Undergrad A version of the twin paradox without accelerations

[Ibix]

Based on how I solved the problems...would it be correct to say that EVERYONE agrees that Alex passes Bob at t=0, but Alex will think his trip is shorter (in time) than Alice and Bob believe it is?

That's a bit vaguely stated; precise would be to say that Alice and Bob experience more proper time during the trip (by their definition of "during") than does Alex.

It seems to me that Alex would still agree that Alice and Bob's clocks are synchronized, but based on this thread that seems to not be the case.

Use the Lorentz transforms to work out what time Bob and Alice's clocks show at $t'=0$, which is what Alex calls "at the same time as I passed Bob". Since Alice and Bob don't have the same $x$ coordinate, it should be immediately obvious from $t'=\gamma(t-vx/c^2)$ that the times aren't the same.

 

[PeterDonis]

would it be correct to say that EVERYONE agrees that Alex passes Bob at t=0

If you make that event the spacetime origin (which I think is the easiest way to set up the two frames), then yes.

It seems to me that Alex would still agree that Alice and Bob's clocks are synchronized

No, he won't. Alex will agree that Bob's clock reads zero at time $t' = 0$ in his frame (since he's passing Bob at that time in his frame). But he will not agree that Alice's clock reads zero at time $t' = 0$ in his frame. He will say that Alice's clock reads some large positive value at $t' = 0$ in his frame. Whereas of course Bob will say that Alice's clock reads zero at $t = 0$ in the Bob-Alice rest frame. This is, of course, just a manifestation of relativity of simultaneity.

 

[Isaac0427]

I drew out the Minkowski diagrams for this situation-- I just wan't to make sure I did this right (note I have color coded the three people on the diagrams)

First, I do know that d₀>d', and t_A=t_B>t₀ (right?). My question is on the time I labeled with a ? on both diagrams. If I am correct, that time will be smaller than t₀, and will be the time that Alex thinks Bob measures when Alex crosses Alice (i.e. if someone a distance d' behind Alex in Alex's frame, whose clock was zeroed at the same time as Bob's clock, looks at Bob's clock when Alex passes Alice). This would show how Alex believes that Bob and Alice both experienced time dilation, while still maintaining that Alice's clock reads higher than his when they pass. It would also seem like if Alice took a picture of her and Alex's clocks when Alex passes her, her pictures would be the same as Alex's pictures--right?

 

[George Plousos]

Is there any problem: All the answers claim that the photos show that Alex will be younger than Alice. But we can assume that there is a new person in Alex's reference system, whom we will name Helen (assume that Helen was present in the original experiment but was not declared). The distance that separates Alex from Eleni is equal to the distance that separates Bob from Alice. Alex and Helen clocks are synchronized with Einstein's method. When Alex stops where Alice is, Helen will stop where Bob is. In addition, Bob photographs his own clock and Helen's clock before Helen stops.

Based on the answers given, Bob's photos will show that Bob is younger than Helen. Therefore, photographs by Alex and Bob show that these two observers are still in sync. This fact is contradictory if we consider the relationships that connect the four photographed watches. However, I suspect the answer will be that the photos of Alex and Bob were not taken at the same time. But then a big problem arises: For Relativity to be consistent, the real age differences between Alex and Bob will depend on the following choices.

1. Alex slows down and stops near Alice.
2. Bob slows down and stops near Helen.
3. Bob and Alex both slow down in the same way to stop next to the girls.

So, since choices 1 and 2 mean that Alex and Bob do not stop at the same time, then choice 3 seems unfeasible, as the third choice ensures the synchronization of Alex and Bob's watches.

I believe that these problems are due to the inability to explain the elementary problem that follows:

Let t' be the time measured by the clock moving at speed u and t is the time measured by the immovable clock. If Alex moves with speed u = 0.6c the relevant formula becomes

t' = t * sqrt{1-0.6^2} = 0.8t

The point that creates the most problems, based on the above, is located in the possibility to consider alternatively Alex as immobile and Bob moving towards him. So, Alex would think that Bob clock are left behind. But according to the initial view, Bob would think something like that about Alex's clock. They do not constitute these things a paradox? The explanation given is that they will not be able to compare the clocks without Alex starting to slow down to reverse his course and slow down again until he stops. The above relationship does not explain what happens to clocks that undergo such changes and I'm not going to get into that.

Since I could not give a physical explanation to the above problems of the four observers, I tried to give at least an artificial interpretation with the help of Minkowski diagrams, with speed u = 0.6c and distance x = 1 light year for each pair of observers, but I was confused. I watched the above discussions and saw Isaac0427's diagram, but I find it difficult to do something similar.

 

[Dale]

All the answers claim that the photos show that Alex will be younger than Alice. But we can assume that there is a new person

I am confused now. Please provide a diagram (in any chosen reference frame) showing who is where and when.

