Does Length Contraction Affect the Wavelength of Light from a Moving Source?

[Zero-G]

Hey, sorry if this is a stupid question but I'm getting really confused...
If I am standing on the Earth and watch a spaceship moving past at say 0.8*c from left to right, I will observe the spaceship to contract in its direction of motion. If a person on the spaceship shines a torch also from left to right, what happens to the wavelength of the light as seen from my reference frame? Does it also contract?

 

[JesseM]

Hey, sorry if this is a stupid question but I'm getting really confused...
If I am standing on the Earth and watch a spaceship moving past at say 0.8*c from left to right, I will observe the spaceship to contract in its direction of motion. If a person on the spaceship shines a torch also from left to right, what happens to the wavelength of the light as seen from my reference frame? Does it also contract?

In the ship's frame, the wave peaks move at c, so the distance between peaks is just c/f, where f is the frequency that the peaks are being emitted by the torch as seen in the ship's frame. In your frame, the peaks still move at c, but the frequency becomes $$f \sqrt{1 - v^2/c^2}$$ due to time dilation, and meanwhile the torch is moving at velocity v so it will have moved a distance of $$v / (f \sqrt{1 - v^2/c^2})$$ between emitting successive peaks, so the the distance between peaks should either be $$c / (f \sqrt{1 - v^2/c^2}) - v / (f \sqrt{1 - v^2/c^2})$$ or $$c / (f \sqrt{1 - v^2/c^2}) + v / (f \sqrt{1 - v^2/c^2})$$ depending on whether the torch is shining in the direction of the ship's motion or in the opposite direction. This simplifies to $$(c \pm v)/(f \sqrt{1 - v^2/c^2})$$. You could also get this from the relativistic doppler shift equation, $$f_{observed} = f_{emitted} \sqrt{1 - v^2/c^2} / (1 - v/c)$$, keeping in mind that the wavelength you observe is $$c / f_{observed}$$.

To compare the wavelength you see with the wavelength seen by the ship-observer, just divide $$(c \pm v)/(f \sqrt{1 - v^2/c^2})$$ (the wavelength seen by you) by c/f (the wavelength seen by the ship-observer), which gives $$(1 \pm v/c) / \sqrt{1 - v^2/c^2}$$ for the factor that the wavelength changes in your frame. This is not the same as the Lorentz contraction factor $$\sqrt{1 - v^2/c^2}$$.

 

[Entropia]

No need to apologize, this is actually a very interesting question! The concept of length contraction can be confusing, but let's break it down.

First, let's clarify what length contraction is. According to Einstein's theory of special relativity, as an object moves at high speeds, its length in the direction of motion appears to shorten when observed by an outside observer. This is due to the fact that as an object moves faster, time slows down for that object, causing it to appear shorter to an outside observer.

Now, let's apply this concept to your scenario. As you observe the spaceship moving past at 0.8*c, you will indeed see it contract in its direction of motion. This means that the length of the spaceship will appear shorter to you than it actually is. However, this does not apply to the wavelength of light.

The wavelength of light is a property of the light itself and is not affected by the motion of the source. So, from your reference frame, the wavelength of the light emitted by the person on the spaceship will remain the same. This is because the speed of light is constant, regardless of the relative motion of the source and observer.

In fact, according to special relativity, the speed of light is the same for all observers, regardless of their relative motion. This means that no matter how fast the spaceship is moving, the speed of light will always be measured as c (the speed of light in a vacuum).

So, to sum up, the length of the spaceship will appear shorter to you due to length contraction, but the wavelength of light will remain the same. I hope this helps clarify things for you!