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February 26[edit]

Relativistic doppler effect[edit]

Is the doppler effect effected by any acceleration of the reciever and/or the source, according to special relativity. In other words, let's say you have two point objects S and R (for source and reciever) which will only move along a straight line. Let's say that at t=0 (from the perspective of R) S emits a photon of wavelength l. So, when the photon will reach R, will the photon be of wavelength l(1+ v/c)γ where γ is the lorentz factor, and v is the velocity between the inertial frame of refrence in which R is stationary when it recieves the photon and the inertial frame of reference in which S was stationary when it released the photon. Taabibtaza (talk) 19:43, 26 February 2024 (UTC)[reply]

When it receives the photon. So, when there's a stream of photons being emitted from S, their wavelengths as observed by R will change as the velocity of R changes. Also: "Is the Doppler effect affected...". --Wrongfilter (talk) 20:27, 26 February 2024 (UTC)[reply]

@Wrongfilter

I'm only talking about one photon, can we use the same formula for it, as what we use for the relativistic doppler effect if there is 0 acceleration. Taabibtaza (talk) 02:13, 27 February 2024 (UTC)[reply]

Our article on the relativistic Doppler effect has a section on the relativistic longitudinal Doppler effect, which corresponds to your scenario. --Lambiam 21:54, 26 February 2024 (UTC)[reply]

I literally learned the effect and the formula I presented from that article, I'm just wanting to know if we can use the same formula if there is acceleration (and what would v mean in the formula in that case). Taabibtaza (talk) 02:17, 27 February 2024 (UTC)[reply]

Same formula, with v being the instantaneous velocity of R with respect to S at the moment that the photon is being detected by R. --Wrongfilter (talk) 12:09, 27 February 2024 (UTC)[reply]

But S will accelerate after the the release of the photon, so shouldn't "v" be the velocity of R with respect to the inertial frame of reference in which S was stationary when it released the photon? Taabibtaza (talk) 20:06, 2 March 2024 (UTC)[reply]

Another response that I missed, sorry. I don't think you specified that S was moving/accelerating as well. So, to be clear, you need the relative velocity of the momentary rest frame of R at the event of reception with respect to the momentary rest frame of S at the event of emission. --Wrongfilter (talk) 21:30, 10 March 2024 (UTC)[reply]