# A Discussion About Wave Momentum From the PHYS-L Listserver - Mar. 1999

Under what conditions can a wave be said to carry momentum? Defining momentum as the time integral of a force, light carries momentum, which manifests itself as an electromagnetic force on a surface it reflects off of. We can understand this purely classically by considering the electric and magnetic fields interacting with a surface electron.

We can also get the momentum quantum mechanically (QM) from p = h/lambda.

Finally, from special relativity (SR) we could invoke "relativistic mass" m and correctly get it from E = mc^{2} and p = mc so that p = E/c for photons.

The QM and SR relations also give the correct momentum for a massive particle such as an electron, where v is now its speed and lambda its de Broglie wavelength.

But what about mechanical waves? For phonons, h/lambda gives the "crystal" momentum, but that's generally not considered a true momentum in the sense of exerting a force. Is there any meaning to the quantity E/v where v is the phonon speed of sound and E = h*nu?

Intuitively it seems a mechanical wave doesn't carry momentum since there's no mass flow. A cork on the surface of the sea just bobs up and down (okay, the bobbing is slightly circular, but never mind that). I assume surfing requires interaction with the drag force from the sea bottom near the beach as the wave breaks, so that still doesn't necessarily mean a water wave carries momentum.

Comments? Carl Mungan

Carl, I suggest that you look at p. 359 of Griffiths' Intro to Electrodynamics. The momentum density p' is equal to S/c^{2} where S is the Poynting vector. There is more on this in the book and it is impossible to type it all here. Oh yeah, this is for an electromagnetic wave so I am not quite sure this answers your question. But I assume you make analogies to the mechanical waves by the fact that a mechanical wave is energy propagating by a medium disturbance (EM waves are only different in that they have a fixed velocity in vacuum of c and need no medium).

Samuel Held <held@rhip.phys.utk.edu>

> But what about mechanical waves?

> Intuitively it seems a mechanical wave doesn't carry momentum since there's

> no mass flow. A cork on the surface of the sea just bobs up and down (okay,

> the bobbing is slightly circular, but never mind that).

Carl is undoubtedly teasing us. We would undoubtedly be more comfortable with the oscillatory motion of individual balls in the Newton's Balls demonstration, I'd think. A liquid may not be able to sustain a transverse body wave, but it most certainly propagates a longitudinal wave quite nicely.

On this topic, there was a fascinating insight in an engineering newsgroup which I am not convinced was entirely an urban fable. It goes like this: Once upon a time in a country far far away (I hear it was an Arabian Peninsula country) the engineers of that place built a pipe fit to transport large quantities of precious liquid - perhaps oil, or it might even have been ... water! They felt it was important to test this pipe which stretched across many miles of desert. And so they pressurized it to a moderate overpressure in the customary way. But it happened that they used compressed air (water being so precious) and what is worse, there was a miscalculation concerning the over pressure to be used for the test. And so, sad to tell, the pipe developed a little crack which stretched a little way along the pipe. Now it happens that the speed of sound in air is about 1100 feet per sec but the speed of sound (at which some cracks propagate) is about 5 times faster in a steel pipe. In an unexpected development, the crack once started, leapt ahead, mile on mile, followed not long after by escaping air from the progressing crack. The crack propagated sooner than the air could escape to relieve it, you see. And so, alone and abandoned across the sand dunes sits a mighty pipeline with a crack longer than any other crack on earth....

Brian Whatcott <inet@intellisys.net>

I don't see that a longitudinal wave carries any momentum either, the famous "Maxell tape" poster notwithstanding. That's the poster/advertisement with a speaker (presumably at high volume) producing a wind which blows on a guy in a easy chair, in case I'm being too obtuse. I'm tempted to suggest that a wave carries momentum only when there is no medium.

James McLean <jmclean@chem.ucsd.edu>

> I don't see that a longitudinal wave carries any momentum either,

> the famous "Maxell tape" poster notwithstanding.

When I see an 8-ounce chrome-plate ball lurch forward, I see a movement I can associate with momentum. Perhaps the issue is the oscillatory nature of mechanical vibrations? Rather like the oscillatory nature of the line voltage in the wall socket?

> (That's the poster/advertisement with a speaker (presumably at high

> volume) producing a wind which blows on a guy in a easy chair, in case

> I'm being too obtuse.)

I think 'abstruse' is the word you had in mind. Nobody claims you are less than acute! :-)

Brian Whatcott <inet@intellisys.net>

But the Maxell tape ad is about a longitudinal wave that carries enough momentum to drive e.g. a pick-up microphone.

James W. Wheeler <jwheeler@eagle.lhup.edu>

Quick calculation for a transverse wave on a string:

Let T = tension on the string. Let mu = mass per unit length of the string. Let the string lie along the x axis, and let the displacement of the wave be in the y direction.

