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The Lagrangian approach to simple waves  several common waves that lack momentum
Summary
In this article we demonstrate use of the Lagrangian for a mass on a spring and then for wave propagation in one dimension in several different media. For most of the wave media, the generalized coordinate s is used, attached to the mass elements in place of the usual Cartesian coordinate, x. For waves on an elastic cord and for sound waves, we found that the average momentum transported by the waves is zero, in contrast to the usual belief that these waves do possess net momentum. We verified our results with computer simulations which relied on extremely basic equations of physics: F = ma for motion of the mass elements and F = −k Δℓ for the springs (Newton's second law and Hooke's law). This article has several Flash animations as listed above.Lagrangian is a mathematical construct used to solve classical mechanics problems in a very general way. It allows use of a very odd coordinate systems chosen for convenience for a particular problem at hand. We demonstrate its use here for several wave media.
Flash animation of mass on a spring 

Fig. 1.1. Mass on a spring. Mouse over to see the action. Mouse off it to suspend it and click on it to restart it. If your computer does not support flash click on the video to the left. 
1. Lagrangian for a mass on a spring
The Lagrangian is defined to be the difference between the kinetic energy and potential energy in a system. For example for a mass supported on the end of a spring (a classical resonator  see Fig. 1.1) we write:where η is the vertical displacement from equilibrium of the mass and k is the spring constant. Note the dot over the η in the first term, standing for a time derivative of η , i.e.
. One oddity of using the Lagrange equation is that we are to treat η and
as separate, distinct and unrelated variables for the purpose of differentiation.
We insert our L into the standard Lagrange equation:
Another oddity of this equation is that it specifies that we take derivatives with respect to the unknowns. Most differential equations used in physics are written in terms of derivatives with respect to time and position, not in terms of derivatives with respect to the unknowns such as η and . Plugging ahead and inserting L we get:
which is the standard differential equation governing the motion of a mass on a spring. It has standard oscillatory sinusoidal solutions. For example see this reference.
2. Lagrangian for one dimensional waves in a medium of masses and linear springs
Flash animation of longitudinal waves  mouse over to activate. Mouse off it to suspend it and click on it to restart it. 

Fig. 2.1. Animation of longitudinal waves in an array of masses interconnected by linear springs. This is a good model for a continuous stretchy string or spring as long as there are many masses in a wavelength of the wave being observed. Color is used to enhance the visibility of the regions of spring compression and stretching. 
Video of longitudinal waves  click on to start 

The first type of wave we discuss is a longitudinal wave on a stretchy cord or slinky (see Figs. 2.1 and 2.2 above as well as a nice video of longitudinal and transverse waves). As we start out, the math below assumes that there is zero tension on the string of masses and springs when they are in equilibrium. Observing this with a normal slinky would require permentantly prestretching it so that it is not collapsed on itself at zero tension.
We assume that this system can be modeled as masses connected by linear springs as shown in Fig. 2.2. We need to solve for the displacement from equilibrium of each mass.
We use as our ‘‘generalized'' coordinate system a number s equal to the mass number as indicated in Fig. 2.2. As shown in the figure, we see that s = 0 is the location of the left most mass, s = 1 is the location of the 2nd mass, s = 2 is the third mass, etc. Note that this coordinate system is dynamic, the point s = 2 will move around as a passing wave moves the masses around. Furthermore, note that the distortion, ξ, at s = 2 will be the movement from equilibrium of the second mass from the left. In general, ξ will be a function of generalized coordinate s and time t, so we can write ξ(s,t). One of the big advantages of using this generalized coordinate system is that the mass per unit s remains unchanged by passing waves. If we are using a stretchy string, this coordinate could be marked with dots on the string itself. It is also true that the spring constant for a unit s is unaffected by wave motion.
In the Fig. 2.2 we have set s equal to the mass number; however if desired, s could be merely proportional to the mass number. In the following work we assume that the proportionality constant is chosen such that when the string is unstretched s and x are identical. That is for any point along the string, the generalized coordinate s equals the Cartesian coordinate x when the string is in a relaxed state.
