Their applications, physics, and math. -- Peter Ceperley
There are all sorts of resonances around us, in the world, in our culture, and in our technology. A tidal resonance causes the 55 foot tides in the Bay of Fundy. Mechanical and acoustical resonances and their control are at the center of practically every musical instrument that ever existed. Even our voices and speech are based on controlling the resonances in our throat and mouth. Technology is also a heavy user of resonance. All clocks, radios, televisions, and gps navigating systems use electronic resonators at their very core. Doctors use magnetic resonance imaging or MRI to sense the resonances in atomic nuclei to map the insides of their patients. In spite of the great diversity of resonators, they all share many common properties. In this blog, we will delve into their various aspects. It is hoped that this will serve both the students and professionals who would like to understand more about resonators. I hope all will enjoy the animations.
This posting includes flash animations showing the physics discussed.
Most computers have a flash player already installed, but if yours does
not, download the free Adobe flash player here.
Flash animations:
The Lagrangian approach to simple waves - several common waves that lack momentum
revised June 2018
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 presented, the generalized coordinates is used, which is simply the x position of each mass element in its equilibrium position. In spite of the fact that the mass elements move around during the passage of a wave, we identify each by its equilibrium position s. We also use the variable ξ which is the displacement of a mass element from its equilibrium position. When a wave is present ξ is a function of s and time, t.
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 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.
The Lagrangian is a mathematical construct used to solve classical mechanics problems in a very general way. It allows use of 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:
, (1.1)
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.
.
We insert our L into the standard Lagrange equation:
. (1.2)
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.
Another oddity of this equation is that it specifies that we take derivatives with respect to the unknowns η and
. 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 from (1.1) we get:
, (1.3)
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
Lumped parameter model for longitudinal waves
Fig. 2.2. Mass and spring model for waves on a slinky (the brand name of a long spring sold as a toy). The springs are considered massless and linear, i.e. spring force = −k Δx where k is the spring constant which does not change with stretching of the spring. The variable Δx is the amount of stretching of the spring.
The s - ruler shows the x coordinates that the masses will rest above when there are no waves. In this illustration we show the effects of a wave passing through the system of masses and springs. The dashed lines connecting the masses to numbers on the s - ruler are purely to show which mass would rest on which number when at rest. There are no forces associated with the dashed lines.
We use the coordinate s to be our generalized coordinate. For a particular mass, s does not change with time even when waves are present. The displacement ξ is the horizontal distance a mass is from its resting place on the s axis. This displacement will change due to waves as a function of s and time t, i.e. ξ(s, t). The x position of a mass also changes with time and is given by x = s + ξ.
The distinction in mass locations is similar to the distinction between Lagrangian and Eulerian coordinates encountered in fluid dynamics where use of our s would be considered Lagrangian.
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 permanently prestretching it so that it is not collapsed on itself at zero tension. How to support the slinky to counter the effects of gravity can also be a problem. We will ignore the gravity problem in this analysis.
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 equilibrium x position that a particular mass has when no waves are present (as illustrated in Fig. 2.2). Furthermore, note that the distortion, ξ, of a particular mass is the distance that mass is from its equilibrium position, from its position when no waves are present. 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. It is also true that the spring constant for a unit s is unaffected by wave motion.
Note that the variable η for the previously solved mass-on-a-spring problem was only a function of the time coordinate t. We wrote the solution for that as η(t).
Because the linear wave system we are now considering has two independent coordinates, s and t, a solution for this is written as ξ(s,t).
We need to broaden the Lagrange equation eqn (1.2) to include derivatives with respect to both coordinates s and t [reference]:
, (2.1)
where
is defined to be
(i.e with respect to time) and ξ ' is similarly defined to be the other derivative,
, i.e. the derivative with respect to the generalized coordinate s. Another change we've made is to replace the simple Lagrangian L with the Lagrangian densityL_{1} which is just the Lagrangian per unit length, as is necessary for a continuum wave medium.
