Follow up question to Doppler redshift in special relativity
A source of light pulses moves with speed v directly away from an observer at rest in an inertial frame. How will the time of the emission of a light pulse as measured by the emitter be different to the time as measured by the observer (how will it be different than the time of the pulse reaching the observer)?
edit: to elaborate slightly, let us say tE1 is the time of emission as measured by the emitter, tE2 is the time of emission as measured by the observer, and tO is the time of observation of the pulse by the observer. How are these three times different, how would they be measured in practice?
Answer
How are these three times different, how would they be measured in practice?
First, let's see if we agree on what these times are. From what I've read, I believe that:
tE1 is the reading of a clock, co-located and co-moving with the emitter, when the pulse of light is emitted.
tE2 is the reading of a synchronized clock, at rest in the frame of the observer, co-located with the emitter at the time of the emission of the pulse.
tO is the reading of the observer's "wrist watch", i.e., it is the reading of a clock at rest at the origin of the observer's frame of reference, when the pulse reaches the origin.
If so, then:
γvtE1=tE2
(assuming standard configuration etc.)
Intuitively, the moving emitter clock runs slow according to the observer so tE1 is smaller than tE2 by the factor γv.
How is this measured? The observer has a synchronized clock at the spatial location of the emission of the pulse. This clock reads tE2 when the emitter is co-located (and emits the pulse). The emitter's clock reads tE1
tO=tE2+dc=tE2(1+vc)=tE1γv(1+vc)
Here, d is the displacement, from the origin of the observer's frame, of the emitter when the pulse is emitted at time tE2.
The coordinate time required for the pulse to travel to the origin is dc but d is just v⋅tE2
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