© *Matt Strassler [August 29, 2012]*

*This is article 4 in the sequence entitled Fields and Particles: with Math. Here is the previous article.*

**Equation for the Ball on a Spring (review)**

In the earlier article about the ball on a spring, we first figured out the formula for the oscillatory motion of the ball,

- z(t) = z
_{0}+ A cos [ 2 π ν t ]

and then found an **equation of motion** for which that formula was a solution,

- d
^{2}z/dt^{2}= – K/M (z – z_{0})

Here

- d
^{2}z/dt^{2 }means the change in time**of**the change in time of z(t) - K is the strength of the spring, M is the mass of the ball, and z
_{0}is the equilibrium position of the ball - ν = √ K/M / 2π

A key step in obtaining the last equation for the frequency in terms of K and M was to compute d^{2}z/dt^{2} for the ball’s oscillatory motion z(t) = z_{0} + A cos [ 2 π ν t ] ; we found

- d
^{2}z/dt^{2}= – (2 π ν)^{2}(z – z_{0})

If at first reading you skipped these calculations that lie at the end of the Ball on a Spring article, you may want to go back and read them now (or at least skim through them) before attempting to read what lies ahead.

**Equation of Motion for Waves**

Now we want to do the same here for waves as we did for the ball and spring. We’ve found a formula for the shape and motion of a wave that is oscillatory **both** in space **and** in time.

- Z(x,t) = Z
_{0}+ A cos (2π [ν t – x/λ]) = Z_{0}+ A cos (2π [t/T – x/λ])

What equation of motion has such a formula among its solutions? We can guess our way to the answer; surely it involves **both**

- d
^{2}Z/dt^{2}, the change in time of the change in time of Z(x,t) ,**and** - d
^{2}Z/dx^{2}, the change in space of the change in space of Z(x,t)

And we would therefore naturally guess the equation should be something like

- C
_{t}d^{2}Z/dt^{2}+ C_{x}d^{2}Z/dx^{2}= C_{0}(Z-Z_{0})

where C_{t} , C_{x} and C_{0} are constants. Note that if C_{t} = 1, C_{x} = 0 and C_{0 }= -K/M we get back the equation for the oscillating ball on the spring. What are the constants in this case?

Well, first, we can always take C_{t} = 1; if instead you wanted, perhaps, to take C_{t} = 5, I would just tell you to divide the whole equation by 5, which would give you an equivalent equation with C_{t} = 1 and new values for the other constants.

But now that we’ve done that, it turns out that the values for C_{x} and C_{0 }take different values for the waves in different physical systems. [Other physical systems actually can have more complicated-looking wave equations, too; but we won’t consider them here.] We’re going to study two different classes of waves that differ qualitatively in what’s happening with these constants.

For both classes we’ll take C_{x} negative, with C_{x} = -c_{w}^{2} ; here c_{w} [short for **c**elerity of the **w**ave] represents the speed at which very high-frequency waves travel.

But the two classes will differ in that in one class, Class 1, we’ll take C_{0 } to be negative, C_{0} = – (2 π μ)^{2}, while in Class 0 we’ll take C_{0} to be zero.

Now let’s explore the properties of the waves for these two classes of equations. Before we start, there’s one calculation to do, though we’ve already done it before.

**A Quick Calculation**

For our infinite wave,

- Z(x,t) = Z
_{0}+ A cos (2π [ν t – x/λ]) = Z_{0}+ A cos (2π [t/T – x/λ])

we’re going to need to know d^{2}Z/dt^{2} and d^{2}Z/dx^{2}. Well, we already showed carefully in an earlier article that for the ball on the spring, whose motion is z(t) = z_{0} + A cos [ 2 π ν t ], we have d^{2}z/dt^{2} = – (2 π ν)^{2} (z – z_{0}). The change in time gives us one factor of 2 π ν, and the change in time ** of** the change in time gives us two factors; and in addition there’s an overall minus sign. It is no surprise, then, that

- d
^{2}Z/dt^{2}= – (2 π ν)^{2}(Z – Z_{0}) - d
^{2}Z/dx^{2}= – (2 π / λ)^{2}(Z – Z_{0})

every change in time gives us a factor of ν = 1/T (the longer the **period**, the slower the change in **time**), while every change in space gives us a factor of 1/λ (the longer the **wavelength**, the slower the change in **space [**see Figure 3 of the previous article].) If you want to see the little bit of calculus that leads to these equations, click here*. *

