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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.

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  • $\begingroup$ What do you mean by the time an observer will "see"? Do you mean the amount of time they calculate to have taken place in the reference frame that they are observing between two events (Alice, observing Bob, sees Bob's clock tick 100 times), or do you mean the time they measure in their own frame between those two events (Alice, observing Bob, sees Alice's clock tick 100 times)? $\endgroup$
    – g s
    Jan 22 at 21:59
  • $\begingroup$ I mean the time it took in their own reference frame. $\Delta t_A$ is the time measured with Alice's clock in Alice's reference frame. $\Delta t_B$ is the time measured with Bob's clock in Bob's reference frame. $\endgroup$
    – Nico G.
    Jan 22 at 22:07

3 Answers 3

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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.

Which $\Delta t$ is to be used to calculate the time the journey takes for Bob?

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.

Would Alice see Bobs time run slower or faster? (suppose there is a clock on board)

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.

In an opposite scenario where Bob travels from the earth to the planet, would Alice see Bobs time run slower or faster?

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.

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  • $\begingroup$ How does the doppler effect relate to this problem? Everything I read about the doppler effect just talks about a shift in frequency and wavelength. Is there any transformation, that accounts for the time that light travels? (unlike the Lorentz Transformation) $\endgroup$
    – Nico G.
    Jan 23 at 9:43
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    $\begingroup$ @NicoG. "Everything I read about the doppler effect just talks about a shift in frequency and wavelength" - the ticking of a clock is also a frequency (1 Hz in your bed-side clock. 9192 MHz in atomic clocks). See edit of the answer. $\endgroup$
    – fishinear
    Jan 23 at 11:10
  • $\begingroup$ This makes it clear. Just to be sure: If Alices clock ticks at $f_A = 1/s$. Then Bob's clock observed by Alice ticks at $f_B = f_A \cdot \gamma \approx 2,29/s$ if he is traveling towards earth with $\beta = 0.9$. And $f_B = f_A / \gamma \approx 0.45/s$ if he is traveling away from earth with $\beta = 0.9$. Correct? $\endgroup$
    – Nico G.
    Jan 23 at 11:38
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    $\begingroup$ No, $\frac{f_B}{f_A} = \sqrt{\frac{1 + \beta}{1 - \beta}}$, see the wikipage en.wikipedia.org/wiki/Relativistic_Doppler_effect. You need to take both the Doppler shift and the time dilation into account, as explained on the wiki page. So when $\beta = 0.9$ then $f_A = 4.35 f_B$ when traveling to the earth, and $f_A = 0.22 f_B$ when traveling away. Here $f_A$ is the received frequency as measured by A, and $f_B$ the frequency of the clock as measured on board B (proper time). $\endgroup$
    – fishinear
    Jan 23 at 12:18
  • $\begingroup$ Important note: $\beta < 0$ if A and B are moving towards each other and $\beta > 0$ if they are moving away from each other. Thank you! Now I somewhat understand it. $\endgroup$
    – Nico G.
    Jan 23 at 14:27
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When we talk about Alice's or Bob's frame, that includes all the spacetime. So, when Bob passes the planet at $t_0$ in Alice's frame, we can postulate there a synchronized clock with Alice's one. Alice herself will receive the information 10 years later. But this planet, that is in her (A) frame receives the information at $t_0$.

The time travel according to Bob's clock is shorter, when he compares it with the difference between the Alices's clock when they meet, and the planet's clock when he passed by there. As the clocks in Alice's frame are synchronized, that is the time travel in her frame.

If Bob was travelling from the Earth to the planet, the time difference would be the same.

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  • $\begingroup$ If Alice is observing Bob's clock during the journey, would it tick faster than her own clock? Is this "observed ticking speed" dependent on the direction of the journey? $\endgroup$
    – Nico G.
    Jan 22 at 22:31
  • $\begingroup$ Yes, and it is an indication for her that Bob is approaching, and at great relative velocity. $\endgroup$ Jan 22 at 23:05
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You are confusing time dilation with the Doppler effect.

In SR, time dilation refers to the fact that the time between two events that occur at the same place in one frame is always less than the time between the same events in another inertial frame moving relative to them. In your example, Bob passing the planet and Bob arriving at Earth are two events that occur at the same place in Bob's frame, so the time between them in Bob's frame will be less than the time between them in Alison's frame.

The Doppler effect means that the interval between light arriving from two-colocated events will be reduced for a person moving towards the events and increased for someone moving away from them. It is caused by the fact that the distance between the observer and the first event is not the same as the distance between the observer and the second event, which means that light has different distances to travel from each event which takes different times.

And yes, as you mention in your comment on Claudio's answer, the Doppler effect is dependent on direction, whereas time dilation is not. If Alison could look at Bob's clock through a telescope she would see it speeded up.

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