Transformation of Doppler shift

harrylin

Yes of course, from your OP this has been about basic or "classical" Doppler, which is also valid for SR within a single reference system. Adding a frame transformation (or two, if you like), one has to multiply with gamma, as usual. For the complicated case of c-sound and c-light mathpages provides a good discussion.

The common relativistic Doppler equation transforms from one to the other's rest frame with only c-light which simplifies the resulting equation.

Thus, starting from basic Doppler:
[tex] \frac{f_r}{f_s} = \frac{c - v_r}{c - v_s} [/tex]
Taking the rest frame of the source v_s=0 =>
[tex] \frac{f_r}{f_s} = \frac{c - v_r}{c} [/tex]
For example v_r=+0.5c => f_r/f_s=0.5

Next transforming to the rest frame S' of the source:
[tex] \frac{f'_r}{f_s} = \frac{c - v_r}{\sqrt{c^2 - v_r^2}} [/tex] so that
[tex] \frac{f'_r}{f_s} = \sqrt{\frac{c - v_r}{c + v_r}} [/tex]
For example v_r=+0.5c => f'_r/f_s=0.577 = 0.5γ.

I notice that the usual relativistic equation follows directly from my notation of basic Doppler, without any messing around with the signs.

Then for both emitter and receiver moving, as already discussed in posts #5 and #6 that gives, with S'' the rest frame of the source:

[tex] \frac{f'_r}{f''_s} = \sqrt{\frac{(c+v_s)(c-v_r)}{(c-v_s)(c+v_r)}} [/tex]

vin300

EDIT: This post is a mess, skip it.
If the wave and receiver and emitter all move along +x, use the formula mentioned in post #5. The same formula is to be used if all these velocities are reversed. If the wave moves along +x, the receiver moves along -x and the source moves along +x, reverse the signs of v_r in the equation. This form of the equation is to be used also if the wave moves along -x, receiver along +x and source along -x. If the wave moves along +x, receiver along +x and source along -x, reverse the signs of v_s in the equation in post #5. This form of the equation is to be used also if the wave moves along -x, receiver along -x and source along +x. All cases have been covered.
The moral of the story is to simply reverse the signs of velocity of the one (the wave, receiver or emitter) which travels in the -x direction.

vin300

My previous post is messy and wrong. This will work: Use the formula given by Harrylin in post #15 with his convention. For the one(the wave, receiver or emitter) that travels in the -x direction, reverse the sign of its velocity. Then bring in the two time dilation factors. The direction of velocity is immaterial in these factors.