In this wave medium of masses and linear springs the Lagrangian density is:
, (2.2)
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 non-stretched length x_{0} , per non-stretched length (i.e. per unit s). We can relate this to the distortion in an incremental system as:
, (2.3)
or in a continuum system we can write this as:
, (2.4)
where, again, ξ ' is defined to be
. This makes our Lagrangian density become:
. (2.5)
The Lagrange equation now becomes:
. (2.6)
Note that we could have derived (2.6) using F = ma in terms of the s coordinate.
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 specifying the resting positions of the masses, instead of the more usual dynamic positions of the masses x . The best known solutions of (2.7) are sinusoidal, such as:
. (2.8)
where κ is the wavenumber (radians per meter of s) and ω is the angular frequency (radians per second) of the wave. We can relate κ and ω to m_{1} and k_{1} with the simple relation:
, (2.8')
which gives the phase velocityv_{p}, 2nd ref. This equation is derivable from a simple but careful examination of (2.6) and (2.8).
Note that (2.8) is an exact solution, not just a small amplitude, linearized solution. In contrast the versions of (2.7) and (2.8) written in terms of x instead of s are only valid in the small amplitude regime.
Energy and momentum
Using (2.8) and the first term of (2.5) we get the kinetic energy in the wave:
, (2.8a)
which we see has an average value of
. We have used the trig identity sin^{2}x = ½ − ½cos2x.
Similarly, using (2.8) with the second term of (2.5), the potential energy becomes:
, (2.9)
which has an average value of
. A little algebra will show that the kinetic energy equals the potential energy (use v = √k_{1}/m_{1} = ω/κ for the wave velocity with respect to the coordinate s). The generalized momentum density in our sinusoidal wave (eqn. (2.8) ) is given by:
. (2.10)
Because the coordinate s is attached to the masses themselves, the s density, the mass per unit s, does not change due to a passing wave. The generalized momentum is thus sinusoidal and averages to zero. That is
.
We might ask if this means that the regular momentum averages to zero.
1. The regular momentum density is the product of mass density m_{1} times velocity v of the masses.
2. First,
is the change in displacement of a mass with respect to time and thus is just the regular velocity.
3. The mass density is mass per length of string.
4. The average momentum density is given by:
, (2.10a)
because the average s-momentum density is zero.
So we conclude: yes, the regular momentum density averages to zero. We verified this finding numerically in an Octave simulation. The results are shown in Table 2.1 below.
Effect of changing from s generalized coordinate to x Cartesian coordinate
Fig. 2.3. Graph showing velocities of the masses as a function of the two horizontal coordinate systems: of the generalized coordinate s (in blue) and of the Cartesian coordinate x (shown in green). While the blue trace is a normal sinusoid, the green trace is a distorted sinusoid. The green trace is narrower in the positive half cycle (where the density is greater) and broader in the negative half cycle where the density is less. With respect to average momentum the density variations cancel the distortion effect and produce zero average momentum in these waves.
The density in terms of x equals m_{1}/[1 - dξ/dx] which is plotted in red. While the density versus x varies over the wave cycle, the density versus s is constant and equals m_{1}.
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 spatial coordinate x. If we have a nice sinusoidal solution in terms of s, a graph of it will appear sinusoidal 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, i.e. somewhat warped by horizontal displacements. 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 ξ is the displacement from equilibrium.
In terms of momentum, we have equal masses in the half cycles labeled a and b. In terms of the graph having x as its axis (the blue one), we see that the upward half cycle is compacted and has a width of "b". This decreased width is compensated for by an increase in mass density during this half cycle shown by the red line. The opposite is true for the lower half cycle: it has increased width and decreased density. Thus the two effects, the width and the density, cancel leaving equal and opposite momentums in the two half cycles.