Note that the variable for the previously solved massonaspring η , was only a function of time t, i.e. we might write η(t). Because the linear wave system we are now considering has two independent variables, s and t, we need to broaden Equation (1.2) to include derivatives with respect to both variables [reference]:
where
is defined to be
(i.e with respect to time) and ξ' is similarly defined to be the other derivative,
, i.e. it is the derivative with respect to the generalized coordinate s.
In this wave medium of masses and linear springs the Lagrangian density is:
where m_{1} is the mass per unit length (in coordinate s) of the string assembly, and k_{1} is the spring constant of a unit length (in s) of the assembly.
Also Δℓ(s,t)/Δs is the change in length of the spring at position s at time t from its nonstretched length x_{0} , per nonstretched length (i.e. per unit s). We can relate this to the distortion in an incremental system as:
or in a continuum system we can write this as:
where, again, η' is defined to be . This makes our Lagrangian density become:
In (2.2) we measure the displacement η of the masses in regular Cartesean cooordinate increments, instead of the generalized coordinate s. We do this because we know the energy formulas needed in (2.2) in Cartesian form. What is important is that we accurately formulate how the two energies (T_{1} and V_{1}), and vary with time and s, which we have done.
The Lagrange equation now becomes:
This is the simple wave equation having d'Alembert's solutions:
Equation (2.7) is the standard (i.e. well accepted) solution to the wave equation with one exception: it is written in terms of s, the coordinate attached to the masses, instead of the more usual coordinate x . The best known solutions are sinusoidal, such as:
Note that this is an exact solution, not just a small amplitude, linearized solution.
Using (2.8) we get the kinetic energy in the wave:
which we see has an average value of .
Similarly,using (2.7), the potential energy is given by:
which has an average value of . A little algebra will show that the kinetic energy equals the potential energy (use v = vk_{1}/m_{1} = ω/κ for the wave velocity with respect to the coordinate s). The momentum density in a sinusoidal wave is given by:
Because the coordinate s is attached to the masses themselves, the density per unit s does not change due to a passing wave. The momentum is thus sinusoidal and averages to zero. That is .
We might take a moment to ponder the difference between a solution in terms of the generalized coordinate s and one in terms of the regular spacial coordinate x. If we have a nice sinusoidal solution in terms of s, we would expect a graph of it to appear as shown in blue in Fig. 2.3. However because s is in general displaced from x by a passing wave, we might expect the wave, when plotted versus x, to appear as plotted in green in the same graph. The equation that relates x to s and ξ is:
x = s + ξ , (2.11)
where for a particular mass, s equals the unstretched x position of that mass and xi; is the amount of stretching.
In terms of momentum, we have an equal mass in the half cycles labeled a and b. In terms of the graph having x as its axis ("b"), we see that the upward half cycle is compacted. Thus, we would expect its mass density to be greater than in the lower half cycle; however at the same time its width is less. The mass in each half cycle are still equal in spite of the mass having been pushed around. Thus the two effects cancel leaving equal and opposite momentums in the two half cycles.
A reader might also note that a given physical point, at a specific time, has a displacement ξ and velocity v = dξ/dt independent of which coordinate system (s or x) we work in.
It would be interesting to show algebraically that the momentum averages to zero in Cartesian coordinates. Such showing involves converting (2.11) from x = f(s) = s + A cos(κs  ωt) into s = f_{inv}(x) . Such inversion is very difficult algebraically. The most straightforward way to accomplish this is numerically. We do this in the Octave simulation section below.
The cancelation of the positive and negative momentum cycles is more obvious in s space where equal Δs always mean equal masses, since the coordinate s is anchored to the mass itself. And since we have perfect sinusoids as solutions in the s space, we can expect exactly zero average momentum.
With static tension: We might wonder how things change if there is a static tension, like when a slinky is stretched between two posts. In this case we would expect a solution (for longitudinal waves) for ξ(s,t) to be something like:
ξ(s,t) = A cos(κs  ωt) + T_{0}s/k_{1} , (2.12)
where T_{0} is the time average (or static) tension.