Fine point: all of the above derivatives are really partial derivatives

**Class 0: Waves of any frequency, all with equal speed.**

In this class of waves, the equation of motion is

- d
^{2}Z/dt^{2}– c_{w}^{2}d^{2}Z/dx^{2}= 0

If we plug in the formula Z(x,t) for the infinite wave and use the formulas from the calculation we just did, we find

- – (2 π ν)
^{2}(Z – Z_{0}) – (-c_{w}^{2}) (2 π / λ)^{2}(Z – Z_{0}) = 0

Dividing the equation by – (2 π)^{2} (Z – Z_{0}) we then get

- ν
^{2}– c_{w}^{2}/ λ^{2}= 0

Since frequencies, speeds and wavelengths are all positive, we can take the square root to get

- ν = c
_{w}/λ , or if you prefer, λ = c_{w}/ν = c_{w}T

From this formula we learn several things.

- Originally our wave, as we wrote it down initially, could have had any frequency and any wavelength; but the equation of motion forces the two to be related. You can pick any frequency you want for Class 0 waves, but once you do, the wavelength is determined by λ = c
_{w}/ν. - All of these Class 0 waves, no matter what their frequency, travel with speed c
_{w}. This follows from the formula λ = c_{w}T, and from watching Figure 3 in the previous article. Watch the wave as it goes through one oscillation cycle during the time of a single period T; what happens? The wave looks the same after T, but each crest has shifted over to where the adjacent crest used to be — a distance λ away. This tells us that each crest moves by the distance λ during the time T — one wavelength for each period of oscillation — and that means the crests are moving with speed λ/T = c_{w}. Note this is true for all frequencies and their corresponding periods and wavelengths! - Just as with the ball on the spring, the amplitude A of these waves can be of any size, large or small. And this is true no matter what the frequency of the wave is.

**Class 1: Waves above a minimum frequency, with unequal speed.**

For this class of waves, our equation of motion is

- d
^{2}Z/dt^{2}-c_{w}^{2}d^{2}Z/dx^{2}= – (2 π μ)^{2}(Z-Z_{0})

If we plug in the formula Z(x,t) for the infinite wave and use the quick calculation above, we find

- – (2 π ν)
^{2}(Z – Z_{0}) – (-c_{w}^{2}) (2 π / λ)^{2}(Z – Z_{0}) = – (2 π μ)^{2}(Z-Z_{0})

Dividing the equation by – (2 π)^{2} (Z – Z_{0}) we then get

- ν
^{2}– c_{w}^{2}/ λ^{2}= μ^{2}

Since frequencies, speeds and wavelengths are all positive, we can take the square root to get

- ν = [ (c
_{w}/ λ)^{2}+ μ^{2}]^{1/2}*(Recall y*^{1/2}means the same as √y)

Now this formula is quite different than the one for the Class 0 waves, and its implications are different too.

First, the equation of motion says there is **a minimum allowed frequency. **Because (c_{w}/λ)^{2 }is always positive,

- ν = [ (c
_{w}/ λ)^{2}+ μ^{2}]^{1/2}≥ μ !

You can approach ν = μ by taking λ larger and larger; so for very long wavelengths, the frequency approaches μ, but it never gets smaller than that. That wasn’t true for Class 0 waves; they had ν = c_{w} / λ, so for them, as you take λ larger and larger toward infinity, ν gets smaller and smaller toward zero. Even for Class 1 waves, though, you can have any ν larger than μ*.*

Second, we found an argument that all Class 0 waves have the same speed; but that argument doesn’t work for Class 1. The only place it (approximately) works is if we take ν much, much greater than μ, which we can make happen by taking λ very small (and so 1/λ very big.) In this case,

- ν = [ (c
_{w}/ λ)^{2}+ μ^{2}]^{1/2 }≈ c_{w}/ λ

Thus *at very high frequency and small wavelength*, Class 1 waves have (approximately) the same relation between frequency and wavelength as Class 0 waves, and so by the same argument we used for Class 0 waves, such waves travel (approximately) at the speed c_{w}.

One thing that is true of Class 1 as well as of Class 0 is that the amplitude A can be anything you like, large or small, independent of the frequency.