A reader might also note that a given physical mass, at a specific time, has a displacement ξ and
velocity v = dξ/dt independent of which coordinate system (s or x) we work in.
Computer simulations … energy and momentum
The 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 Hooke's 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.
In order that we test for sinusoidal waves it is important that we try to achieve single frequency results in the simulations. After all, a wave that is a pulse will transport net momentum (and mass) since it only has a positive part and no compensating negative half cycle. We minimize the non-sinsuoidal waves by exciting the setups using a pure sinusoidal excitation. The ring resonator is somewhat better in this regard because the excitation used there is an extremely long sinusoid. On the other hand, use of the ring resonator requires us to determine and use the exact resonant frequency of the ring resonator for the excitation signal.
Computer simulation results
Progression of longitudinal waves in a linear array
Slow build up of longitudinal rotating waves in a ring resonator
⇑ Fig. 2.4. In the first simulation, a burst of waves were introduced on the left side and observed to propagate towards the right. The simulation ceased when the waves reached the right side. The calculations showed that the potential and kinetic energies were equal. They also demonstrated that the momentum was orders of magnitude smaller than these energies divided by their velocity (the commonly held relationship) and certainly consistent with zero as derived above. Some numbers from this simulation are reported below in Table 2.1.
⇑ Fig. 2.5. In the second simulation, we used a ring resonator setup, where waves traveled round and round the ring, slowly building up. This simulation also showed the potential and kinetic energies to be equal while the momentum was orders of magnitude smaller than the energies divided by the wave velocity. The actual simulation used hundreds of masses around the ring. Some of the results from this simulation are reported below in Table 2.2.
This is a flash animation. To see the action mouse over the image, off to suspend it and click on it to restart it. This animation does not show the slow build up of rotating waves but only the propagation of constant amplitude rotating waves around the ring resonator.
⇑ Fig. 2.6. Progression of small amplitude longitudinal waves in linear simulation setup (shown above). This graph shows the wave at a progression of times, during and after its launch. The length of the simulation medium is 10. An animation of this is shown below in Fig. 2.8.
⇑ Fig. 2.7. Results for the ring resonator. The above graph shows snap shots of longitudinal waves traveling around the ring taken at equal time intervals after a long time interval of build up. The horizontal axis is labeled in terms of distance around the circumference (total circumference equals 10) while the y axis is the wave amplitude. These waves propagate similar to traveling waves and should have similar momentum. The advantage of this simulation setup is that it avoids the effect of short sinusoidal bursts and instead uses continuous sinusoidal excitation. An animation of this is shown below in Fig. 2.9.
Fig. 2.8.⇑ Flash animation of a wave burst propagating on a linear medium, such as shown in Fig. 2.4 above. The ξ displacement is plotted on the y axis and does not mean to imply that the displacement is in the y direction. Mouse over the graph to see the wave propagation. Mouse off it to suspend it and click on it to restart it.
Fig. 2.9.⇑ Flash animation of a rotating wave mode in a ring resonator, shown in Fig. 2.5 above. The ξ displacement (longitudinal - in the direction around the ring) is plotted on the y axis. Mouse over the graph to see the wave propagation. Mouse off it to suspend it and click on it to restart it.
Note that the wave at 2π radians has the same value as that at 0 radians as is required for a circular resonator (since 2π radians is physically the same place as 0 radians). Note also that the waves propagate in the positive direction around the resonator, much the same as traveling waves would. Note also the slow steady build up of the waves due to the exciting source. In the actual Octave simulation, the build up took even more cycles of the resonance.
Table 2.1. Octave calculated values of energy and momentum at four different times during the wave propagation in a linear array. Wave velocity was set equal to 1. These results show that the average momentum is much less than the wave energy divided by the wave velocity, consistent with zero momentum. As in many numerical simulations, the results are unitless, i.e. normalized to some arbitrary value.