The energy density of the wave is then 0.5*mu*(dy/dt)^{2} + 0.5*T*(dy/dx)^{2}.

The current of energy (flux) is -T*(dy/dt)*(dy/dx).

Dividing this by the speed of the wave gives -sqrt(mu*T)*(dy/dt)*(dx/dt) which would be a momentum flux for the wave.

James W. Wheeler <jwheeler@eagle.lhup.edu>

You are confusing me mightily about these waves and their momentum. I will stipulate that phonons in solids and photons in the electromagnetic field are concentrations of energy, and that when a concentration of energy moves there is a momentum associated with that motion. That said, I must admit that the only way I see a momentum flux to be associated with the traveling waves that may be moving down a stretched string is relativistic and is again associated with the motion of a bundle of energy down the string. So far as I can see no rest mass is transported and no momentum moves down the string.

Barlow Newbolt <newboltw@madison.acad.wlu.edu>

Howdy. Maybe in a sound wave thinking of the energy and momentum transport as successive collisions it gets easier to picture. The original source transfers energy and momentum to adjacent molecules which collide with the next layer, etc.

Good luck, Herb Schulz <herbs@interaccess.com>

This applies to a pulse traveling along the "domino wave." We must be clear about what is not a wave, what is a wave, and what is more than a wave. Calling these three things waves may be the cause of our difficulties.

Ludwik Kowalski <kowalskil@mail.montclair.edu>

Our simplistic model (redundancy?) ascribing purely vertical motion to a string supporting a wave has inherent contradictions. If you haven't thought carefully about it, see the eye-opener: Reuben Benumof, "Simple harmonic motion in harmonic waves," AJP **48**, 387-392 (May 1980).

Bob Sciamanda <trebor@velocity.net>

One way to see the momentum carried by a wave is to use a long rod of some metal, say aluminum or steel. Use a bit of vaseline to stick a small plug or weight on the end of the rod. Then hit the other end of the rod with a hammer. The momentum contained in the wave will be trapped in the plug which will pop off the end of the rod with some velocity. Experiment with the mass of the plug to maximize the transfer of momentum, basically impedance matching.

Eric T. Lane <elane@cecasun.utc.edu>

I must say I've been enjoying the responses to my query about whether mechanical waves carry momentum. I hope people will continue to post, because as far as I can tell at present, some folks think mechanical waves clearly do carry momentum, and others (including me) are not so sure. I'm with Barlow: confused!

I should clarify that I am thinking of a net time-averaged momentum. Of course I grant that Newton's balls swing. But then they swing back!

Ludwik asks what is a wave? Maybe that's a good question to start with - anyone care to take a crack? In my mind, I was counting pulses as waves - after all, you could Fourier compose many sinusoids to get a pulse. I am however ruling out standing waves - there does have to be a net energy transport, hence a traveling wave.

Brian's Arabian story is interesting. But I'm not sure what conclusion to draw from it. I'm not teasing, really, so do give me the moral of it.

I actually heard someone else recently say that light wouldn't carry momentum if the ether actually existed. I don't know what that means, but it does sound like James McLean's comment, "I'm tempted to suggest that a wave carries momentum only when there is no medium." Please elaborate - sounds intriguing! I appreciated your example of the Maxell tape ad. But what about Eric's example of striking the end of a metal bar and seeing what happens at the other end? (Presumably, the bar is clamped loosely at its midpoint - is that analogous to managing to surf because of a sea bottom effect?)

Bob's point that purely vertical motion of a string wave is over-simplified is definitely related to my question. Again, please elaborate and I'll go look up the AJP article next time I'm at the library.

Carl Mungan

> But the Maxell tape ad is about a longitudinal wave that carries enough

> momentum to drive e.g. a pick-up microphone.

It carries enough *energy* to drive a microphone. I don't think there is a net change in the velocity of the microphone (and hence in it's momentum). First increased pressure pushes it one way, then decreased pressure pulls it the other.

James McLean <jmclean@chem.ucsd.edu>

> When I see an 8-ounce chrome-plate ball lurch forward, I see a movement

> I can associate with momentum.

You must be referring to a demo that I'm not familiar with. Could you describe it in more detail? Does the ball actually obtain a net momentum change, or does it only lurch forward, then backward again?

James McLean <jmclean@chem.ucsd.edu>

> One way to see the momentum carried by a wave is to use a long rod of

> some metal, say aluminum or steel. Use a bit of vaseline to stick a

> small plug or weight on the end of the rod. Then hit the other end of

> the rod with a hammer. The momentum contained in the wave will be

> trapped in the plug which will pop off the end of the rod with some

> velocity. Experiment with the mass of the plug to maximize the transfer

> of momentum, basically impedance matching.

In this case, haven't you basically constructed a momentum rectifier? The rod can push the plug much better than it can pull.