 

[Janus]

Is there any problem: All the answers claim that the photos show that Alex will be younger than Alice. But we can assume that there is a new person in Alex's reference system, whom we will name Helen (assume that Helen was present in the original experiment but was not declared). The distance that separates Alex from Eleni is equal to the distance that separates Bob from Alice.

As measured by Bob and Alice or Alex and Helen?

Alex and Helen clocks are synchronized with Einstein's method. When Alex stops where Alice is, Helen will stop where Bob is.

Again, according to who? Does Bob and Alice say that that Helen stops at Bob when Alex stops at Alice, or does Alex and Helen claim this.

In addition, Bob photographs his own clock and Helen's clock before Helen stops.Based on the answers given, Bob's photos will show that Bob is younger than Helen.
Therefore, photographs by Alex and Bob show that these two observers are still in sync.

If we assume that the distance between Helen and Alex is 0.866 ly as measured by Helen or Alex, and their clocks are synced according to this pair. This allows Alex to reach Alice and Helen to reach Bob at the same time, according to Helen and Alex and with both of their clocks reading the same time, of 1.732 yrs .(Assuming a 0.5c relative speed), During that 1.732 yrs, Bob's clock will tick off 1.5 yrs. So that is the time his clock shows when Helen with her clock reading 1.732 yr.
If we consider this from Bob's rest frame. Helen and Bob are 0.75 ly apart, So Helen starts off that far away from Bob. Due to the relativity of simultaneity, Helen's clock already reads 0.433 yrs. It takes 1.5 yrs on Bob's clock for Helen to reach him, during which time Helen's clock is time dilated and ticks off 1.299 yrs. Since it started at 0.433 yr, it will read 1.732 years upon reaching Bob. This is exactly what Alex and Helen would say.
However, Alex's clock only reads 1.299 yrs. But since the distance between Alex and Helen is only 0.75 ly, he is still 0.25 ly short of reaching Alice at this moment. He has to travel for another 0.5 yrs, accumulating another 0.433 yrs to reach Alice.
Ergo, according to Bob and Alice, Helen and Alex to not stop "at the same time" Alice would stop at Bob and Alex would continue on until he reaches Alice.
So, according to Bob and Alice, for the majority of the time Helen's and Alex's clocks are not in sync with each other, and only become so after Alex finally comes to a stop.*

This fact is contradictory if we consider the relationships that connect the four photographed watches. However, I suspect the answer will be that the photos of Alex and Bob were not taken at the same time. But then a big problem arises: For Relativity to be consistent, the real age differences between Alex and Bob will depend on the following choices.

1. Alex slows down and stops near Alice.
2. Bob slows down and stops near Helen.
3. Bob and Alex both slow down in the same way to stop next to the girls.

Whichever of the choices you make will not change the final results. It will only change how any particular observer determines how these results were achieved.

So, since choices 1 and 2 mean that Alex and Bob do not stop at the same time, then choice 3 seems unfeasible, as the third choice ensures the synchronization of Alex and Bob's watches.

I believe that these problems are due to the inability to explain the elementary problem that follows:

Let t' be the time measured by the clock moving at speed u and t is the time measured by the immovable clock. If Alex moves with speed u = 0.6c the relevant formula becomes

t' = t * sqrt{1-0.6^2} = 0.8t

The point that creates the most problems, based on the above, is located in the possibility to consider alternatively Alex as immobile and Bob moving towards him. So, Alex would think that Bob clock are left behind. But according to the initial view, Bob would think something like that about Alex's clock. They do not constitute these things a paradox? The explanation given is that they will not be able to compare the clocks without Alex starting to slow down to reverse his course and slow down again until he stops. The above relationship does not explain what happens to clocks that undergo such changes and I'm not going to get into that.

Since I could not give a physical explanation to the above problems of the four observers, I tried to give at least an artificial interpretation with the help of Minkowski diagrams, with speed u = 0.6c and distance x = 1 light year for each pair of observers, but I was confused. I watched the above discussions and saw Isaac0427's diagram, but I find it difficult to do something similar.

There are no contradictions involved, as long as you are careful to apply all the Relativistic effects correctly.

* This is one of those instances where you have really be careful if you try to assume instantaneous velocity changes, as they can introduce contradictions (contradictions that don't arise in reality, because you can't actually have instantaneous velocity changes.)[/QUOTE]

 

[Ibix]

Before I start, @George Plousos, a quick warning: you've taken a fairly simple scenario, listened to our explanations, and added a new element to it, leading to more confusion. This is a fairly common pattern here (I call it "yes, but...") and tends to repeat - you add a complication, get an explanation, go "yes, but" and add another complication. It never ends well. The problem just gets messier without adding to your understanding, and we get frustrated repeating variants on the same thing (usually, "you don't seem to understand the relativity of simultaneity"). With that in mind, I strongly suggest that you study the maths, rather than adding more elements to your scenario - it's only yourself you are tripping up with the added complexity.

