- #26

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Corrected, thanks.Um - I think you made a small but important typo...

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- #26

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Corrected, thanks.Um - I think you made a small but important typo...

- #27

Isaac0427

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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? 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.I don't think it needs to be a new thread, since a more detailed working of the problem might help the OP in this thread.

I would recommend picking a speed for Alex relative to Alice and Bob (@Ibix recommended two good choices in post #2), working in Alice's and Bob's common rest frame, and choosing the spacetime origin ##(x, t) = (0, 0)## to be the event where Alex passes Bob. Then work out the coordinates of the event where Alex passes Alice, based on Alex's speed and the distance from Bob to Alice in the chosen frame, and calculate the spacetime interval between those events, which will be the time elapsed on Alex's clock between them, using the standard interval formula.

Once you have coordinates of the relevant events in the Bob-Alice rest frame, drawing a spacetime diagram should be easy. If you want to then transform into Alex's rest frame to see how things look there, the standard Lorentz transformation formulas will give you the coordinates of the relevant events in that frame so you can draw another diagram.

- #28

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Alex will also say that the distance between Bob and Alice is shorter in distance than Bob and Alice measure it as being.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? 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.

As far as Bob's and Alice's clocks being in sync. let's assume that Alex has been traveling for some time at 0.5c when he passes Bob. At the moment they pass each other, they both are pointing very powerful telescopes at Alice and reading what they see on her clock. Bob and Alice are 1 ly apart in their frame and the image of Alice's clock will read as being 1 year behind Bob's clock reading. Because Bob knows that Alice's distance from him is a constant, he (and anyone at rest with respect to Bob and Alice) will conclude that Bob's and Alice's clocks are in sync.

Alex will see the same time reading for Alice; 1 yr behind Bob's clock reading. However, for him, Alice has not maintained a constant distance from him. That means that the distance between Alice and himself has not been a constant.

As I mentioned above, the distance between Bob and Alice is less than 1 ly according to Alex (0.866 ly). But that is the distance to Alice as he passed Bob. Since it took time for the image to reach him, He and Alice had to have been further apart when the light carrying that image left Alice. That light also was traveling at c relative to Alex. This means that the time between the light leaving Alice and Bob seeing it is longer than 1 yr ( 1.732 yrs) by Alex's clock. Alice's clock will be time dilated according to Alex and and 1.5 yrs will pass on it.

Thus, when Alex sees a reading of 1 yr behind Bob's clock in Alice;s clock as he passes Bob, he knows that 1.5 yrs has passed for Alice since that light left. This is 0.5 yr more than what Bob says passed on Alice's clock.

From this, alex has to conclude that. at the moment he passes Bob, Alice's clock reads 0.5 yr later than Bob's clock.

- #29

Isaac0427

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This seems to be an argument based on how the clocks were synchronized. What if Alex starts going at .5c right as the light signal from Alice reaches Bob? Wouldn't Alex see that Alice's clock reads .866 years?Alex will also say that the distance between Bob and Alice is shorter in distance than Bob and Alice measure it as being.

As far as Bob's and Alice's clocks being in sync. let's assume that Alex has been traveling for some time at 0.5c when he passes Bob. At the moment they pass each other, they both are pointing very powerful telescopes at Alice and reading what they see on her clock. Bob and Alice are 1 ly apart in their frame and the image of Alice's clock will read as being 1 year behind Bob's clock reading. Because Bob knows that Alice's distance from him is a constant, he (and anyone at rest with respect to Bob and Alice) will conclude that Bob's and Alice's clocks are in sync.

Alex will see the same time reading for Alice; 1 yr behind Bob's clock reading. However, for him, Alice has not maintained a constant distance from him. That means that the distance between Alice and himself has not been a constant.

As I mentioned above, the distance between Bob and Alice is less than 1 ly according to Alex (0.866 ly). But that is the distance to Alice as he passed Bob. Since it took time for the image to reach him, He and Alice had to have been further apart when the light carrying that image left Alice. That light also was traveling at c relative to Alex. This means that the time between the light leaving Alice and Bob seeing it is longer than 1 yr ( 1.732 yrs) by Alex's clock. Alice's clock will be time dilated according to Alex and and 1.5 yrs will pass on it.

Thus, when Alex sees a reading of 1 yr behind Bob's clock in Alice;s clock as he passes Bob, he knows that 1.5 yrs has passed for Alice since that light left. This is 0.5 yr more than what Bob says passed on Alice's clock.

From this, alex has to conclude that. at the moment he passes Bob, Alice's clock reads 0.5 yr later than Bob's clock.

- #30

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It has nothing to do with how the clocks were synced to each other. We just start with the assumption that according to Bob and Alice their clocks are synchronized. Alex however, has to conclude that Alice's and Bob's clocks can't be synchronized.This seems to be an argument based on how the clocks were synchronized. What if Alex starts going at .5c right as the light signal from Alice reaches Bob? Wouldn't Alex see that Alice's clock reads .866 years?

When Alex and Bob pass each other, they have to see the same image of Alice's clock. You can't have Alex and Bob being right next to each other and seeing different readings for Alice's clock.

If we assume that Alex's acceleration is instantaneous*, he will see Alice's clock reading 1 yr behind Bob's clock both before and after the acceleration. What will change is what time he would conclude is actually on Alice's clock.

*you have to be cautious when assuming "instantaneous" changes in velocity, This equates to a non zero change in velocity in 0 time, or a = dV/0. Since division by zero is undefined, you risk producing contradictory results if you are not careful. You can use a "near-instantaneous" velocity change instead, But then you have to accept that Alex will be in an non-inertial frame at some point, and account for how that effects what he says is happening to Alice's clock.

- #31

Ibix

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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.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?

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.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.

- #32

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If you make that event the spacetime origin (which I think is the easiest way to set up the two frames), then yes.would it be correct to say that EVERYONE agrees that Alex passes Bob at t=0

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

- #33

Isaac0427

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First, I do know that d

- #34

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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.

- #35

Dale

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

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- #36

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As measured by Bob and Alice or Alex and Helen?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.

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

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.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 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.*

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.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.

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

* 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]

- #37

Ibix

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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:

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:

\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:

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:

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.

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

\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:

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.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.

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:

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

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- #38

Ibix

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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.

- #39

Ibix

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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##.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 thatAlex thinksBob 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 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).

- #40

Isaac0427

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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.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.

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?

- #41

Nugatory

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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).Is that correct or nonsense?

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.

- #42

Isaac0427

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Assuming throughout their lives they have only gone at nonrelativistic velocities, then there appear to be two cases: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.

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?

- #43

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

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 travelled different distances and that it’s not surprising to find different steering inputs on different paths.

- #44

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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: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?

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.

- #45

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Here is a concrete simple example in Schwarzschild spacetime: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.

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

- #46

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

- #47

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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.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 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?

- #48

Nugatory

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

- #49

Ibix

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

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.

- #50

Ibix

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