Wave Mechanics: Wave Intensity and Flux

WM18 introduced the energy carried by a one-dimensional wave, and WM19–WM20 developed the instantaneous and average power transported through a fixed point on an ideal string. The next step is to distinguish total power from power distributed over an area.

For a wave whose average power ⟨P⟩ is distributed uniformly across an area A, the wave intensity is

|---------| I = ⟨P⟩. | ------A----

(1)

The SI unit of intensity is

|------------| [I] = W ∕m2. | --------------

(2)

Intensity is therefore an areal energy-flux density: it tells us how much energy per unit time crosses each unit area perpendicular to the direction of transport [3, 4, 1, 2].

This distinction becomes essential when waves spread in two or three dimensions. A source may emit the same total power while that power is distributed over an ever larger wavefront. In that situation the total power can remain constant even while the intensity decreases.

1 Power is not the same as intensity

Power measures the rate of energy transfer:

P = dE-. dt

(3)

Its unit is the watt:

[P] = W = J ∕s.

(4)

Intensity divides that transported power by area:

I = ⟨P⟩. A

(5)

Thus two waves can carry the same total power but have very different intensities if they occupy different cross-sectional areas.

Figure. Intensity is average wave power per unit area measured through a surface perpendicular to the direction of energy transport.

For a uniform intensity over the surface,

|----------| |⟨P ⟩ = IA. | ------------

(6)

This simple relation will appear repeatedly in acoustics, optics, electromagnetic waves, and other wave systems.

2 Geometric spreading

Suppose the same average power passes through two perpendicular surfaces with areas A₁ and A₂. If there is no absorption or other loss between the surfaces,

⟨P ⟩ = I1A1 = I2A2.

(7)

Therefore

|----------| |I2 A1- | |I1 = A2. | -----------

(8)

If the area increases, the intensity decreases even though the total transported power remains the same.

Figure. The same total power distributed over a larger area produces a smaller intensity. This is geometric spreading, not necessarily dissipative loss.

This distinction is important:

geometric spreading redistributes the same power over a larger area;
dissipation or absorption converts some wave energy into other forms, reducing the power that remains in the wave.

A measured decrease in intensity can result from either effect, so the changing wavefront area must be considered before concluding that energy has been dissipated.

3 Energy flux as a vector quantity

In more than one spatial dimension, energy transport has a direction as well as a magnitude. Introduce an energy-flux vector

JE,

(9)

with units

[JE ] = W ∕m2.

(10)

The instantaneous power crossing an oriented surface S is

|------------------| | ∫ | |P(t) = JE ⋅ dA.| ---------S----------

(11)

Here

dA = ˆn dA

(12)

points normal to the surface.

If the flux is uniform over a flat area and makes an angle α with the surface normal, then

P = JEA cosα.

(13)

The largest power crosses when the surface is perpendicular to the direction of propagation, so α = 0 and

P = JEA.

(14)

For a steady or periodic wave, the scalar intensity is commonly the magnitude of the time-averaged energy flux in the propagation direction:

|-----------| I = |⟨JE ⟩|. | -------------

(15)

When direction matters, keeping the vector or signed flux is more informative than using only the positive scalar intensity.

4 Local conservation of wave energy

WM19 obtained the one-dimensional conservation law

|------------| |ℰt + Px = 0,| --------------

(16)

where ℰ is energy per unit length and P is signed power along the string.

The corresponding three-dimensional local conservation law is

|------------------| |∂w- | | ∂t + ∇ ⋅ JE = 0,| -------------------

(17)

where w is energy per unit volume:

3 [w ] = J∕m .

(18)

The divergence term measures the net outward energy flux from a small volume. If more energy flows out than flows in, the energy stored inside must decrease.

This is the same bookkeeping principle used in WM19, now written in a form appropriate to multidimensional waves [2, 6].

5 Connecting the 1D string to a 3D flux density

Suppose, only for the purpose of connecting the dimensions, that a one-dimensional wave model represents a uniform wave field across a constant area A. Then

ℰ = wA

(19)

and

P = JEA.

(20)

Dividing the 1D conservation equation by A gives the corresponding 3D density form.

Figure. The one-dimensional quantities ℰ and P correspond to volume energy density w and areal energy flux density I when a uniform cross-sectional area is introduced.

For the ideal string, P is the natural flux quantity. The string model does not require an areal intensity because its energy density was defined per unit length rather than per unit volume. Introducing I = P∕A is therefore a bridge to higher-dimensional wave fields, not a replacement for the 1D string power.

6 Intensity and energy density for a progressive wave

WM19 showed that for a pure right-moving nondispersive wave on an ideal string,

P = cℰ .

(21)

ℰ = wA

(22)

and

P = IA,

(23)

then

IA = cwA.

