Length contraction & light

1. Nov 17, 2005

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?

2. Nov 17, 2005

JesseM

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

Last edited: Nov 17, 2005