Let's look at the maths. Your original scenario was:

• Bob, at rest.
• Alice, at rest with respect to Bob, 10 light years distant as measured in their rest frame.
• Alex, moving from Bob to Alice at 0.6c as measured by them.

And the question was what time does each person's watch read when Alex meets Alice?

The easiest frame to work in is Bob and Alice's rest frame, since most of the information is specified in that frame. What we need to write down is the position (call this $x$) and time (call this $t$) of every event of interest. Events of interest are:

• Alex passes Bob and both start their clocks
• Alice starts her clock
• Alex reaches Alice and they both stop their clocks
• Bob stops his clock

Let's measure distance in light years and time in years. Further, let's say that Bob is at rest at the origin ($x=0$) of the spatial coordinates. That means that Alice is at $x=10$. We agreed that they start their clocks simultaneously in this frame at $t=0$. We don't know when Alex reaches Alice, but we can calculate it from Alex's velocity - to cross 10 light years at 0.6c takes 10/0.6=50/3=16.7 years. Neither Alice nor Bob has moved, so their $x$ coordinates remain the same, and we agreed that both stop their clocks simultaneously in this frame at $t=16.7$. So, to summarise:$$
\begin{array}{|r|c|}
\hline
&x,t\\
\hline
\textsf{Alex passes Bob}&0,0\\
\textsf{Alice starts her clock}&10,0\\
\textsf{Alex reaches Alice}&10,16.7\\
\textsf{Bob stops his clock}&0,16.7\\
\hline
\end{array}$$OK so far?

We now need to determine the position and time of these events in the frame where Alex is at rest. These quantities are usually denoted with a prime, $x'$ and $t'$, to distinguish them from the first set. To find these, you use the Lorentz transforms:$$\begin{eqnarray*}
x'&=&\gamma(x-vt)\\
t'&=&\gamma(t-vx/c^2)
\end{eqnarray*}$$where $\gamma=1/\sqrt{1-v^2/c^2}$. The velocity, $v$, is Alex's velocity, 0.6c, giving $\gamma=1.25$. We just plug in the $x,t$ values from the table above to generate $x',t'$ values. They are:$$\begin{array}{|r|c|c|}
\hline
&x,t&x',t'\\
\hline
\textsf{Alex passes Bob}&0,0&0,0\\
\textsf{Alice starts her clock}&10,0&12.5,-7.5\\
\textsf{Alex reaches Alice}&10,16.7&0,13.3\\
\textsf{Bob stops his clock}&0,16.7&-12.5,20.8\\
\hline
\end{array}$$You can check my results on a calculator (or spreadsheet, which makes this easier) if you like. Remember that $c$ is one light year per year in these units. Now we can simply read off the time Alex's clock reads when he reaches Alice: 13.3 years (or, more generally, the difference between the start event and end event, but the start event is at $t'=0$ in this case). Note that Alex is unsurprised to find that Alice's clock is ahead of his: she started it at -7.5 years in this frame, so it's been ticking for 20.8 years (125/6 years, to be precise) at 0.8 of the rate of his, so he expects it to read 0.8×125/6=16.7 years, as it does. Furthermore, Bob's clock ticks slowly but he doesn't stop his clock until 20.8 years (again, actually 125/6) have passed, which explains why he stops his clock at 16.7 years. I stress that, in Alex's $x',t'$ coordinates, Alice and Bob did not start or stop their clocks simultaneously. This is the bit everyone forgets when they're thinking about time dilation and length contraction.

If you wish to draw a Minkowski diagram, now you can - draw one using the $t$ coordinate as the vertical axis and $x$ as the horizontal, and another with $t'$ as the vertical and $x'$ as the horizontal.

Finally, let's add your new complication, Helen. You say:

The distance that separates Alex from Eleni is equal to the distance that separates Bob from Alice. Alex and Helen clocks are synchronized with Einstein's method. When Alex stops where Alice is, Helen will stop where Bob is.

I presume Eleni is a misprint for Helen. Unfortunately, it's not clear which frame you are using when you say they are the same distance apart as Bob and Alice and they stop at the same time. I'm going to assume that you mean in Alex's rest frame. If you didn't mean that, you'll need to modify the following.

