# Doppler effect for light wave

1. Apr 7, 2009

### loup

How to prove the doppler effect for light wave?

Isn't that the speed of light is constant for both the observer and source?
If the wavelength for both the source and observer is the same, then the frequency is the same too, so there is no doppler effect?

If the wavelength is different, but if we use Lorren's transfer, the new wave obtained will be the same if the observe is moving towards or leaving the source!

How to deal with this?

2. Apr 7, 2009

### cragar

the speed is the same but the frequency's are different

3. Apr 7, 2009

### pixel01

In sound wave, the sound frequency depends on speed, why not in this case?

4. Apr 7, 2009

### cragar

i guess the same reason you don’t just add velocities like you do two moving cars
with light you cant add the velocities so the light is always perceived to the observer at c and all frequency's travel at c. i guess when you travel towards the light source you see a blue shift the distance between the crest's is shorter as your eye perceives it but the time between the crest's is shorter so the speed still comes out c .
Dont know if this is 100% accurate.

5. Apr 7, 2009

### Bob S

How come the wavelength of light is shortened after it falls from the top of a 3 story building to the basement? The famous Mossbauer Effect experiment at Harvard by Pound and Rebka in 1959 showed that the 14 keV photons from an iron-57 source gained energy (wavelength shortened) when they fell from the top of the physics building.

6. Apr 7, 2009

7. Apr 7, 2009

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9. Apr 7, 2009

### yelram

Are there any substantial experiments with visible light, and the doppler effect, that dont rely on large celestial bodies? Or the travel of light through space?

10. Apr 7, 2009

### Staff: Mentor

11. Apr 7, 2009

### loup

Proof for sound effect I clearly understand. But light cannot be treated as relative velocity, it is absolute, isn't it?

12. Apr 7, 2009

### JesseM

The Doppler effect is just a consequence of the fact that if an object emitting regular signals (or peaks of a continuous wave) is in motion relative to you, then each signal (or peak) has a different distance to travel to reach your eyes...the relativistic Doppler effect also factors in time dilation, but that's the only difference. For example, suppose a clock is traveling away from me at 0.6c, and it's programmed to send out a flash every 20 seconds in its own rest frame. In my frame, because of time dilation the clock is slowed down by a factor of $$1/\sqrt{1 - 0.6^2}$$ = 1.25, so it only flashes every 1.25*20 = 25 seconds in my frame. But that doesn't mean I see the flashes every 25 seconds, the gap between my seeing flashes is longer since each flash happens at a greater distance. For example, suppose one flash is emitted when the clock is at a distance of 10 light-seconds from me, at time t=50 seconds in my frame. Because we assume the light travels at c in my frame, if the flash happens 10 light-seconds away the flash will take 10 seconds to reach me, arriving at my eyes at t=60 seconds. Then, 25 seconds after t=50, at t=75, the clock emits another flash. But since it was moving away from me at 0.6c that whole time, it's increased its distance from me by 0.6*25 = 15 light-seconds from the distance it was at the first flash (10 light-seconds away), so it's now at a distance of 10 + 15 = 25 light-seconds from me, so again assuming the light travels at c in my frame, the light will take 25 seconds to travel from the clock to my eyes, and since this second flash happens at t=75 in my frame, that means I'll see it at t=100 seconds. So, to sum up, the clock flashes every 20 seconds in its own rest frame, and once every 25 seconds in my frame due to time dilation, but I see the first flash at t=60 seconds and the second at t=100 seconds, a separation of 40 seconds. This means the frequency that I see the flashes (1 every 40 seconds) is half that of the frequency the clock emits flashes in its own frame (1 every 20 seconds), which is exactly what you predict from the relativistic Doppler equation if you plug in v=-0.6c (negative because the clock is moving away from me): $$\sqrt{\frac{1 - 0.6^2}{1 + 0.6^2}} = \sqrt{0.25} = 0.5$$. And you can see from the italics above that I specifically assumed the light from each flash traveled at exactly c between the clock and my eyes.

Last edited: Apr 8, 2009
13. Apr 8, 2009

### Staff: Mentor

See for example the link in post #8.

Yes.

No, it isn't, if you mean "wavelength in the source's rest frame" and "wavelength in the observer's rest frame."

The speed of the wave is the same for both observers, so if the wavelength is different, then the frequency must be different, also.

The relativistic Doppler effect actually involves three different frequencies, and you need to be clear about which one you are talking about:

1. The frequency of the source, in its own rest frame. An observer riding along with the source would see this frequency. I call it the proper frequency of the source.

2. The frequency of the (moving) source, in the observer's rest frame. This is different from the proper frequency because of time dilation.

3. The frequency of the wave emitted by the source, as received by the observer. This is different from #2 for exactly the same reason as in the classical Doppler effect: the source is moving, so successive "peaks" of the wave are emitted from different locations in the observer's reference frame. If the source is moving towards the observer, each successive peak needs to travel a shorter distance to reach the observer. It arrives at the observer earlier than if the source had remained stationary, which increases the frequency at the observer. Similarly, if the source is moving away from the observer, each successive peak needs to travel a larger distance, so it arrives at the observer later than if the source had remained stationary.