James McLean <jmclean@chem.ucsd.edu>

> This applies to a pulse traveling along the "domino wave." We must be clear

> about what is not a wave, what is a wave, and what is more than a wave.

> Calling these three things waves may be the cause of our difficulties.

Howdy. I guess I don't understand this statement. Given the random motion of the gas molecules with an additional momentum superimposed on the molecules, why would you expect that the "dominoes" couldn't rebound? The collisions don't all have to be head-on elastic collisions with one molecule initially stationary.

Herb Schulz <herbs@interaccess.com>

> You must be referring to a demo that I'm not familiar with. Could you

> describe it in more detail? Does the ball actually obtain a net

> momentum change, or does it only lurch forward, then backward again?

As normally constituted, the big balls in coffee table Newton's Balls demonstrations bounce to and fro. If one chooses a suitably light suspension string for the last ball, one can arrange for it to pop off and roll away.

I think we are into "dc" vs "ac" momentum here.

Brian Whatcott <inet@intellisys.net>

> Dividing this by the speed of the wave gives -sqrt(mu*T)*(dy/dt)*(dx/dt)

> which would be a momentum flux for the wave.

Is this a typo? Did you mean (using P for the second quantity) P/v = P/sqrt(T/mu) = -sqrt(mu*T)*(dy/dt)*(dy/dx)? Why did you choose to divide by the velocity of the wave, rather than the velocity of a point on the rope?

James McLean <jmclean@chem.ucsd.edu>

> Dividing this by the speed of the wave gives -sqrt(mu*T)*(dy/dt)*(dx/dt)

> which would be a momentum flux for the wave.

Sorry I don't think this last step gives the momentum flux of the wave. It gives the flux of the *magnitude* (i.e. absolute value) of the momentum which is merely a scaled version of the energy flux. The E vs. p dispersion relation for the wave equation is E = |p|*c (actually this is the relation between the energy and the momentum of the quantized progagating phonons along the string, and this is equivalent to the relation omega = |k|*c relating frequency to wave number) where c = sqrt(T/mu). If the absolute value function was not taken in this relationship then you might be nearly correct (up to a factor of c) for the momentum density. Because the momentum is a signed quantity (or vector if you will) it is possible to show that the total momentum for the excitations of the string (assuming the endpoints are fixed) is zero. All momentum concentrations and fluxes are mere local imbalances that both total to zero and average out to zero over multiple cycles of the waves. This is not necessarily true of the energy flux which can propagate unidirectionally with running waves until they reflect off of the end points and the energy returns to its starting place. But this time scale is given by the length of the string divided by the wave speed, not by the wave frequency like the momentum reversals are.

If you thought that you could get the momentum *density* from the energy *flux* by dividing by c^{2} like you can from the Poynting flux for EM waves, then I think you you were misled in this string case. Waves for a scalar-valued disturbance field propagating in a 1-dimensional medium are different than 4-vector-potential-valued waves propagating in 3 dimensions. BTW, the momentum flux is given by the stress tensor, not c times the momentum density anyway. Of course, since I have not worked it all out, I'm not quite sure at the moment just what the expression for the 1 x 1 stress tensor actually happens to be for this string system, but I kind of doubt that it would come out to your final expression.

David Bowman <dbowman@georgetowncollege.edu >

I just realized that I probably need to clarify a point in my last post. I had said:

> BTW, the momentum flux is given by the stress tensor, not c times the momentum density....

I should point out for the engineering dimension that this symmetric 2nd rank (stress) tensor that comes in depends on the spatial dimensionality of the medium. If the medium is 3-dimensional then its dimension is that of force/area. If the medium is 2-dimensional then its dimension is force/length. If (as is the case for James' string example) the medium is 1-dimensional then its dimension is just plain force and the tensor then only has 1^{2} = 1 component.

David Bowman <dbowman@georgetowncollege.edu>

> Ludwik asks what is a wave? Maybe that's a good question to start with -

> anyone care to take a crack? In my mind, I was counting pulses as waves -

> after all, you could Fourier compose many sinusoids to get a pulse. I am

> however ruling out standing waves - there does have to be a net energy

> transport, hence a traveling wave.

A wave is a pattern of movement in some medium in which the pattern moves in a different way than the medium does. In light, the medium is electromagnetic fields, which do not in themselves move at all; instead there is a time-varying value for the field at each point. The wave is in the pattern of variation of the fields at each point.

In some cases the medium undergoes a net displacement as a result of the wave, but the displacement is different than the movement of the pattern. A classic example of this kind of wave is a crowd startled by a sudden noise. The people nearest to the noise step away from it, forcing the next line of people to move back a step, which forces the next line of people to step back, and so forth.

A shock wave is a pattern of changes in the form of motion of the particles in a medium, where the location of the change may move at a different speed or direction than the particles themselves. (This definition may not be general enough.) Clearly a shock wave can carry momentum. In an oscillatory wave, the particles of the medium undergo cyclical motion with no net displacement. Most but not all contributions to this thread had this kind of wave in mind.

