Loudness, Intensity and Decibels

Beginner

Hold a can of spray paint close to a wall and you get a small, thick patch. Step back to twice the distance and the same paint covers a patch twice as wide and twice as tall, four times the area, so the layer is four times thinner. Sound from a loudspeaker spreads in the same way, in all directions, like a growing bubble. The blue rings in the animation are that bubble, and they fade as they grow.

The speaker sends out the same sound energy every second. The intensity is how much of it arrives on each square metre every second, measured in watts per square metre. Look at the gold squares: the sound passing through one square near the speaker is shared by four equal squares twice as far away, so each of them gets only a quarter.

Our ears cope with an enormous range, from the faintest sound they can detect to noise a million million times more intense. To keep the numbers friendly we use decibels (dB), shown on the chart: every step of 10 dB means ten times the intensity. So 60 dB is a hundred times as intense as 40 dB. Moving twice as far away costs about 6 dB.

Our sense of loudness is more modest than the numbers: a sound needs roughly ten times the intensity, 10 dB more, to seem twice as loud. And loudness is not pitch: high or low depends on frequency, so a soft whistle can be high and a loud drum low.

Look after your hearing: hours of sound above about 85 dB can damage it for good.

Check yourself: a radio plays at 70 dB and another sound reaches 90 dB. How many times more intense is it? Answer: 100 times, two steps of 10 dB.

Intermediate

Intensity: power through each square metre

The acoustic power $P$ of a source is the sound energy it sends out each second, in watts. Sound carries little energy: a typical loudspeaker turns 1 % or less of its electrical power into sound, and the one in the animation radiates just $P=10\mathrm{mW}$.

Your ear collects only the share of that power that falls on its own small area. The intensity is the power crossing each square metre of a surface facing the sound:
$I=\frac{P}{A}$

It is measured in W/m² and describes the sound at one place, not the source. Intensity is not loudness, which is how strong the sound seems to a listener. Nor is it pitch, high or low, which is set mainly by the frequency.

The inverse-square law

Picture a small source sending sound out evenly in every direction. At a distance $r$ the energy it emits each second is spread over a sphere of area $4\pi r^2$, so the intensity of a point source is
$I=\frac{P}{4\pi r^2}$

Doubling the distance makes the sphere four times larger, so the intensity drops to a quarter; tripling it divides the intensity by 9. When you compare two distances, $P$ and $4\pi$ cancel:
$\frac{I_2}{I_1}={\left(\frac{r_1}{r_2}\right)}^2$

The gold pyramid in the animation shows the geometry. Rays through the corners of one square at $r=2\mathrm{m}$ enclose, at 4 m, a square twice as wide and twice as tall: four equal squares, labelled $I/4$. The energy that crossed one square is shared among four: none is lost, it is only spread thinner.

The decibel scale

The ear handles an enormous range, from about $10^{-12}W/{\mathrm{m}}^2$, the faintest 1 kHz tone a young listener can hear, to about 10 W/m², where sound becomes painful. Instead of numbers spread over thirteen powers of ten, we count the powers of ten. The sound intensity level is
$\beta =10{\log }_{10}\frac{I}{I_0},I_0=10^{-12}W/{\mathrm{m}}^2$
in decibels (dB). A factor of 10 in intensity adds 10 dB and a factor of 100 adds 20 dB; since ${\log }_{10}2=0.301$, doubling adds 3.0 dB and quadrupling 6.0 dB.

Two working rules follow. Because $I_2/I_1=(r_1/r_2{)}^2$, the logarithm turns the square into a factor 2, and $10\cdot 2=20$: the level falls by 20 dB per tenfold distance and by 6 dB per doubling. Because independent sources add their intensities, $N$ equal sources raise the level by $10{\log }_{10}N$:

• ${\beta }_2={\beta }_1-20{\log }_{10}\left(\frac{r_2}{r_1}\right)$
• ${\beta }_N={\beta }_1+10{\log }_{10}N\text{(}N\text{ equal incoherent sources)}$

Worked example: the meter in the animation

The speaker radiates $P=10\mathrm{mW}=0.010\mathrm{W}$. For the near listener, at $r=2\mathrm{m}$:
$A=4\pi r^2=4\pi \cdot 2^2=50.3{\mathrm{m}}^2$
$I=\frac{0.010\mathrm{W}}{50.3{\mathrm{m}}^2}=1.99\times 10^{-4}W/{\mathrm{m}}^2$
so $I/I_0=1.99\times 10^8$ and
$\beta =10{\log }_{10}\left(1.99\times 10^8\right)=10\cdot 8.30=83.0\mathrm{dB}$

