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Twin A leaves Earth in a spaceship while Twin B stays on Earth. They both agree that at the one year anniversary of their separation, according to their local clock, they will take a photo of themselves and send it by space-carrier-pigeon to a pre-arranged location.

What will the photos show?

Remember, there is no acceleration, no turning around, none of the usual answers apply. It doesn't matter if it takes a million years for the carrier-pigeon to deliver the photos. The question remains, what will the photos show when they arrive at the common destination?

Update: To avoid the initial acceleration, let's amend the scenario as follows:

Two spaceships pass each other and they have a physical sensor sticking out that triggers a clock when the two ship sensors touch each other. The photos are taken at the one year anniversary of that trigger.

Update 2: Let's change the stop-trigger event to something external. Spaceship B will take a photo of themselves when they arrive at Alpha Centauri. Spaceship A will calculate how long (according to Spaceship A's clock) it would take for Spaceship B to get to Alpha Centauri, and take a photo at when Spaceship A's clock reads that time.

Níckolas Alves
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Jay
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  • But if the twins separate and never meet again this isn't the twin paradox. Are you asking why their time dilation isn't symmetric?. – John Rennie Mar 10 '23 at 05:33
  • @JohnRennie I have amended the question and, yeah, it is no longer a Twins paradox but a similar question of time dilation. It cannot be physically symmetric. What would the photos show? – Jay Mar 10 '23 at 05:37
  • So your question comes down to this:: If, on my 20th burthday, I commit to taking a photo of myself exactly one year from today, what will the photo show? Hint: You don't need to know anything about relativity to answer this question. – WillO Mar 10 '23 at 05:39
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    Does this answer your question? How can time dilation be symmetric? – Golden_Hawk Mar 10 '23 at 05:40
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    @Jay But you haven't said who decides when the one year anniversary is due. If both twins photograph their clock when their clock says one year, then both photos will be of a clock showing one year. This is WillO's point above. If we say either twin A's clock determines the anniversary or twin B's clock determines the anniversary than that introduces an asymmetry because one twin has the anniversary determined by their own clock while the other has it determined by someone else's clock, and now the pictures will be different. – John Rennie Mar 10 '23 at 05:45
  • But now it isn't a paradox because by preferring one twin over the other you have broken the symmetry. So can you clarify how exactly the one year anniversary is determined? – John Rennie Mar 10 '23 at 05:46
  • BTW a smoke and mirrors free explanation of the paradox is given in the answer to What is the proper way to explain the twin paradox?. – John Rennie Mar 10 '23 at 05:49
  • @JohnRennie I have updated the question to use an external stop trigger. The photos are taken when Spaceship B arrives at a predetermined destination. Spaceship B looks out the window. Spaceship A uses its clock to decide when that happens. – Jay Mar 10 '23 at 05:52
  • @Jay Your latest edit makes the question the same as How long would it take me to travel to a distant star?. – John Rennie Mar 10 '23 at 05:52
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    @JohnRennie : The OP did in fact specify that the twins each decide when to snap the photos by measuring one yeear "according to their local clock". So you are answering a much more interesting question than was asked. – WillO Mar 10 '23 at 05:53
  • @JohnRennie but here's the problem. When I reach the distant star, I will claim that the 'other' people (on Earth or Spaceship A) have aged very little bc they are the ones moving. Those people will say I have aged little. When both of us take the photos, who will be proven right? – Jay Mar 10 '23 at 05:55
  • "Who will be proven right?"???? How can taking a picture of yourself prove anything at all about the age of someone a million miles away? – WillO Mar 10 '23 at 05:58
  • @Jay but now the problem is that events that are simultaneous in one frame are not simultaneous in other frames. In your frame you photographing yourself and me passing Alpha Centauri are simultaneous, but in my frame they are not. So they cannot be used as a universal timer. This is known as the breakdown of simultaneity. So your question is still ill defined. – John Rennie Mar 10 '23 at 06:05
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    What you are discovering is that the paradox appears to exist only because it is much harder to ask the question precisely than you think. – John Rennie Mar 10 '23 at 06:09
  • @JohnRennie you are absolutely right and I thank everyone because your comments are helping me refine my question. – Jay Mar 10 '23 at 07:06

3 Answers3

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Since the photos are taken after one year with respect to each of their local clocks the photos look identical. The answer is still the same if the traveling twin undergoes acceleration or in the presence of strong gravitational fields, etc. He takes the picture after aging one year and hence will be a single year older than when they departed. The same holds for the other twin. Therefore, the photos are identical.

