Celinamg8

2022-09-25

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)? To elaborate slightly, let us say ${t}_{{E}_{1}}$ is the time of emission as measured by the emitter, ${t}_{{E}_{2}}$ is the time of emission as measured by the observer, and ${t}_{O}$ is the time of observation of the pulse by the observer. How are these three times different, how would they be measured in practice?

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berzamauw

Expert

${t}_{{E}_{1}}$ is the reading of a clock, co-located and co-moving with the emitter, when the pulse of light is emitted.
${t}_{{E}_{2}}$ 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.
${t}_{O}$ 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:
${\gamma }_{v}{t}_{{E}_{1}}={t}_{{E}_{2}}$
(assuming standard configuration etc.)
Intuitively, the moving emitter clock runs slow according to the observer so ${t}_{{E}_{2}}$ is smaller than ${t}_{{E}_{1}}$ by the factor ${\gamma }_{v}$.How is this measured? The observer has a synchronized clock at the spatial location of the emission of the pulse. This clock reads ${t}_{{E}_{2}}$ when the emitter is co-located (and emits the pulse). The emitter's clock reads ${t}_{{E}_{1}}$
${t}_{O}={t}_{{E}_{2}}+\frac{d}{c}={t}_{{E}_{2}}\left(1+\frac{v}{c}\right)={t}_{{E}_{1}}{\gamma }_{v}\left(1+\frac{v}{c}\right)$
Here, $d$ is the displacement, from the origin of the observer's frame, of the emitter when the pulse is emitted at time ${t}_{{E}_{2}}$.
The coordinate time required for the pulse to travel to the origin is $\frac{d}{c}$ but $d$ is just $v\cdot {t}_{{E}_{2}}$

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