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The Twin Paradox is undoubtedly one of the most discussed things in special relativity and have a tendency to confuse most of us.

Classically, it's resolved by either stick to one of the three reference frames in question – one with the Earth at rest, one following the ship as it goes outwards and one following the ship as it comes back – or by invoking general relativity equalizing acceleration being in a strong gravitational field.

I, however, would like to find a solution where the ship is in rest all the time, tracking its "view" on the world.

Let's take real numbers. The binary star system Alpha Centauri is about 4.39 light years away making it the perfect candidate for this thought experiment in special relativity. To make calculations more simple we also choose the velocity of the ship such that $\gamma = 2$ is true: we find that $v = 0.87c$.

Now, the ship takes off from Earth heading towards Alpha Centauri with $v = 0.87c$. Since we are aboard on the ship, we instead see Alpha Centauri closing in on us with $v = -0.87c$ from a distance of $\frac{4.39 \; \text{ly}}{\gamma} = 2.20 \; \text{ly}$. This makes the duration of our journey only $\frac{2.20 \; \text{ly}}{0.87c} = 2.53 \; \text{years}$, while $1.26 \; \text{years}$ have passed on Earth.

We have reached our destination and look back at Earth, concluding that in a coordinate frame with our present ship as the origin $(0, 0, 0, 0)$ the Earth has the coordinates $(0, -2.20 \; \text{ly}, 0, 0)$.

We then instantaneously change our Earthly relative velocity from $v=0.87c$ to $v=-0.87c$ but we still want a reference frame in which we are at rest. We find that a such coordinate system has a relative velocity of $v = -0.99c$ to our reference frame making $\gamma = 7.09$.

Where is the Earth we looked back at in this new reference system?

Well, we had the event $(0, -2.20 \; \text{ly}, 0, 0)$, which we write $\begin{cases} t = 0 \\ x = 2.20 \; \text{ly} \end{cases}$

The coordinates are then transformed according to $\begin{cases} t' = \gamma(t - \frac{vx}{c^2}) \\ x' = \gamma(x-vt) \end{cases}$ yielding

$\begin{cases} t' = -15.44 \; \text{years} \\ x' = -15.60 \; \text{ly}\end{cases}$

We therefore conclude that our instantaneous turn have resulted in $15.44 \; \text{years}$ passed on Earth. But where is the simultaneous Earth?

Since the simultaneous Earth now has the relative velocity $v=0.87c$ its coordinates must therefore be

$\begin{cases} t' = -15.44 \; \text{years} + \Delta t\\ x' = -15.60 \; \text{ly} + 0.87 c \cdot \Delta t \end{cases}$

For $t' = 0$ to be satisfied we find $x' = -15.60 \; \text{ly} + 0.87 c \cdot 15.44 \; \text{years} = -2.2 \; \text{ly}$.

Perfect! That's the result we expected. Our ride back to Earth takes $\frac{2.20 \; \text{ly}}{0.87c} = 2.53 \; \text{years}$.

But here's my question.

For a stationary person on Earth we travelled a total distance of 8.78 light years with the speed of 0.87c. This translates into 10.1 years of Earth time.

But just our turn back caused the axis of simultaneously to shift the time on Earth 15.44 years in the future.

How is this possible?

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  • $\begingroup$ @JahanClaes But why does axis of simultaneity shift 15 years when the whole journey only took 10 Earth proper time years? $\endgroup$ – Markus Klyver Aug 27 '15 at 14:05
  • $\begingroup$ @JahanClaes No it's not, it's going to take 10 years only. $\endgroup$ – Omar Nagib Aug 27 '15 at 15:30
  • $\begingroup$ @JahanClaes How? The total distance (both ways) is 8.78 light years and the velocity is 0.87c. How do you yield 20 years from that? $\endgroup$ – Markus Klyver Aug 27 '15 at 15:44
  • $\begingroup$ Sorry, misread the problem! $\endgroup$ – Jahan Claes Aug 27 '15 at 17:30
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Perhaps it's more illuminating to look at the whole thing in a spacetime diagram. we have the earth frame with coordinates $(t,x)$, and its trajectory through spacetime is the blue line. The trajectory of the spaceship is the red one.

Straight worldlines are inertial frames of reference, curved or non-straight worldlines are non-inertial frames of reference. It's very clear that the spaceship is non-inertial frame, since it has a curved worldline, hence it's impossible to find an inertial frame attached to the ship, in which it can claim a state of rest all along the journey, as you wanted your solution to be.

However, What you considered in your calculations is the following idea:In outbound journey, the ship has an inertial frame $(t',x')$ attached to it, at the point of turnaround, the ship switches to another inertial frame $(t'',x'')$ which moves in the opposite direction with the same speed as in the outbound journey.

This act of switching inertial frames is your turning back. Now since $S'$ and $S''$ are two inertial frames in relative motion(they move opposite to each other), they have different conceptions of simultaneity. In the turnaround point, $S'$ claims events on his violet time axis are simultaneous. After switching to $S''$, you switch to another time and space axes, where the events(or points) on the green line are what is simultaneous for $S''$. It's very clear from the diagram why this act of switching creates this shift in time which is so puzzling to you.