This is all off the top of my head, so I expect some corrections, but I think that I got the gist of the definitions correct.

Maurice Barnhill <mvb@udel.edu>

> Bob's point that purely vertical motion of a string wave is over-simplified

> is definitely related to my question. Again, please elaborate and I'll go

> look up the AJP article next time I'm at the library.

This sadly neglected paper is MUST reading. By constructing a more realistic model, this paper develops a picture of a string wave which departs significantly from our usual view of simple harmonic motion of each particle. Besides answering the problems raised in this thread, quite different conclusions are reached on other matters, e.g. "At any point x on the string [supporting a traveling wave] the potential and kinetic energies of a length dx are always equal in magnitude and in phase." Elsewhere, "In the case of standing waves, the energy surges from the nodes, where it is entirely potential, to the antinodes, where it is entirely kinetic, and back again." There's lots more, including similar conclusions about longitudinal mechanical waves.

A follow-up note - Mathur & Sastry, AJP **51**, 276-277 (March 1983) confirms the above and draws analogous parallels in E&M waves. Don't wait - read them both now.

Bob Sciamanda <trebor@velocity.net>

A mathematician would say that any function of x and t in which these two variables appear together as z=x-vt is a one-dimensional wave propagating to the right along the x-axis. For example: y=A*sin(x-vt), y=A*(x-vt)^{2}, or y=exp(x-vt). Each of them satisfies the differential wave equation. Neither energy nor momentum appears in the definition. Note that y can be any quantity: the displacement, pressure, E, B, or even the probability of finding a particle.

Ludwik Kowalski <kowalskil@mail.montclair.edu>

The so-called "electromagnetic wave" is actually a pair of waves supporting each other. This is very different from the waves of displacement y and velocity v=dy/dt, of particles along a stretched rope. Unlike for y and v, the planes of polarization of E and B are mutually perpendicular. That is a significant factor for transporting or not transporting linear momentum. A surfer pushed by an ocean wave, near the shore, will gain linear momentum, but this is not a transverse wave traveling with constant v.

Ludwik Kowalski <kowalskil@mail.montclair.edu>

To drive a microphone, e.g. a variable capacitor type, it must exert a force, and hence it must carry momentum! The average momentum over a cycle may well average to zero, but it is not zero at every point in the cycle.

James W. Wheeler <jwheeler@eagle.lhup.edu>

Thank you! It was indeed a typo.

The derivation for the energy flux is straightforward. The extension to momentum was oversimplified and conjectural. It brought forth comments of suitable complexity from David Bowman.

It is interesting to think about a traveling triangular wave moving left to right. During the rise of the string (wave arriving) dy/dt is positive, but dy/dx is negative, so the flux is positive. During the fall of the wave (after the wave passes) dy/dt is negative, but dy/dx is positive, so the flux is still positive. This makes good sense for the flux of energy, but makes much less sense for a flow of momentum. The (canonical) momentum density of any point in the wave is mu*dy/dt and is perpendicular to the direction of propagation of the wave. The stress energy tensor is probably the right way to approach the momentum density. I'm getting mu*(dy/dt)*(dy/dx) for the momentum per unit length. I still have questions about the interpretation of this, so I will do a little looking.

James W. Wheeler <jwheeler@eagle.lhup.edu>

> To drive a microphone, e.g. a variable capacitor type, it must exert a

> force, hence it must carry momentum! The average momentum over a cycle

> may well average to zero, but it is not zero at every point in the cycle.

We are in complete agreement. Which brings us back to Carl's question: a (continuous) sound wave does not carry (time averaged) momentum, but an EM wave does. What's the crucial difference? How does a shock wave fit in?

James McLean <jmclean@chem.ucsd.edu>

In this example or the previously posted example, if you choose a small enough dt I think you would ascribe a momentum change. Or assume the last ball is not in a gravitational field, starts at rest, and is not connected to anything. (Never mind how you achieve this: buy the necessary equipment at the massless string and pulley store.) Then that ball will achieve a net change in momentum and I would ascribe it as momentum transferred. And BTW, I don't believe phonons do not involve transport of momentum.

Joel Rauber <rauberj@mg.sdstate.edu>

> Which brings us back to Carl's question: a (continuous) sound

> wave does not carry (time averaged) momentum, but an EM

> wave does. What's the crucial difference?

The sound wave also differs from an EM wave by its longitudinal nature. Ask the same question in the context of an EM versus a mechanical wave along the stretched rope. In this way we compare transverse waves only. The crucial difference is that dB/dt vector (which is proportional to E) is perpendicular to B, while for the rope wave the dx/dt vector (v of particles) is in the same plane as their displacements x.

