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 $ 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?

  • $\begingroup$ The coordinate time used by each observer differs by a time dilation factor. $\endgroup$
    – Will
    Jul 14, 2013 at 15:53
  • $\begingroup$ I am looking for a bit more detail in order to understand it intuitively $\endgroup$
    – user997712
    Jul 14, 2013 at 17:51

1 Answer 1


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:

$t_{E_1}$ 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 $t_{E_1}$ is smaller than $t_{E_2}$ 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 $t_{E_1}$

$t_O = t_{E_2} + \frac{d}{c} = t_{E_2}(1 + \frac{v}{c}) = t_{E_1}\gamma_v(1 + \frac{v}{c})$

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 $v\cdot t_{E_2}$

  • $\begingroup$ If $ t_{E_2} $ were to be the time of emission as measured by a clock co-located (and at rest) with the observer, would it be same as $ t_O $? This is really the source of my confusion. Please see the linked question for the background discussion. $\endgroup$
    – user997712
    Jul 14, 2013 at 19:48
  • $\begingroup$ @user997712, the clock co-located with the observer can only record the time of events co-located with the observer. The time of events spatially separated from the origin must be recorded by synchronized clocks co-located with those events. This is in fact, the operational meaning of the time coordinate of an event. The measurement of the one-way time of flight requires two clocks - one at the origination and one at the destination. We would say that the emission of the pulse is simultaneous, according to the observer, with the origin's clock reading $t_{E2}$. $\endgroup$ Jul 14, 2013 at 19:55
  • 1
    $\begingroup$ Thanks for the answer and the link to Einstein Synchronization. That makes it much more clear. $\endgroup$
    – user997712
    Jul 14, 2013 at 20:12

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