Table 2.2. Octave calculated values of energy and momentum at four different times during the wave propagation in a circular array. Wave velocity was set equal to 1. These results show that the average momentum is much less than the wave energy divided by the wave velocity. The wave amplitude was different than in Table 2.1.
Energy in the wave
5.53, 5.59, 5.61, 5.77
Energy over wave velocity
5.53, 5.59, 5.61, 5.77
Momentum in the wave
1.40e-014, 1.40e-014, 1.40e-014, 1.40e-014
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.
With static tension added to the spring
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 displacement from equilibrium.
Assuming a reasonably small sinusoidal term for the purposes of converting (2.13) between the two coordinate systems s and x, we can approximate ξ as being dominated by the last static stretching term in (2.12) and ignore the wave induced stretching:
, (2.14)
which means the relation between s and x becomes:
. (2.15)
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:
. (2.16)
We can find the wave velocity in the x coordinate system by transforming an x interval into an s interval using (2.15):
. (2.17)
This is the same result we would get using v_{x} = √k_{x}/m_{x} and (2.16).
In the limit where T_{0} » k_{1} these become:
and
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 deal with in Section 4 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 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.
3. Mixed longitudinal and transverse waves in a linear medium
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:
. (3.1)
The length change of the springs from a horizontal unstretched length is:
. (3.1a)
This makes the potential energy density:
, (3.2)
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 spatial 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:
. (3.3)
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:
. (3.4a)
The second Lagrange equation (for the y direction) is:
. (3.4b)
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) ):
, (3.5)
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) ):
. (3.6)
The Lagrange equation for the vertical motion becomes:
. (3.7)
For small amplitude waves, the last term can be ignored, leaving the standard wave equation with a wave velocity of
, (3.8)
which is the commonly derived velocity for transverse waves on a string.
3. Strings with a zero unstretched length are weird in that they represent a setup where all the mass elements occupy the same spot before any waves are created and oscillate through each other when waves are present. We can model these in our math but physically making such a setup would require some creativity. Such a system is of interest because of its particularly simple wave solutions.
One computational complication is that s for all masses is the same, all have s = 0 . We can avoid this by letting s for the various masses increase in extremely tiny increments, not actually all 0 but close enough to be mathematically distinct from each other.
The two Lagrange equations for x and y directions become:
. (3.11)
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. One way to avoid this is 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).
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.
References
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 non-linear 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.
If the string cannot be stretched there is no potential energy in stretching but because these are transverse waves there will be extra string length required by the wave. This extra length will presumably come from flexing of the end supports of the string or some sort of pulley and mass at the ends. This means that there must be longitudinal motion which we will address farther down.
The general expression where there is both longitudinal and transverse motion is:
, (4.1)
where
is the amount of extra string needed because of the transverse wave per unit s as we derived in (3.1a).
If we assume that the wave is mostly transverse then η'≫ξ ' and the potential energy becomes:
Modifying (3.3) with the above approximations we have:
. (4.4)
This is basically the Lagrangian that leads to the famous “wave equation” as we saw in (2.5) with η exchanged for ξ and T_{0} exchanged for k_{1} . So we should expect the same solution as shown in (2.8) after making these exchanges:
. (4.5)
The constants κ and ω are related to m_{1} and T_{0} as
, (4.6)
where there is a remaining adjustable constant which we can take as the exciting frequency ω. These ratios give the phase velocity v_{p} of the waves.
Horizontal motion
As stated above, a transverse wave in this medium will have a horizontal component as well as mass transfer and horizontal momentum. To show this we start with calculating ∂ξ/∂s picking from the above equations (4.2)-(4.3):
. (4.7)
Our solution, i.e. (4.7), is in the form of a d'Alembert solution of the wave equation, i.e. the solution is a function of κx−ωt. Using standard d'Alembert trickery we note that (4.7) is in the form of a function f(u) where u(κ,ω) = κs−ωt. This allows us to use the chain rule of derivatives to do the following:
. (4.8)
Combining these to eliminate ∂f/∂u and replacing f with ξ we get:
. (4.9)
But ∂ξ/∂t is just the horizontal velocity of the mass units and we can see from the right hand expression has a constant term. The constant term represents a drifting of the masses in the positive x direction caused by the wave. Thus in this wave system “pure” transverse waves have a horizontal double frequency component and this component carries mass. The carrying of mass at a certain velocity means the waves carry momentum.