Ludwik Kowalski <kowalskil@mail.montclair.edu>

> A mathematician would say that any function of x and t in which these two variables appear

> together as z=x-vt is a one-dimensional wave propagating to the right along the x-axis. For

> example, y=A*sin(x-vt), y=A*(x-vt)^{2}, or y=exp(x-vt). Each of them satisfies the differential

> wave equation.

A small quibble: this is a description of a traveling wave that retains its shape (no dispersion), so shouldn't be taken as completely exhausting the types of wave phenomena.

Joel Rauber <rauberj@mg.sdstate.edu>

> Because the momentum is a signed quantity it is possible to show that the

> total momentum for the excitations of the string (assuming the endpoints are fixed) is zero.

Assuming that this snippet can be regarded as applying to the classical vibrating wave on a one-dimensional string, I view the parenthetical comment as crucial. It is not surprising that if both endpoints are fixed we conclude that there is zero momentum transport (flux) as we may describe such a situation as two traveling harmonic waves of equal amplitude (a standing wave) which would be equal momentum transport in opposite directions, summing to zero since momentum is a "signed" quantity.

On the other hand, what happens if the string is very long (infinitely long shall we say) and we attach an oscillator at one end, which we turn on at t=0, and the other end is not fixed (or we look at times less than the propagation time for the leading edge of the disturbance to reach the far end). What then?

Joel Rauber <rauberj@mg.sdstate.edu>

I am a bit mystified by the comments on this thread. There is no real EM wave -- we have given up the idea of an aether long ago have we not. There _is_ real motion in a sound or, to stretch things, water wave -- ie. molecules do move and thus one can more easily imagine a momentum flow, but I am a bit nervous about this as well. Am I being too "classical" -- or worse, just ancient?

Thus an EM wave doesn't "carry" mass nor momentum -- there is a charge at the "originating" site which imposes a force on the distant site and thus it may cause some momentum at that site. We just imagine a wave momentum which we hope will make calculations easier -- just as we talk about EM "fields" which are not "real" either but are sometimes helpful to imagine. Of course I readily admit that action at a distance is not much more "real."

Jim Green <jmgreen@sisna.com>

> There is no real EM wave.

I wouldn't have said that. By any reasonable definition, EM waves really exist.

> We have given up the idea of an aether long ago have we not.

We have indeed. But we needn't throw out the EM baby with the bath-water (or bath-ether).

> Of course I readily admit that action at a distance is not much more "real."

That gets directly to the point. Conservation-at-a-distance is almost as useless as no conservation at all. It is precisely to ensure *local* conservation of momentum that we need expressions for the momentum content of an EM field. These points (and most of the other issues raised in this thread) are pretty well covered in _The Feynman Lectures on Physics_ chapter 27, "Field Energy and Field Momentum."

John S. Denker <jsd@monmouth.com>

> Am I being too "classical" or worse, just ancient?

No. You are just being incorrect. Hertz discovered real EM waves well over a century ago. They were predicted by Maxwell who also identified light an an example of them.

Denying the EM field degrees of freedom and including only Coulomb-esque actions at a distance between charged particle degrees of freedom is not equivalent to the Maxwell theory and is inconsistent with experiment and special relativity. Any attempt (even at the purely classical, nonquantum level) to "integrate out" the field degrees of freedom from the charged matter + EM field + their mutual interactions Lagrangian results in a very nonlocal particle-only theory which can only be solved perturbatively in inverse powers of c^{2}. The lowest order theory is Coulombic electrostatics tied to Newtonian dynamics. The next order includes initial lowest order magnetic field effects and leading order relativistic corrections (for moving particles) to the Coulombic interaction between particles as well as the leading order relativistic corrections to the particles' dynamics. The resulting Lagrangian for this interacting particle-only theory is the Darwin Lagrangian. The next order correction to this theory is much more complicated and I don't know what the name of its Lagrangian is, or even if it has a name. It is only if an infinite-order expansion is made whose dynamical equations of motion result in an infinite set of coupled *infinite-order* differential equations that the Maxwell field + particles theory is asymptotically reproduced as a particle-only theory.

Where did you get the idea that EM fields are not "real", or are somehow less "real" than particles? After all, we consider the so-called 'particles' to really be the classical limit (in terms of wave function localization in position and momentum) of quantum excitations of underlying dynamical *fields*. In the case of electrons (and positrons) it is the lowest mass lepton generation of a Dirac field coupled to the Maxwell (i.e. EM) field via a local connection that happen to make a global U(1) gauge invariance local.

> Of course I readily admit that action at a distance is not much more "real."

It's not any more real. I think it is less real. Somehow, I'm getting a sense of deja vu about this thread. Am I the only one?

John S. Denker <jsd@monmouth.com>

> The crucial difference is that the dB/dt vector (which is proportional

> to E) is perpendicular to B, while for the rope wave the dx/dt vector

> (v of particles) is in the same plane as their displacements x.

