0
$\begingroup$

I've seen many questions answered on the twin paradox but they hand-wave the problem away using non-inertial frames. "you must accelerate away and come back therefore we use the Rindler metric". But I'm not interested in that, I'm interested in this inertial situation:

I am in my spaceship, you are on the ISS. We are twins. I am already moving at $86.6\%$ of the speed of light. As I whizz past you, right next to the ISS, we both start our clocks.

I travel to andromeda. I do not accelerate, I do not decelerate.

We both wait $1$ year according to our clocks, then we take a selfie, and we both fire out a radio signal containing our selfie and the timestamp towards each other.

By happenstance, an alien is in a UFO and at the midpoint and is static relative to you when the two radio packets both arrive at the same position. He intercepts both simultaneously, and decrypts them, and looks at both selfies.

Who is younger in the selfie? In your frame of reference I was moving at $0.866c$, so:

$$t' = \frac{t}{\sqrt{1-0.866^2}} = 2t$$

But my frame of reference you were moving at $-0.866c$, so

$$t' = t' =\frac{t}{\sqrt{1-(-0.866)^2}} = 2t$$

So in my frame you should've aged at half speed but in your frame I should've aged at half speed.

So what does the alien see? whose selfie is older?

(I would've liked to have used the example that as soon as I reached andromeda I collapsed a quantum entangled particle and told you to take a selfie upon collapse and then we saw who was older but I guess someone would just say "you can't transfer information through entanglement" so I had to use the alien example).

$\endgroup$
4
  • 5
    $\begingroup$ All twin paradox questions and confusions get resolved when you draw a spacetime diagram. Have you drawn a spacetime diagram? $\endgroup$ – Prof. Legolasov May 28 at 5:22
  • $\begingroup$ We both wait 1 year according to our clocks, then we take a selfie...Who is younger in the selfie?" Seriously? If you were both 35 when you passed each other, you both waited 1 year according to your own clocks, then you took the selfies --- then you are both 36 in the selfies. Sometimes it pays to stop and think for a moment before you start writing down formulas. (In this case you wrote down the wrong formula for the Lorentz transformation, which depends on both $t$ and $x$, not just $t$.) $\endgroup$ – WillO May 28 at 5:30
  • 2
    $\begingroup$ PS: @Prof.Legolasov gave you good advice re spacetime diagrams in general. But in this specific case, 30 seconds of thought would have sufficed. $\endgroup$ – WillO May 28 at 5:34
  • $\begingroup$ " We both wait 1 year according to our clocks, then we take a selfie, and we both fire out a radio signal " Do you wait 1 year, from the moment of when clocks were set to 0 , or 1 year from the moment of passing andromeda ? If you are waiting 1 year from the moment of passing ISS, then why is andromeda part of the question ? Isnt it better to remove the andromeda part entirely ? $\endgroup$ – silverrahul May 28 at 6:24
3
$\begingroup$

There is only one correct resolution to the twin "paradox" regardless of your setup. There is no need for hand-waving or use a special set of coordinates (such as the Rindler coordinates OP mentions). No. The only real definition for the time elapsed on someone's person clock is the proper time of their world-line. I would go as far as to argue that the only real way to avoid confusing oneself with relativity is to try and think exclusively in terms of world lines and what they are doing.

To make this precise, suppose that in the coordinates $x^\mu=\langle t,\vec x\rangle$, the observer on Earth is simply sitting at the origin and time is elapsing for them. That is, the world-line (path through spacetime) of the person on Earth may be written $x^\mu_E=\langle t,0\rangle$, the subscript $E$ for Earth. We can also write the path traveled by the spaceship person (in the same coordinates!) as $x^\mu_S=\langle t,\vec vt\rangle$. Normally, physicist would say that this is the path through spacetime taken by the spaceship as seen by the Earth observer.

We then need to define the proper time, and to do so we must introduce the Minkowski metric: $$ ds^2 = -dt^2+d\vec x^2=dx^\mu\eta_{\mu\nu}dx^\nu. $$ I am working in units where $c=1$ since the factors of $c$ aren't important for determining who's older. Note $\eta_{\mu\nu}$ is itself the Minkowski metric here, which is $\text{diag}(-1,+1,+1,+1)$. In this metric signature, a timelike curve will have negative length, so we define the proper time to be $d\tau=\sqrt{- ds^2}$, so $$ \tau = \int_C\sqrt{-d s^2} $$ is the time elapsed on the personal clock of someone traveling along the curve $C$ through spacetime.