All we need to do is add a row to our summary table for the events "Helen starts her clock" and "Helen reaches Bob and stops her clock". This is actually easiest to do in Alex's rest frame, because we can simply copy the time coordinates from the "Alex passes Bob" and "Alex reaches Alice" events, and the distance between Bob and Alice is 8 light years in this frame so the position coordinates just need to be -8. Adding this to our table:$$\begin{array}{|r|c|c|}
\hline
&x,t&x',t'\\
\hline
\textsf{Alex passes Bob}&0,0&0,0\\
\textsf{Alice starts her clock}&10,0&12.5,-7.5\\
\textsf{Alex reaches Alice}&10,16.7&0,13.3\\
\textsf{Bob stops his clock}&0,16.7&-12.5,20.8\\
\textsf{Helen starts her clock}&&-8,0\\
\textsf{Helen reaches Bob}&&-8,13.3\\
\hline
\end{array}$$Then you just need to rearrange the Lorentz transforms to calculate $x$ and $t$ from $x'$ and $t'$ instead of the other way around. You can do some algebra, look up the inverse Lorentz transforms, or just observe from symmetry that the result is:$$\begin{eqnarray*}
x&=&\gamma(x'+vt')\\
t&=&\gamma(t'+vx'/c^2)
\end{eqnarray*}$$Then you can use these to fill in the blanks in the table:$$\begin{array}{|r|c|c|}
\hline
&x,t&x',t'\\
\hline
\textsf{Alex passes Bob}&0,0&0,0\\
\textsf{Alice starts her clock}&10,0&12.5,-7.5\\
\textsf{Alex reaches Alice}&10,16.7&0,13.3\\
\textsf{Bob stops his clock}&0,16.7&-12.5,20.8\\
\textsf{Helen starts her clock}&-10,-6&-8,0\\
\textsf{Helen reaches Bob}&0,10.7&-8,13.3\\
\hline
\end{array}$$And now you have everything you need to understand Helen: in Alex's frame she started her clock simultaneously with him and met Bob 13.3 years later. In Bob and Alice's frame, she started her clock six years before them and met Bob 16.7 years later (same journey time as Alex) when Bob's clock reads 10.7 years, six years before Alex reaches Alice.

There cannot be any contradictions or ambiguities here. The whole process is simply an affine transform. All of the problems people have arise because they try to use time dilation and length contraction formulae, which are special cases of the Lorentz transforms and do not always apply. The first step in learning to understand relativity is the recipe I've used above:

1. Write down the coordinates of events of interest in one frame or another
2. Apply the Lorentz transforms and/or their inverse
3. Read off the results

I do recommend drawing Minkowski diagrams because they help to build intuition, so you don't really need to do the maths after a while. But it's nearly dinner time so I'll put that off until later.

 

[Ibix]

And here are Minkowki diagrams. I suggest that you compare the locations of the crosses to the coordinates given in my last table in the previous post. First, here's one drawn in the rest frame of Alice and Bob (using numbers from the left hand column of the table):

Bob's worldline is drawn in red and Alice's in green. Neither is moving so their lines are vertical. Alex's worldline is blue dashed, and moves from Bob to Alice. Helen's worldline is drawn in purple dashed and meets Bob's worldline. I've added crosses at the end of each worldline, to indicate that these are the events where clocks are started and stopped. You can see immediately that Alice, Bob and Alex all start and stop their clocks simultaneously in this frame, but Helen starts and stops at completely different times.

Now look at the diagram in the rest frame of Alex and Helen (numbers from the right hand column in the table):

The colours are the same as before. You can see that the dashed lines (Alex and Helen) start and stop simultaneously, but in this frame the solid lines (Alice and Bob) do not. It's this difference of simultaneity that makes all the direct observables ("what my clock and yours show as we pass each other") consistent in every frame. Unfortunately, it's also this difference of simultaneity that is left out of a lot of pop-sci accounts of relativity. Don't make the same mistake.

 

[Ibix]

My question is on the time I labeled with a ? on both diagrams. If I am correct, that time will be smaller than t0, and will be the time that Alex thinks Bob measures when Alex crosses Alice (i.e. if someone a distance d' behind Alex in Alex's frame, whose clock was zeroed at the same time as Bob's clock, looks at Bob's clock when Alex passes Alice).

I'd phrase it as Bob's coordinate time for the event on Bob's worldline that is simultaneous in Alex's frame with Alex and Alice's meeting. Despite the complexity of the sentence, you can calculate it easily. You know that $x=0$ (Bob's position) and $t'=t_0$ (Alex's arrival time), so the Lorentz transform for $t'$ is $t_0=\gamma(t-v×0/c^2)$ and hence your question mark time is just $t_0/\gamma$.

This is, of course, the "paradoxical" result of both frames saying the other's clock ticks slowly. But from the diagrams it's easy to see how to resolve the paradox: Bob stops his clock simultaneously with Alex's arrival using his frame's simultaneity, not Alex's frame's simultaneity. In other words, the frames have different definitions of what "during Alex's trip" means for Bob, so naturally they disagree about how much time elapsed on Bob's clock "during Alex's trip". They do not disagree what time Bob's clock shows when he stops it (or photographs it, as the OP says).

 

[Isaac0427]

I'd phrase it as Bob's coordinate time for the event on Bob's worldline that is simultaneous in Alex's frame with Alex and Alice's meeting.