The far listener, at 4 m, receives a quarter of that intensity, $4.97\times 10^{-5}W/{\mathrm{m}}^2$, at a level of $83.0-20{\log }_{10}2=77.0\mathrm{dB}$. These are the $I$ and $\beta$ readouts as the red meter reaches the listeners' distances: at the near listener about 199 µW/m² and 83.0 dB, and at the far listener 49.7 µW/m² and 77.0 dB.

The meter starts at 1 m (795.8 µW/m², 89.0 dB) and ends at 8 m (12.4 µW/m², 70.9 dB). Its distance doubles every 2 s, so each step divides the intensity by 4 and lowers the level by the same 6.0 dB, and the pointer slides down the chart in equal steps. Over the whole walk the intensity shrinks 64 times, yet the level falls by only 18 dB.

Worked example: a safe distance from a drill

A meter 2 m from a pneumatic drill reads 100 dB. How far away must you be for the level to fall to 85 dB, the hearing-risk mark on the chart? At 2 m, $I=I_0\cdot 10^{100/10}=0.01W/{\mathrm{m}}^2$; if the drill radiated evenly in all directions, its acoustic power would be $P=4\pi \cdot 2^2\cdot 0.01\approx 0.50\mathrm{W}$. The level must drop by 15 dB, so
$20{\log }_{10}\left(\frac{r}{2\mathrm{m}}\right)=15$
$r=2\mathrm{m}\cdot 10^{15/20}=2\mathrm{m}\cdot 10^{0.75}=11.2\mathrm{m}$

A second, identical drill raises the level at 2 m to $100+10{\log }_{10}2=103.0\mathrm{dB}$, and the 85 dB distance grows to $2\cdot 10^{18/20}=15.9\mathrm{m}$: $\sqrt{2}$ times farther, not twice as far. Hard ground adds roughly the same 3 dB at every distance: it halves the power estimate but leaves these distances unchanged.

A common occupational guideline allows 8 hours a day at 85 dB and halves the time for every 3 dB more. A worker 2 m from one drill is 15 dB above that, five halvings, so the limit is $8\mathrm{h}/2^5=15\min$.

Reading the decibel chart

The chart beside the speaker is a ruler in decibels, so equal distances on it are equal factors of intensity: from a 30 dB whisper to a 60 dB conversation is a factor of 1000, and from 0 dB to the threshold of pain at 130 dB a factor of $10^{13}$. Its coloured bar is green up to 70 dB, gold up to 85 dB and red above, where long exposure harms hearing. The pointer starts in the red at 89.0 dB, crosses 85 dB when the meter is about 1.6 m from the speaker, and ends at 70.9 dB, just above the green.

Loudness grows far more gently than intensity. A sound must gain about 10 dB, ten times the intensity, to seem twice as loud. So a 110 dB rock concert, 50 dB above a conversation, carries $10^5$ times the intensity but sounds only about $2^5=32$ times as loud.

Common mistakes

• Decibels do not add. Two identical, independent 80 dB machines give 83 dB, not 160 dB: the intensities add, and doubling the intensity adds 3 dB. Reaching 86 dB takes four machines.
• Doubling the distance does not halve the intensity but quarters it, a drop of 6 dB; only a long line of sources, such as a busy road, loses just 3 dB per doubling.
• 0 dB is not silence but the reference intensity $I_0$, about the faintest sound a young ear can detect; fainter sounds have negative levels, and $10^{-13}W/{\mathrm{m}}^2$ is −10 dB.
• Four times the intensity does not sound four times as loud: the 6 dB gained by halving your distance sounds only about 1.5 times as loud, since loudness doubles roughly every 10 dB.
• Small slips cost whole factors: dividing by $r^2$ instead of $4\pi r^2$ makes an intensity 12.6 times too large, and dropping $I_0$ or the factor 10 turns 70 dB, the level of $10^{-5}W/{\mathrm{m}}^2$, into −50 dB or 7 dB.