The differential equation (2.6) as derived above for this medium involves second derivatives with respect to s and time, t. The new stretching term (T_{0}s/k_{1}) in the above solution does not survive the second derivative and so has no effect on the waves in terms of the generalized coordinate s and we can take the above solution (2.12) as an exact solution in generalized coordinates.
At the same time there is an important effect once we convert back to the Cartesian coordinate x. In this converting back we shall used a small amplitude approximation.
The relationship between x and s (shown earlier in (2.11) ) is:
x = s + ξ , (2.13)
where for a particular mass, s equals the unstretched x position of that mass and ξ is the amount of stretching.
Assuming a reasonably small sinusoidal term for the purposes of (2.13), converting between the two coordinate systems s and x, we can approximate ξ as being dominated by the last (static stretching) term in (2.12):
which means the relation between s and x becomes:
The two media parameters are the spring constant for a unit length in s and the mass m_{1} in a unit length of s. These equal the spring constant for an unstretched unit length of string and the mass in an unstretched length.
What about these constants in Cartesian coordinates? We will call the spring constant for a unit length in the Cartesian coordinates k_{x} and the mass in a unit length in x as m_{x} . When the string is stretched k_{x} and m_{x} change while k_{1} and m_{1} do not change. The relationships between the above parameters are:
The wave velocity in the x coordinate system equals:
In the limit where T_{0} » k_{1} these become:
where in the last step of v_{x} we have used . Note that this result, that , is the standard cited velocity for transverse waves on a string (remembering that we are discussing longitudinal waves here).
When is T_{0} » k_{1} ? The stretched length of the string is so the condition that T_{0} » k_{1} will hold if the stretched length is multiple times larger than the unstretched length. If there is preset tension in the string (like many elastic cords have) we must plot the tension versus string length and linearly interpolate back to zero tension to find the mathematical "unstretched length".
Another similar condition (that we will not mathematically deal with here) is where the spring constant k_{1} is very large (as in the case of using a piano wire or a very stiff fishing line) but the setup has some sort of "elasticity" in the end conditions of the ‘‘string''. This would be the case where a pulley and weight constitute one of the ends or one of the end posts is quite flexible. In either case active potential energy is stored in the weight or flexible end post.
If T_{0} « k_{1} then from the equations just above m_{x} ≈ m_{1} , k_{x} ≈ k_{1} and v_{x} ≈ v_{p} . That is to say, the parameters see little change from their unstretched values.
Octave simulation … energy and momentumThe figures below show the setups for two Octave simulations. The springs between the masses (black rectangles) were assumed to be linear. In both Octave simulations we use Newton's second law and Hooks law for a linear spring to calculate the dynamics of wave propagation for the particular setup. Because the simulations rely on extremely basic equations of physics, we believe the simulations are good alternate checks on the results of the Lagrange algebraic methods above. It is important that we try to achieve single frequency results in the simulations. We do this by exciting the setups in a simple way and watching the resulting waves. The ring resonator is somewhat better because the excitation used there is single frequency and gradual and really lets the medium determines the wave pattern. 

Agreement with other references:
Elmore and Heald and other references imply that all waves transport momentum equal to the wave energy divided by their velocities. This is in disagreement with our findings.
3. Mixed longitudinal and transverse waves on a linear media
Flash animation of mixed wave propagation involving both x and y movement of the masses. Mouse over the figure to activate it. Mouse off it to suspend it and click on it to restart it. 

Fig. 3.1. Animation of mixed waves in an array of masses interconnected by linear springs. This is a good model for a continuous stretchy string or spring as long as there are many masses in a wavelength of the wave being observed. If your browser does not support Flash, use the video just below. 
Video of mixed wave propagation  click to start and move the cursor off the video. 

We add to the above work the possibility that the waves are both longitudinal and transverse. One example of mixed waves are water waves, although they are somewhat more complicated than waves on a stretchy string.
