# Relativistic Doppler effect

If i have 2 coordinate systems (CS) which are travelling one towards another. CS $xy$ with an observer in its origin and CS $x'y'$ with a source in its origin. Correct me if I am wrong, but i think that I have to use this variation of an equation for a relativistic Doppler effect:

$$\nu = \nu' \sqrt{\frac{c+u}{c-u}}$$

Here $\nu$ is a frequency that observer receives and $\nu'$ is a frequency which source transmits. Correct me if I am wrong.

Question: Do I have to swap $\nu$ and $\nu'$ in the equation above if I put observer in CS $x'y'$ and source in CS $xy$? A brief explanation will do just fine.

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u-c is negative – C.R. Sep 20 '12 at 12:20
Woops i did fix that. – 71GA Sep 20 '12 at 12:29

The equation is correct, though whether it's $(c + u)/(c - u)$ or $(c - u)/(c + u)$ depends on your sign convention. As it stands your equation is correct if $u$ is positive when the source is approaching you and negative when the source is moving away from you. The Wikipedia article you cite actually uses the opposite convention.
Anyhow, provided $\nu^'$ is always the moving frame and $\nu$ is the stationary frame the equation applies to both sets of observers. To make this clearer let's take an example. Suppose both frames have a source emitting the same frequency of 1KHz and their relative speed is 0.6$c$ then both observers will see the other source to be 2KHz when approaching and 0.5KHz when receding (I chose 0.6$c$ to give a factor of 2).