I knew it had to be using the Doppler Effect somehow. You know when you are driving along and you hear a fire engine or police siren behind you, and how the pitch shifts as the emergency vehicle passes you? That's the Doppler effect on sound waves.

What happens is, the wavelength of the sound gets compressed as the siren approaches you. Conceptually, the siren is producing a sound wave that peaks every so often (hundreds or thousands of times each second). In the case of a stationary siren, these wavefronts would be equally spaced, but in the case of a moving siren, the wavefronts are going to be squished together in the direction the siren is going, and spaced farther apart in the opposite direction.

Mathematically speaking, we know the wave equation: v = f, where v represents the velocity of the wave, is the wavelength, and f is the frequency. Let's suppose, to keep things simple, that you (the observer) are stationary, and the siren is moving. The frequency of the wave that you observe is proportional to the velocity of the observed wave divided by the frequency of the observed wave (i.e., f

_{o}= v

_{o}/

_{o}). The velocity of the wave is constant; the siren isn't pushing the air faster, it's altering the location and frequency of the wavefronts. So v

_{o}= v (a constant). How does the movement of the siren change the wavelength of the wave? Well, there is some amount by which the wavelength

_{o}gets changed from the at-rest wavelength ; let's call that a. So

_{o}= -a, making our equation

f

_{o}= v/(-a).

The shift in wavelength, a, has to be proportional to the velocity of the siren, v

_{s}(i.e., v

_{s}= af). We can substitute v/f for , and v

_{s}/f for a, by solving for the wavelength in the wave equation. Thus we obtain

f

_{o}= v/(v/f - v

_{s}/f)

f

_{o}= (v/(v-v

_{s})) f.

We can go through this exercise again, with the siren stationary and the observer moving, and obtain the equation

f

_{o}= ((v-v

_{o})/v) f,

and then combine the two equations into the more general

f

_{o}= (v-v

_{o})/(v-v

_{s}) f.

Of course, a radar gun uses electromagnetic waves, not sound waves. When v is much, much bigger than v

_{o}or v

_{s}, then we can simplify this equation somewhat. Suppose we multiply the right-hand side, ((v-v

_{o})/(v-v

_{s})) f, by (v+v

_{s})/(v+v

_{s}), or in other words, by 1. Then we obtain

f

_{o}= f (v-v

_{o})(v+v

_{s})/((v-v

_{s})(v+v

_{s}))

= f (v

^{2}- (v

_{s}-v

_{o})v - v

_{o}v

_{s})/(v

^{2}-v

_{s}

^{2}).

This doesn't look very nice or helpful, but since the velocities of the source and the observer are so tiny compared to the velocity of the wave, then we can cancel out any second-order (i.e., squared) terms in v

_{o}and v

_{s}:

f

_{o}= f (v

^{2}- (v

_{s}-v

_{o})v -

_{o}v

_{s}

^{2}-

_{s}

^{2}

^{2}- (v

_{s}-v

_{o})v)/v

^{2}

So finally, we end up with the much simpler equation

f

_{o}= f (1-(v

_{s}-v

_{o})/v).

This is the equation that is used in a radar gun. The radar gun shoots out some radio waves, which bounce off your speeding car. The frequency shift is used to determine the speed of your vehicle.

It is slightly more complicated, however, when faced with identifying the actual culprit who is speeding, as well as correcting for certain types of errors that crop up when the radar is not trained in the exact direction of the car's velocity, etc. But combining the radar gun readings with other techniques, such as observation, speed matching, etc., police are able to catch speeders with better and better accuracy.

Sources:

How Radar Guns Work

Doppler Effect (Wikipedia)

Doppler Shift (Eric Weisstein's World of Physics)

## 2 comments:

There is a double shift in reflection, one when the wave is "receieved" by the car and a second when it is sent back to the radar gun, but otherwise this is correct.

The one interesting detail you left out is that they "beat" the shifted frequency against the transmitted frequency, and use the beat frequency to determine the speed. (The link is about the more familiar case of beating two sound waves against each other, like when tuning an instrument.) Your last equation shows that the frequency difference is directly proportional to the speed.

Interestingly, because the frequency (10^9 Hz) and speed of light (3x10^8 m/s) are of the same order of magnitude, the beat frequency is in the audio range of a few hundred Hz. Cops can listen to the beat frequency and hear the guilty ones slowing down.

Also, FYI, highway patrol have both forward and backward looking radar. They can catch speeders coming up from behind.

Thanks for the additional info, Dr. Pion. I didn't realize they can catch you coming and going... I am now doubly glad I don't have a lead foot!

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