The reason I phrased it as I did is that if Alex took a picture of Bob's watch, accounting for the time it took light to get there, he would conclude that Bob's watch hit the "?" time simultaneously with Alex crossing Alice. Bob would disagree with this, but that's okay, as spacelike events need not be simultaneous in all frames of reference.One other point that I want to make sure I am correct on:
Say Alex and Alice are twins. This would mean their "biological clocks" were "synchronized" in some mutual rest frame. It seems as though the question of who is older would still require acceleration to resolve, right? If they were born at rest wrt. Bob and Alice's rest frame (in the current situation), then it seems Alex would be younger, but if they were born at rest wrt. Alex's rest frame (in the current situation), then Alice would be younger. This is still consistent with the photographs, as Alice zeroed her clock when she was biologically younger than Alex when he zeroed his clock (i.e. in Alex's frame, the "birth frame," she zeroed her clock before Alex).

Is that correct or nonsense?

 

[Nugatory]

Is that correct or nonsense?

It seems reasonable to think that the twins were colocated with one another and their mother at the moment of their birth, so their biological clocks were synchronized at time zero. (Note that colocation is all that's necessary; I'd expect that they were also at rest reative to one and their mother, but that is not a requirement).

If they separate, then reunite we can calculate the amount each one ages just by looking at the length of their respective paths through spacetime. That either or both accelerated is irrelevant, although we can infer that at least one of them was accelerated if they followed different paths.

 

[Isaac0427]

It seems reasonable to think that the twins were colocated with one another and their mother at the moment of their birth, so their biological clocks were synchronized at time zero. (Note that colocation is all that's necessary; I'd expect that they were also at rest reative to one and their mother, but that is not a requirement).

If they separate, then reunite we can calculate the amount each one ages just by looking at the length of their respective paths through spacetime. That either or both accelerated is irrelevant, although we can infer that at least one of them was accelerated if they followed different paths.

Assuming throughout their lives they have only gone at nonrelativistic velocities, then there appear to be two cases:
1. The birth frame is at rest wrt. Bob and Alice, in which case Alex is older.
2. The birth frame is at rest wrt. Alex, in which case Alice is older.

Is it even possible to construct a true twin paradox without acceleration being relevant? Otherwise, how could one twin be moving wrt. the other and still end up in the same place to compare ages?

 

[Nugatory]

Is it even possible to construct a true twin paradox without acceleration being relevant?

In flat spacetime it will always be the case that at least one of the twins was subject to proper acceleration. However, in a curved spacetime there can be multiple acceleration-free paths between the separation and reunion events, so we need the different paths explanation for these cases. But that explanation also works in the flat spacetime case... so there’s no reason to think of acceleration as anything more than one of the ways of setting the twins on different paths through spacetime. It’s the only way that works in flat spacetime, but that’s just because our options are limited there.

Consider also the case in which the outbound traveller meets another traveller moving inbound and the inbound one sets their clock to whatever the outbound clock read at their meeting. Now when inbound arrives back at Earth their clock will tell us exactly the same thing as the biological clock of a twin making the round trip, and the result will be just as “paradoxical” (travelers say that the stay-at-home clock is slower at all times, yet also say that it ends up ahead). But there’s no acceleration anywhere,

Treating the acceleration as the cause of the differential aging, as opposed to a corollary of the having different paths, is like saying that turning the steering wheel of a car is what causes its odometer to disagree with the odometer of a car moving in a straight line to the destination. It’s simpler to say that the odometers disagree because the cars traveled different distances and that it’s not surprising to find different steering inputs on different paths.

 

[PeroK]

Is it even possible to construct a true twin paradox without acceleration being relevant? Otherwise, how could one twin be moving wrt. the other and still end up in the same place to compare ages?

Fundamentally, this is one problem with using twins rather than clocks to measure time. Geometrically, the twin paradox is simply the hyperbolic geometry equivalent of the triangle inequality in Euclidean geometry:

https://www.physicsforums.com/threads/twin-paradox-explained-for-laymen.991504/page-4#post-6370123

It is, of course, possible to synchonise clocks using the Einstein convention. You could, I suppose, synchronise twins by moving them about locally at high relativistic speed!

The concept of twins in the paradox is an artifact aimed at obsuring the nature of spacetime geometry. To see the physics of it clearly, in my view, you must replace the twins by clocks and not be distracted by inessentials, which are irrelevant to the phsyics.

 

[vanhees71]

In flat spacetime it will always be the case that at least one of the twins was subject to proper acceleration. However, in a curved spacetime there can be multiple acceleration-free paths between the separation and reunion events, so we need the different paths explanation for these cases.

Here is a concrete simple example in Schwarzschild spacetime:

https://www.physicsforums.com/threads/twin-paradox-for-freely-falling-observers.991709/

 

[PeterDonis]

It is, of course, possible to synchonise clocks using the Einstein convention.

Yes, but they have to be at rest relative to each other during the entire synchronization process, and if they ever move relative to each other, the synchronization is lost. So this method is actually quite limited in practical usefulness.

