# 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:

• 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. – PM 2Ring Aug 30 '19 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). – x-vision Aug 30 '19 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 $c$. – PM 2Ring Aug 30 '19 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. – Paradoxy Aug 30 '19 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 '19 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.