The dormitive virtue of root-power quantities
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One of the concepts that comes up in the Introduction to Phonetics course that I'm teaching this semester — first meeting yesterday — is SNR ("Signal to Noise Ratio"). This is the ratio between the power of the "signal" (defined as the stuff you care about, essentially) and the power of the "noise" (the stuff that you aren't interested in).
And at this point, there are a few things that students need to learn. Since SNR is a ratio of power to power, it's a dimensionless quantity. Similar ratios of physical quantities come up elsewhere in acoustics, like "sound pressure level" (SPL), defined as the ratio of sound pressure to the some reference level, usually taken to be the nominal threshold of human hearing. Because additive scales are more intuitive (and because psychophysical scaling is roughly logarithmic), we generally take the log of such ratios. And because powers of ten are inconveniently far apart, we generally multiply log10(whatever ratio) by 10 to get "decibels".
Now comes the part that I'm interested in this morning: the power of a sound wave is proportional to the square of its amplitude. And I'm looking for a simple and correct way to justify this statement, and to explain why we generally quantify "levels" of physical signals as ratios of powers rather than as ratios of amplitudes.
The Wikipedia article on Amplitude tells us,
The amplitude of a periodic variable is a measure of its change over a single period (such as time or spatial period). There are various definitions of amplitude (see below), which are all functions of the magnitude of the difference between the variable's extreme values. […]
Root mean square (RMS) amplitude is used especially in electrical engineering: the RMS is defined as the square root of the mean over time of the square of the vertical distance of the graph from the rest state.
For complex waveforms, especially non-repeating signals like noise, the RMS amplitude is usually used because it is both unambiguous and has physical significance. For example, the average power transmitted by an acoustic or electromagnetic wave or by an electrical signal is proportional to the square of the RMS amplitude (and not, in general, to the square of the peak amplitude)
At this point, it's natural to ask why "power" is proportional to "amplitude" squared. And why do we take the ratio of powers rather then the ratio of amplitudes?
Of course, after we've taken the log of the ratio, this distinction is just a scaling factor, since log(x^2) = 2*log(x). But still, it's natural to ask why.
If you look this issue up on the internet, you mostly find answers like those that I got when I first asked this question at the age of 16 or so, back in neolithic times. For instance, a student asks in the Physics Forums asks
My book states: "Power is proportional to amplitude squared"
How can this be derived?
He gets a bunch of answers that mostly just offer more elaborated versions of the statement he is trying to understand, e.g.
… the power of the electricity in a resistor is the current-squared times R (watts), or the voltage-squared divided by R (watts). We also talk about the product of voltage and current as power. So current or voltage are amplitudes, and the power is watts.
This is also true in electromagnetic waves. We talk about the transverse E vector or the H vector (amplitudes) in an electromagnetic wave. [Note that E is volts per meter and H is amp-turns per meter]. E and H are amplitudes, while the Poynting vector, which is E x H , is power (per square meter).
Or
The power of a signal or waveform is the signal or waveform multiplied by its conjugate.
Power = U x U*, where U is the function describing the signal.
For example; U = Ae(t/b) Where U is the signal function, A is the amplitude, t is time (or what ever you want it to be), and b is just a coefficient.
Since P = U x U*
P = Ae(t/b) x Ae(-t/b) = A2
None of this really explains why the power of a signal is proportional to the square of its amplitude — it just persuades the student to stop asking and respect the authority of those responding.
Poking around in Wikipedia is not much more helpful. Thus the link on power tells us that
In physics, power (symbol: P) is defined as the amount of energy consumed per unit time. […] The integral of power over time defines the work done.
And later in the same article we learn that
This tells us again that power is related to amplitude squared, adding references to energy and work. But why isn't power related to amplitude cubed? or the square root of amplitude? or the square root of amplitude cubed? or just plain amplitude?
Additional levels of authoritative mystification can be derived from ISO-80000-1, where Wikipedia tells us that:
According to Annex A, "[t]he logarithm of the ratio of a quantity, Q, and a reference value of that quantity, Q0, is called a level". For example LP = ln(P/P0) is the level of a power quantity P.
Annex C introduces the concepts of power quantities and root-power quantities.
So to justify the statement that power is proportional to amplitude squared, we could explain that amplitude (like pressure) is a "root-power quantity". As an explanation, this would rival the insight of the Sçavantissimi doctores in Moliere's Le Malade Imaginaire, who answer the question "quare Opium facit dormire ?" ("Why does opium cause sleep?") by explaining that "Quia est in eo Virtus dormitiva" ("Because it contains a dormitive property").
There are no doubt many simple explanations for the basic distinction between "power quantities" and "root-power quantities" (previously "field quantities"), and for the fact that the amplitude of a sound waveform is a "root-power quantity", and for the practice of defining levels in terms of the ratio of "power quantities". One story might depend on the fact that this all makes the conservation of energy work out. Another might point to sound as pressure variation considered over a surface rather that at a point.
But I'm having trouble coming up with a simple, memorable, and correct story to put into my lecture notes.
In the end, of course, people just have to remember the factor of 20 (rather than 10) in the expression for SNR in dB:
SNR = 20*log10(RMS(signal)/RMS(noise))
Still, it would be nice to explain the factor of 2 in terms of something other than its virtus dormitiva. So I'm offering a lifetime subscription to Language Log as a prize to the commenter telling the best story about this.
One of the interesting-but-not-helpful things that I learned, in the course of poking around on the internet for the right way to explain why amplitude is a root-power quantity, was the story of where the decibel came from. I previously knew the vague outlines, but not the details.
From the Wikipedia article on the decibel:
The decibel originates from methods used to quantify reductions in audio levels in telephone circuits. These losses were originally measured in units of Miles of Standard Cable (MSC), where 1 MSC corresponded to the loss of power over a 1 mile (approximately 1.6 km) length of standard telephone cable at a frequency of 5000 radians per second (795.8 Hz), and roughly matched the smallest attenuation detectable to the average listener. Standard telephone cable was defined as "a cable having uniformly distributed resistance of 88 ohms per loop mile and uniformly distributed shunt capacitance of .054 microfarad per mile" (approximately 19 gauge).