 

[George Plousos]

It seems reasonable to think that the twins were colocated with one another and their mother at the moment of their birth, so their biological clocks were synchronized at time zero. (Note that colocation is all that's necessary; I'd expect that they were also at rest reative to one and their mother, but that is not a requirement).

If they separate, then reunite we can calculate the amount each one ages just by looking at the length of their respective paths through spacetime. That either or both accelerated is irrelevant, although we can infer that at least one of them was accelerated if they followed different paths.

Ibix's explanations helped me understand how Minkowski diagrams are made - I feel grateful for that. I think I have become accustomed to this and my confidence in Relativity has increased somewhat, mainly in terms of the mathematical consequence of the theory. Whatever modifications I tried in the original experiment and with any parameter values (eg with the prices suggested by Janus, and with the participation of Helen), these diagrams always gave a clear mathematical interpretation. But when I try to understand - from a physical point of view - what is happening in the watches of the four observers, I am forced to use acceleration, which makes the explanation more difficult if not impossible. I have stayed at this point.

Isaac0427 suggests looking at the case from the point of view where Bob and Alice are considered twins. Then a problem arises related to the mother who gave birth to the twins and their separation at a great distance. I think this problem can be overcome with the following argument: There is no law forbidding the existence of two identical people at any distance at the same time - in the sense that the twins are synchronized with Einstein's convention, and that should be enough for this case. That is, twins do not have to be born to a single mother and then separate. If I'm right, how should Isaac0427's question be answered?

 

[Nugatory]

That is, twins do not have to be born to a single mother and then separate.

If the twins don’t synchronize their clocks (declaring them twins just means that we consider their biological clocks to be synchronized, but we could use their wristwatches instead) while colocated, separate, and then reunite so that we can compare their clocks again while colocated then we don’t have the twin paradox, we just have a routine application of the relative of simultaneity.

 

[Ibix]

But when I try to understand - from a physical point of view - what is happening in the watches of the four observers

If you write down any two events, $x_1,t_1$ and $x_2,t_2$ you can calculate $c^2(t_2-t_1)^2-(x_2-x_1)^2$. If you then use the Lorentz transforms to calculate $x'_1,t'_1$ and $x'_2,t'_2$ for any $v$ you like and then calculate $c^2(t'_2-t'_1)^2-(x'_2-x'_1)^2$ you will find that the answer is the same (you can prove this straightforwardly by algebra, or just pick random numbers for the unprimed quantities and repeat until you get bored). This quantity is called the interval. Furthermore, if you pick events that happen in the same place in one frame (say the primed one) then you will find that the interval is just $c^2(t'_2-t'_1)^2$ - i.e., the square of the elapsed time for a clock that's at rest, multiplied by $c^2$.

The interval is the spacetime equivalent of length in space, which we are familiar with. The length of a straight line between space coordinates $x_1,y_1$ and $x_2,y_2$ is given by $l^2=(x_2-x_1)^2+(y_2-y_1)^2$. The interval along a straight line between spacetime coordinates $x_1,t_1$ and $x_2,t_2$ is given by $s^2=c^2(t_2-t_1)^2-(x_2-x_1)^2$. What this means is simple: the reason that the twins' watches read different elapsed times is that their watches measure the interval (the "length") along the paths they have followed through spacetime, and they followed paths with different intervals. It's like two cars zeroing their odometers , going on different journeys, and meeting up again. You wouldn't be surprised that their odometers read different things. With the insight that elapsed time is just "length" through spacetime (give or take factors of $c$), you shouldn't be surprised that their clocks read different things too.

You can easily calculate the interval along the worldlines of Alice, Bob et al in your thought experiment, or in a normal twin paradox scenario. Acceleration isn't relevant, except that in special relativity at least one of the twins must have accelerated if they are to meet up again. You can easily create scenarios where one twin does more acceleration but ends up the same age, or younger or older.

 

[Ibix]

Is it even possible to construct a true twin paradox without acceleration being relevant?

What do you mean by a "true" twin paradox? Just the standard example? Yes, but only in general relativity where you can have one or both twins in freefall paths that meet multiple times. Post #45 has a link.

 

[George Plousos]

I will study your notes, Ibix. In the meantime I have prepared this thought experiment:

I think I can show that the answer to the original question may be different. To achieve this I will use two additional people, Helen and Tom, but in the end I will remove these two people as if they never existed, so only Bob, Alice and Alex will remain on stage, as the original question wants. The image helps a lot.

Fig 1: The distance that separates Bob from Alice in the yellow reference frame is the same as the distance that separates Alex from Helen in the pink reference frame. The clocks of the observers in each separate reference frame are synchronized with the Einstein method. Tom is considered immobile, while Bob and Alex move at the same speed in Tom's view. At the meeting point of the three boys, their three clocks are synchronized.

