All of our discussions of waves have been ones with stationary sources and stationary observers. We had no reason to have these things in motion, but as it turns out, the frequency of sound perceived will change if one or the other is in motion. [br][br]The same is true for light, but the derivation is harder and involves concepts from relativity, so we will stick to sound for now. It is perhaps serendipitous, however, that for speeds of relative motion that are somewhat smaller than around c/10, the results for light are the same as those for sound. And there are very few objects in nature that move even a tenth the speed of light.[br][br]As a general rule about the Doppler shift, when objects approach one another the pitch (frequency) is higher than when stationary, and when objects are separating, the pitch (frequency) is lower. Depending on whether the source or the observer (or both) is doing the moving, the derivation is different. Before we go into the mathematics, however, please take a look at the animation below to get a sense of the phenomenon.

When a source of sound is moving, it will literally chase after its own sound waves in front, and run away from the sound waves it's producing behind itself, as you can see in the animation above. While this doesn't affect the period of the sound from the source's perspective, it will affect the wavelength. Certainly there is a shortening of the wavelength in front of the object and a lengthening behind the object. The wavelength would be the distance between the rings in the animation. [br][br]An observer measures waves unaltered in speed. In front of the source they would hear a higher frequency or pitch than with a stationary source, and a lower frequency behind the object due to the altered wavelength. As expected, from the perspective of the observer, the period is altered as well.[br][br]We should also ask what happens if the observer is off to the side at some angle [math]\phi_s[/math] with respect to the forward direction of the source. In this case, nothing changes conceptually. What I mean is all that matters is how quickly is the source approaching the observer. You can see in the interactive diagram below that the rate of approach is just [math]v_S\cos\phi_s[/math]. Move the point representing the observer to a different location and see the effect. We will use this now to derive the frequency that the observer will hear.

I will denote the observed frequency [math]f_O[/math] and the frequency produced by the source [math]f_S[/math]. We discussed above that the source is chasing after its own sound in the forward direction, but runs away from it in the backward direction. At the sides (right angle to the velocity of source) it will do neither. This notion of chasing after or running away from its sound is dependent on angle, and the speed at which it "chases" is the expression in the diagram [math]v_S\cos\phi_s[/math] that we just determined. [br][br]The period T[sub]S[/sub], or time between pulses of sound going out from the source remains fixed whether the source is moving or not. The speed of sound also remains fixed for a stationary observer. Therefore [b]the change in pitch the stationary observer measures is really due to a change in the wavelength of the sound[/b]. The wavelength for a stationary source and observer is given by [math]\lambda_S=\frac{v}{f_S}=vT_S[/math]. Notice here that the velocity without a subscript is the speed of sound in air, not the speed of the source of sound - two entirely different things! [br][br]If the source is chasing after its own sound at the rate described above, then during the interval of one period [i]T[sub]S[/sub] [/i]the source will have traveled by a distance equal to [math]v_S\cos\phi_sT_S[/math]. This means that the wavelength will be reduced by this amount as compared to that for a stationary source. So the observed wavelength is [math]\lambda_O=vT_s-v_S\cos\phi_sT_S=\left(v-v_s\cos\phi_s\right)T_s[/math]. Combining this equation with the expression for wavelength for a stationary source above gives us[br][br][center][math]\lambda_O=\frac{\lambda_S\left(v-v_S\cos\phi_s\right)}{v}=\lambda_S\left(1-\frac{v_S\cos\phi_s}{v}\right).[/math][/center]Written in terms of frequency by just inverting the equations, this gives us:[br][br][center] [math]f_O_{ }=f_s\left(\frac{1}{1-\frac{v_S\cos\phi_s}{v}}\right)[/math].[/center]

Let us now consider the situation with a stationary source S, but a moving observer. The speed at which the observer approaches the source is [math]v_O\cos\phi_o[/math] where [math]\phi_o=0[/math] when the observer is moving directly toward the source and [math]\phi_o=\pi[/math] when moving directly away from the source. What changes in this case versus the prior one is that the wave speed in the observer's frame of reference is higher. At the beach, the waves come at you faster when you are running at them than when you are just standing still. However, nothing changes about the wavelength of the ocean waves. Nor will it for the sound waves. Please notice that this is different than the case with the moving source in which the speed of the sound was unchanged and the wavelength was altered.[br][br]The easiest way to approach this derivation is to look at frequency from the start. With a stationary source and observer, we find [math]f_S=\frac{v}{\lambda_S}[/math]. The moving observer will measure a different frequency [i]f[sub]o [/sub][/i]on account of an increase in speed when moving toward the source. This gives [math]f_O=\frac{v+v_O\cos\phi_o}{\lambda_S}[/math]. Combining these expressions so that we can relate the observed frequency to the source frequency gives us [br][br][center][math]f_O=f_S(1+\frac{v_O\cos\phi_o}{v}).[/math][/center][br]As mentioned above, when the angle [math]\phi_o=0[/math], the observer is understood to be moving directly toward the source of sound, and when [math]\phi_o=\pi[/math] the observer is moving directly away from the source. Clearly this implies an increase in frequency when moving toward a sound source and a decrease when moving away.