The transmission unit (TU) was devised by engineers of the Bell Telephone Laboratories in the 1920s to replace the MSC. 1 TU was defined as ten times the base-10 logarithm of the ratio of measured power to a reference power level. The definitions were conveniently chosen such that 1 TU approximately equaled 1 MSC (specifically, 1.056 TU = 1 MSC). In 1928, the Bell system renamed the TU the decibel. Along with the decibel, the Bell System defined the bel, the base-10 logarithm of the power ratio, in honor of their founder and telecommunications pioneer Alexander Graham Bell. The bel is seldom used, as the decibel was the proposed working unit.
"Miles of Standard Cable" has a certain steampunk charm that almost makes me sorry that we lost it. But from the same article, I learned that the decibel itself has been challenged as a confusing residue of slide-rule thinking:
According to several articles published in Electrical Engineering and the Journal of the Acoustical Society of America, the decibel suffers from the following disadvantages:
* The decibel creates confusion.
* The logarithmic form obscures reasoning.
* Decibels are more related to the era of slide rules than that of modern digital processing.
* They are cumbersome and difficult to interpret.
Hickling concludes Decibels are a useless affectation, which is impeding the development of noise control as an engineering discipline.
And of course, the issue of choosing an appropriate power of ten for everyday applications of standard units brings up the Jargon File entries for microlenat and microfortnight.
Michael Bench-Capon said,
August 29, 2013 @ 8:17 am
Mark,
Sorry if I'm either flaunting my ignorance or giving more dormitive virtue, but I'll have a guess.
The distance the particles are going backwards and forwards is proportional to the amplitude, so the speed of this movement is as well, so the kinetic energy is proportional to the square of the amplitude, and the power is proportional to the kinetic energy.
Any use?
Josh said,
August 29, 2013 @ 8:44 am
I had this same question many years ago and I imagined it like this (I have to admit, I have no idea if this is right):
You have a rope in your hand, and the other end is affixed to the wall some distance away. You want to make a wave in the rope (like cracking a whip). To do that, you raise your hand up and down quickly. That makes a sin wave that travels the length of the rope.
Now, to get a hump A meters high, you need to put in power equal to P. And we're saying that for power to be the square of amplitude, to get a hump 2A meters high, you would need to put in power equal to 4P.
This could probably be measured experimentally, but I believe the reason for this is that you have to go both twice as far with your hands, but you also need to go twice as fast to cover the extra distance in the same time, otherwise you end up with a lower frequency.
And if you work through all the physics equations (P=W/T; W=FD; F=MA, etc.), you get that power is proportional to both distance and acceleration, and it's this product of the extra distance and the extra acceleration that makes it A^2.
And I think the same thing should apply to any kind of wave.
Mark P said,
August 29, 2013 @ 8:50 am
Maybe the first step would be to tell everyone to forget about their intuitive understanding of the word "amplitude" (it's how high the wave is) and think about it as a physical quantity like power or energy. It won't make it any easier to understand the why of it, but then it's not any easier to understand why the energy of a given mass should be related to the square of the speed of light.
[(myl) I've been teaching people in that mode for several decades. This year, I thought I'd look for a more contentful explanation. I figured it would be easy — just a matter of refreshing some first-year physics — but it seems to be surprisingly hard to find a first-year physics explanation that doesn't just restate the relationship, perhaps in a different domain.]
Graeme Williams said,
August 29, 2013 @ 8:50 am
If you could freeze time, you could build a wave at the beach by moving water from the trough to the peak of the wave you were building. If you built one wave and wanted to build one twice as high, you'd have to move twice as much water, but you'd also have to move it twice as far, on average.
The same is true of a sound wave. You could move atoms of "air" from a low pressure area to a high pressure area, but each atom you move increases the pressure in the high pressure area and so the amount of work you have to to do for the next atom. To double the amplitude, you have to move twice as many atoms, *and* work twice as hard, on average, to do it.
"On average" refers to the root-mean-square calculation you need to do to get the numbers to work out.
[(myl) This is the best story that I've heard so far. The only thing that it leaves out is why we want to define "level" in terms of the ratio of the squared quantities rather than the unsquared ones.]
Tim Silverman said,
August 29, 2013 @ 8:51 am
Perhaps it would be easiest to start with the observation that the energy stored in a spring is proportional to the displacement squared. This can be seen from the fact that the increments of energy added (or "work done") as you compress the spring are the product of the incremental displacement and the resisting force, but the force is proportional to the displacement. So the displacement enters in twice—once directly and once via its effect on the force.
In the case of compression of a gas, although the pressure is not proportional to the density (in the case of adiabatic compression), we can make a linear approximation in the case of small deviations, so that the "restoring force" (change of pressure) is nearly proportional to the "displacement" (change in volume) and then the same argument applies to the work done to compress the volume as above.
In the case of electric fields, it's not quite so clear, but we have a plausibility argument along the same lines: we can think of building up an electric field in a region by moving electric charges into that region, and then the resistance to bringing further increments of charge in is proportional to the size of the electric field already built up. Hence, e.g. the energy stored in a capacitor is proportional to the square of the charge on it, and generally the energy density of an electric field is proportional to the square of the magnitude of the field (ignoring various complications which don't affect the general idea).
Then, in a wave, we have to supply energy, proportional to amplitude squared, to build the amplitude up at the source, which is then passed along through space as the wave relaxes back to the undisplaced state, and then we need to supply the energy again to build the amplitude up, and so on.
That's a bit handwavey but hopefully not completely useless.
Brett said,
August 29, 2013 @ 8:52 am
Yesterday, I introduced energy in my senior-level classical mechanics class. I told them (and I will tell them again next semester in E&M) that any discussion related to energy ultimately has to come back to a calculation of the amount of work being done. Because work is expressed as an integral of a product (force times displacement), any energy-related quantity such as power is also going to be a second-order function—a product like Fv.
