Dopper Effect in Special Relativity

AI Thread Summary
The discussion centers on a hypothetical scenario involving a driver who claims to see a traffic light in the X-ray region instead of the infrared region, raising questions about the implications of the Doppler effect in special relativity. The original poster calculates the car's speed using the Doppler effect formula and concludes that it would need to exceed the speed of light, approximately 600c, which is deemed impossible. A participant points out that the Doppler formula involves frequency rather than wavelength and suggests that the original poster's expansion of the formula may be incorrect. They emphasize the need to properly convert between frequency and wavelength to accurately apply the Doppler effect. The conversation highlights the complexities of applying relativistic physics to practical scenarios.
yukcream
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Suppose there is a planet, the traffic law there is~
car should stop if the traffic light in infrared region,i.e wavelenght ~10^-5m, otherwise the driver will be fined.
There is a driver claimed that he saw the traffic light still in the X-ray region i.e wavelenght ~10^-10m when passing the traffic light so he hadn't violated the law
If possible, what will be the speed of the car~

However, I found it is impossible for the driver to have this dopper shift, as the speed of the car will be about 600c ~

I did in following step:
by the formula of dopper effect for SR
[(1-v/c)/(1+v/c)]^1/2 = L0/L = 10^-5/10^-10where L is the wavelenght
L0/L = 10^5
the RHS value is so large~
so I expand the LHS as:

(0.5 - 0.5V/c) (0.5+ 0.5v/c) than I get the value as v~600c!

What's going wrong, can anyone help me!

yukyuk
 
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For one thing, the doppler formula is:
f = f' \sqrt{\frac{1 + v/c}{1 - v/c}}

As far as your expansion, I don't understand what you are doing. (Realize that v/c is not necessarily small compared to 1.)
 
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f is stand for frequency here I use wavelenght instead~
 
yukcream said:
f is stand for frequency here I use wavelenght instead~
You can convert from one to the other: f = c/\lambda.
 
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