Where you meet it

Noise limits at work and near airports, the ratings of ear defenders (good earmuffs cut 20 to 30 dB, a factor of 100 to 1000 in intensity) and the noise maps of cities all rest on these rules. The decibel serves wherever ratios span many powers of ten: amplifier gains, the light lost in optical fibres, the strength of phone signals.

The inverse-square law is not special to sound: light, gravity and the electric field of a point charge also spread over spheres, as Gravitation and Orbits, Coulomb's Law and Electromagnetic Waves show. Coherent sources, which add amplitudes rather than intensities, lead to Interference and Superposition, and sound trapped between walls leads to Standing Waves.

Advanced

What intensity measures

Intensity tells us how much sound energy flows through a surface: the power crossing each square metre held square-on to the flow, in watts per square metre. It is what a microphone or an eardrum responds to, and energy conservation alone fixes how it changes with distance.

Where does the flow come from? In a sound wave the air pressure rises and falls by a small excess pressure $p$ about its usual value, while the air itself moves to and fro with velocity $u$. The pressure pushes on the moving air in front of it, so it does work at the rate $pu$ per square metre: force per area times velocity, the acoustic form of force times velocity. This product swings with the wave, so the intensity $\vec{I}$ is defined as its average over one cycle. It is a vector that points the way the energy flows.

Now surround a source by any closed surface and cut the surface into small patches. Write each patch as a vector $d\vec{A}$ whose length is the patch's area and which points outward. Only the part of the flow that crosses a patch carries energy through it, so the power leaving through one patch is $\vec{I}\cdot d\vec{A}$. If the source runs steadily and nothing absorbs sound, energy cannot pile up inside, so the whole power $P$ of the source must leave through the surface (the circled integral sign means a sum over the entire closed surface):
$\oint \vec{I}\cdot d\vec{A}=P$

Spreading through a cone: the inverse-square law

The easiest way to see what that equation says about distance is to follow one narrow bundle of the flow.

Legend

$\Delta P$	Power carried by the cone: the share of the source's output that crosses each patch.	$\mathrm{W}$
$r_1$	Distance from the source to the first patch.	$\mathrm{m}$
$r_2$	Distance from the source to the second patch, twice $r_1$.	$\mathrm{m}$
$A_1$	Area of the first patch, facing the source.	${\mathrm{m}}^2$
$A_2$	Area of the second patch: four times $A_1$, one ray in each of its four cells.	${\mathrm{m}}^2$
$I_1$	Intensity at the first patch: the power per square metre there, $\Delta P/A_1$.	$W/{\mathrm{m}}^2$
$I_2$	Intensity at the second patch, $\Delta P/A_2$.	$W/{\mathrm{m}}^2$

Figure 1 follows the energy inside one narrow cone with its tip at the source. Its four rays stand for the power $\Delta P$ that the cone carries. They cross a square patch of area $A_1$ at distance $r_1$ and then a second patch of area $A_2$ at $r_2=2r_1$; both patches face the source. The sides of the cone are straight lines from its tip, so every side of the square grows in proportion to the distance: at twice the distance the patch is twice as wide and twice as high, and $A_2=4A_1$. The cross lines on $A_2$ divide it into four cells, each the size of $A_1$.

Now look at the dots where the rays cross. On $A_1$ all four rays pass through one cell's worth of area; on $A_2$ there is one ray in each cell. The same power is spread over four times the area, so the power per square metre has fallen to a quarter, $I_2=I_1/4$. In symbols: nothing is lost between the patches, so $I_1{A}_1=I_2{A}_2=\Delta P$. Along a cone the cross-section grows as $r^2$, so the intensity falls as $1/r^2$, and the cone carries the same power at every distance.

Fill all directions with such cones. For a source that radiates evenly, on a sphere of radius $r$ centred on it, $\vec{I}$ points straight outward and has the same size $I$ everywhere. Each patch then passes $IdA$, and the patches add up to the area of the sphere, $4\pi r^2$. The closed-surface equation becomes $4\pi r^{2}I=P$, which is the inverse-square law:
$I=\frac{P}{4\pi r^2}$
The live check above the first expert graph shows it for the animation's 10 mW loudspeaker: from 1 m to 8 m, $I$ plunges from 796 to 12.4 µW/m², yet $4\pi r^{2}I$ stays at 10.0 mW.

