# Reference frames symmetry in Special Relativity

I have a couple of questions related to reference frames in STR.

Let's consider a rocket that is inertially moving towards a star with a relative velocity 0.9c.

I'd like to look at this example from both the rocket's and the star's perspectives.

In the reference frame of the rocket:

• The rocket is at rest and the star is moving towards the rocket.
• At time t(0), the distance between the rocket and the star is 10 light years.
• Since the distance between the two is 10 ly - and their relative velocity is 0.9c - the star will reach the rocket in 11.1 years.
• From the rocket's perspective, time is slowing down for the star, so only 4.8 years will have passed in the star's reference frame.

In the reference frame of the star:

• The star is at rest and the rocket is moving towards the star.
• At time t(0), the distance between the rocket and the star is 10 light years.
• Since the distance between the two is 10 ly - and their relative velocity is 0.9c - the rocket will reach the star in 11.1 years.
• From the star's perspective, time is slowing down for the rocket, so only 4.8 years will have passed in the rocket's reference frame.

I have calculated the 4.8 years interval using the time dilation formula:

So, my questions/comments are:

• Is my math correct ;)
• Given that there is no acceleration involved in this example, can we safely assume that the two reference frames are fully symmetrical?
• When we switch the roles of "stationary" and "moving" between the star and the rocket, the proper distance between them doesn't change.
• The proper distance in this example is always in the reference frame of the stationary observer.
• The distance to the star is contracted in the rocket frame. Aug 30, 2019 at 4:20
• Interesting. I automatically assumed that if the roles of "stationary" and "moving" are interchangeable, then both observers will see the same distance at time t(0) - provided that their clocks are synchronized at t(0) (which was another automatic assumption on my part). Aug 30, 2019 at 5:10
• No, but they will agree on the magnitude of the relative velocity between the two frames. The initial distance from the station to the star is 10 light-years in the star frame, but it's only 4.359 light-years in the rocket frame. If the rocket's length is 100 m in the rocket's rest frame, it's only 43.59 m long in the star frame. Without length contraction, the speed of the star in the rocket frame would be greater than $c$. Aug 30, 2019 at 5:34
• Observers can't compare their clock in the end. Because they haven't synchronised their clock at beginning. So it'd be meaningless to compare 4.8y with 11.1y. and also they can't synchronize their clock as well, because they are at different positions with different velocities. Aug 30, 2019 at 10:31
• What exactly is the question here? You have two separate and different scenarios. They cannot both hold, it is one or the other. The numbers you have chosen suggest that you think that this is just one case from two different perspectives. Well, it isn't!
– MBN
Sep 4, 2019 at 11:21

Any two inertial reference frames in Special Relativity are completely symmetrical. So all the logic, math, and numbers in your question are correct, except for the following:

1. In the reference frame of the rocket: At time $$t(0)$$, the distance between the rocket and the star is $$10$$ light years.
2. In the reference frame of the star: At time $$t(0)$$, the distance between the rocket and the star is $$10$$ light years.

These two statements cannot be both correct, because the moment of time $$t(0)$$ when the trip begins is not the same for the spaceship and the star due to Relativity of Simultainety.

1. When we switch the roles of "stationary" and "moving" between the star and the rocket, the proper distance between them doesn't change.

This is not a rigorous statemenet, because distance and time are relative concepts while it is unclear exactly what distance at exactly what (and whose) time this statement describes.

Consider the captain of the spaceship blinked his eyes when he measured the distance to the star to be $$10$$ light years. Let's call this the spacetime Event A. Similarly, lets assume a momentary solar flare happened on the star just when the spaceship was $$10$$ light years away in the frame of the star. Let's call this flare the spacetime Event B.

Based on your correct math, we know that $$4.8$$ years have passed on the star between the Event A and the arrival. Thus in the frame of the star, the Event A happened long after the Event B and when the spaceship was already almost a half way through.

You can completely reverse this argument by symmetry as follows.