In a problem with, say, a sound wave, both the force exerted on a surface and the velocity at which the surface is moving end up being proportional to the wave amplitude; hence the energetics of the wave are determined by the amplitude squared. This doesn't mean much without all the intermediate calculations, but it's the reason that the power is a second order quantity.
===Dan said,
August 29, 2013 @ 8:56 am
"…and don't get me started on the Richter scale!"
John Spevacek said,
August 29, 2013 @ 9:01 am
The amplitude can fluctuate to both positive and negative values. If you simply add these, you get a cancellation that isn't really there. (The sum of a sine wave over a complete period is zero, even though the power certainly is not). By squaring all the values, the negative values are now positive and you have a positive power. If you were to go with the third power (as you asked about), then you can have negative values again.
Other options can be used – take the absolute value of the amplitude at all times, but that is mathematically difficult to use in equations. Additionally, the RMS shows up in multiple situations, particularly in statistics, and so there is something fundamental about it, much like the way that pi keeps showing up in applications far removed from circular geometry.
david said,
August 29, 2013 @ 9:02 am
Perhaps it would be good to avoid using the word power mathematically and reserve it for the physical property. Or avoid using it all together, most non-physics majors have only the vaguest ideas about power. Your working with definitions 10-12 in this dictionary http://www.thefreedictionary.com/power
Sound is a variation of air pressure, the pressure increases above and decreases below atmospheric pressure, which could be measured with a barometer. The average of this variation over a suitably long interval of time is zero, there is an equal amount of increase and decrease in pressure. When discussing the amount of sound it makes more sense to consider the square of the variation which is always positive. (A negative number times itself is a positive number.) The positive square root of the average of the square of the variation is called the root mean square (RMS) variation. People like RMS values because they have the same units as the variation, Pascals in this case.
You could quit there and not mention power at all. The 20 (or the 2) is in the equation because we used the RMS instead of the amount of sound. To get to power you'll need to distinguish sound pressure and sound intensity and define acoustic impedance. This is laid out densely here http://www.sengpielaudio.com/calculator-soundlevel.htm
David L said,
August 29, 2013 @ 9:04 am
I'm not sure I understand your question, but if it's about the relationship between amplitude and power, a good place to start is the pendulum, or any similar example of simple harmonic motion. Elementary mechanics explains why the energy goes as the amplitude of the motion. For other linear wave motions, such as sound or electromagnetism, the same general idea applies.
But maybe that's not what you're asking…
[(myl) That should be "…as the square of the amplitude of the motion", if it's going to be an explanation of the relationship between amplitude and power/intensity, right?]
David L said,
August 29, 2013 @ 9:05 am
(others made the same point while I was pondering…)
Eugene van der Pijll said,
August 29, 2013 @ 9:05 am
Pressure waves exist when the pressure at a point oscillates around a fixed equilibrium pressure. The potential energy of the wave at that point depends on the difference between the actual pressure and the equilibrium pressure, and is given by a potential energy curve that just happens to be quadratic in the pressure difference: U = c (p – p_0)^2.
The total energy of the system is a mixture of kinetic energy and potential energy; at a point in the wave where the pressure is maximal, the total energy consists of potential energy, and so the total energy of the wave (at that point) is given by E = c (p_max – p0)^2. Here, p_max – p_0 is of course the amplitude of the wave.
So the quadratic nature of power (which is a derived quantity of energy) comes from the quadratic energy potential. The reason that most physical systems are described by a quadratic function is that most models (including the one for air waves) describe small disturbances of a system around an equilibrium.
The actual energy potential (as a function of e.g. pressure) is likely to be a complicated but smooth function. As such, it can be written as a Taylor series, U = c0 + c1 * (p-p0) + c2 * (p-p0)^2 + c3 * (p-p0)^3 + … .
Now the constant term in this expansion is not important; it does not influence the dynamics of the system, and can in fact be arbitrarily chosen. As the system we want to describe is in equilibrium at p0, the linear term has to be zero; a system in equilibrium is at a (local) minimum in the energy potential curve. There are no restrictions on c2 (except that it cannot be negative; if it was, the equilibrium would be unstable), and in most systems, it's non-zero. So for most systems, the potential energy is approximately quadratic for small distances to the equilibrium value.
If power was equal to amplitute^3 for a system, the energy curve would be U(p) = |p-p0|^3, which is not smooth at p0 (the third derivative is non-continuous). Amplitude^3/2 is even more problematic. Physical models with power = amplitude^4 would in theory be possible, but are rare (or non-existant, perhaps).
Faldone said,
August 29, 2013 @ 9:11 am
I'm not sure what the problem is in understanding why power is expressed in "squared" terms. The amplitude measured is either the amplitude of the current or the amplitude of the voltage. Power is defined in terms of current squared, voltage squared, or current times voltage. In any of these power comes out as a square of amplitude.
[(myl) I'm looking for an explanation, not a definition. Definitions we have plenty of.]
Theo Vosse said,
August 29, 2013 @ 9:11 am
In electrical circuits, it's simple. The power is V * I, as it represents the amount of work. I is proportional to the number of electrons that move per time, and V is proportional to the potential energy for each one. Hence, V*I is proportional to the amount of work that has to be done. Consequently, when I is constant, the power can be derived as V * V / R, and is proportional to V^2.
Probably, the amount of work needed to make a spring move proportional to a periodic signal of amplitude a is also proportional to a^2, since the power is the integral over time of the force needed to move the spring and that force is linear to the amplitude. If a loudspeaker cone behaves like a spring, then the relation would hold there too, but I'm no physicist, so I'm not sure about that.
KevinM said,
August 29, 2013 @ 9:13 am
This plagued in high school: For gravity, e.g., how does nature KNOW about the inverse square law? Like most, I accepted it and moved on. There is a wrong-end-of-the-telescope quality to the question, I suppose; our mathematics is designed to fit our observations, not vice versa. But still, I have a why-do-pine-cones-observe-the-Fibonacci-sequence freakout when I try to delve too deep.