Fig 2: At some point along the way, Alex will meet Alice and they will be photographed with each other. At some point the same thing will happen between Bob and Helen. We will return to figure 2 later.

Fig 3: Tom meets Alice and Helen. Because of the symmetry, Alice and Helen's clocks remain synchronized in Tom's view. If you want, the two girls are photographed with each other. Immediately after this observation, Tom sees Alice and Eleni slowing down at the same speed and stopping. Obviously, in the short time between Tom's last observation and the immobilization of the girls, there is no dramatic change in their clocks, but this is true anyway because of the symmetry of the system with respect to Tom.

Now all observers are motionless with each other - they are in the same frame of reference. Alice is in sync with Bob and Helen, who is in sync with Alex. So Bob, Alice, Alex and Helen are all in sync with each other. But then all four observers must have been in sync at all stages of the journey between Figures 1 and 3. Consequently, in the events described above for Fig 2 these four observers would also be in sync. We can now forget the involvement of Helen and Tom in the experiment, to find that the same conclusions apply to the original question - we now refer to the facts of figure 2 without Helen and Tom.

The answer to the original question was that the photos between Alex and Alice would show that Alex's watch was going backwards relative to Alice's watch before Alex stopped, but based on the new arguments the same photos must show that their clocks have remained in sync.

I can not see any error in the above arguments, unless I am blind because of my prejudice, or I have not learned everything I should.

 

[Nugatory]

or I have not learned everything I should.

You haven’t.
Go back and read post #37 by @Ibix in this thread. Then follow the recipe he outlined there, writing down the coordinates assigned to each relevant event using one frame and then applying the Lorentz transformations to find the coordinates using the other frame.

 

[PeterDonis]

To achieve this I will use two additional people, Helen and Tom, but in the end I will remove these two people as if they never existed, so only Bob, Alice and Alex will remain on stage, as the original question wants.

It has already been pointed out that, if you don't have the basic scenario with just three people (Bob, Alice, and Alex) in it straight on its own, adding more people won't help, it will just confuse you.

I strongly suggest that you take a step back and really consider this advice.

 

[Ibix]

In the meantime I have prepared this thought experiment:

You are playing the "yes, but..." game I mentioned in #37. As I said in that post, it confuses you, frustrates us, and you don't learn anything. Focus on applying the maths instead of verbal reasoning about ever more complex scenarios.

The answer to the original question was that the photos between Alex and Alice would show that Alex's watch was going backwards relative to Alice's watch before Alex stopped, but based on the new arguments the same photos must show that their clocks have remained in sync.

There cannot be contradictions in this type of scenario because the Lorentz transforms are analogous to rotations in the same way interval is analogous to length. Claiming a contradiction here is like drawing a triangle on a piece of paper, turning it round, and claiming that the edges no longer connect - clearly you've done something wrong if you predict such a result.

SR may or may not be an accurate model of reality, but it is not self-contradictory.

 

[Janus]

I will study your notes, Ibix. In the meantime I have prepared this thought experiment:

I think I can show that the answer to the original question may be different. To achieve this I will use two additional people, Helen and Tom, but in the end I will remove these two people as if they never existed, so only Bob, Alice and Alex will remain on stage, as the original question wants. The image helps a lot.

View attachment 267159

Fig 1: The distance that separates Bob from Alice in the yellow reference frame is the same as the distance that separates Alex from Helen in the pink reference frame. The clocks of the observers in each separate reference frame are synchronized with the Einstein method.

According to which frame? From what you say later, I assume you mean according to the frame in which Tom is considered at rest.

Tom is considered immobile, while Bob and Alex move at the same speed in Tom's view. At the meeting point of the three boys, their three clocks are synchronized.

Fig 2: At some point along the way, Alex will meet Alice and they will be photographed with each other. At some point the same thing will happen between Bob and Helen. We will return to figure 2 later.

Fig 3: Tom meets Alice and Helen. Because of the symmetry, Alice and Helen's clocks remain synchronized in Tom's view. If you want, the two girls are photographed with each other. Immediately after this observation, Tom sees Alice and Eleni slowing down at the same speed and stopping. Obviously, in the short time between Tom's last observation and the immobilization of the girls, there is no dramatic change in their clocks, but this is true anyway because of the symmetry of the system with respect to Tom.

Now all observers are motionless with each other - they are in the same frame of reference. Alice is in sync with Bob and Helen, who is in sync with Alex. So Bob, Alice, Alex and Helen are all in sync with each other. But then all four observers must have been in sync at all stages of the journey between Figures 1 and 3.