(24)

Canceling the area gives

|--------| -I-=-cw.-|

(25)

For a periodic wave the corresponding average relation is

|----------| -I-=-c⟨w-⟩.|

(26)

This result has a simple interpretation: energy density tells us how much energy occupies a unit volume, while multiplication by the propagation speed tells us how rapidly that energy sweeps through a unit area.

The relation I = cw is especially natural for a single progressive nondispersive wave. It should not be applied blindly to Standing Waves, arbitrary multidirectional fields, or dispersive media without reconsidering the appropriate energy-transport velocity.

7 Spherical spreading and the inverse-square law

Consider an ideal isotropic point source emitting average power

P . source

(27)

At distance r, the power is distributed uniformly over a spherical surface with area

A (r) = 4πr2.

(28)

If no power is lost between the source and the sphere, then

Psource = I(r)4πr2.

(29)

Therefore

|--------------| | Psource | |I(r) = ----2-.| ---------4πr----

(30)

This gives the inverse-square law

|--------| | -1 | -I-∝-r2-.|

(31)

For two radii,

|--------------| |I2 ( r1 )2 | |-- = -- . | -I1-----r2-----|

(32)

Figure. For an ideal isotropic spherical wave, the same source power crosses every sphere. Intensity decreases as 1∕r² because the area grows as 4πr² [3, 4].

Doubling the distance gives

I (r) I(2r) = --4- .

(33)

Tripling the distance gives

I(3r) = I-(r) . 9

(34)

Again, this decrease does not by itself imply energy loss. It can occur entirely because of geometric spreading.

8 Different spreading geometries

The dependence of intensity on distance is controlled by the area of the wavefront.

For an ideal plane wave with constant cross-sectional area,

I ≈ constant

(35)

in the absence of loss.

For a cylindrically spreading wave, the wavefront area per fixed axial length grows proportional to r, so

I ∝ 1. r

(36)

For a spherical wave,

-1 I ∝ r2 .

(37)

Thus the familiar inverse-square law is not a universal property of all waves. It is a consequence of spherical geometry.

9 Intensity and wave amplitude

For a linear sinusoidal wave in a fixed medium, average transported power is proportional to the square of the wave amplitude. WM20 found for the ideal string

2 ⟨P⟩ ∝ A .

(38)

The same square-law structure occurs for many other linear wave systems, although the proportionality constant depends on the medium and on the physical meaning of the amplitude.

If a spherically spreading linear wave satisfies

I ∝ A2

(39)

and

-1 I ∝ r2 ,

(40)

then its far-field amplitude scales as

|-------| | 1 | A ∝ -. | -----r---

(41)

This is another way to understand why a wave can become weaker with distance even in a lossless medium.

10 Worked Example 1: Convert power to intensity

A wave carries an average power

⟨P ⟩ = 4.0W

(42)

uniformly through an area

A = 0.020 m2.

(43)

Find the intensity.

Solution

Use

⟨P⟩ I = ---. A

(44)

Then

I	= W∕m²	(45)
	= 200 W∕m².	(46)

Therefore

|---------------| I-=-200-W-∕m2.---

(47)

11 Worked Example 2: Spherical spreading

An ideal isotropic source radiates

Psource = 18 W.

(48)

Find the intensity at r = 3.0 m and at r = 6.0 m.

Solution

At 3.0 m,

I₁	= W∕m²	(49)
	≃ 0.159 W∕m².	(50)

At twice the radius, the inverse-square law gives

( )2 3.0 I2 = I1 6.0 .

(51)

Therefore

I = I1 ≃ 0.0398 W ∕m2. 2 4

(52)

Thus

|-----------------------| | 2 | I(3.0-m)-≃-0.159-W-∕m----

(53)

and

|----------------------2-| I-(6.0-m-) ≃-0.0398-W-∕m-.-

(54)

The total power is unchanged; only the area over which it is distributed has increased.

12 Worked Example 3: Intensity from volume energy density

A progressive wave has average volume energy density

⟨w ⟩ = 0.015 J∕m3

(55)

and propagation speed

c = 340 m/s.

(56)

Find the intensity. Then find the average power through a perpendicular area of 0.40 m².

Solution

For a progressive nondispersive wave,

I = c⟨w ⟩.

(57)

Thus

I	= (340)(0.015) W∕m²	(58)
	= 5.10 W∕m².	(59)

The average power is

⟨P⟩	= IA	(60)
	= (5.10)(0.40) W	(61)
	= 2.04 W.	(62)

Therefore

|----------------| |--------------| |I = 5.10 W ∕m2, | ⟨P ⟩ = 2.04W. | ----------------- ----------------

(63)

13 Worked Example 4: Same power through a larger area

A wave has intensity

I1 = 0.25W ∕m2

(64)

through a uniform area

A1 = 2.0m2.

(65)

The same total power later occupies an area

A2 = 5.0m2.

(66)

Find the new intensity.