As to a field (I know, I know) like electricity or gravity: I found it helpful to consider that it was not pulling like a rope; it was spreading like spilt milk. That is, the earth and the sun are not at either end of a stick; it's more helpful to think of the orbit as a dinner plate with the earth on the edge. Viewed that way, the force is spreading over an area, not just diminishing with distance. Hence the squared quality of the diminution. This isn't science, I know, but it is an intuitive point of entry.
Theo Vosse said,
August 29, 2013 @ 9:13 am
PS I wouldn't know where the assumption that I is constant comes from, nor in what applications it is correct, but it is a common assumption in electronics.
David L said,
August 29, 2013 @ 9:14 am
Oops, and now I see I omitted "square of" the amplitude when referring to the energy…
To add to what E van der P just said, the fundamental point is that when the restoring force is linear in the displacement, the energy goes as the square.
[(myl) Right. But again, this is a restatement in slightly different terms of the concept for which we want an explanation.]
Ollyver said,
August 29, 2013 @ 9:22 am
To add to Brett's answer:
Let's describe power as the amount of energy added in a certain period of time.
Imagine you have a box that the wave is going through. It contains a certain quantity of energy E, that is proportional to the amplitude of the wave.
Now every time the box contains an entirely new section of wave, it gets another E.
How fast it gets new sections is also proportional to the amplitude.
So power is proportional to amplitude twice.
[(myl) Huh? I think I must be misunderstanding you, since the speed of sound is independent of amplitude (or power, for that matter), so "how fast it gets new sections", which depends on the velocity of propagation, is also basically constant, and thus independent of all of the quantities under discussion.]
KeithB said,
August 29, 2013 @ 9:45 am
I was going to do the electrical analog bit, but does that apply to acoustics? Could it be that the current is analogous to the density of a medium? And it is more convenient to ignore that and just focus on the amplitude?
After all, a pressure wave of a certain amplitude in water will carry much more power than one through air, and I can believe that it increases as the square of the amplitude.
sarang said,
August 29, 2013 @ 9:56 am
Isn't a "root-power quantity" just something that is not a scalar? An oscillator has a phase, a current or a displacement has a direction, etc. The simplest well-behaved (in the sense of the relevant inner-product space structure) scalar you can construct from an object like this is the square of its modulus. (In other words, (a.a) is a natural object, whereas Sqrt[(a.a)] is not. I guess the intensities of independent sources also add, as there's no specific phase relation between them.) You can express all this in terms of cons. of energy, of course, but that reduces to asking why the energy is a squared quantity…
kureshii said,
August 29, 2013 @ 9:59 am
The story I give my students in high school physics goes as follows.
Sound waves are a propagation of energy through a medium, typically air. This energy mainly manifests itself in the form of particle motion (kinetic energy). Intuitively we can try to measure the energy of a wave as the collective kinetic energy of its particles.
Sound waves are a form of simple harmonic motion, with displacement of particles from their equilibrium position in the direction of propagation. This displacement varies with time sinusoidally. Since velocity is the time-derivative of displacement, this means that the velocity varies sinusoidally as well. And the maximum velocity is proportional to the maximum displacement, related by omega term, the "angular velocity" of the wave.
From earlier mechanics topics we know that kinetic energy is proportional to the square of velocity. However, we run into a minor hiccup here. The sinusoidal variations of displacement and velocity are actually out of phase by a quarter of a full wave; when the displacement is maximum, the velocity is zero, and when the displacement is zero, the velocity is maximum. (This does not contradict the earlier statement, which states that _maximum_ velocity is proportional to the _maximum_ displacement)
Thus, we cannot simply conclude that _each_particle's_ velocity is proportional to its displacement and hence its kinetic energy is proportional to displacement. But what we can do is observe that the average kinetic energy per particle for the entire wavetrain of particles would be equal to the maximum kinetic energy of each particle (draw the graph of kinetic energy against time and see). This lets us link amplitude (maximum displacement) and average kinetic energy of a wave: amplitude is proportional to _maximum_ velocity, from which we can calculate the _maximum_ kinetic energy of a particle, and halving that gives us the average kinetic energy.
All this takes place in a single snapshot of the wave. Get the wave moving again, and what we have is a propagation of energy. The rate of this energy propagation gives us the power. Since the speed of the wave is not affected by the amplitude, we can say that power is proportional to the square of the amplitude.
Hope this helps.
Jerry Friedman said,
August 29, 2013 @ 11:42 am
If they know some physics, I think the best method, as people have suggested, is to remind them that W = Fd or P = Fv, so in a medium with a linear response to force (an elastic medium, with F = -kd), work is proportional to the displacement squared and thus to the force squared. There's an analogous argument in electronics for voltage and current.
If you want a non-rigorous real-life argument for P = Fv, changing gears in a car or bike shows how, for a given amount of power, you have a trade-off between F and v (or torque and angular velocity, which are proportional to F and v), which is at least consistent with P = Fv. Levers, wrenches, and the like also make good examples.
If you want to show why W = Fd is important, I think you have to show that it's one way to transfer energy, which is a conserved quantity, so it constrains all the interactions. This gets you into the theorem about work and kinetic energy, and such things (ultimately, I suppose, to Noether's Theorem, which I don't want anyone to think I ever understood mathematically). I think The Feynman Lectures might help with this kind of approach. It would definitely be contentful, but it might well be more than you and your students want.
If you want to go with Graeme Williams's explanation, your question was why level is the important quantity. That might be easiest to understand in terms of power and the conservation of energy. It's the rate of energy delivery that determines how loud the produced sound is and how strong an effect the received sound has an an eardrum or a mike.
Jerry Friedman said,
August 29, 2013 @ 11:58 am
By the way, the reason we take the log isn't just that we have better intuition for addition. Our perception of sound is approximately logarithmic—Wikipedia tells me it's called a "near miss" of logarithmic perception. I tell my students this allows us to hear differences in loudness between soft sounds while being able to accommodate loud ones. All this is subject to correction by those who understand this field, and I'm hoping my mention of it is interesting to at least some people here.