True for circularly polarized EM waves, but not linearly polarized. How about this: take any wave supported by a (massive) medium and which leaves the medium unchanged by its passage. That is, any piece of the medium ends up in the same position where it started, so that it's average velocity (and hence average momentum) is zero. This in turn means that the momentum flux (which passed through the piece) must time average out to zero. Thus, an oscillatory wave through a massive medium cannot transmit net momentum. (Sound transmits momentum *fluctuations*, but not *net* momentum.) A domino or shock wave does transmit net momentum because it doesn't leave the medium undisturbed.

I'm guessing that a wave in a 'field' (a non-massive medium?) will always carry momentum. Can a field-theory-type person verify that?

James McLean <jmclean@chem.ucsd.edu>

> Any attempt (even at the purely classical, nonquantum level) to "integrate out" the

> field degrees of freedom from the charged matter + EM field + their

> mutual interactions Lagrangian results in a very nonlocal particle-only

> theory which can only be solved perturbatively in inverse powers of c^{2}.

Although I'm inclined to agree, let me play devil's advocate. What your message mostly says is that it is possible, but extremely inconvenient, to avoid the field degrees of freedom. I don't suppose that nature is required to be convenient. So are you basically saying that your notion of reality is based on obtaining simpler equations, at the expense of having more ethereal things (pun intended) be considered 'real?' Or do you have a deeper reason to lump electric fields into the same category as, say, my desk?

James McLean <jmclean@chem.ucsd.edu>

I have long since given up on Feynman as a reliable source. He does a fine job of illuminating the physics myths, but a poor job of overthrowing them. There will always be a discussion about "reality" and another about, "Well of course it (any current subject) isn't real, but treating it as real is so handy that we should all agree to use ‘reality’ language." The problem here is that the myths are not accurate (though visually handy) and we should all keep the fact that they are myths in mind when we use physics jargon -- lest we fall into useless semantic arguments as in this thread. "Waves" and "fields" (and, yes, "heat") are among such myths.

Jim Green <jmgreen@sisna.com>

I certainly agree that there is a question about "reality" and just what should be considered "real." And another discussion about just what is "invention" and what is "discovery." I guess what I have to say is that we should all keep this problem in mind during discussions like this thread and NOT assume anything absolute one way or the other. After all there is convincing "proof" that "light" is particulate and other equally convincing "proof" that it is a wave. It is likely neither. Just as English is neither German nor French but sometimes smacks of each -- I don't recall who first offered this analogy -- I can't claim it to be mine, but I wish it were.

And it is also true that the fall of one pet (even nearly universally accepted) "belief" will unstabilize perhaps a major structure of physics. But this happens from time to time -- we all should be mentally and emotionally prepared for it. As "scientists" we should not adamantly hold on to any given idea lest we fall with the old theory. Yes, one might go to any region in space and measure varying electric potentials etc, but we should not loose sight of the fact that the effect is due to a moving charged particle somewhere -- not to something independent of such a charge -- if in fact there "really" are such things as charged particles. (:-)

Jim Green <jmgreen@sisna.com>

Did someone claim somewhere in this thread that an EM wave is independent of an accelerating charged particle? I believe that one person's reality is another person's model of the universe. As a natural philosopher, I try to understand and to explain and to exploit the relationships between measurable properties. Whether or not my understanding of nature is "real," I'll leave to a humanist.

Paul O. Johnson <pojhome@flash.net>

> A domino or shock wave does transmit net momentum, because it doesn't leave the medium

> undisturbed.

After the shock wave has passed, isn't the medium undisturbed? This seems to be the case as the sonic boom passes through the air. Ditto for a wave pulse.

Joel Rauber <rauberj@mg.sdstate.edu>

> Although I'm inclined to agree, let me play devil's advocate. What your

> message mostly says is that it is possible, but extremely inconvenient,

> to avoid the field degrees of freedom. I don't suppose that nature is

> required to be convenient.

This is true. But a similar construction can be made for *any* interacting system where some of the system's degrees of freedom can be "integrated out." When the equation of motion for a given degree of freedom is linear then it can be formally solved to describe the time dependence of that degree of freedom in terms of integrals over the entire past history of the other degrees of freedom with which it interacts. This formal expression can be substituted into the other equations of motion for all the other degrees of freedom and the resulting equations of motion become complicated nonlocal integro-differential equations involving only the remaining degrees of freedom without any reference to the formally solved ones (except for possibly reference to their initial conditions). Similarly, the "solved" expressions for the "solvable" degrees of freedom can be substituted into the original Lagrangian resulting in the Lagrangian taking on a nonlocal character but depending only on the remaining degrees of freedom. The extremely nonlocal character of the theory (involving integro-differential equations or coupled sets of infinite order diff-eqs) betrays the fact that the formally appearing degrees of freedom are not all of the story. The system simplifies tremendously when the theory is written in terms of all of the legitimate dynamical degrees of freedom. In that case the theory becomes local and all of the equations of motion for all of the degrees of freedom are 2nd order (in the Lagrangian formulation, and 1st order in the Hamiltonian formulation). Such is the situation for classical electrodynamics when the EM degrees of freedom are integrated out.

