξ(0,t) ξ(−Δx,t)

F = −kS ξ(0,t) − ξ(−Δx,t)⎡⎣ ⎤⎦

Creating Waves

Multiply this by the velocity

Time average of sin2ωt is 1/2

This matches the energy transfer rate, as expected

F = −kS ξ(0,t) − ξ(−Δx,t)⎡⎣ ⎤⎦ = −kS

∂ξ(x,t)∂x

⎛⎝⎜

⎞⎠⎟ x=0

Δx = −K∂ξ(x,t)∂x

⎛⎝⎜

⎞⎠⎟ x=0

= −Kξ0k sinω t

ξ(x,t) = ξ0 cos(kx −ω t)

−ξ0ω sinω t

P = Kξ02kω sin2 ω t = Kρlω

2ξ02 sin2 ω t

k =ω

ρl

K

The average power needed to create the waves

P =12

Kρlω2ξ0

2

Example: Audio Speakers Sound is an example of longitudinal waves Will be discussed in the next lecture

Imagine a round speaker attached to a pipe Air has volume mass density

1.29 kg/m3 ρl = 1.29π r2 kg/m Sound velocity cw is 330 m/s

2-inch speaker moving ±1 mm at 1 kHz 16.5 Watts 12-inch speaker moving ±5 mm at 20 Hz 6.0 Watts

The ω2 factor makes it much more difficult to generate low frequency sound

r

cw =

Kρl

dEdt

=12

Kρlω2ξ0

2 =12

cwρlω2ξ0

2 = 670r 2ω 2ξ02 (Watts)

Momentum Density Momentum is just (mass) x (velocity) Consider a mass m at position x

Time average of sinωt is 0 No net momentum

This is wrong The problem: mass density variation

p = m ×

∂ξ(x,t)∂t

= mξ0ω sin(kx −ω t)

Density Wave We are studying longitudinal wave The direction of movement = the direction of transmission

The movement can be seen as changing density

To calculate the momentum, we must take this density variation into account

wave

Density Wave Consider the piece between x and x + Δx Thickness is changed by the wave

For small Δx,

This changes the density

x x + Δx

x + ξ(x,t)

x + Δx + ξ(x + Δx,t)

Δx → Δx + ξ(x + Δx,t) − ξ(x,t)

ξ(x + Δx,t) − ξ(x,t) = ∂ξ

∂xΔx

Δx → Δx +

∂ξ∂x

Δx = Δx 1+ ∂ξ∂x

⎛⎝⎜

⎞⎠⎟

Density Wave

Assume small waves

The density should not change too much…

For

ρl →ρl

1+ ∂ξ∂x

small

ρl → ρl 1− ∂ξ(x,t)

∂x⎛⎝⎜

⎞⎠⎟

ξ(x,t) = ξ0 cos(kx −ω t)

ρl → ρl 1+ ξ0k sin(kx −ω t)( )

Δx → Δx 1+ ∂ξ

∂x⎛⎝⎜

⎞⎠⎟

How It Looks Like

ξ(x,t) = ξ0 cos(kx −ω t)

Δρl = ρlξ0k sin(kx −ω t)

ρl + Δρl

velocity = ξ0ω sin(kx −ω t)

momentum density = mass density × velocity

Momentum Density

Momentum density (in kg/s) is

Average over time sin(kx – wt) 0 sin2(kx – wt) ½

The rate of momentum transfer (in newtons) is

Δρl = ρlξ0k sin(kx −ω t) v = ξ0ω sin(kx −ω t)

dpdx

= ρl + Δρl( )v = (ρl + ρlξ0k sin(kx −ω t))ξ0ω sin(kx −ω t)

dpdx

=12ρlωkξ0

2 =12ρl

cw

ω 2ξ02

cw =

ωk

dpdt

= cw

dpdx

=12ρlω

2ξ02

Energy vs. Momentum Energy density and momentum density are related

This is true for any waveform on this medium This is true even for different kinds of waves as long as it follows the

non-dispersive wave equation Electromagnetic waves is one such example e.g. a 100 MW laser pulse produces a force of

dEdx

=12ρlω

2ξ02

dpdx

=12ρlω

2ξ02

cw

energy densitypropagation velocity

= momentum density

1×108 (W)/3 ×108 (m/s) = 0.3(N)

Force and Wave Creation

Waves carry momentum Motor must give net force to create waves Let’s see if we can confirm this

Motor movement is ξ0coswt Warning: Tricky math ahead!

Force From the Motor We have already calculated that the force motor must produce was

Time-averaging this gives 0 (again!)

What’s wrong this time? Density variation is changing the wavenumber k

This makes the whole thing non-linear!

F = −K

∂ξ∂x

⎛⎝⎜

⎞⎠⎟ x=0

= −Kξ0k sinω t

cw =

ωk=

Kρl

=kS

mΔx

Δx → Δx 1+ ∂ξ

∂x⎛⎝⎜

⎞⎠⎟

A Bit of Reflection We did not take the effect of density variation into account when we calculated the energy Why was it allowed?

Whole calculation is based on linear approximation We always take the first-order in Taylor series Valid if the wave amplitude x0 is small

If the first-order terms cancel out to 0, we can no longer ignore the higher-order term This happens when you try to calculate wave momentum

Back to the Force

The force is (to the 2nd-order approximation)

Average over time

F =

12

Kk 2ξ02 =

12ρlω

2ξ02 = momentum transfer rate

k →k

1+ ∂ξ∂x

≈ k 1− ∂ξ∂x

⎛⎝⎜

⎞⎠⎟

cw =

ωk=

Kρl

=kS

mΔx

Δx → Δx 1+ ∂ξ

∂x⎛⎝⎜

⎞⎠⎟

F = −Kξ0k 1−∂(ξ0 cos(kx −ω t))

∂x⎛

⎝⎜⎞

⎠⎟ x=0

⎛

⎝⎜

⎞

⎠⎟ sinω t

= −Kξ0k sinω t(1− ξ0k sinω t)

Summary Studied the energy and momentum carried by waves Energy is distributed non-uniformly over space

Kinetic energy = spring potential energy It travels at the wave velocity Momentum behaves similarly Energy density / velocity = momentum density

Calculated what it takes to create the waves Power needed = energy transfer rate Force needed = momentum transfer rate

We know all about longitudinal waves now