What happens when twins synchronize clocks and one teleports?

[Pyter]

TL;DR Summary

A variant of the twin paradox where the reference frames are symmetrical, as opposed to the classical one

Hello all.

Recently this twin paradox variant occurred to me, and I can't wrap my head around it:

Alice and Bob are in the same (roughly inertial, for our purposes) reference frame, separated by a sideral distance. Let's say Alice is on Earth and Bob on Pluto.
They synchronize their clocks at 0 through a radio signal from Alice to Bob.

The very moment Bob resets its clock to 0, Charlie flies by at constant relativistic speed, headed towards Alice, reads Bob's clock and resets its one accordingly. So that Alice, Bob and Charlie clocks are all synchronized.

Now the situation of Alice and Charlie is perfectly symmetrical: in Alice's RF Charlie is headed towards her at velocity -v, and in Charlie's, Alice is headed towards him at velocity -v.
When they finally meet, what are their clocks' readings? it can't be the same because for Alice, Charlie's clock runs slow due to the time dilation, and the same is true for Charlie and Alice's clock.
Who's "younger" and who's "older" when they meet?

I call it "twins' paradox with teleportation" because the outbound leg of the traveling "twin" is missing, there's only the inbound leg, no deceleration/acceleration is involved, and Charlie never switches his IRF.

Note: I did a quick check on the forum to see if something similar was already posted, the one coming closest is here, but in that case the second twin accelerates to reach the first one so their situations are not mirrored.

 

[Ibix]

Clock synchronisation is frame dependent. If Alice and Bob synchronise their clocks using their mutual inertial rest frame then Charlie's clock will show less time elapsed. If they synchronise using Charlie's rest frame, Alice's clock will show less time elapsed. There is even a frame where the two show the same elapsed time because they move with equal and opposite velocities.

The discussion in Will Flannery's thread applies here too. Acceleration at the start and/or end of the journey is irrelevant to the result.

 

[hutchphd]

Summary:: A variant of the twin paradox where the reference frames are symmetrical, as opposed to the classical one

So that Alice, Bob and Charlie clocks are all synchronized.

This phrase by itself has no meaning. And in your scenario Charlie will likely not agree that Bob and Alice's clocks have been properly synchronized.

 

[PeterDonis]

So that Alice, Bob and Charlie clocks are all synchronized.

No, they're not. Clocks that are not at rest relative to each other cannot be synchronized. So Charlie's clock cannot be synchronized with Alice's and Bob's. (Also, as @hutchphd points out, Charlie will not agree that Alice's and Bob's clocks are synchronized, since "synchronized" is frame-dependent.)

Now the situation of Alice and Charlie is perfectly symmetrical

No, it isn't. Alice is at rest relative to Bob, and Charlie isn't. And the time "zero" on Alice's clock is only simultaneous with the time "zero" on Bob's clock in Alice's and Bob's common rest frame. It is not simultaneous with the time "zero" on Bob's clock in Charlie's rest frame.

I call it "twins' paradox with teleportation" because the outbound leg of the traveling "twin" is missing

This is a misnomer; there cannot be any "twin paradox" with two observers who only meet once. They have to meet twice for any "twin paradox" to be present. The reason is that the fact of two observers meeting is invariant--it's independent of any choice of reference frame. But simultaneity--which pairs of spatially separated events happen "at the same time"--is not invariant; it depends on your choice of reference frame. You have failed to take that relativity of simultaneity into account in your analysis.

 

[Dale]

Now the situation of Alice and Charlie is perfectly symmetrical: in Alice's RF Charlie is headed towards her at velocity -v, and in Charlie's, Alice is headed towards him at velocity -v.

In Alice’s frame their clocks are synchronized when Charlie passes Bob. In Charlie’s frame they are not. They are not symmetrical.

 

[Janus]

Summary:: A variant of the twin paradox where the reference frames are symmetrical, as opposed to the classical one

Hello all.

Recently this twin paradox variant occurred to me, and I can't wrap my head around it:

Alice and Bob are in the same (roughly inertial, for our purposes) reference frame, separated by a sideral distance. Let's say Alice is on Earth and Bob on Pluto.
They synchronize their clocks at 0 through a radio signal from Alice to Bob.

The very moment Bob resets its clock to 0, Charlie flies by at constant relativistic speed, headed towards Alice, reads Bob's clock and resets its one accordingly. So that Alice, Bob and Charlie clocks are all synchronized.