Why the amplitude falls as one over r

The inverse-square law is about energy. What happens to the wave itself, the swing of pressure a microphone records? It must shrink too; the question is how fast.

Legend

$p$	Excess pressure of the sound wave at the instant shown.	$\mathrm{Pa}$
$r$	Distance from the source along the ray.	$\mathrm{m}$
$p_1$	Height of the crest at $r_1$: the pressure amplitude there.	$\mathrm{Pa}$
$p_2$	Height of the crest at $r_2$: half of $p_1$.	$\mathrm{Pa}$
$r_1$	Distance from the source to the first marked crest.	$\mathrm{m}$
$r_2$	Distance to the second marked crest, twice $r_1$.	$\mathrm{m}$
$\lambda$	Wavelength: the distance between neighbouring crests, the same at every $r$.	$\mathrm{m}$
$C/r$	Envelope of the wave, with $C$ a constant set by the source: crests touch $+C/r$, troughs $-C/r$.	$\mathrm{Pa}$

Figure 2 is a snapshot of the excess pressure $p$ along one ray leaving a loudspeaker at $r=0$. Look first at the spacing: neighbouring crests are always one wavelength $\lambda$ apart, so distance does not change the pitch. Then look at the heights. The crest at $r_1$ reaches $p_1$; the crest at $r_2=2r_1$ reaches only $p_2=p_1/2$. Every crest touches the dashed curve $+C/r$ and every trough the curve $-C/r$, where $C$ is a constant set by the strength of the source. Twice the distance, half the amplitude: the pressure amplitude goes as $1/r$.

The wave equation confirms this. Let $c$ be the speed of sound. When a point source sends sound evenly in all directions, $p$ depends only on the distance $r$ and the time $t$. In that case the three-dimensional wave equation can be rewritten, exactly, as a one-dimensional wave equation for the product $rp$:
$\frac{{\partial }^2(rp)}{\partial r^2}=\frac{1}{c^2}\frac{{\partial }^2(rp)}{\partial t^2}$
This is the equation of a wave on a string, whose solutions are shapes that travel at speed $c$ without changing. The outgoing one is $rp=f(t-r/c)$ for any function $f$: whatever the source does at a given moment is repeated at distance $r$ a time $r/c$ later. Dividing by $r$,
$p(r,t)=\frac{f(t-r/c)}{r}$

Read this against Figure 2: the function $f$ draws the wiggles and the divisor $r$ draws their envelope $\pm C/r$. It is also why the animation draws its wavefronts thinner and paler as they grow.

Why does an amplitude falling as $1/r$ mean an intensity falling as $1/r^2$? Think of what receives the sound, an eardrum or the membrane of a microphone. Such a receiver is linear: driven at a given frequency, it moves with a velocity proportional to the force on it. The power it absorbs is force times velocity, so it goes as the square of the amplitude. Amplitude as $1/r$ and power as amplitude squared give intensity as $1/r^2$. Energy conservation and the wave equation agree, and the same two facts hold for the electric field of light.

Pressure and intensity in a plane wave

To turn a pressure reading into an intensity we need the air velocity $u$ that goes with a given $p$. Far from the source a small patch of the spherical wave is practically flat: a plane wave travelling along $x$. Newton's second law for a thin slab of air of density $\rho$ says that its mass per unit volume times its acceleration equals the net push, which comes from the pressure being higher on one face than on the other:
$\rho \frac{\partial u}{\partial t}=-\frac{\partial p}{\partial x}$
For a wave travelling towards $+x$, $p=F(t-x/c)$ for some function $F$, so the right-hand side is $F^{\prime }/c$. Integrating over time then gives
$u=\frac{p}{\rho c}$
The air moves exactly in step with the pressure: where the pressure is high, the air surges forward. The power per square metre $pu$ is then $p^2/(\rho c)$. For a sine wave of amplitude $p_0$ the average of $p^2$ over a cycle is $p_0^2/2$, so
$I=\frac{p_0^2}{2\rho c}$
The product $\rho c$ is the acoustic impedance of the medium: it says how much pressure it takes to make the air move at a given speed. For air at 20 °C, $\rho c\approx 413kg/({\mathrm{m}}^2\mathrm{s})$.

