tl;dr: A process takes time $\Delta t_B$ in some frame of reference B.
Is the time an actual observer $A$ would see $\Delta t_A = \Delta t_B / \gamma$?
Or is the time that the light travels to the observer not included in $\Delta t_A$?
A little thought experiment to make my question clear:
There are two people Alice and Bob. The two reference frames are denoted with $A$ and $B$ as indices.
Alice is on the earths surface (or in orbit) the entire time.
Bob is in a spaceship travelling near the speed of light towards earth.
Suppose Alice had a telescope on earth, that is precise enough to see exactly what is going on in Bobs spaceship.
At time $t_0$ Bob passes a planet that is $d_A = 10 ly$ (lightyears) away from earth.
Alice sees this event happening at time $t_{1,A} = t_{0,A} + 10 y$ (years), because the light takes 10 years to travel to her.
Bob flies at such a speed that it takes $\Delta t_A = 11y$ until he reaches earth.
Then Bob would pass earth at $t_{2,A} = t_{0,A} + \Delta t_A = t_{1,A} + 1y$ and Alice would see the entire journey of Bob in $\Delta t'_A = 1y$.
What is the difference between $\Delta t_A$ and $\Delta t'_A$?
Which $\Delta t$ is to be used to calculate the time the journey takes for Bob?
Would Alice see Bobs time run slower or faster than her own? (suppose there is a clock on board)
In an opposite scenario where Bob travels from the earth to the planet, would Alice see Bobs time run slower or faster?
I think time dilation would say time always runs slower in a moving frame of reference, but including the time that light travels to the observer would suggest a dependence on direction.
Best Answer
The time that the light takes to travel to the observer is not included. The time an event happens in a reference frame is defined as the time that a clock at the location of the event would show. To tell time in a reference frame, Special Relativity imagines clocks at all relevant locations, and all those clocks properly synchronised within the reference frame. It is important to keep that definition in mind, because it also leads to Relativity of Simultaneity, which many novices don't count with.
In reference frame A, Bob reaches $d_A$ at time $t_{0,A}$ and Alice at time $t_{0,A} + \Delta t_A$, so the journey would take 11 years in reference frame A. In reference frame B, the journey takes $\Delta t_A /\gamma$ years.
As perceived from reference frame A, the clocks in reference frame B run slower. If with "see", you mean what Alice sees through the telescope, then you need to take the Doppler effect into account, and Alice would see the clocks of Bob run faster.
The Doppler shift is caused because each tick of the moving clock occurs at a point a bit closer to Alice than the previous tick. The light from the previous tick had to travel a bit of time to get to that same point, and therefore the two ticks appear closer together to Alice when viewed through the telescope. That is, the clock appears through the telescope to run faster than it actually does.
Perceived from reference frame A, Bobs time runs just as slow as on the way towards the Earth. Through the telescope, Alice would see an additional slow-down due to the receding Doppler effect.