Prior Alex, Bob, Helen and Alice coming to a rest with respect to Tom, Alice and Bob would not say that their clocks were in sync. Neither would Alex and Helen say that their clocks were in sync.
So while Tom might say that all four of these clocks started in sync, started their deceleration at the same time, decelerated at the same rate, and thus remain in snyc the whole time, Non of the other observers would say the same. Alice, for example, would say that Bob's clock reads ahead of hers at the start. He will begin his "deceleration" before she does, and thus after that his clock will run slow compared to hers.
Then when she finally reaches Tom, and begins her "deceleration", the effects of being in an accelerated frame causes here to say that Bob's clock starts running even slower. By the time she comes to a rest with respect to Bob, she will say that both their clocks do read the same. But that does not mean that She and Bob agree that their clocks read the same during the whole exercise.
If we look at figure 2, it shows Alex and Alice lining up at the same time that Bob and Helen do. We also assume that all four clocks read the same time as this happens. But all this is according to Tom's rest frame.
In Alice and Bob's frame, The distance between Alex and Helen is length contracted, and thus is shorter than the distance between Alice and Bob. Thus Bob has to line up with Helen before Alice reaches Alex. The one thing all frame will agree on is that The times on Bob's, Alice's, Helen's and Alex's clock all read the same value during these passings ( Alices clock reads the same as Alex's when they meet and so do Bob's and Helen's, and in addition if Alice's and Alex's clock read 2:00 when they pass, Bob's and Helen's Clocks read 2:00 when they pass. However, since Bob's and Helens's clock meet before Alice and Alex do, according to Alice and Bob, Bob's and Helen's clocks has to read 2:00 before Alice's and Alex's do.

Here's an animation which is similar to what you are trying to do here:
We have two rows of clocks, in motion with respect to each other. We have arranged things, so that from the rest frame of the lower row of clocks, the clocks in the upper row are spaced the same distance apart as those in the bottom row, run at the same rate as those in the bottom row and are snychronized with the bottom rows clocks, and themselves. Thus, as each clock passes the one below it, they read the same time.

However, if we switch to being at rest with respect to the upper row, we see this:

The clock's in the lower row run slower than the upper row clocks, the distances, are different, and none of the clocks are in sync in either row.
However, every time a upper and lower row clock pass each other, they still read the same time.
Thus the two frames remain in agreement about what happens as any two clocks pass each other, they just don't agree on how this comes about.

 

[kent davidge]

View attachment 267167

However, if we switch to being at rest with respect to the upper row, we see this:

View attachment 267166

why is the second clock in the first row brighter than all others? does this have something to do with the physics?

 

[Janus]

why is the second clock in the first row brighter than all others? does this have something to do with the physics?

That's a lighting artifact. This animation was done using ray-tracing software, So it simulates the appearance of a real object being lit by a real light. Basically, you are just seeing a specular highlight due to the relative postions of the light, object, and "camera" in the scene.

 

[Ibix]

why is the second clock in the first row brighter than all others?

@Janus is pretty good at 3d modelling, if his avatar didn't give that away.

 

[George Plousos]

Indeed, Janus's moving clocks show what is happening, and a few simplified calculations can actually show that the diagrams for Figures 1 - 3 work properly, however, the impression is created that the mathematical consequence is due to the inherent impossibility of synchronizing clocks in moving reference frames - ie the theory slips like an eel to the point where a clearer explanation is needed. On the other hand, Einstein's two axioms seem to make it necessary to modify the concepts of space and time. I have no objection to that, but perhaps nature hides surprises as is often the case from the point of view of quantum physics, and a (reliable?) experiment that combines the two theories will test the "twin paradox" in a few years, as long as it does not have the luck of similar experiments that have been abandoned in the past:

https://scitechdaily.com/physicists-put-einstein-to-the-test-with-a-quantum-mechanical-twin-paradox/

 

[mitochan]

I think I can show that the answer to the original question may be different. To achieve this I will use two additional people, Helen and Tom, but in the end I will remove these two people as if they never existed, so only Bob, Alice and Alex will remain on stage, as the original question wants. The image helps a lot.

Say Alice and Bob have synchronized clocks in their IFR and Alex and Helen have synchronized clodks in their IFR. In the simultaneous snapshots in the IFR of Tom, the reading of the clocks according to their owners are ;

Figure 1
Alice = $T(1-\frac{1}{\Gamma})$
Alex =0
Bob =0
Helen = $T(1-\frac{1}{\Gamma})$
Tom =0

Figure 2
Alice = T
Alex =$\frac{T}{\Gamma}$
Bob =$\frac{T}{\Gamma}$
Helen = T
Tom = $\frac{t}{2}$

Figure 3
Alice = $T(1+\frac{1}{\Gamma})$
Alex =2$\frac{T}{\Gamma}$
Bob =2$\frac{T}{\Gamma}$
Helen = $T(1+\frac{1}{\Gamma})$
Tom = t

where
T=\frac{L_0}{V}
t=\frac{L_0}{\gamma v}
V=\frac{2v}{1+\frac{v^2}{c^2}}
\Gamma=\frac{1}{\sqrt{1-\frac{V^2}{c^2}}}
\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
$L_0$ : proper length of the bars
$\pm v$ : speed of the bars