The kinetic energy density for waves having both longitudinal and transverse components is:
The length change of the springs from a horizontal unstretched length is:
This makes the potential energy density:
where ξ and η are the x and y displacements of the masses (x and y distances from their equilibrium positions), and are their time derivatives, their velocities, and ξ' and η' are their spacial derivates ( and . The length change Δℓ/Δs is the stretching of the springs per unit s of the springs.
Using (3.1) and (3.2) our Lagrangian density can be written:
Because the displacement is now a vector with x and y components, our Lagrange equation is also vectorial (having an equation for each direction). The first Lagrange equation is:
The second Lagrange equation (for the y direction) is:
We see that the two components of the Lagrange equation are both rather difficult to unravel and separate out the x and y components (ξ and η) of the motion due to waves.
Two special cases:
1. Pure longitudinal waves will mean (see Equation (3.1a) ):
and we get the same potential energy density as in Section 2 above ... see (2.4) above for example. The results of calculations with this potential energy density are shown in Section 2 above.
2. Pure transverse waves will mean (again see Equation (3.1a) ):
The Lagrange equation for the vertical motion becomes:
For small amplitude waves, the last term can be ignored, leaving the standard waves equation with a wave velocity of
which is the commonly derived velocity for transverse waves on a string.
3. Strings with a zero unstretched length have particularly simple wave equations. For these (3.2) becomes:
which makes the Lagrangian:
The two Lagrange equations for x and y directions become:
Note that in contrast to (3.4a) and (3.4b), in Equation (3.11) the x and y components are nicely separated. That is, the first equation has only the x component of the distortion (ξ) as an unknown, while the second equation has only the y component η as an unknown. Thus the longitudinal waves and transverse waves are independent of each other. Furthermore, each of these two equations is a simple wave equation, the same as (2.6) above. As we stated for (2.6) these equations have simple sinusoidal solutions and convey zero average momentum. Just as for (2.6) this equation is written in terms of the generalized coordinate s instead of the usual Cartesian coordinate x.
In order for "zero unstretched length" to have a physical meaning the masses would have to all coexist at the same spot without a rest separation distance. To avoid this we are required to having a static tension, to separate the masses from each other. As discussed in Section 2 above we would use an x solution given by (2.12).
With zero unstretched length we need to alter (2.13) which relates x to s and ξ. In all the above work we assumed that the s coordinate system was set up so that s and x were identical when the string was unstretched, without waves, and at rest. However with the zero unstretched length condition, all the masses have the same x (call it x = 0 ) location when the string is without tension. At the same time the s coordinate must vary through all its range for the various masses. We cannot accomodate this in (2.13). To fix this problem we change (2.13) into:
x = ξ . (3.12)
Using the solution (2.12) we see that the coordinate x will take on the form:
x = ξ = A cos(κs  ωt) + T_{0}s/k_{1} ~ T_{0}s/k_{1} . (3.13)
So x will pretty much monotonically increase with s, plus a small oscillating component. The time varying part of the distortion will vary as a normal wave as ξ(s,t) = A cos(κs  ωt) . The component of the distortion η (distortion in the y direction, i.e. transverse waves) will have a similar wave function with amplitude, phase and frequency which are independent of those for the x component. These transverse waves will be strongly governed by the static tension (in the x direction) and weaky affected by the longitudinal waves. The longitudinal waves will weakly modulate their x coordinate unless the longitudinal waves have large amplitude creating oscillating tension comparable to the static tension.
As to the momentum content of these waves, the issue is fairly complicated as discussed at great length by Rowland and Pask (scroll down a page to see the text). As the text explains, momentum depends on a lot of factors and detailed constraints. Purely transverse waves have no x momentum, but in some setups transverse waves will be accompanied by longitudinal waves which may bring momentum. Rowland and Pask maintain that there is a missing factor of 2 in the common assertion that the momentum equals the energy in the wave divided by its velocity. (See Elmore and Heald Section 1.11 as an example of the common formula.) Peskin derives the usual relationship that the momentum for a nonlinear vibrating string equals the energy over the velocity (see Equation (100) in his reference) but surprisingly he concludes that the momentum is zero for the linear case which is consistent with my findings just above. He says in the linear case, wave motions in the three cases are easily separable, a fact that I find is true when the equilibrium length is zero.