*Fine point — actually, more than just fine. I have just pulled some wool over your eyes; perhaps you’ve noticed. I never actually calculated the speed of the Class 1 waves. That’s because there is a very tricky subtlety lurking here. For Class 0 waves I calculated their speed by looking at how fast their crests move. That works only because for Class 0 waves, waves of all frequencies travel at the same speed. But for Class 1, or any other class where the waves of different frequencies travel at different speeds, the speed of a realistic wave is not given by the speed at which its crests move! It turns out the crests move faster than c _{w}, but the speed of the wave is actually slower than c_{w}. To see this involves some logic that is not readily obvious and involves the difference between “group” and “phase” velocities. I am going to skirt this subtlety for now; all I am doing here is alerting you that it exists, so that you aren’t left with a misconception. [At some point I’ll explain it with a short article, but not soon.]*

**Some Final Comments About Classical Waves**

There are many familiar examples of Class 0 waves, including sound in air or water or metal (where c_{w} is the speed of all sound waves in that material), light and other electromagnetic waves (for which c_{w} = c in vacuum), and the waves on a rope or string, as shown in Figure 2 of the previous article. For this reason, Class 0 waves are the ones that are taught in beginning physics classes. I know of no readily available examples of Class 1 waves in daily life, but we’ll soon see that Class 1 waves are just as crucial in the universe.

We had a nice formula E = 2 π^{2} ν^{2} A^{2} M for the energy of a ball of mass M on a spring, and the formula for other oscillators depends in detail on what they are but has a similar form. But we haven’t discussed energy at all for waves. That’s partly because we’ve been looking at waves with an infinite number of crests to keep the math simple. Intuitively there should be some energy stored in the motion and shape of each crest and trough of the wave, so with an infinite number of crests and troughs, the amount of energy in such a wave is infinite. However, there are two ways around this. Again, the exact formulas depend on what type of wave we’re looking at, but let’s specifically consider the Class 0 waves on a rope.

- The amount of energy
of the wave (stored or carried in the region between the point x and the point x + λ) is finite, and equal to 2 π**per wavelength**^{2}ν^{2}A^{2}M_{λ}, where M_{λ}is the mass of a rope segment of length λ . - No realistic wave is infinite in length; like the pulse of crests and troughs shown in Figure 2 of the previous article, any wave will actually be finite, with a finite number of crests and troughs. If the wave extends over a region of length L, so that it has L/λ crests, then the energy carried by this wave will be about 2 π
^{2}ν^{2}A^{2}M_{L}, where M_{L}is the mass of a rope segment of length L; this is just L/λ times the energy per wavelength above.

For waves that aren’t on ropes, the details of these formulas will be different, but the energy per wavelength is always proportional to ν^{2} A^{2}, as for any simple oscillating system.

There’s a very special and interesting Class 1 wave that doesn’t exist for Class 0, and that’s the case where ν = μ , its minimum value, and λ = infinity. In this case the wave takes the form

- Z(x,t) = Z
_{0}+ A cos (2π μ t )

a wave which is independent of x — at all times, Z(x,t) is a constant across all space — and Z oscillates with time exactly the same way as a ball on a spring with frequency μ. This stationary wave, illustrated in Figure 2, will turn out to be quite important in what follows.

**On to Quantum Waves**

For the ball on the spring, the difference between the classical and quantum systems was that in the former case the amplitude could be anything large or small, and the energy too; but in the quantum case, the amplitude and energy were quantized. This is true for any similar oscillating system. Perhaps, then, we can anticipate that the same is true of waves…

The equation for the Class 0 is traditionally called the wave equation or d’Alembert equation; does the equation for Class 1 have a proper name as well?

Hey,

Is there any possible way your website could be modified to use LaTeX? I love what you are doing, and I’ve been following you for quite a while, but in an article like this it’s hard to follow all the math sometimes.

Regardless, keep up the great work!

Why would LaTeX help you? I’m not understanding your request.

i could be wrong, but i think aidan finds the equations difficult to read. using the square root symbol and a differential notation (especially of the second order) that isn’t constrained to a single line of text are the places i see this as particularly useful.

I like these articles and the animations are a lot of fun 🙂

However, I agree with Aidan that LaTex equations are much easier to read.

The particular notation used here, with many 2\pi s still floating around for example instead of using wavenumbers and angular frequencies, makes it sometimes a bit harder to connect what I read here whit what I’ve read or seen elsewhere.

On the 2 pi’s: This was a judgment call. Angular frequencies and h-bars are convenient for physicists, not so much for non-physicists. I decided not to use h-bar omega and h-bar k ; h nu and h/lambda are more intuitive. You could make a good argument either way.

On the formatting: also a judgment call about how my time would be spent…

I wonder if different browsers ( or missing plugins ) are causing it to display differently.

