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For time dilation, both observers see the other as experiencing a time dilation symmetrically based on relative velocity to each other. In the twin paradox, however, one must be careful of which twin is the "proper" or rest time, and this is based off of acceleration to know who is truly moving. I know this, yet my question is: does the Relativistic Doppler Effect follow these same rules?

Does the doppler effect take into account who is the proper time or is the rest frequency based solely on the velocity of the two frames, irregardless of "proper time"? Because for example I would think that in the twin paradox, both twins would see each other's signals as red shifted since they are moving away from each other. But how could this be the case if their time dilation aren't symmetrical due to only one being in an inertial reference frame, considering that the relativistic doppler shift is originally based off of Lorentz Transformations / Time Dilation in the form of

λ0=(c+u)T0

Is this T0 a different proper time than in the normal time dilation?

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The flaw is in this part of your argument:

In the twin paradox, however, one must be careful of which twin is the "proper" or rest time, and this is based off of acceleration to know who is truly moving.

While there is no acceleration yet, both twins are "truly moving" and each one sees the time of the other running slower. There is no paradox in the fact that A sees the time of B running slower while B sees the time of A running slower.

With the Doppler effect, the same principle applies. In fact, this is true not only for light, but even for sound. Imagine two trains on parallel tracks going away from each other. If they sound a signal of the same pitch, people on each train would hear its pitch higher than the pitch of the other train. Similarly, each twin would see the light coming from the other twin red shifted.

Relativistic Doppler effect

$$\dfrac{\lambda_o}{\lambda_s}=\dfrac{f_s}{f_o}=\sqrt{\dfrac{1+\beta}{1-\beta}}$$

Where

$$\beta=\dfrac{v}{c}$$

There also is a transverse efect, which is beyond the scope of your question.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Oct 21 '17 at 17:48

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