How big are these quantities? At the animation's near listener, 2 m from the source, $I=1.99\times 10^{-4}W/{\mathrm{m}}^2$. Then $p_0=\sqrt{2\cdot 413\cdot 1.99\times 10^{-4}}\approx 0.41\mathrm{Pa}$, and the velocity amplitude of the air, $p_0/(\rho c)$, is about 1 mm/s. Dividing a velocity amplitude by $2\pi$ times the frequency gives the displacement: at 1 kHz the air moves only 0.16 µm back and forth, and at the threshold of hearing about $10^{-11}\mathrm{m}$, a tenth of an atom's diameter.

Two decibel scales

Hearing spans about thirteen powers of ten in intensity, so levels are given on a logarithmic scale. The sound intensity level is $\beta =10{\log }_{10}(I/I_0)$, with $I_0=10^{-12}W/{\mathrm{m}}^2$, roughly the faintest 1 kHz sound a young ear can hear. Each factor of ten in intensity adds 10 dB: the near listener's $1.99\times 10^{-4}W/{\mathrm{m}}^2$ is 83.0 dB.

A meter, however, senses pressure, not intensity. It reports the sound pressure level $L_p=20{\log }_{10}(p_{\mathrm{rms}}/p_{\mathrm{ref}})$, where $p_{\mathrm{rms}}=p_0/\sqrt{2}$ is the root-mean-square pressure and $p_{\mathrm{ref}}=20\mu \mathrm{Pa}$. The factor is 20, not 10, because intensity goes as pressure squared, and $20{\log }_{10}$ of a ratio equals $10{\log }_{10}$ of its square. Since $I=p_{\mathrm{rms}}^2/(\rho c)$, the two scales differ only by a constant, $L_p-\beta =10{\log }_{10}(\rho c{I}_0/p_{\mathrm{ref}}^2)\approx 0.14\mathrm{dB}$ in air, so in practice they are interchangeable: the near listener's 83.0 dB reads 83.1 dB on a meter.

The pressures involved are tiny. A 60 dB conversation has $p_{\mathrm{rms}}=0.02\mathrm{Pa}$, two ten-millionths of an atmosphere, and a painful 120 dB is only 20 Pa. That smallness is what makes the wave equation linear to excellent accuracy, and linearity is what lets the waves of several sources simply add, as a later section uses.

Straight lines on logarithmic axes

Logarithms do more than compress a huge range: they turn power laws into straight lines. Between two distances from a point source $I_2/I_1=(r_1/r_2{)}^2$, and taking $10{\log }_{10}$ of both sides gives
${\beta }_2={\beta }_1-20{\log }_{10}\left(\frac{r_2}{r_1}\right)$
Against $\log r$ the level therefore falls along a straight line: 20 dB for every factor of ten in distance (a decade), and $20{\log }_{10}2=6.02$ dB for every doubling.

The two expert graphs show both views. On the first, $I(r)$ drops from 796 µW/m² at 1 m to 199 at 2 m and then creeps towards 12.4 µW/m² at 8 m, a curve on which equal ratios are hard to see. On the second, plotted against $\log r$, the distances 1, 2, 4 and 8 m are equally spaced and the level runs down a straight line through 89.0, 83.0, 77.0 and 70.9 dB. The meter, whose distance doubles every 2 s, moves along that line at a steady pace.

The slope measures the exponent. Any power law $I\propto r^{-n}$ gives a straight line of slope $-10n$ dB per decade on these axes: −20 for a point source, −10 for a long line source (the worked problem below), 0 for a plane wave or for sound guided along a tube. Anechoic rooms, whose walls absorb almost all sound, are tested this way: wherever the measured level strays from the −20 dB-per-decade line, reflections are adding to it.

Adding sources: phase decides

When two sources sound together, the levels in decibels do not add: two 80 dB machines do not make 160 dB. Because the waves are small, they superpose, and at each instant their pressures add, $p=p_1+p_2$. The intensity follows the mean square of the pressure; with a bar for the average over time,
$\overline{p^2}=\overline{p_1^2}+\overline{p_2^2}+2\overline{p_1{p}_2}$
The first two terms are the two sources' own contributions. The last one, the cross term, depends on how the two waves keep in step, and it decides the result.