It looks fine to me. i.e. I see a single Greek letter and not the two letter combination “pi”.

Talking about technicalities of the site itself, I would appreciate it a lot if the comments could be made editable by their author to fix embarassing typos later for example …

maybe there’s a way to do that… it *is* a problem…

In the ball and spring analogy I fail to see what causes the sideways motion – it would appear to be only an illusion and if a wave starts from a single point, what in the physical world determines if this movement should be to the left or right?

Another thing that has always puzzled me is when 2 waves interfere and “cancel each other out” apparently resulting in “emptiness” what happens to all this information that describes them? – it seems to contradict the first law of thermodynamics?!

Thanks for your time.

Zbynek

In the case of Figure 2, the wave moves to the right because it was the left end of the rope that was wiggled.

If instead you had wiggled the middle of the rope, you’d get two waves moving away from the center of the rope; one to the right of center moving to the right, and one to the left of center moving to the left. Try it!!

Two waves that aren’t traveling in exactly the same direction (so that they can really be identified as two different waves) never cancel each other out perfectly. They may cancel in all places some of the time, and they may cancel in some places all of the time, but they never cancel in all places all of the time. Cancellation is destructive interference. There will always be constructive interference somewhere else. For example: radio stations will arrange their antennas so that the waves from the antennas have destructive interference in areas of very low population and constructive interference near cities. The power emitted gets redirected from the areas of destructive interference and instead gets spent in the areas of constructive interference. In this way they can reach more people with a stronger signal without wasting power on unpopulated areas.

The only way to have perfect destructive interference is to take one wave that travels in the a certain direction and take another wave that travels in the same direction with its troughs lined up so as to exactly cancel the others crests. But in that case nothing ever moves; there’s no wave. You don’t have cancellation of two physical things, you have no physical thing at all! the cancellation is just there in the math.

Does that implay that unless confined by boundries, waves are always created in pairs moving in opposite directions?

Concerning the destructive interference: yes they have to be precisely aligned and perfectly out of phase but do not have to travel the same distance – so in a thought experiment you could have a light beam traveling from alpha centauri encounter the second wave created in your lab – does it then vanish?

1) If you make a ripple in a pond, the waves go out in all allowed directions. If you burn a match, it can be seen from all directions. However the constraints may not all be boundaries; there may be physical absorption or reflection of waves by materials.

2) Your thought-experiment isn’t well-defined; you have to consider how you did this. The answer depends on how you do it. But energy will be conserved somehow — either your apparatus will absorb the wave from alpha centauri and become a bit hotter, or it will reflect the wave back the way it came, recoiling as it does so. There’s no magical way to make the energy and momentum of the wave disappear.

Nice article – looking forward to next one.

It is clear that Class 1 waves are the ones that can explain particles in quantum physics. Class 1 waves represent wave packets. If we want to represent a particle (like say, an electron) with a Fourier series expansion, what we get is the expression of a wave packet, with the group velocity of the packet being equal to the speed of the particle.

I undestand that Prof. Strassler will get to explain all this, in time (and space!).

Kind regards, Gastón

I fact, with the equation for the packet and a little math, it is very simple to derive the usual expression for the Uncertainty Principle.

Hi Matt,

As you say “there are no readily available examples of Class 1 waves in daily life” unless you happen to be a microwave engineer. I think your class 1 waves can be found in (for example) rectangular waveguides. Such as here in

Kevin Brown’s math pages

http://www.mathpages.com/home/kmath210/kmath210.htm

and here in Dr. Pat Donohoe’s Waveguides and Cavity Resonators course notes (ece3323waveguides.pdf) from

http://www.ece.msstate.edu/~donohoe/ece3323.html

I have a feeling that when I stick a conducting plate on either end of my waveguide then that particular real life example is going to continue to exhibit the sort of physics your next article or two will be discussing? Can’t wait…..

The part that confuses me is the “And we would therefore naturally guess the equation should be something like Ct d2Z/dt2 + Cx d2Z/dx2 = C0(Z-Z0)” I don’t see where the ability comes from to feel free to add those two terms. I look at the x and t as two totally different variables and I wouldn’t have guessed or rather assumed that I can put them together in a linear combination. Makes it easier for sure but where does that come guess from?

Well, this is not a derivation! It is more of a inspired suggestion, and it isn’t unique.

