# The Doppler Effect and Electromagnetic Waves

1. Mar 23, 2006

### BoogieL80

I had a question. I did a problem where a distant galaxy emitted light that had a wavelength of A. On earth, the light had a wavelength of B. B ended up being greater than A. The question asked if the galaxy was receding from the earth or approaching the earth. The answer turned out to be the galaxy is receding from the earth. I'm not sure I understand why though. I do understand that the frequency of light is less on earth than in the galaxy, but get a little lost after that.

2. Mar 23, 2006

### Tide

HINT: Wavelength varies inversely with frequency.

3. Mar 23, 2006

### BoogieL80

I understand that, but I still don't understand why that would mean the galaxy is receding from earth.

4. Mar 23, 2006

### ranger

Take a plane for example, as the distance from the plane increases, the wavelenght of the sound wave increases. Since there is more distance for the wave to cover, it "spreads out". If you were to superimpose wave B over A, you would notice that the frequency (of B) has decreased. and the wavelength has increased.

So if wavelength B>A, this can only mean that the source of the light is receding.

Last edited: Mar 23, 2006
5. Mar 23, 2006

### Tide

You said you understood why the frequency is lower so I assumed that you did. If so, then all you need is a relationship between frequency and wavelength to complete the picture.

6. Mar 23, 2006

### lightgrav

Tide, he said he understood THAT the frequency was lower ...
since wavelength was longer.

How a difference in frequency or wavelength arises from relative velocity
is simple for waves that travel in a material (like sound or water waves).
It is a bit trickier for "waves" that have no frame to propagate relative to.

While a car moves away from us through the air , the sound waves back
(that eventually reach our ears) travel through the air at the same speed
as the sound waves forward. So each sound wave back is farther from the sound source (car) than its simultaneously-made sound wave forward.

But I can't explain this for EM waves without knowing WHAT BoogieL knows
about reference frames and relativity.

7. Mar 23, 2006

### Tide

lightgray,

Thanks. I missed that on first reading.

Boogie,

Suppose a source of radiation moves at constant speed. Over the course of one full period of oscillation an interval of time will have elapsed and an observer will see on full wavelength of the light emitted. The start of the wave will have travelled a distance $c \Delta t$ while the source will have moved $v \Delta t$ during that period so the wavelength an observer sees will be

$$\lambda ' = (c - v) \Delta t$$

Classically, $\Delta t$ is just the period of oscillation or $\lambda / c[/tex] so $$\lambda ' = \frac {c-v}{c} \lambda$$ which gives you the basic idea behind the Doppler shift. However, Einstein tells us that since the source and observer are moving relative to each other, their measures of time are different so we have to adjust for that and replace [itex]\Delta t$ with

$$\frac {\lambda}{c \sqrt {1-v^2/c^2}}$$

from which

$$\lambda ' = \lambda \sqrt{ \frac {1-v/c}{1+v/c}}$$

Note that this is 1-dimensional and a simple geometric adjustment is necessary if the source is approaching or receding at an angle with repect to the observer.

8. Mar 26, 2006

### BoogieL80

Thank you.

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