Legend

$p_1$	Pressure amplitude of the first wave at the listener: the length of its arrow.	$\mathrm{Pa}$
$p_2$	Pressure amplitude of the second wave, here equal to $p_1$; its direction shows its phase relative to the first.	$\mathrm{Pa}$
$p$	Amplitude of the combined wave: the arrow sum $p_1+p_2$.	$\mathrm{Pa}$
$\phi$	Phase difference: the angle through which $p_2$ is turned from the direction of $p_1$; $0$ in step, $\pi$ half a cycle apart.	$\mathrm{rad}$

Figure 3 works out the sum at one listener for two sine waves of the same frequency. From here on $p_1$, $p_2$ and $p$ denote amplitudes. Each wave is drawn as an arrow whose length is its pressure amplitude and whose direction is its phase, the point it has reached in its cycle. The arrow $p_1$ lies along the horizontal axis. The arrow $p_2$ starts at its tip and is turned through the phase difference $\phi$, the angle by which the second wave runs ahead of the first. The combined wave is the arrow $p$ from the start of $p_1$ to the tip of $p_2$, which closes the triangle. The angle of the triangle opposite $p$ is $\pi -\phi$, so the law of cosines gives
$p^2=p_1^2+p_2^2+2p_1{p}_2\cos \phi$
and the last term is the cross term of the mean square, now written through the phase. In the figure the two sources are equal, $p_1=p_2$, so as $\phi$ changes the tip of $p$ runs round the dashed circle centred on the tip of $p_1$.

Independent sources, such as two machines or two voices, have no fixed phase relation. Their phase difference drifts at random, the tip of $p$ wanders all round the circle, and $\cos \phi$ averages to zero. The cross term disappears and the intensities simply add, so $N$ equal sources give $N$ times the intensity of one:
${\beta }_N={\beta }_1+10{\log }_{10}N\text{(}N\text{ equal incoherent sources)}$

Two sources add 3.0 dB, ten add 10 dB. Unequal sources are dominated by the louder: 80 dB and 70 dB are $10^8$ and $10^7$ times $I_0$, so together they give $10{\log }_{10}(10^8+10^7)=80.4\mathrm{dB}$. Run backwards, the rule removes background noise from a measurement: subtract intensities, never decibels.

Coherent sources, such as two loudspeakers fed the same tone, keep a fixed phase difference, so the cross term survives. Its value depends on where the listener stands, because $\phi$ depends on the difference between the two path lengths. Where the waves arrive in step, $\phi =0$ and the tip of $p$ sits at the right-hand point of the circle: the arrows line up, $p=2p_1$, and the intensity is four times that of one source, +6.0 dB. Where they arrive half a cycle apart, $\phi =\pi$ and the tip of $p$ sits at the left-hand point, back at the start of $p_1$: the waves cancel. Averaged over the whole pattern the intensity is still twice that of one source. Interference moves energy around; it does not create it.

Worked problem: the noise of a straight road

Everyday noise often comes from many sources in a row. Model a long straight road as a row of independent sources, each like the animation's loudspeaker, $P=10\mathrm{mW}$, one every $d=20\mathrm{m}$. What does a meter read at $r=40\mathrm{m}$ from the road, and how fast does the level fall as you walk away?

Legend

$P$	Acoustic power of each source.	$\mathrm{W}$
$d$	Spacing between neighbouring sources.	$\mathrm{m}$
$M$	The sound-level meter.
$r$	Distance from the meter to the road, which is also its distance to the nearest source.	$\mathrm{m}$
$x$	Position of one source along the road, measured from the foot of the perpendicular from $M$.	$\mathrm{m}$
$s$	Distance from that source to the meter, $\sqrt{r^2+x^2}$.	$\mathrm{m}$
$dx$	Length of a thin slice of road; it holds $dx/d$ sources.	$\mathrm{m}$

Figure 4 shows the road from above, drawn with $r=2d$ as in the numbers. The meter $M$ stands a distance $r$ from the road, level with the nearest source. Another source sits a distance $x$ along the road, measured from the foot of the perpendicular from $M$. By Pythagoras it is $s=\sqrt{r^2+x^2}$ from the meter, so on its own it gives $P/(4\pi s^2)$ there.