4. Unstretchable string, no longitudinal mass effect, constant tension for transverse waves:
In this case the potential energy is:
where
is the amount of extra string needed because of the transverse wave per unit x.
The kinetic energy in terms of x is complicated by the change in density (per x) because of the transverse waves:
The expression for in terms of x and y is:
Taking the derivative we will need for the Lagrangian analysis:
The Lagrange equation:
with L_{1} = K_{1}  V_{1} becomes:
Taking the limit (small amplitude approximation) where , and also the above becomes:
which is a wave equation with the wave velocity given by as is the usual result.
One interesting fact about a transverse wave traveling on an unstretchable string is that such wave transports mass. To see this consider a burst of such waves (similar to that in Fig. 2.8 above) containing perhaps 10 cycles of the wave with no waves before and after. Inside the burst, there is a longer length of string per Δx (as per Equation(4.3) above) than on either side of the burst. Presumably the source of the waves inserted this extra length (and string mass) during the creation of the burst. Since the burst is traveling at velocity v with extra mass only in the burst region where the waves are, the extra mass must also be traveling at the same velocity. The momentum of this wave is just this extra mass times its velocity. We could easy make the argument that a wave with average momentum along its direction of propagation is always accompanied with a corresponding mass transport. See Rowland and Pask Section VI for more on this.
5. Sound waves
Flash animation of model of masses and massless but spongy gas parcels for sound propagation. Mouse over the figure to activate it. Mouse off it to suspend it and click on it to restart it. 

Fig. 5.1. Animation of sound propagation through model of masses (black rectangles) and massless spongy gas parcels (in red and blue). This model is ok as long as there are many masses in a wavelength of the wave being observed. Color has been added to indicate the amount of compression in the gas parcels and make the wave more visible. This animation assumes that there is an invisible source of waves on the left and an absorber of waves on the right. 
Sound through a gas has the same kinetic energy term as above, but the potential energy is different. For adiabatic sound transmission we have the following equation:
where V = Ax is the volume of gas under consideration and γ = C_{p}/C_{V} is the gas constant. Also, A
is the cross sectional area of the sound duct (assumed constant) and x_{0} is the length of a gas parcel.
At an ambient pressure P_{0} this becomes:
where V_{0} and x_{0} are the volume and length of a gas parcel under ambient pressure and temperature. Putting these equations together yields
that is to say, the pressure is inversely proportional to the volume to the gamma power or alternatively, inversely proportional to the length of the gas parcel to the gamma power.
To calculate the potential energy in a gas parcel, we consider a gas parcel in isolation where we have allowed its length x to become very long. Then we calculate the energy required to compress the gas to make its length equal l:
We could define the potential energy we use here as shown in (5.4) minus a reference energy, such as the potential energy when the parcel's pressure equals P_{0} . On the other hand this reference energy term would not survive the differentiation step in the Lagrange equation and so in the end makes no difference to the final equation. This is typical of potential energies: we can usually choose our reference, i.e. add an arbitrary constant, without affecting the outcome of our calculations.
We need the potential energy density per unit s, similar to m_{1} and k_{1} . This potential energy density is given by:
where is the length of a gas parcel whose length at ambient pressure is x_{0} . Also, ξ' is defined as dξ/ds . So our Lagranian becomes
We substitute this into
to get
where
Rearranged this becomes:
This equation has the feature that if ξ has small amplitude, the ξ' in the denominator in the right hand side can be ignored, i.e. the small signal linearization. This equation says that the small signal wave velocity is , consistent with Elmore and Heald.
Next we investigate second order behavior of the waves predicted by (5.8a) using a power series solution of (5.8a).
The general binomial equation is:
or when x = 1 and y = e and e is much smaller than 1, this becomes:
to the second power in e.
