Doppler

Peter Fisher

11/26/08

8.033

You will see this in lecture Wednesday, but I promised I would post this after running out of time in recitation today. Think about a clock at rest (Frame S) ticking every τ seconds. Now suppose you are in a frame S’ and the clock is moving away from your position at the origin with velocity v. At t=t’=0, the clock is at the origin. When the moving clock ticks, the sound has to travel to your ear through the air, at rest with respect to you. We can think of the clock ticks as events. In the clock rest frame, the first tick occurs a t=0, the second at t=τ and so on, all at the origin. This is shown in column 2 of the table. Next, we want to find where the ticks occur in our frame S’. Since this is classical, we use the Galilean transformation:

t'=t x ' = x + vt The third column shows the locations of the tick events in our frame S’. Now that we know when and where in our frame each tick occurs, we can compute the time for the sound to travel back to our ear at the origin. The sound travels with velocity vs and the transit time is shown in the fourth column. Finally, the absolute time we hear each tick is the emission time from the third column plus the transit time from the fourth column. The fifth column give the time we hear each tick. Tick

(t,x)

(t’x’)

Transit time

Time heard

1

(0,0)

(0,0)

0

0

2

(τ,0)

(τ, vτ)

vτ/ vs

τ+vτ/ vs

3

(2τ,0)

(2τ,2 vτ)

2vτ/ vs

2τ+2vτ/ vs

From the table, it as apparent that we hear a tick every τ(1+v/ vs)= τ’ sec. Now instead of ticks, we could call each tick the peak of a sine wave of frequency ν and then ν=1/τ. Then, ν’=ν/ (1+v/ vs). So, the frequency from a source moving away is lower than when the source is at rest. As was pointed out in recitation, this is a much easier way of working it out than I was doing. The nice thing is that to go to relativity, just change the Galilean transformation to the Lorentz transformation and fill in the table. You will get

ν′ = ν

1 − v / vs 1+ v / vs

If v/vs<<1, then 1 − v / vs ν  ν 1 − (v / vs )2 = ν 1 − 2v / vs + v 2 / vs2  ν 1 − 2v / vs  ν (1 − v / vs )  1+ v / vs 1+ v / vs which is just the classical expression. I used three Taylor expansions; see if you can follow them. v′ = ν