As we know, $$4.8$$ years have passed on the spaceship between the Event B and the arrival. Thus in the frame of the spaceship, the Event B happened long after the Event A and when the star was already almost a half way through.

Based on this logic you can easily see and calculate the length contraction in both cases, but the effect will remain fully symmetrical.

The accepted answer is not correct. As summarized there, your assumptions are

1. In the reference frame of the rocket: At time t(0) , the distance between the rocket and the star is 10 light years.

2. In the reference frame of the star: At time t(0) , the distance between the rocket and the star is 10 light years.

According to the accepted answer, it's not possible for both these things to be true. But it is. Here is an accounting of the history, as recounted by the captain of the rocket ship (who considers himself always stationary):

Time -14.1: The star is 22.7 light-years away, though the folks living there say it is 10 light-years from me. The star-clock has just been set to time 0.

Time 0: The star is now 10 light-years away and I am setting my clock to 0. The clock on the star has been running for 14.1 years at speed .44, and now reads 6.2.

Time 11.1: The star has arrived, with its clock reading 11.1. [Since time 0 when it said 6.2, it has gained $$11.1 \times .44 = 4.9$$ ticks.]

(I've done some rounding off.)

An observer on the star (considering himself stationary) tells exactly the same story, with "the star" everywhere replaced by "the rocket ship". So everything is perfectly symmetric.

1. In setting the conditions for your thought experiment, you wrote: “At time 0, the distance between the star and the rocket is 10 light years”. It is unclear how the clock on the star and the clock in the rocket are synchronised to define the time 0, and how the distance was measured at that instance? I suggest the following addition to your experiment: there are two distant stars, the distance between them is known and does not change with time (it is an idealisation, but all thought experiments require some idealisation). The clocks at both stars can be synchronised (for example, by Einstein’s synchronisation). The rocket moves with constant speed along the straight line which connects the two stars. When the rocket flies over the first star close to its surface, it exchanges the light-speed signals with the star so they both register the reading of each other’s clock. The rocket moves on towards the second star. The clock on the second star is synchronised with the clock on the first star, so the time 0 is the same for the second star, the first star, and the rocket. Note, that we synchronise only the moment when we start counting time in both frames, and after that the time is measured separately in two frames. Now, if the reference frames of the second star and the rocket are symmetrical, as required by SRT (meaning that each of them observes itself being at rest, and the other one being in motion), then, in the reference frame of the star, the clock in the rocket must show less time than the clock on the star; and, in the RF of the rocket, the clock on the star must show less time than the clock on the rocket. When the rocket flies over the second star, they exchange the signals carrying information about the readings of their clocks. It is quite obvious that the readings will be the same, because both will show the proper time in each RF. This time will be less, though, than that calculated in Newtonian world, because the distance between the moving objects was contracted in both RF by the Lorentz factor. But this result will contradict the fact that in the system of the two stars, the distance covered by the rocket was not contracted, so the synchronised clock on the stars must register a longer time than the clock in the rocket. Obviously, a thought experiment which involves only binary relation between moving objects contradicts a thought experiment which involves a fixed system of distant objects and a particle in motion between them.
2. The question then arises, why in the real world the clock of a particle which moves at about the speed of light runs slower than the clock of the same particle at rest on Earth? And why the clock on the satellite runs slower than the clock on Earth? If the time were slowing down both ways, the clocks on the satellite and on Earth would show the same time when they exchange information, as described above, but this is not the case (take GPS for example).
• As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.
– Community Bot
Apr 13 at 20:52
• Asymmetry of SRT: We can say that a single clock always moves slowly relative to a set of synchronously running clocks if a single clock moves relative to this set. The readings of many clocks flying past individual clocks, on the contrary, always change at an accelerated rate in relation to individual clocks. In this regard, the term “time dilation” is meaningless without specifying whether this dilation refers to a single clock or a set of clocks synchronized and at rest relative to each other Apr 28 at 7:32