So to sum up, the shift in time is caused by switching between two inertial frames in relative motion. When you switch between two different time axes(and space axes but this is not our concern here), this result in this shift in time of earth into the future.

Edit: I got the OP's question wrong. He wanted to ask how the shift in time for earth time(that is about $15$ years) is greater than the whole journey which is $10$ years. Well although I haven't checked your calculations, you must have done something wrong. It's pretty straight forward how to calculate the shift in time $t_\text{shift}$ from the diagram. The blue world line is made up of three parts, the first part that is before the intersection of violet line with the blue line which I'll call $\Delta t$, and the shift in time $t_\text{shift}$, and the last part which is after the intersection between the green line with the blue line which I call $\Delta t$ since by symmetry from the diagram, it's equal to the first time interval.

So that the whole time of earth journey is: $t_\text{total}= \Delta t+ t_\text{shift}+ \Delta t=t_\text{shift}+2\Delta t$.

So that: $t_\text{shift}=t_\text{total}-2\Delta t$.

$\Delta t$ is simply the time on earth clock when the ship reached the turn point(from the ship's perspective), therefore: $\Delta t= 2.53/\gamma=2.53/2=1.26$

Therefore: $t_\text{shift}=10-2(1.26)=7.48$.

So the shift in time should be $7.48$ assuming the whole journey on earth is $10$ years and that it takes $2.53$ years on the ship to reach the turn point. Redo you calculations.

Further Edit: I'm going to analytically calculate what the shift in time is.

It should be noted that at the turn point, the spaceship(which claims a state of rest) will have coordinates $(t,x)=(2.53,0)$, Owing to time dilation it will claim that $\dfrac{2.53}{2}=1.26$ years has passed on the clocks in the earth frame.

However this is different from what the clocks in the earth frame gonna read. The problem was set up in such a way that the coordinates of the earth and the spaceship coincide that is $(t,x)=(t',x')=(0,0)$.

So that according to the spaceship at the turn around point, the earth clock is gonna read $1.26$ years, and its position will be given by $-2.2$ light years. However, owing to the relativity of simultaneity, according to the ship frame, clocks in the earth frame are out of sync,so that the clock in the earth frame located at Alpha Centauri will read $1.26+l_0v/c^2$ years, not $1.26$ years, where $l_0$ is the rest length between earth and alpha.

Then the space ship switch to another frame with velocity $-0.87c$. You talked about the fact that such a frame will be moving with $0.99c$ with respect the old space frame and will have $\gamma=7.09$ but this is totally irrelevant here. What matters to calculations is that the ship has now just switched to another frame that is moving with $-0.87c$ with respect to earth, or in other words, it claims a state of rest, while the earth moves with $0.87c$ with respect to it.

We previously calculated that the clock at Alpha Centauri will read $1.26+l_0v/c^2$, this is still the case and has not changed. However owing to this switch of frames, clock at earth is gonna no longer read $1.26$ years, it will read $1.26+l_0v/c^2+l_0v/c^2$ years, that is the time at alpha centuri $1.26+l_0v/c^2$ plus the time difference $l_0v/c^2$ owing to relativity of simultaneity.

The shift in time on earth clock is the difference between what the clock read when the spaceship was moving with $0.87c$ and then switched to $-0.87c$, that is $t_\text{shift}= 1.26+l_0v/c^2+l_0v/c^2 -1.26=2l_0v=2(4.39)(0.87)=7.6$ years as calculated before.

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    $\begingroup$ Great example of clear technical writing. $\endgroup$ – WetSavannaAnimal Aug 27 '15 at 3:41
  • $\begingroup$ Great answer, but unfortunately you did not answer the question I asked. I did not ask why the slice of simultaneously changed but rather how the the shift in axis could cause a change of 15.4 years on Earth when the whole journey for someone on Earth should only take 10 years. $\endgroup$ – Markus Klyver Aug 27 '15 at 13:41
  • $\begingroup$ @MarkusKlyver Check my edit. $\endgroup$ – Omar Nagib Aug 27 '15 at 15:29
  • $\begingroup$ @OmarNagib Thank you. However, I still don't know what's wrong with my calculations. t' and x' gives the exact distance to Earth as we would expect. Changing those would violate length contraction as described by the Lorentz transformation. $\endgroup$ – Markus Klyver Aug 27 '15 at 15:43
  • $\begingroup$ @MarkusKlyver I think I figured out what's wrong with your calculatiuons, I'll further edit my answer after I return home. $\endgroup$ – Omar Nagib Aug 27 '15 at 15:59
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The axis of simultaneity, or in other words, the set of events which are simultaneous as measured in the rest frame of the ship, does indeed change suddenly when we turn back. This is because it depends on your reference frame. There isn't a single inertial frame that stays with the ship for the whole journey; you can either accept that the frame is non-inertial, which makes it much harder to define simultaneity, or you can switch frames at the same time you turn back. But if you switch frames the axis of simultaneity also changes, because it's not a real thing: it depends on your frame.

This kind of "paradox" is best handled if you forget about trying to make any kind of sense of simultaneity, because as you can see it's a tricky notion. It's better to think in terms of events and spacetime diagrams. In fact, drawing a diagram can help you understand why simultaneity changes suddenly when you turn back.

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