Eugene van der Pijll said,
August 29, 2013 @ 12:45 pm
Maybe this would work?
Imagine a closed volume of air, e.g. in a piston, at the pressure of the surrounding air. To increase the pressure by a certain amount, you have to put a certain amount of work into it,to move the piston for a certain distance.
If you want to increase the pressure by double the amount, you not only have to move the piston for twice the distance, but also you have to push against twice the pressure difference, so ir costs four times the effort . So the energy you put into the system is quadratic with the pressure.
rhr said,
August 29, 2013 @ 12:46 pm
The answer to this lies in the units we use to measure the various quantities. Power –
or power density in this case – has basically non-arbitrary units dictated by the laws
pf physics as commonly defined (power density is mass length^-1 time^-3). Amplitude is kind of arbitrary though, and you can choose to represent it in various different units. But when you make a specific choice, you determine a mathematical relationship between amplitude and power density.
So the first answer to your question is that you square amplitude to get power density
because we choose to measure amplitude in units where … you have to square it to get
power density. You could measure amplitude in units proportional to power density, and then of course you wouldn't have to square it.
So why do we use the units we use? I think the answer is a combination of practicality
and reverence for the simple harmonic oscillator. The standard setup involves a test
mass connected to a Hooke's law spring. A force F on the mass produces a displacement x according to F = -kx. Now it's a general law of mechanics that potential energy U and force are related by F = -∂U/∂x, so if F is proportional to x then U must be proportional to x^2.
What ends up happening is that amplitude is defined as a quantity that's proportional to x and the power density is (necessarily, by mechanics) proportional to U and therefore x^2. Think of a condenser microphone, where you have a flexible conductive membrane forming part of a capacitor. The membrane is the test mass and also the spring. Sound pressure exerts a force. This produces a displacement proportional to the force, a capacitance proportional to the displacement, and ultimately a voltage proportional to that. The usual practice is to define amplitude as a quantity proportional to all of the above. The power density of the sound wave is proportional to the potential energy stored in the membrane at maximum displacement, and therefore to the square of displacement, voltage, amplitude etc.
For other kinds of waves the setup with a test mass and a spring isn't as natural, but
amplitude is still defined the same way. It can even end up making sense, e.g. ocean waves have a power density proportional to the square of their height.
Theodore said,
August 29, 2013 @ 1:24 pm
Power quantity is important quite simply because it represents the (incredibly tiny) amount of heating that occurs on your ear's sensing structures (or inanimate objects, for that matter) when they are impinged upon by sounds.
Other commenters' statements of P=Fv, P=V²/R, etc. explain well enough why the square relationship exists.
Bob Ayers said,
August 29, 2013 @ 1:31 pm
The statement of Graeme Williams is excellent. Here is another that uses a water analogy: You have a reservoir with water in it, at a certain altitude ("head"). You let the water run out through a turbine. If you have twice the "head", you get twice the water pressure on the turbine blades. But the water also runs out twice as fast. So you get four times the power *per unit time*. (This is the "you double the volts and you also double the amps" statement converted to the water analogy.)
Faldone said,
August 29, 2013 @ 1:40 pm
Well, since the definition is the explanation I guess I can't be of any help. I will not, however, ask to have my subscription cancelled. Meanwhile, I wish you the best of luck in understanding this question.
Layra said,
August 29, 2013 @ 3:25 pm
In terms of why power is the relevant quantity rather than amplitude, it could be that amplitude isn't transferable between different mediums in anything resembling a simple fashion, while power is. If vibrating air hits a membrane, the amplitude of the vibrations in the air will be different from the amplitudes of the vibrations in the membrane since air and solid membranes have different tensions and modes. But the power in the air will transfer to the membrane linearly, so it's easy to keep track of it without having to consider the exactly material properties of the media too much.
Nor is amplitude really that useful a measure anyway; a vibrating string that's fixed at both ends has a different amplitude at every point along the string, while for power we just integrate over the whole thing to get a function of time. We could do a similar integration for amplitude but as pointed out before that often just averages out to 0, and integrating the magnitude without squaring becomes hairy mathematically.
D.O. said,
August 29, 2013 @ 4:31 pm
The reason to use power (as in square amplitude) is because energy is conserved. Thus, if you have 2 independent sources their powers will add up, but what happens to the amplitude depends on definition. and precise situation. Another reason is that power (or energy) is something that you might be interested in. For a sound wave power will be proportional not only to the square of the amplitude, but also to the frequency etc.
Basically, in many applications you can transform energy (that is power) to various forms, with different amplitudes etc. and it might be nicer to have a unit of accounting which conserves (now, don't tell it to Mr. Bernanke, he should keep printing money).
Rubrick said,
August 29, 2013 @ 4:36 pm
I am embarrassed that it had never dawned on me that "decibel" must historically have consisted of "deci-" + "bel".
Stefano said,
August 29, 2013 @ 4:58 pm
If I correctly understand, you are asking two separate questions:
1. Why the power of a sound wave is proportional to the square of its amplitude.
2. Why we generally quantify "levels" of physical signals as ratios of powers rather than as ratios of amplitudes.
I'll address first the second question. The main reason is that power is a quantity that is portable across physical domains: for instance, you can compare the power of an electrical signal reaching a microphone with the power of the sound wave leaving it, since they are both powers and their value is expressed in comparable units of measure. On the contrary, the amplitude of an electrical signal (a voltage expressed in Volt) and of a sound wave (a pressure, expressed in Pascal or a multiple thereof) cannot really be compared in any meaningful way.
As for question one, I'll split in two parts. First, you can characterize any physical phenomenon by the quantifiable change on a system brought about by a quantifiable stimulus. In the case of a sound wave the stimulus is a pressure change, which results in a volume of air to be displaced. By definition, power is the product of the applied instantaneous stimulus with the rate of affected change. What makes this definition useful is that, as it happens, there is conservation principles that states that any change of system's energy per unit of time must equal the net power inflow.
