Let's say for example:

Two people sing the same note (frequency) and volume (amplitude) together.

Why is it that the two persons sound louder than they would individually?

I would imagine that since the traverse waves generated by the two persons would randomly be louder or softer, based on the phase shift between the two sound waves. But that hypothesis is obviously untrue since in the real world a choir is clearly louder than a soloist.


2 Answers 2


This is a neat question. Did you know that adding two Sine waves of the same frequency but different phase together always produces another Sine wave? Of course you can imagine two perfectly out-of-phase Sine waves that "cancel" by adding to a line but in that case you can just imagine the result as a Sine wave with 0 amplitude.

Using gnuplot with the following commands I plotted the sum of 4 Sine waves:

a = rand(0)
b = rand(0)
c = rand(0)
d = rand(0)
plot sin(x + a * (2 * pi)) + sin(x + b * (2 * pi)) + sin(x + c * (2 * pi)) + sin(x + d * (2 * pi))

And on the first go I got this:

add 4 sine waves together

Notice the shape and frequency is the same but the amplitude has increased. If you add them randomly together again you'd get something like:

add 4 sine waves together again

Notice that the amplitude is still greater than 1 but the phase has shifted.

What's happening is that each Sine wave you add increases the maximum possible amplitude so the more you add the better the chance of getting a wave with amplitude greater than 1.

Chris White posted a fantastic answer about the brightness of the sun where he goes into a lot more details about the math of adding waves and the statistics of it.

  • $\begingroup$ I would bet that the reason they seem louder (extremely few singers can produce a sine wave and even fewer could probably sing one together) is because there are now two power sources for the audio instead of one (2x more volume). The end result is better audio spreading and on average more power delivered to audience. $\endgroup$
    – user6972
    Apr 7, 2014 at 9:35

Chris White's answer using an analogy to incoherent light pretty much answers the question; it's fundamentally a question of the statistics of how wave sources add.

Here's a slightly different but equivalent rephrasing of Chris White's answer using matrices:

Given $N$ wave sources, incoherent waves add "diagonally" ($I\propto N)$, ie, additively. Coherent waves add "off-diagonally" ($I\propto N^2$).

From reasonably far away, a choir of people can be approximated as a set of $N$ spherical wave sources located at points $\mathbf{q}_k$. At a point $\mathbf{p}$ (observer), the amplitude/pressure from the $k$th source becomes $A_k(\mathbf{p},t)=\frac{K_k}{|\mathbf{p}-\mathbf{q}_k|}V_k(t-|\mathbf{p}-\mathbf{q}_k|/c)$ where $K_k$ is some amplitude constant and where $V_k(t)$ is the time-dependent part. The total pressure at time $t$ becomes $\sum_{k=1}^NA_k(\mathbf{p},t)$.

Since acoustic intensity is the square of the modulus of the pressure, the expectation value of the intensity at $\langle I(\mathbf{p})\rangle$ becomes $$\langle I(\mathbf{p})\rangle=\left<\overline{\left(\sum_{j=1}^N\frac{K_jV_j\left(t-\frac{|\mathbf{p}-\mathbf{q}_j|}{c}\right)}{|\mathbf{p}-\mathbf{q}_j|}\right)}\left(\sum_{j=1}^N\frac{K_jV_j\left(t-\frac{|\mathbf{p}-\mathbf{q}_j|}{c}\right)}{|\mathbf{p}-\mathbf{q}_j|}\right)\right>$$ $$=\sum_{j=1}^N\sum_{k=1}^N\frac{\overline{K_j}K_k}{|\mathbf{p}-\mathbf{q}_j||\mathbf{p}-\mathbf{q}_k|}\left<\overline{V_j\left(t-\frac{|\mathbf{p}-\mathbf{q}_j|}{c}\right)}V_k\left(t-\frac{|\mathbf{p}-\mathbf{q}_k|}{c}\right)\right>$$ $$=\sum_{j=1}^N\sum_{k=1}^N\frac{\overline{K_j}K_k}{|\mathbf{p}-\mathbf{q}_j||\mathbf{p}-\mathbf{q}_k|}\left<\overline{V_j\left(t\right)}V_k\left(t+\frac{|\mathbf{p}-\mathbf{q}_j|-|\mathbf{p}-\mathbf{q}_k|}{c}\right)\right>$$ $$=\mathbf{a}^\dagger\boldsymbol{\Gamma}\mathbf{a}$$ where $$\mathbf{a}_j=\frac{K_j}{|\mathbf{p}-\mathbf{q}_j|}$$ is a vector of amplitude coefficients and $$\boldsymbol{\Gamma}_{jk}=\left<\overline{V_j\left(t\right)}V_k\left(t+\tau_{jk}\right)\right>$$ is the correlation matrix element between the fields at $V_j$ and $V_k$ with time delay $\tau_{jk}=\frac{|\mathbf{p}-\mathbf{q}_j|-|\mathbf{p}-\mathbf{q}_k|}{c}$.

Without loss of generality we can assume the $V_k$ are normalized, ie, $\boldsymbol{\Gamma}_{kk}=1$, by absorbing any excess amplitude into the intensity factor $K_k$. Then the off-diagonal elements $\boldsymbol{\Gamma}_{jk}\leq1$ by elementary geometry, and $\boldsymbol{\Gamma}$ is called the normalized coherence matrix.

Note $\mathbf{a}^\dagger\boldsymbol{\Gamma}\mathbf{a}$ is a matrix quadratic form, and in the absence of correlation between the sources $A_k$ (in the time-averaged sense) the off-diagonal elements vanish and $\boldsymbol{\Gamma}=I$, the identity matrix, giving $$\langle I(\mathbf{p})\rangle=\mathbf{a}^\dagger\mathbf{a}=\sum_{k=1}^N\left|\frac{K_k}{|\mathbf{p}-\mathbf{q}_k|}\right|^2,$$ which is just the sum of the individual intensities of the voices. This is what is meant by "incoherent waves add diagonally"; their intensities just add, because only the on-diagonal elements of the normalized coherence matrix come into play, and $I\propto N$.

Meanwhile, if there are long-term correlations between the sources $A_k$, the off-diagonal elements will not be zero, and additional intensity will come from the $N^2-N$ off-diagonal entries of $\boldsymbol{\Gamma}$. In this case, the intensities add "off-diagonally", and in the best-case scenario they can add as $I\propto N^2$.


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