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If we have a periodic motion happening in frame A and if a moving frame B was to observe the event, B would see the time period reduced to $\frac{T_A}{1+v/c}$. This happens because of the finite amount of time light takes to deliver the message to B that change occurred in A

Apart from this, there is time dilation from SR that affects the mechanism of periodic motion itself. That is if the period of motion is $T_A$, (time between ticks of any clock), the period as observed by B would be

$T_B = T_A\sqrt{1-\frac{v^2}{c^2}}$

So if two frames are in relative motion, shouldn't we consider both effects? I mean in common physics questions like : If I were to look at the clock in a moving spacecraft, And if my clock ticks once every second, then at what rate would his clock tick? I believe it should be :

$\frac{1}{1+v/c} \times\text{dilated time between ticks} $ $=\frac{1}{1+v/c} \times \ \sqrt{1-v^2/c^2}={\sqrt{\frac{1-v/c}{1+v/c}}}$

but textbooks just mention the dilation factor only.

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If I were to look at the clock in a moving spacecraft, And if my clock ticks once every second, then at what rate would his clock tick?

Physics textbooks only consider the time dilation factor here and not the light propagation delay or doppler effect factor because the question is at what rate is his clock ticking, not at what rate does his clock "appears" to tick to my eyes.

These are 2 different questions. How fast the clock ticks in my frame, is just a matter of time dilation. How fast it "appears" to tick to my eye would necessitate accounting for the doppler effect or light propagation delay as well.

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  • $\begingroup$ thank you for the answer... New to SR and didn't have a strong Idea about doppler shift. Thanks a lot :) $\endgroup$ – Rishab Navaneet May 19 at 6:53
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There are two separate effects at work. One is time dilation, the other is the Doppler effect. They tend to be treated separately in textbooks to avoid confusing the underlying principles.

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  • $\begingroup$ thank you for the answer... New to SR and didn't have a strong Idea about doppler shift. Thanks a lot :) $\endgroup$ – Rishab Navaneet May 19 at 6:53

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