If the curve $C$ is parametrized by a function $X(t) = \langle t,\vec x(t)\rangle$, then the above integral can be computed in the same way we would compute the arclength of a curve in a calculus 3 class: $$ \tau = \int_C\sqrt{-d x^\mu\eta_{\mu\nu} dx^\nu} = \int_{t_1}^{t_2}\sqrt{-\frac{dx^\mu}{dt}\eta_{\mu\nu}\frac{dx^\nu}{dt}}dt. $$ This last expression is just a regular 1D integral. Note that this makes no assumptions about the particular coordinates we are using (if you change to, for example, polar coordinates, you would need to also change the components of $\eta$, but it is what it is) and certainly makes no assumptions about what kind of path the person is taking through spacetime...so there is absolutely zero inhibition to thinking about accelerating observers.

So, let's see how this calculation works out for the Earth observer, starting their path at $t=0$ for sake of convenience. $$ \tau_E=\int_0^T\sqrt{1+0}dt = T. $$ This should not be a surprise, they were just sitting there till their clock hit $T$. What about the spaceship person? $$ \tau_S=\int_0^T\sqrt{1-v^2}dt = T\sqrt{1-v^2} = T/\gamma. $$ Again, this should not be surprising...we have two inertial observers and when the Earth observer finally sees the ship person reach their destination, they had to wait longer for the light of the spaceship at its destination to get back to Earth.

In the particular situation OP has described with an alien, I will first point out that saying the alien is "static" would seem to suggest OP has a preferential reference frame in mind. This of course cannot happen. But even that aside, OP has said that both people wait 1 year to take their selfie...so assuming they have followed their own instructions they will be indeed 1 year older in the selfie. The difference would come up in which selfie the alien receives first, and the answer to that would depend on exactly what the alien is doing. Again, there is no sense in which the alien can be "static."

$\endgroup$
1
$\begingroup$

When they take their selfies, both individuals will have aged by 1 year according to their local clocks. So the selfies will show them at the same age.

However, you have posed the question with a built in asymmetry. The alien observer is stationary with respect to the ISS and will see time elapse at the same rate as the twin on the ISS. He/she will however see time going at half that rate for the twin traveling to Andromeda. As a result, the alien will not receive the selfies simultaneously. The selfie from the ISS twin will arrive first and one year later the selfie from the Andromeda twin will arrive.

The asymmetry can be fixed by having the alien traveling towards Andromeda at half the speed. So he is always midway between the two twins. In this case he/she will simultaneously receive two selfies showing (as before) both twins at the same age.

$\endgroup$
0
$\begingroup$

You will each appear to have aged by one year. However, if you were to stop at Andromeda (I mean reduce your speed to 0 relative to your twin back on Earth) and ask someone there the date and time, you will find it to be much further ahead than your own watch and calendar would suggest. In other words, the people there will believe that your watch has been running slow since you synchronised it with your twin back on Earth.

The secret to all these supposed time dilation paradoxes is that a moving clock must be compared to two stationary clocks which the moving one passes in turn. The differences in the times shown by the two stationary clocks when the moving clock passes each of them is always less than the time that seems to have elapsed during the same period according to the moving clock. The discrepancy appears to be time dilation to the stationary clocks, but to the moving clock it seems to be due to the stationary clocks being out of synch.

In your example, suppose you had been trailed by another space ship moving at the same distance as you, and whose clock was synchronised with yours, and which was so far behind you that it arrived at Earth just at the same time as you arrive at Andromeda according to your clocks. From the perspective of you and your companion ship, you have both been stationary and it is your brother who has moved between you. The people on the companion ship will see that you brother has aged only one year, which will surprise them because their clocks will suggest it is a much longer time since your brother started his journey from you.

What happens is that the plane of simultaneity in your frame of reference is tilted from that in your brother's, so even if you synchronise your frame's clocks when you and your brother are together, the further out in your respective frames, time becomes increasingly out of synch.

To make that concrete, suppose you had also been accompanied by a third ship travelling at the same speed as you but so far ahead that they were already at Andromeda when you were just passing Earth (according to your clocks). For the forward companion ship the time would be the same as it is for you, but if they were to ask the Andromeda locals (in the same frame as your brother) they would have an entirely different view of the date and time.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.