In a medium like air and for a stimulus not too violent, experiments show that the relation between rate of change (y) and instantaneous stimulus (x) is linear (y = k * x). Hence, when I multiply the two of them to compute the power (P= x * y), I obtain a quadratic function of the instantaneous stimulus (P = k * x^2).
KWillets said,
August 29, 2013 @ 5:17 pm
Engineers tend to repeat these definitions circularly, perhaps after one painful derivation in college. The basics of wave motion in a fluid (basically diffusion of pressure) can be reduced to potential and kinetic energy (both equal) as others have noted.
I keep a copy of Elmore and Heald's Physics of Waves to remind myself of the derivations of various parameters. They mention that acoustic power/intensity corresponds to the "subjective notion of loudness", so for instance a signal varying in frequency will be perceived as the same loudness so long as power is kept constant.
One thing to note is that power is proportional to both amplitude squared and frequency squared, so if you have a mix of frequencies, or unknown frequencies, you have to use RMS to sum them correctly. In the case of a single sinusoid, amplitude and RMS are identical, and you can go on about Hilbert spaces if you want.
Linguistics note: c, the usual parameter for phase velocity and the speed of light, derives from Latin celeritas, speed.
Chris Waigl said,
August 29, 2013 @ 5:19 pm
I'm not in the ideal place right now to write this up (and clarify in my own head) completely properly, being stuck with a flight delayed by 6h on the northern tip of the US (Barrow, AK) and having only a slow satellite internet connection, but I love this kind of question. Some have already said pretty much what I'm going to say, but maybe in more complicated forms.
The central point here is that all the processes you're talking about involve traveling waves. That is, we're looking for power transported in harmonic propagating processes, all of which can be reduced to the concept of the harmonic oscillator. In its simplest form, we have Hooke's law that the force exerted to extend the harmonic oscillator (think, spring) by a distance x is proportional to x (F = – k x), where x is the (instantaneous) amplitude. (This is either the definition of a harmonic oscillator or an observation about those real entities we chose to call (mechanical) harmonic oscillators — but there you have your amplitude/square-of-power quantity.) Now the elastic potential energy stored in this system at a given instant is the work done to stretch the "spring" to distance x, which is the integral of F from 0 to x, that is 1/2 * k * x^2. Now there's a larger step involved moving from a simple spring to a propagating wave and looking at the energy stored in one whole wavelength worth of propagation. x gets replaced by A (the amplitude of the *wave*) * omega (the angular frequency). Also, the whole thing needs to be considered as the infinitesimal potential energy change *along* the wave, and integrated over a whole wave. Be that as it may, the square-of-x survives this process. In the end, to get from energy to power, you divide by a wave period, which means multiply by the frequency. However, the whole square-of-linear-extent (Amplitude) factor survives the manipulation, as it does for other harmonic propagating waves, such as the electromagnetic wave. (Where again, you have a field-type quantity, which is proportional to the force exerted on charged particles or magnetic dipoles, and the power comes out as proportional to the square of that quantity.)
Breffni said,
August 29, 2013 @ 5:26 pm
I don't know how far Eugene's piston example is from the original problem, but certainly it should evoke concrete sensorimotor experience for anyone who's ever manually inflated a bicycle tyre.
Ø said,
August 29, 2013 @ 8:39 pm
Why is power (in the physicist's sense) the key quantity here? Because it is the rate at which energy is transferred, and that's what matters most. (See others' comments above about conservation of energy. Energy is a pretty non-obvious idea, and it is the conservation of idea that makes it an important idea.)
Why is power (meaning the rate at which energy is transferred) proportional to the square of amplitude? Basically because kinetic energy is proportional to the square of velocity.
Why is kinetic energy (which is Fx, force times distance, or rather the integral of Fdx, therefore the integral of Fvdt) proportional to the square of velocity? By Newton's equation F=mv' the derivative of the integral of kinetic energy is Fv=mvv', one half the derivative of mvv.
Dr. Decay said,
August 30, 2013 @ 6:40 am
1st. I'm amazed at the number of physics geeks who seem to read this blog assiduously and are ready jump in with their two cents.
2nd. I really love this kind of question too and so here are my two cents.
I hope you're satisfied with some of the arguments appealing to harmonic motion to explain why energy is usually proportional to the square of an amplitude. But in case its not obvious, I want to say something about why energy (or power) is such an important quantity. As other comments have mentioned, energy is conserved and because of that fact, this quantity allows you to easily compare waves in different media. Consider a sound wave encountering a boundary, going from air to a solid for example. There will be a reflected and transmitted wave. The amplitudes of these waves will have no simple relationship, but their energies do.
Conservation of energy leads to another important insight when discussing noise. If you have several independent noise sources, the total noise corresponds to the sum of the noise powers of the individual sources. So, say you're standing on a train platform and there is a fan and an idling engine nearby both of which are preventing you from understanding a conversation, then the total signal to noise ratio is the signal power divided by the sum of noise powers from the fan and the engine separately. You can discuss signal to noise in terms of amplitude ratios if you like, but you will have to add noise sources in quadrature.
On the other hand, maybe its pedagogically better to present the idea of noise power in terms of stocastic processes. If you combine several independent, Gaussian processes, the total variance (which is proportional the square of some typical amplitude) is the sum of the variances. You could make a few diagrams, or even animations, that show how this works. It gets to the point more quickly and might make a stronger impression on students who don't want to think about springs and masses and waves and Watts and Pascals and Volts…
Haamu said,
August 30, 2013 @ 10:56 am
From a linguistic standpoint, one of the most interesting things about this thread is the inability of the commenters to agree on a single sense of the word "why."
"Why [mathematically]?" is essentially an epistemological question (isn't it?) while "Why [physically]?" is ontological, i.e., "What is going on?"
It's intriguing that this gap is, for some, not easily perceived. Maybe those who have thought more deeply in this area will argue that the gap doesn't actually exist.
Plane said,
August 30, 2013 @ 2:41 pm
KWillets, "a signal varying in frequency will be perceived as the same loudness so long as power is kept constant" is not quite correct. It's important to distinguish power and loudness. For a graphic illustration that the two are different, you might consider looking up "equal-loudness contours" (often referred to as Fletcher-Munson curves)–though I believe the actual relationship between the two is more complicated than looking at the charts might suggest.