When we apply (5.9a) to the right hand side of Eqn. (5.8a), we get:
This makes (5.8a) become:
A power series solution of sinusoids up through square terms is:
or
where X = κs  ωt . We only have the cosine term (to the first power) and not the sine because we are free to pick the phase of our solution, at least the phase of the first order sinusoids.
We now substitute (5.12) into (5.10a) one side at a time. On the left hand side of (5.10a), the a_{0} term disappears while the a_{1} term stays linear in cosX. All the other terms come through as second order in sinusoids of X and so are important (remember we are doing this to second order). The left side of (5.10a) now looks like:
Next we do the right hand side of (5.10a) remembering that we are only keeping terms of power two or less. The rightmost factor automatically gives us one power (or more), so we only need to use the linear term in . The right most factor (not the whole right hand side) is:
which is essentially the same as the previous formula (5.13).
The important part of the other factor on the right hand side is:
After multiplying two factors together, (5.14) and (5.15), the right hand side of (5.10a) becomes:
Equating the linear terms of the two sides (in (5.13) and (5.16) ) we get:
which gives the relationship between frequency and wavenumber:
which is the wave velocity.
Development of shocking 

Fig. 5.3. We show the addition of higher order terms to the basic sinusoid (in blue) to create shock waves. The green line shows the result of adding a sinXcosX = 0.5sin2X term to the basic sinusoid as discussed in the text to the left. The red line shows the eventual fully developed shock with a great number of higher order terms. See an early blog for the complete infinite Fourier series of a saw tooth wave. One can also see this general reference on shocking of sound waves. 
Equating the cos^{2}X terms, we have:
which yields:
and is the same as before and doesn't tell us anything new.
The sin^{2}X terms (in (5.13) and (5.16) ) give us:
again yielding the same relationship for the third time.
Equating the sinXcosX terms yields:
which can be solved:
The addition of a sinXcosX term is the beginnings of the formation of shock waves. This behavior of sound is illustrated in Fig. 5.3. Shocking is a distortion of the sinusoid acoustic wave that results from the fluid parcels in the forward moving part of the sinusoid moving faster than the rest of the wave. This phenomenon takes place at high acoustic power levels.
The cos2X factor averages to zero, implying that the average momentum in the s coordinate system is zero. As we saw with respect to (2.10) above, this means the average momentum is also zero in the Cartesian x coordinate system. This result disagrees with Peskin under "sound waves" Equation (40). Our results do agree with the findings of Rowland and Pask Section VI. Rowland and Pask also discuss pseudomomentum which has applications in quantum mechanics. We need to remember that the above algebra is only been done to second order in the sinusoids. Higher order calculation, applicable to very intense sound, might yield a nonzero result. On the other hand, we have shown that momentum will not be proportional to the energy over the wave velocity.
We next discuss a computer simulation to confirm the zero result.
Octave simulation for sound … energy and momentum
To confirm some of the algebraic results of this section, we ran Octave simulations similar to those discussed at the end of Section 2 above. Because the circular or ring configuration was shown to be more accurate (see Section 2 above), we restricted the simulation to the circular configuration. Two of the figures and animations above in Section 2 (Fig. 2.5 and Fig. 2.9) are appropriate for this section. Compared with the simulations used for Section 2 we only changed the interaction force between adjacent masses to be appropriate for gas parcels, Eqn (5.3), instead of that appropriate for linear springs.
The results of the calculations of the energies, energy over waves velocity and momentum are shown in Table 5.1. We see that the momentum is orders of magnitude smaller than the energy over velocity, implying that the momentum is zero.
Table 5.1. Octave calculated values of energy and momentum at four different times during the wave propagation on a circular array. The wave velocity was set equal to 1.18 . The amplitude was different than that in Section 2 above. These results show that the average momentum is much much less than the wave energy divided by the wave velocity.  
Energy in the wave  7.6814e006, 3.0705e005, 6.8912e005, 1.2211e004 
Energy over wave velocity  6.4920e006, 2.5951e005, 5.8241e005, 1.0321e004 
Momentum in the wave  2.5324e015, 2.5307e015, 2.5278e015. 2.5322e015 
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