So the question is how we generalize the ball-and-spring case. The form I chose is both the simplest possible form and the most symmetric,

ifwe want it to be similar to the ball-and-spring example.What you should then ask is: what would have happened if we’d combined the time and space derivatives differently? For example, you could have tried

Ct d2Z/dt2 + Cx d4Z/dx4 = C0(Z-Z0)

or

Ct d2Z/dt2 + Cx d2Z/dt2 d2Z/dx2 = C0(Z-Z0)

Or if we didn’t insist on being similar to the ball and spring,

d2Z/dt2 d2Z/dx2 = C0(Z-Z0)

These also have wave solutions with different relations between frequency and wavelength. As you say, they are more complex, but they’re not wrong. However, now more sophisticated considerations have to come in to tell you that the latter isn’t generally a good equation of motion for technical reasons … technical details follow … beyond the scope … etc.

Even more to the point, most of these alternatives will turn out to be inconsistent with special relativity, which is what we need to impose when we get to article 7. The forms I wrote, with one special choice, will turn out to be the simplest wave equations consistent with Einstein’s rules.

In the end, the justification for studying the types of equations I wrote down is that they turn out to be essential in particle physics. They aren’t unique, but they are the simplest.

Dr. Strassler,

I’m a grad student in theoretical physics and am a big fan of your site.

Regarding the above request for LaTex equations, it’s reasonably straightforward in WordPress-based sites (which I believe your site might be?). Enclosing math in tags seems to render things nicely on the site I have for myself. (More info: http://en.support.wordpress.com/latex/)

As hinted above, you probably do not have any interest in formatting things this way, but I thought the information might possibly be interesting or even useful.

Thanks for all the time and effort you put into the site!

Best,

Will

thanks for the info…

Great article! Looking forward to reading the rest.

Although the equations are different (i.e. there is no minimum allowed frequency for one), another example of a Class-1-“like” (which by “like” I mean at least they are both dispersive!) wave would be large scale transverse waves in the atmosphere (e.g., ripples in the jetstream) which in the midlatitudes are responsible for our general day-to-day variability in the weather. The phase speed relation for the simplest type of such a wave (called a Rossby wave) is c – u_bar = -Beta/K^2, where c is the phase speed of the wave, u_bar is the mean west-to-east wind speed, Beta is the gradient of the Coriolis force with latitude (increases with increasing latitude toward the poles), and K^2 is the square of the horizontal wavenumber. Thus, for these types of waves, shorter waves move more quickly, while larger ones move more slowly.

Dan — there are many types of dispersive waves, as you note. Class 1 is a specific example, and is not intended to refer to the more general case, which would include the case you mentioned, as well as many examples in solid state physics, fluid dynamics, etc.

Argh! Quick clarification: the phase speed c for a Rossby wave is actually faster (toward the west) for longer waves *relative to the mean wind* u_bar. But, relative to the ground, because the mean wind u_bar is usually almost always from the west in the midlatitudes, longer and longer waves move more slowly to the east, and may even “retrograde” to the west if they are sufficiently long. Sorry for the mixup!

I have learned a lot from your web site and look forward to more threads. Regarding a real world, lay person available example of Class 1 waves (which have different velocities depending on wave length); it is my understanding that a diffractive prism functions because the different wavelengths of light travel at different velocities while in the prism. Could light traveling in prism be an example (at least for electromagnetic waves in the visible spectrum) of a class 1 wave?

I can’t think of a common Type 1 example either, but I think one could be made.

Start with a Shive wave machine (like the one shown here http://www.youtube.com/watch?v=qUohelhrtl8), but add a weak spring from each bar to the table (set so that the spring is unextended when the bar is horizontal).

Now the ν=μ case is simply the whole machine twisting in unison against the new springs, without ever putting any torsion into the central wire. The ν≅c/λ case is when the wavelength is sufficiently short that the torsion in the central wire completely dominates over the new springs.

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Matt,

Just re-reading this group are articles. They are fantastic and would make a great book.

I was just trying to get a simple intuition for class 0 and class 1 waves. I can imagine with class 0 waves that the rope just wants to be straight and so it changes in proportion to how curved the rope is at a particular point. Thinking about class 1 waves it seems like the rope would also have little springs attached at every point.

Would a lightly sprung mattress exhibit class 1 waves ?

In fact I think I could now give a hand waving explanation to my friends of how the higgs field gives rise to mass. The base of the mattress being the higgs field, the surface being the particle field, the springs being the interaction, the fact that in the limit of long wavelength there is still an irreducible frequency of vibration due to the springs, and so an irreducible quantum of energy/mass even for a stationary particle. ( or something along those lines )