In the case that both the observer and the source of sound are moving, we will have both of the above derivations in operation. For both effects at the same time, we need to multiply the effects. Think about it like this: If a source is moving toward you such that you hear a higher frequency on account of the shortened wavelength, and then you decide to run toward the waves such that the wave speed is increased, then both of our derivations apply. The result for both in motion, and therefore the most general expression is: [br][br][center][math]f_O=f_s\frac{1+\frac{v_o}{v}\cos\phi_o}{1-\frac{v_s}{v}\cos\phi_s}.[/math][/center]

In the case of a moving source that exceeds the speed of the waves that it is producing, a phenomenon called a [b]sonic boom [/b](shock waves) will be produced. This is a consequence of waves building up in front of an object without an ability to escape. The interactive animation of the rings will allow you to see this if you set [math]v\ge c[/math]. When [math]v>c[/math], a [b]Mach cone[/b] of superposed sound waves is produced. This is a 3D analog to a 2D wake produced behind a boat, which forms when the boat travels faster than surface water waves on a lake or ocean. It turns out that boat wakes don't follow the same physics due to strong dispersion (wave speed being dependent on wavelength), but the appearance of such wakes is familiar for the sake of analogy.[br][br]The [b]Mach number, M[/b] is the ratio of the speed of the source to the speed of the sound wave: [math]M=\frac{v_S}{v}[/math]. The angle of the Mach cone can be calculated very easily if you consider that the sound wave will propagate in all directions a distance vT, while the airplane will travel parallel to its own velocity a distance v[sub]S[/sub]T. Simple geometry tell us that the half angle of the cone is given by [br][br][center][math]\sin\phi=\frac{v}{v_S}.[/math][/center] [br][br]The same equation will not be true for the wake behind a boat. The angle of a boat wake (with rare exceptions) is fixed at a half angle around 19.5 degrees. This is due to the interference effects of many wavelengths governed by dispersion - the boat acting like the source of many different wavelengths, the longest of which travel fastest. We will not go into the physics of this phenomenon. Here is a good article on the nuance of boat wakes: [url=http://www.fast.u-psud.fr/~moisy/papers/2013_rabaud_prl.pdf]Kelvin or Mach angle?[/url] The only condition for which the Mach cone is seen is the shallow water limit - shallow relative to the wavelengths.[br][br]If we now go back to the Mach cone, in the animation above it looks like the waves build up on the Mach cone in a fashion that should make their pressure go infinite. The problem with this is that nature does not, as far as we know, contain any infinities. What really happens to the air on the Mach cone is that under the large pressure it will typically condense - at least the moisture content - to a liquid. This can be seen beautifully in the photo of the plane below. It is worth noticing that there is an obvious Mach cone at the location of the jets (where the majority of the sound is produced), but also a little half-cone is produced from turbulent air right above the cockpit. The sound of rushing air over the pilot's canopy is playing the role of a supersonic sound source in addition to the sound of the engines.

The sound of a gunshot is familiar. If you think it sounds that way because it's a little explosion of gun powder, then in a sense you have it backwards. The reason that an explosion sounds like an explosion is that projectiles or shrapnel from the explosion exceed the speed of sound. A bullet has the sound it does only because it is traveling faster than sound and therefore produces a sonic boom.[br][br]If you shop for rifle ammunition you will occasionally see ammo like subsonic 22LR bullets. These are 22 caliber long rifle subsonic bullets. They travel a few hundred feet per second slower than regular 22LR ammo, but are still very lethal. The difference is that when fired they don't make the familiar loud gunshot sound. Rather they sound like a low-powered air rifle. Sounds can be deceiving.[br][br]If you ever have a wet towel and snap it just right and get the sharp cracking sound, then you managed to make the tip of the towel exceed the speed of sound. The same is obviously true of a bullwhip.

It is possible to travel faster than a light wave. The catch is that this will never happen in vacuum. If you recall from our discussions of optics, light travels slower in any medium by a factor of the medium's refractive index. For instance, in glass, the speed of light is roughly v=c/1.5. [br][br]This fact begs the question: What if a particle exceeds the speed of light in some substance? It turns out that this happens all the time in nuclear reactors and in the upper atmosphere. The result, if the particle is charged, is not a sonic boom, but an optical boom in the form of a flash of light called Cherenkov radiation. This radiation tends to have a characteristic blue glow to it. The light quickly bleeds energy from the particle until it is traveling under v=c/n. Under that speed no more light is emitted.