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.


1 Answer 1


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).

The situation has to be symmetric otherwise it would be possible for the observers to tell which frame was moving and which was standing still.

  • $\begingroup$ If i am correct this comes from 2nd Einstein's postulate that all coordinate systems are equivalent. $\endgroup$
    – 71GA
    Commented Sep 20, 2012 at 14:59
  • 1
    $\begingroup$ I'd say it was really the first postulate "The laws of physics are the same in all inertial frames of reference", though I suppose it depends on how you count the postulates. $\endgroup$ Commented Sep 20, 2012 at 15:30

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