In practice, the build up of mass at one end of a finite string will result in a steady counterflow which will counteract the above effect. On the other hand the mass and momentum effect could be observable if short wave bursts were used.
In fact if we think about a short burst of transverse waves it's clear that they contain extra string length required in order to allow a transverse sinusoidal shape. This extra string length means extra mass in the waves. If we have a burst moving along, it's clear that this extra mass will need to move along with the burst meaning there is mass transfer. Mass in motion also means momentum transfer.
We could easy make the counter 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.
Energy in the waves
All waves transfer energy. The equations for energy flow of both the kinetic and potential kinds for waves on a unstretchable string are:
. (4.10)
One interesting fact is that the kinetic energy density equals the potential energy density. Also, we used (4.6) in the last line of the equations.
. (4.11)
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:
Comparison of adiabatic ideal gas and linear spring
Fig. 5.2. The blue line shows the relative pressure of a gas parcel undergoing adiabatic compression versus x, the motion of a compressing piston. This is compared with the restoring force for an equivalent linear spring (in green), made to be equivalent around the point x = 1. We see that while the first derivatives (the slopes) of the two curves are the same around this point, their second derivatives (their curvatures) differ greatly.
, (5.1)
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:
, (5.2)
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
, (5.3)
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:
. (5.4)
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:
, (5.5)
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
. (5.6)
We substitute this into
, (5.7)
to get
, (5.8)
where
. (5.8aa)
Rearranged this becomes:
. (5.8a)
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.
Higher power - second order calculations
Next we investigate second order behavior of the waves predicted by (5.8a) using a power series solution of (5.8a).
or when x = 1 and y = ε and ε is much smaller than 1, this becomes:
, (5.9a)
to the second power in ε.
When we apply (5.9a) to the right hand side of Eqn. (5.8a), we get:
. (5.10)
This makes (5.8a) become:
. (5.10a)
A power series solution of sinusoids up through square terms is:
, (5.11)
or
, (5.12)
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:
. (5.13)
Next we do the right hand side of (5.10a) remembering that we are only keeping terms of power two or less. The right-most 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:
, (5.14)
which is essentially the same as the previous formula (5.13).
The important part of the other factor on the right hand side is:
. (5.15)
After multiplying two factors together, (5.14) and (5.15), the right hand side of (5.10a) becomes:
. (5.16)
Equating the linear terms of the two sides (in (5.13) and (5.16) ) we get:
, (5.17)
which gives the relationship between frequency and wavenumber:
, (5.18)
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:
, (5.19)
which yields:
, (5.20)
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:
, (5.21)
again yielding the same relationship for the third time.
Equating the sinXcosX terms yields:
, (5.22)
which can be solved:
. (5.23)
Equation (5.23) indicates that we expect an additional sinXcosX term in proportional to the amplitude of the fundamental cosX term squared. Trig tells us that sinXcosX = 0.5sin2X which is double the freqency of cosX. That means that in the presence of a high power fundamental (i.e. large a_{1}) we expect a double frequency term. Fig. 5.3 shows how this will result in peaks in the wave crests.
This is the beginnings of the formation of shock waves. 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.
Momentum is given by
:
. (5.24)
The cos2X factor averages to zero, implying that the average momentum in the s coordinate system is zero (the a_{1} term also averages to zero as we have seen in previous sections.) 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 has only been done to second order in the sinusoids. Higher order calculation, applicable to very intense sound, might yield a non-zero 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.