> Do you have a deeper reason to lump electric fields into the same category as, say, my desk?

Well, most of what makes your desk is mostly artifacts of electric fields and their interactions with electrons and nuclei. The only non-EM aspect of your desk is contained in the details of the binding of the quarks and nucleons in each of the atomic nuclei of your desk.

Joel Rauber <rauberj@mg.sdstate.edu>

> After the shock wave has passed, isn't the medium undisturbed? This seems

> to be the case as the sonic boom passes through the air. Same comment with

> respect to a wave pulse.

A wave pulse down a rope certainly does leave the medium undisturbed and therefore transmits no net momentum (although, again, it transmits a momentum fluctuation). For a true shock wave, the medium is disturbed. For instance, with a sonic boom there is a region of compressed air, but no region of rarefied air (as there is in a normal sound wave). Thus, as the boom passes through the air, each little chunk of air ends up displaced from its initial position. Although the pressure, average velocity, etc. of the air is the same afterwards as before, the position of each chunk has changed, and hence its average velocity (during passage of the boom) was nonzero.

James McLean <jmclean@chem.ucsd.edu>

> True for circularly polarized EM waves, but not for linearly polarized.

The vectors of E and B are mutually perpendicular even when the EM wave is linearly polarized.

Ludwik Kowalski <kowalskil@mail.montclair.edu>

It’s been a long time since I studied shocks, so I can't provide a lot of detail. As I write this, I am referring to "Astrophysical Formulae" by K.R. Lang. Pressure, velocity, and temperature do change across a shock front. I think a shock is, by definition, a discontinuity. The changes are described by a set of "jump conditions" called the Rankine-Hugoniot equations. Each equation can be expressed in terms of the adiabatic gas constant (gamma) and M1, the Mach number, for the gas behind the shock front. The equations are derived directly from Euler's equation, energy conservation, and the ideal gas law. If subscript 2 represents gas ahead of the shock and subscript 1 the gas behind the shock, the jump conditions are:

Pressure: P2 / P1 = 1 + 2*gamma*(M1^{2} - 1) / (gamma + 1)

Density: rho2 / rho1 = [1 + (gamma + 1)/(gamma - 1)*P2/P1] / [(gamma + 1) /(gamma - 1) + P2/P1]

Temperature: T2 / T1 = P2/P1 * [(gamma + 1)/(gamma - 1) + P2/P1] / [1 +(gamma + 1) / (gamma - 1)*P2/P1]

Philip Zell <zell@act.org>

> The vectors of E and B are mutually perpendicular, even when the EM wave is linearly polarized.

Sorry, I should have been more specific: E and B are always perpendicular, but for linearly polarized EM waves dB/dt is parallel to B. In general, dB/dt = -c curl(E), which may or may not be proportional to E.

James McLean <jmclean@chem.ucsd.edu>

If you connect a wire to the head of the metal swinging hammer, and another to the metal rod itself, and then use a battery and oscilloscope with appropriate triggering so as to show a trace for the contact time of the hammer on the rod, then that contact time corresponds to the length of time it takes the sound wave to travel from the hammer end to the free and and reflect back to the hammer end again. This is because the reflected wave, on arrival back at the hammer in contact with the "source end" bounces the hammer away slightly. The scope trace corresponds to the to-and-fro time for wave travel. It can be captured with a storage scope or the Vernier apparatus (the fast interface, not the old serial one), or even just by repeated trials and marking on the scope face with a water soluble pen. It's neat. So, you can get the speed of sound in the rod from an indirect time-of-flight type measurement.

Allen Brown <brownan@educ.queensu.ca>

I have finally looked up the reference suggested by Bob Sciamanda -- Reuben Benumof, "Simple harmonic motion in harmonic waves," AJP **48**, 387-392 (May 1980) -- and it is definitely an eye-opener!

For light, we all know that the energy-momentum relation is E=cp. Reuben shows that EXACTLY the same relationship holds for mechanical waves, where E and p are now average (over either time or space) densities (e.g. per unit length for a wave on a string, per unit volume for sound, etc.) and c is the wave speed.

So the answer to my original question is: all traveling waves do carry a nonzero average momentum. Hence, the Maxell ad, although wrong in magnitude, is right in principle: a sound wave will blow your hair back. You can calculate from the above relationship that the force should be experimentally verifiable for a decent stereo system - ie., I suggest dropping a feather in front of your speakers when the volume is cranked and there are otherwise no drafts in your room and see if there is a net drift. (I haven't tried this experiment myself.)