I'm not sure whether or not this information relates to Mark Liberman's question of why, I'm afraid, so my apologies if I've strayed off-topic.
Eric P Smith said,
August 30, 2013 @ 7:58 pm
Interesting thread. I would have contributed sooner but I have been away for a couple of days.
A number of contributors have used the concept of a simple harmonic oscillator. I wouldn’t myself do that. There is a connection but not an essential one. There is no simple harmonic oscillator in a sound wave propagating in an infinite medium. It is of the essence of a simple harmonic oscillator that it oscillates at a fixed frequency, whereas a sound wave propagating in an infinite medium can consist of an arbitrary mixture of frequencies.
Here, then, is my story. A sound wave in air propagates through the continuous interchange of two forms of energy: the potential (stored) energy of the pressure differences between volumes of air, and the kinetic energy of the motion of the volumes of air. I will consider the potential energy and the kinetic energy separately.
First the potential energy. Tim Silverman’s analogy of a spring is a good one. The force exerted by a spring is proportional to the displacement. The energy stored in a spring is proportional to the displacement squared, because to compress the spring twice as far we have to apply twice the force over twice the distance, doing four times the work. For exactly the same reason, the energy of the pressure difference between two volumes of air is proportional to the square of the pressure difference. If we double the amplitude of the sound wave, we double all the pressure differences and thus quadruple all the potential energies.
Now the kinetic energy. As several readers have pointed out, the kinetic energy of a moving mass is proportional to the square of the speed. (Why should this be? Because to increase the speed from, say, 10 to 10+dv we have to apply some force over some short period of time, whereas to increase the speed from 20 to 20+dv we have to apply the same force over the same period of time, which means applying it over twice the distance, thus doing twice the work.) If we double the amplitude of the sound wave, we double all the speeds and thus quadruple all the kinetic energies.
The total energy associated with the sound wave in any volume of air is the sum of the potential energy and the kinetic energy. We have shown that each of these is proportional to the square of the amplitude: hence so is their sum.
Finally, we have to remember that we are ultimately concerned not with the energy in a fixed volume of space but with the power (energy per unit time) propagated by the wave across a wavefront. The energy propagated across (say) a circular area of wavefront of area 1m² in 1 second is equal to the energy in a cylinder of cross-section 1m² and length 343m (the speed of sound being 343m/s). As we have seen, that is proportional to the square of the amplitude.
Mathematical detail: when I say the amplitude I should really say the root mean square amplitude over a representative time sample.
Graeme Williams’s comparison with a gravity water wave is intuitively appealing but I think it is of limited explanatory value because the physics is so different. He has shown that the potential energy of a water wave is proportional to the the square of the amplitude: but to show that the potential energy of a sound wave is proportional to the square of the amplitude we really need to do the physics afresh. Also I fancy that he confuses the “kinetic” model of a gas with the “elastic medium” model. A molecule, as such, doesn't do any work when it enters a higher pressure area: it just flies right in.
[(myl) Thanks, Eric! This is very helpful.]
Theo Vosse said,
August 31, 2013 @ 7:47 am
For some reason, this got rejected one or two days ago, but let's try again.
If you want to demonstrate the difference between amplitude and power, you can play sounds with identical amplitude, but different power. It's pretty easy to generate a sine, triangle or square wave of fixed amplitude, but their (perceived) loudness is different. One of the reasons is that complex waves in this case can be seen as constructed of a sum of sine waves of different amplitudes, and therefore must generate more power than a single sine wave (when everything is in phase). So amplitude is a bit deceiving as a measure. For a truly powerful signal, you could also play white noise…
[(myl) This doesn't quite help, because RMS ("root mean square") amplitude is in a monotonic relationship with power, which is (proportional to) RMS squared.
Thus a square wave of amplitude 1 (i.e. values 1 and -1) obviously has RMS of 1 and power of 1; sine waves of whatever frequency have RMS about 0.707 and power of 0.5 (as long as we calculate over an integer number of periods); triangle waves of whatever frequency have RMS about 0.577; and power of 0.333; uniform random numbers on the interval [-1,1] have RMS about 0.58 and power about 0.33; etc.
So this sort of demonstration, in itself, can't help convince us that we should want to look at the ratio of mean squares rather than the ratio of root mean squares.]
Barbara Partee said,
August 31, 2013 @ 3:41 pm
I don't know enough physics to contribute, but I just have to say that I LOVE this thread! I too am impressed to see who-all is reading Language Log and ready to contribute! And the angle about the non-triviality of interpreting "why" questions could potentially be a whole 'nother thread!
a George said,
September 1, 2013 @ 7:52 am
Among the responses to your query there are many that are perfectly correct physically but not really adapted to most modern undergraduates. I find the best coherent presentation yet of this problem complex in "Fundamentals of Musical Acoustics" by Arthur H. Benade (Second Revised Edition 1990, Dover Books, still in print). You will find it on pp. 225-228: "13.2 The Decibel Notation and Its Application to Acousticial Signals". As Benade writes: "We have reached the place in this book where the term needs to be explained and preliminary indication of its usefulness sketched out".
I think I have mentioned it before, but I still believe this book to be essential for anybody engaged — in music making and listening, obviously — but also the sister discipline phonetics. Its presentation of 'formants' is magnificent. Strangely enough, the scientific community seems to be oblivious of its existance, even among anonymous peer reviewers. For instance:
a note in Nature:
"Tuning of vocal tract resonance by sopranos"
by Joliveau, Smith,and Wolfe
NATURE, Vol. 427, 8 January 2004, p. 116
with a subsequent in-depth paper:
"Vocal tract resonances in singing: The soprano voice"
by Joliveau, Smith,and Wolfe in J. Acoust. Soc. Am. 116 (4), Pt. 1, October 2004, pp. 2434–2439.
