Modern Physics/The Doppler Effect

The Doppler Effect

You have probably heard how the pitch of a train horn changes as it passes you. When the train is approaching, the pitch or frequency is higher than when it is moving away from you. This is called the Doppler effect. A similar, but distinct effect occurs if you are moving past a source of sound. If a stationary whistle is blowing, the pitch perceived from a moving car is higher while moving toward the source than when moving away. The first case thus has a moving source, while the second case has a moving observer.

In this section we will compute the Doppler effect as it applies to light moving through a vacuum. The figure below shows the geometry for computing the time between wave fronts of light for a stationary and a moving reference frame.


 * The time between wavefronts for the stationary observer, in the stationary frame, is T.


 * The time between wavefronts for the moving observer, in the stationary frame, is T&prime;.


 * The time between wavefronts for the moving observer, in the stationary frame, is &tau;

Since the world lines of the wave fronts have a slope of unity, the sides of the shaded triangle both have the same value, C. If the observer is moving at speed U, the slope of the observer's world line is c/U, i.e


 * $$\frac{c}{U}=\frac{cT+X}{X}$$

Solving this for X and substituting in give cT&prime;=cT+X gives


 * $$T'=\frac{T}{1-\frac{U}{c}} \quad (1)$$

In classical physics T&prime; and &tau; are the same so this formula, as it stands, leads directly to the classical Doppler shift for a moving observer.

However, in relativity T&prime; and &tau; are different. We can use the Lorentz transformation to correct for this.

The second wavefront passes the moving observer at (UT&prime;,cT&prime;) in the stationary observers frame, but at (0,c&tau;) in its own frame. The Lorentz transform tells us that.


 * $$c\tau = \gamma \left( cT'-\frac{U}{c}UT' \right)

= cT' \gamma \left( 1 - \frac{U^2}{c^2} \right) = cT'/ \gamma $$

Substituting in equation (1) gives


 * $$\begin{matrix}

\tau & = & T \frac{ \sqrt{ 1 - U^2/c^2 }}{1-U/c} \\ & = & T \sqrt{ \frac{ (1-U/c)(1+U/c)}{(1-U/c)^2}} \\ & = & T \sqrt{ \frac{c+U}{c-U} } \end{matrix}$$

From this we infer the relativistic Doppler shift formula for light in a vacuum:
 * $$\omega^{\prime}= \omega \sqrt{ \frac{c-U}{c+U} } $$

since frequency is inversely proportional to time.

We could go on to determine the Doppler shift resulting from a moving source. However, by the principle of relativity, the laws of physics should be the same in the reference frame in which the observer is stationary and the source is moving. Furthermore, the speed of light is still c in that frame. Therefore, the problem of a stationary observer and a moving source is conceptually the same as the problem of a moving observer and a stationary source when the wave is moving at speed c.

This is unlike the case for, say, sound waves, where the stationary observer and the stationary source yield different formulas for the Doppler shift.