This also means that a wave on a string is not simply transverse. There is a forward component. One might be able to think of this as the symmetry breaking phenomenon which drives the wave forward rather than backward, though I'd welcome comments on this interpretation.

I also think we can carry Reuben's analysis a step further. Let's write p=mc where m is the net mass transmitted by the wave. Hence we have E=mc^{2}. I think the reason this didn't come out as 0.5mc^{2} is because the KE and PE of any unit length (or volume) of the displaced medium are equal so that E=2*KE. I would then conclude that typical textbooks which state that there is no bulk mass flow for a wave on a string, water surface, sound, etc are WRONG. (Of course, the wave must be traveling not standing, I hasten to add.) After all, this is what I usually do to elongate my garden hose when I'm too lazy to walk to the other end and tug on it.

Carl Mungan

Recently published is D. Rowland and C. Pask, "The missing wave momentum mystery," AJP **67**, 378-388 (May 1999). I note that the Benumof 1982 paper is cited. Check it out.

Bob Sciamanda <trebor@velocity.net>

I have only begun reading Benumof's article, but my progress has been severely impeded by what I see as a serious error on the very first page on which most of his later results seem to depend. Specifically, Benumof derives the potential energy in the string by implicitly assuming that pieces of the string engage in *no* longitudinal motion. The result of this assumption is that a portion of the string that normally has length dx is stretched by the waveform to a length dx*secant(theta), where theta is the instantaneous angle of the piece of string wrt the "at rest" string. Accordingly he derives a *local* increase in the elastic potential energy that is directly proportional to (tangent(theta))^2.

He provides no justification for this assumption. Indeed, he does not seem even to be aware that he is making it.

Moreover, the assumption goes very strongly against my intuition; I would anticipate that the additional stretch that *is* necessitated by the string's curvature would be accommodated *globally*. For instance, I don't see any reason to assume, as Benumof *does*, that portions of the string that are instantaneously at the tops and bottoms of a large amplitude wave do not share in the overall stretch of the string. I'll go one step further: Since the overall length of the string is strictly *constant* in the presence of an extended, unidirectional harmonic wave, it would be my intuition that the stretch factor should be essentially constant everywhere.

John Mallinckrodt <ajm@csupomona.edu>

**Postscript:** Since this series of postings, I have come across other papers discussing wave momentum.

R.T. Beyer, "Radiation pressure in a sound wave," AJP **18**, 25 (1950). (Summarized in A.P. French's book "Vibrations and Waves," p. 243.) The momentum flow depends on the equation of state of the fluid in which the sound wave is traveling. In particular, there is a nonzero average acoustic pressure for a sound wave in an ideal gas (corresponding to a net momentum transport of approximately E/v) but not in an elastic liquid.

D.W. Juenker, "Energy and momentum transport in string waves," AJP **44**, 94-99 (Jan. 1976). This is a very accessible discussion in which the author emphasizes that the momentum transport depends on two factors: the properties of the medium (viz. whether the total length of the string can stretch or not) and the nature of the source (ie. how the wave is excited). In particular, for a transverse wave on an inextensible string, there is necessarily a net transport of linear momentum equal to E/v (where E is purely kinetic since the string cannot stretch), as well as of angular momentum for a helical wave. However, a "springy" string of variable tension and density can transmit either density or shape wave modes. If a density wave is purely transverse and sinusoidal, as in typical textbook presentations, then there can be no net mass and hence linear momentum transfer.

D.R. Rowland, "Comment on 'What happens to energy and momentum when two oppositely-moving wave pulses overlap?'," AJP **72**, 1425-1429 (Nov. 2004). This is a long comment on a paper by N. Gauthier [also see his paper with P. Rochon on pages 1227-1231 of the Sep. issue] which reviews the topic of longitudinal momentum carried by transverse waves on a string. An extensive bibliography lists many other key papers.

D.R. Rowland, "The potential energy density in transverse string waves depends critically on longitudinal motion," EJP **32**, 1475-1484 (Nov. 2011). In this article, Rowland contrasts three different formulae for potential energy density and shows that those that neglect longitudinal motion are wrong, even though that motion can be neglected when considering kinetic energy density, total integrated energy, or the derivation of the transverse wave equation.

B.C. Denardo, S.G. Freemyers, M.P. Schock, and S.T. Sundem, "Acoustic radiation force due to a diverging wave: Demonstration and theory," AJP **82**, 95 (2014). This paper shows that the sound wave coming out of a loudspeaker can either repel or attract a hanging styrofoam ball, depending on the distance from the ball to the speaker, and on the wavelength and amplitude of the sound.

This page has been translated into Estonian by Catherine Desroches from DoMyWriting.

Happy trails, Carl Mungan