The sources are independent, so by the previous section the meter adds their intensities. Rather than add them one by one, spread them evenly along the road: a slice of length $dx$ holds $dx/d$ sources. Summing the slices along the whole road turns the sum into an integral. Its middle step uses
${\int }_{-\infty }^{\infty }\frac{dx}{r^2+x^2}=\frac{\pi }{r}$
and the result is
$I={\int }_{-\infty }^{\infty }\frac{P}{4\pi (r^2+x^2)}\frac{dx}{d}=\frac{P}{4\pi d}\cdot \frac{\pi }{r}=\frac{P}{4dr}$

At 40 m, $I=0.010/(4\cdot 20\cdot 40)=3.1\times 10^{-6}W/{\mathrm{m}}^2$, so $\beta =64.9\mathrm{dB}$, where the nearest source alone would give 57.0 dB. At 80 m the level is 61.9 dB and at 160 m 58.9 dB: 3.0 dB per doubling, half the fall of a point source, because now $I\propto 1/r$. The reason is visible in the integral: half of the sound comes from the stretch of road within $r$ of the nearest point on either side. Doubling $r$ quarters each source's share but doubles the number of sources in that stretch, so the intensity only halves.

Two checks on the model. The integral matches the exact sum over separate sources to 0.4 % once $r\ge d$; closer in, the nearest source dominates and the road no longer looks continuous. And a road of finite length $L$ looks like a point source again from well beyond $L$, where the level returns to 6 dB per doubling.

A subtler point: is 64.9 dB the energy actually flowing past the meter? Draw a cylinder of radius $r$ around the road. Waves from distant sources cross it at a slant, and only the part of their flow that points straight out leaves the cylinder. Counting only that part gives $P/(2\pi dr)$ per square metre; times the cylinder's area per metre of road, $2\pi r$, it returns the $P/d$ that each metre of road radiates. This is $2/\pi$ of the meter's value, about 2 dB lower, because the meter adds the intensities of all the waves whatever their direction. So $p_0^2/(2\rho c)$ equals the energy flow only for a wave travelling one way.

Limits of validity

The inverse-square law rests on four assumptions: a small source, open space, still air that absorbs nothing, and small amplitudes. Each one fails somewhere.

Close to the source: within about a wavelength of a real source (3.4 m at 100 Hz), or within a few times its size, pressure and air velocity are not yet in step as they are in a plane wave, and the pressure of an extended source does not yet fall as $1/r$. Real sources also beam: a loudspeaker sends wavelengths shorter than its cone mostly forward. A directivity factor $Q$ in $I=QP/(4\pi r^2)$ describes this; a source on hard ground radiates into half the space only, so $Q=2$ (+3 dB). Such factors do not depend on distance, so the 6 dB per doubling survives.

Indoors, sound reflects from the walls. Beyond the critical distance, where the reflected sound becomes as strong as the direct sound (often about a metre in a living room), reflections dominate and moving away barely lowers the level.

Outdoors, air absorbs a fixed number of decibels per kilometre on top of the spreading, and the loss rises steeply with frequency: at 20 °C and 70 % humidity, about 0.3 dB per kilometre at 125 Hz, 5 dB at 1 kHz and nearly 80 dB at 8 kHz, against the 60 dB that spreading costs from 1 m to 1 km. High pitches die away first, hence the low rumble of distant thunder.

Linearity fails at high amplitude. At 150 dB, near a jet engine, $p_{\mathrm{rms}}\approx 630\mathrm{Pa}$ is only 0.6 % of an atmosphere, yet crests travel slightly faster than troughs, and a 1 kHz plane wave would steepen into a shock within about 7 m. No undistorted sine wave exceeds about 191 dB at sea level, where its amplitude would equal the atmospheric pressure and its troughs would reach vacuum.

Finally, $\beta$ measures physics, not loudness: the ear is far less sensitive at low frequencies, so meters usually apply an A-weighting, a filter that turns down the frequencies the ear hears poorly, and quote the result in dB(A).

Where this leads

Flux through closed surfaces also makes gravity and the field of a point charge fall as $1/r^2$, in Gravitation and Orbits and Coulomb's Law; it is the idea behind Gauss's law. Light follows the same arithmetic: in Electromagnetic Waves its intensity is $E_0^2/(2Z_0)$, with the impedance of free space $Z_0\approx 377\Omega$ taking the role of $\rho c$. Coherent sources and their cross term lead to Interference and Superposition, sound confined between walls to Standing Waves, and the eardrum as a driven receiver to Damping and Resonance. Absorption at a fixed number of decibels per kilometre is the same exponential decay as in Radioactivity and Half-Life.