Solution

First compute the conserved power:

⟨P ⟩ = I A = (0.25)(2.0) = 0.50 W. 1 1

(67)

Then

⟨P ⟩ 0.50 2 I2 = -A-- = -5.0 = 0.10 W ∕m . 2

(68)

Therefore

|----------------| -I2 =-0.10-W-∕m2.-

(69)

14 Worked Example 5: Bridge from string power to an areal intensity

A sinusoidal wave on an ideal string has

μ = 0.012 kg/m, c = 100 m/s,

(70)

A = 2.0 mm, f = 40 Hz.

(71)

First find the average string power. Then, purely as a dimensional bridge to a uniform three-dimensional field, suppose that power is distributed over an effective area

A = 5.0 × 10−4m2. eff

(72)

Find the corresponding intensity.

Solution

Use the WM20 result

1 2 2 ⟨P ⟩ = --μA ω c. 2

(73)

The angular frequency is

ω = 2πf = 80 πrad/s.

(74)

With

A = 0.0020 m,

(75)

we obtain

⟨P⟩	= (0.012)(0.0020)²(80π)²(100)	(76)
	≃ 0.152 W.	(77)

If this power is assigned to the stated effective area,

I	= W∕m²	(78)
	≃ 303 W∕m².	(79)

Thus

|---------------| |---------------| |⟨P⟩ ≃ 0.152 W, | |I ≃ 303 W ∕m2. | ----------------- -----------------

(80)

The string calculation itself needs only ⟨P⟩. The areal intensity appears only after an effective area is introduced.

15 Worked Example 6: Separate geometric spreading from true loss

At radius r₁ = 2.0 m from a source, an intensity I₁ is measured. At radius r₂ = 5.0 m, the measured intensity is

I2 = 0.10I1.

(81)

If the wave were lossless and spherically spreading, what ratio would be expected? What fraction of the power crossing the first sphere remains in the wave at the second sphere?

Solution

Pure spherical spreading predicts

	= ²	(82)
	= ²	(83)
	= 0.16.	(84)

Thus geometric spreading alone would give

|----------------| -I2,geom--=-0.16I1.|

(85)

The actual powers through the two spherical surfaces satisfy

P2 I24πr2 ---= -----22-. P1 I14πr1

(86)

Therefore

	= 0.10²	(87)
	= 0.625.	(88)

|------| |62.5% | --------

(89)

of the wave power crossing the inner sphere remains at the outer sphere. The remaining

|------| |37.5% | --------

(90)

has been removed from the propagating wave by effects beyond ideal geometric spreading.

16 Common mistakes

Mistake: treating power and intensity as the same quantity. Power is measured in watts; intensity is power per unit area.
Mistake: assuming any decrease in intensity means energy has been dissipated. Intensity can decrease simply because the wave spreads over a larger area.
Mistake: applying the spherical inverse-square law to every wave geometry. Plane, cylindrical, and spherical spreading have different area laws.
Mistake: forgetting that flux has a direction. The dot product J_E ⋅dA determines the signed power crossing an oriented surface.
Mistake: applying I = cw to arbitrary standing, dispersive, or multidirectional waves without checking the transport physics.
Mistake: assigning a unique areal intensity to the ideal 1D string without first defining an area. The intrinsic 1D transport quantity is the power P.

17 What WM21 adds to the wave-mechanics picture

The sequence of transport quantities is now

|------------------------------------------------------------------| |energy − → energy density − → power −→ intensity or flux density. | -------------------------------------------------------------------

(91)

For the one-dimensional string,

|------------| -ℰt +-Px-=-0.-

(92)

For a multidimensional wave field,

|------------------| |∂w- + ∇ ⋅ JE = 0.| --∂t---------------|

(93)

For a progressive nondispersive wave,

|----------| |I = c⟨w ⟩.| -----------

(94)

And for an ideal isotropic spherical source,

|--------------| | Psource | |I(r) = 4πr2 .| ----------------

(95)

These ideas provide the transport language needed later for acoustic intensity, electromagnetic energy flux, RF propagation, and higher-dimensional wave equations.

References

[1] A. P. French, Vibrations and Waves, M.I.T. Introductory Physics Series, W. W. Norton & Company, 1971.

[2] Frank S. Crawford, Jr., Waves, Berkeley Physics Course, Volume 3, McGraw-Hill, 1968.

[3] William Moebs, Samuel J. Ling, and Jeff Sanny, University Physics, Volume 1, OpenStax, 2016, Section 16.4, “Energy and Power of a Wave.”

[4] William Moebs, Samuel J. Ling, and Jeff Sanny, University Physics, Volume 1, OpenStax, 2016, Section 17.3, “Sound Intensity.”

[5] Massachusetts Institute of Technology, 8.03SC Physics III: Vibrations and Waves, MIT OpenCourseWare, Fall 2016.

[6] Massachusetts Institute of Technology, 2.24 / 13.022 Ocean Wave Interaction with Ships and Offshore Energy Systems, Lecture 4, “Wave Energy Density and Flux,” MIT OpenCourseWare, Spring 2002.

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