Now the situation of Alice and Charlie is perfectly symmetrical: in Alice's RF Charlie is headed towards her at velocity -v, and in Charlie's, Alice is headed towards him at velocity -v.
When they finally meet, what are their clocks' readings? it can't be the same because for Alice, Charlie's clock runs slow due to the time dilation, and the same is true for Charlie and Alice's clock.
Who's "younger" and who's "older" when they meet?

I call it "twins' paradox with teleportation" because the outbound leg of the traveling "twin" is missing, there's only the inbound leg, no deceleration/acceleration is involved, and Charlie never switches his IRF.

Note: I did a quick check on the forum to see if something similar was already posted, the one coming closest is here, but in that case the second twin accelerates to reach the first one so their situations are not mirrored.

A lot of people have already mentioned the clock synchronization issue, and there is another factor to consider, length contraction.
Pluto is roughly 5 light hrs from Earth, so we will assume that this is the distance between Alice and Bob, as measured by either. Let's assume that Charlie is traveling at 0.8c between them.
Thus, by Alice's and Bob's clocks, it will take 6 1/4 hrs for Charlie to make the trip. ( if he passes Bob. when Bob's clock reads 12:00, he will reach Alice when her clock reads 18:15). During which time, They will note that Charlie's clock runs slow by a factor of 0.6 and advances only 3 hrs during the trip. If he set his clock to Bob's as they passed, his will read 15:45 when he gets to Alice.

However according to Charlie, it is Bob and Alice that are traveling at 0.8c. This causes three things.
He will measure their clocks as running slow by a factor of 0.6
He will measure them as length contracted by a factor of 0.6 ( And this includes the distance between them, which he will measure as only being 3 light hrs.
Alice and Bob's clocks will not be sychronized, with Alices clock reading 4 hrs ahead of Bob's.

At 0.8c, it takes 3. 3/4 hrs between Bob passing Charile until Alice crosses the 3 light distance them, and meets up with Charlie by Charlies clock. Charlie's clock,( having been set to 12:00 as Bob passed) reads 15:45 when Alice arrives at his position. This agrees with what Bob and Alice said his clock would read when he and Alice meet.
During this period, Both Alices and Bob's clocks tick off 3 1/4 x 0.6 = 2 1/4 hrs.
But since Alice's clock reads 4 hrs ahead of Bob's according to Charlie, when Charlie passes Bob, and set his clock to 12:00, Alice' clock already reads 16:00. Alice's clock then ticks off 2 1/4 hrs to read 18:15 when she meets up With Bob.
Thus while Alice, Bob, and Charlie disagree as to who's clock ticks slower, They all agree that Both Bob's and Charlie's clock read 12:00 as those two passed, and that Charlies clock reads 15:45 while Alice's clock reads 18:15 when Alice and Bob meet up.

 

[Pyter]

Let's recap the events sequence in Alice and Bob's frame:

a) Alice sends a radio impulse to Bob;
b) she waits for a time $d/c$, where $d$ is the distance between her and Bob, then resets her clock to 0;
c) at the same time, Bob's clock receives the radio impulse and reset itself to 0: Alice and Bob's clocks are now synchronized;
d) simultaneously, Charlie flies by, reads Bob's clock and resets his own to 0.

So Alice and Bob agree that when d) occurs, their clocks and Charlie's all read 0.
But Charlie, having a different "now line" from Bob, at c) sees Bob's clock at 0, but Alice's one in the future, say at t1 > 0.
This means that for Charlie, event b) has happened before c), at time -t1 in his IRF (after he's reset his clock)?

 

[PeroK]

Let's recap the events sequence in Alice and Bob's frame:

a) Alice sends a radio impulse to Bob;
b) she waits for a time $d/c$, where $d$ is the distance between her and Bob, then resets her clock to 0;
c) at the same time, Bob's clock receives the radio impulse and reset itself to 0: Alice and Bob's clocks are now synchronized;
d) simultaneously, Charlie flies by, reads Bob's clock and resets his own to 0.

So Alice and Bob agree that when d) occurs, their clocks and Charlie's all read 0.
But Charlie, having a different "now line" from Bob, at c) sees Bob's clock at 0, but Alice's one in the future, say at t1 > 0.
This means that for Charlie, event b) has happened before c), at time -t1 in his IRF (after he's reset his clock)?