Neither of these papers, nor further papers by the same group on this subject, mention that Benade in "Fundamentals of Musical Acoustics" actually identifies, analyses, and explains precisely this phenomenon on pp. 382-387. And this was even in the first edition of the book, in 1976!
[(myl) Thanks, George. I have a copy of this book somewhere, but it didn't occur to me to look at it for an answer to this question. I'll do so now — as soon as I can figure out which bookcase (in which office) it's in!]
KWillets said,
September 3, 2013 @ 12:43 am
@Plane thanks for the refinement — the example I was paraphrasing was in the 50-100 Hz range, where I believe the equal loudness contours are close to the power curve. This is the range where phonograph records ran out of amplitude range sufficient to support bass notes (required amplitude approaches infinity as frequency approaches zero, for any fixed intensity level), so a compression scheme was used.
In this range a sine wave of fixed amplitude will rapidly decrease in loudness as frequency is lowered — @Theo's example with more complicated waveforms is not needed.
Conversely, if amplitude is held steady, it takes more power to wiggle pressure multiple times in an interval than once (and the gradients are proportionately steeper, which means more energy transfer).
Mark F. said,
September 3, 2013 @ 11:21 am
It sounds like the question of why the power is proportional to the amplitude has been answered more definitively than why the power is the thing you should care about. Two of the answers that people have given are that power behaves better (mathematically) as you go from one medium to the next, and that power levels of noise combine linearly. But I think Mark's audience probably needs a good intuitive explanation for why these behaviors hold for power but not amplitude.
I think the medium-boundary issue could be explained by imagining a concrete example, with a less-dense medium next to a denser one. Somehow each blob of less-dense material won't be able to move its more-dense neighbor as far as it "wants to", but it will do just as much work on it. I may have already drifted off into error here, but I feel like this is the kind of explanation that would be nice.
As for the additivity of noise power, that seems like a tough one. It reminds me of the fact that variances of random variables add, I'm not sure that's helpful even if it's related.
Eric P Smith said,
September 3, 2013 @ 11:46 am
@K WIllets: it depends on what variable you are taking the amplitude of. I think you are talking about the amplitude of the displacement. Normally however sound is regarded as a pressure wave, and "amplitude" means the amplitude of the pressure above or below ambient pressure. It is in that sense that power is proportional to the square of amplitude, the constant of proportionality being the same across all frequencies.
David Stinson said,
January 1, 2014 @ 10:18 am
When you're talking about signal-to-noise, you're really talking about the amount of information you can transmit in a signal. Information is measured as entropy, which is a measure of non-usable energy. Then there's still a time factor to get to power. So basically, you're looking at information over time.
Bringing up entropy can also help explain why you're dealing with logs.
Sampo Syreeni said,
March 30, 2014 @ 3:35 pm
I'm a year late to this one, but let's see if I can make it bit clearer. Originally the idea indeed comes from electrical work, so let's start there. Let's also make a further assumption which people often don't spell out, but which was already made in the original Bell Labs work, since they were working with perfectly compensated transmission lines: at whatever frequency you're dealing with the medium your putting your signals into behaves like a perfectly matched, or resistor-like sink of power.
In those conditions the power expended in a resistor is voltage over it times the current through it. Keeping the resistance constant, doubling the voltage by Ohm's law forces twice the current through the resistor, but at the same time, by assumption, the voltage two is doubled So the total power expended is quadrupled.
The point here is that this is characteristic of all linear, time-invariant systems, be they about masses and springs as in mechanics, electrical circuits, electromagnetical radiation, acoustical radiation/sound, or even the small scale, linearized perturbations of more general, nonlinear systems. All of them can be analysed analogously. In the case of linear acoustics, what is called voltage in electrical work becomes sound pressure variation around the mean atmospheric pressure, current becomes airflow (or in radiative fields, mean field velocity) which work to equalize pressure gradients, and resistance becomes the frequency, temperature, total pressure, medium and whatnot dependent loss term which converts sound, too, gradually into heat.
As such, the sound energy per unit time, or power, works just the same in acoustics as it does in electronics. If the amplitude doubles, it not only means that the amount of diffusion through your dissipative medium was doubled, it also means that your energy source could keep up with the extra loss induced, while maintaining the doubled amplitude. Those terms multiply, so that when you in fact finally measure twice the amplitude given the same proportional amount of dissipation, it took four times as much power to yield the outcome; twice for the doubled amplitude, and twice again to compensate for the fact that in order to maintain the new level, you also have twice the amount of collisions between your air molecules, in addition.
The easiest way to see how the energetics work out is to think of hydropower. There, the water pressure differential over a dam ("head") is the "field quantity" which provides the forcing and can be likened to voltage in electronics or instantaneous pressure differential in acoustics. The flow amount from the bottom in litres per second of the dam equals current, or "wind" (more accurately the differential drift rate) in acoustics, and the cross sectional area of the hole at the bottom of the dam can be roughly likened to resistance. Now there are two equivalent ways of giving twice the power from an ideal turbine affixed to the hole.
You could double the pressure head and spin a single turbine twice as fast. Or you could maintain the head but use two turbines in parallel. The trouble is, as soon as you double the head, but maintain the same size of hole, the water runs out twice as fast, so that you can run two turbines, both twice as fast. The power output will be quadrupled, and so will the steady state input, needed to maintain the water level (head, voltage, whatever) within your dam.
So, finally, the reason why field quantities need to be squared to get power quantities, and so why a factor of two needs to be put in the corresponding decibel counts for field quantities, is that you're only measuring the real energetic outcome of your process when you talk about power measures. There the conservation of energy makes everything simple. If you instead talk about the naive forcing behind your system, you need to know how precisely your system reacts as well. In order to make all that simple, any calculation of decibels having to do with a forcings assumes the simplest kind of, Ohmic system, in steady state (or using ideally matched sinusoidal excitation), where the correspondence between forcing and energetics necessarily takes on the form of a square law. Plus there are then tons of mathematical reasons why this really is double-plus-good, easy, natural, or even the only way, in other regards.
I hope that makes the underlying ideas at least a bit more transparent. :)