Yes. It's simpler to use the Lorentz transformation. We have the original A-B rest frame and the C frame, which is moving (in the direction B to A) at some speed $v$.

We have two events: C and B setting their clocks to 0 can be taken at the common origin; and A setting her clock to zero has coordinates $(0, d)$ in the A-B frame.

Using the Lorentz Transformation we a time coordinate for this second event in the C frame:
$$t' = \gamma(t - \frac{vd}{c^2}) = -\frac{\gamma vd}{c^2}$$
I.e. in C's frame A set her clock before B did, so by the time B sets his clock (along with C) A's clock has already advanced (in C's frame).

 

[Ibix]

a) Alice sends a radio impulse to Bob;
b) she waits for a time $d/c$, where $d$ is the distance between her and Bob, then resets her clock to 0;

It's important to state who measures what every time. So here you should say that $d$ is the distance in Alice and Bob's frame.

c) at the same time, Bob's clock receives the radio impulse and reset itself to 0: Alice and Bob's clocks are now synchronized;

"...are now synchronised in Alice and Bob's frame".

So Alice and Bob agree that when d) occurs, their clocks and Charlie's all read 0.

Yes.

But Charlie, having a different "now line" from Bob, at c) sees Bob's clock at 0, but Alice's one in the future, say at t1 > 0.

Some care with language is needed here - he will always see Alice's clock in the past because of the finite travel time of light. But if he corrects for the travel time then he will calculate that, in his frame, Alice's clock reads after zero, yes.

This means that for Charlie, event b) has happened before c), at time -t1 in his IRF (after he's reset his clock)?

That depends on who is measuring t1. Is it a reading on Alice's slow-ticking clock? If so, then (b) happened at a time Charlie would call $-t_1/\gamma$.

 

[Janus]

Let's recap the events sequence in Alice and Bob's frame:

a) Alice sends a radio impulse to Bob;
b) she waits for a time $d/c$, where $d$ is the distance between her and Bob, then resets her clock to 0;
c) at the same time, Bob's clock receives the radio impulse and reset itself to 0: Alice and Bob's clocks are now synchronized;
d) simultaneously, Charlie flies by, reads Bob's clock and resets his own to 0.

So Alice and Bob agree that when d) occurs, their clocks and Charlie's all read 0.
But Charlie, having a different "now line" from Bob, at c) sees Bob's clock at 0, but Alice's one in the future, say at t1 > 0.
This means that for Charlie, event b) has happened before c), at time -t1 in his IRF (after he's reset his clock)?

As Charlie passes Bob, he and Bob see the same light that originated from Alice. For Both Bob and Charlie, in your scenario, this is light that left Alice sometime before she sets her clock to 0.
For Bob. this light left 5 hrs earlier, and took 5 hrs to cross the 5 light hr distance. Bob also knows ( by prior agreement) that Alice then waits 5 hrs before setting her own clock to zero.

Charlie seeing the same image of Alice, has to figure events like this:
The light I am now seeing from Alice has been traveling at c relative to myself.
Bob and Alice are moving at 0.8c relative to me.
Since I am now next to Bob, Alice is 3 light hrs away from me (length contraction).
However, the light I am seeing now had to have left Alice before I passed Bob, and thus left Alice when the distance between us was greater than 3 light hrs away.
If I backtrack the light I an seeing now( which is traveling at c) and Alice(who is presently 3 light hrs away and moving at 0.8c) they meet up at a point 15 light hrs away. Thus Alice was 15 light hrs from me when the light left, and it took 15 hrs for that light to reach me. In that 15 hrs, Alice's time ran 0.6 as fast as my own. The 5 hrs she waited by her clock was 8 1/3 hrs by my clock.
This means she set her clock to zero, 6 2/3 hrs ago by my clock, and during that remaining time ticked off 6 2/3 * 0.6 = 4 hrs, and reads 4 hrs past 0 at the moment I am passing Bob and his clock reads 0.
So yes, according to Charlie, Alice started her clock ticking from 0 before Bob started his clock from 0.
But because of the speed of light limitations, the effects of this can't produce any causality paradoxes.
For example, While Charlie would "know" that it is 4 hours later for Alice for him than it is for Bob as they pass, Any tangible information they both have about Alice is exactly the same. Charlie has no more idea about what really happened after Alice sent that signal than Bob does. Both have to assume that everything went by plan when they determine what time is on Alice's clock as Charlie passes.