To clarify the apparent paradox, let me allow the travelling twin to return. Imagine they agree to do this every year (according to their own clocks) until they meet again on Earth. In this scenario, the first photos look identical, and so do the second ones, and so do the thirds, etc. However, the twin that stayed on Earth took more pictures than the twin who traveled. If the traveling twin took, say, 17 photos, then the first 17 photos they took look exactly the same, but the twin who stayed on Earth has more photos. He did age more.

Níckolas Alves
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  • I think you were unwise to answer before the OP had decided what they were actually asking. The repeated edits have somewhat invalidated your answer. – John Rennie Mar 10 '23 at 06:07
  • The comments to the question have helped me to refine the question into using an external stop-trigger. – Jay Mar 10 '23 at 06:25
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Alice stays on earth. Bob passes earth en route to a star. They are both celebrating their 20th birthdays.

Alice calculates that it takes Bob four years to reach the star, whereupon she takes a picture of herself. Bob takes a picture of himself just as he reaches the star.

Alice's picture shows her as 24 years old. Bob's picture shows him as 22 years old.

Alice's story: I took my picture on my 24th birthday, just as Bob was arriving at the star. Bob aged slowly, so he was just turning 22 at the time.

Bob's story: I took my picture on my 22nd birthday, just as I was arriving at the star. Alice took her picture six years later, on my 28th birthday, long after I had passed that star. But she aged slowly, so she was only 24 in the picture.

Who has been "proved wrong"?

WillO
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  • Sorry, why did Alice wait eight years according to Bob? According to him, he is sitting still. Alpha Centauri and Earth are travelling past him. According to Bob, it should take Earth only two years to cover the distance to Alpha Centauri in the opposite direction. According to Bob, Alice should have taken her picture at age 22. Actually even less bc her clock should have sowed. – Jay Mar 10 '23 at 06:33
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    @Jay: Yes, it would be better to say "just as the star was passing me", not "just as I was arriving at the star". As to "why did Alice wait eight years according to Bob", the answer is because this is what relativity predicts, as you'll see if you're willing to do the calculations. – WillO Mar 10 '23 at 06:36
  • There is no acceleration involved here, only uniform motion. If we replace Alice/Earth, Bob, and Alpha Centauri with three equal sized spaceships in deep space, there is no reason to say who waits 2 years who who waits 8. – Jay Mar 10 '23 at 06:41
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    Nobody said anything about acceleration. @JohnRennie has already explained the source of the asymmetry: It's Alice, not Bob, who thinks the two pictures are snapped simultaneously. I think it would be a very good idea for you to go back and digest those comments before you post any further. – WillO Mar 10 '23 at 06:41
  • Let me rephrase your scenario. Bob is on Earth and Alice passes by him leading a giant armada of spaceships. Bob takes his picture when the last spaceship passes him by. Alice does the same bc she knows exactly how large her armada is and when the last ship will pass Earth. – Jay Mar 10 '23 at 06:44
  • All the 'answers' implicitly assume that one of the participants is 'really' moving. That defeats the whole entire point of SRT. According to SRT, an electron by itself can claim it is at rest and the rest of the universe is doing all the moving. – Jay Mar 10 '23 at 06:46
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    You've gotten good answers. They do not implicitly assume any of the things you claim they do. You can either try to digest the answers [ideally with a good textbook in front of you] or you can keep posting additional proof that you haven't understood them. Either way, I don't think there's any point in continuing to respond to you. – WillO Mar 10 '23 at 06:49
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Let's clarify the situation by defining exactly the events that interest us. We'll take you to be on Earth and me speeding past you in a rocket travelling at a speed $v$. We could take $v = 0.866c$ (i.e. $\sqrt{3}/2$) as a concrete example as this gives a nice round value of $\gamma=2$, and let's take a star $0.866$ light years away so that in your frame it takes me $1$ year to reach the star. So we have three events at times $t$ and distances $x$ where the time $t$ is what you measure on your clock and the distance $x$ is the distance you measure from Earth:

  1. at $t=0, x=0$ I pass you and we synchronise our clocks

  2. at $t=1$ year, $x=0.866$ light years I reach the star and photograph my clock

  3. at $t=1$ year, $x=0$ you photograph your clock

So you photograph your clock on Earth at the exact instant that I pass the star and photograph my clock i.e. in your frame events 2 and 3 are simultaneous.

The paradox arises because in our photographs our clocks will show different times, and this seems like a paradox because there appears to be a symmetry i.e. we both see the other person moving at $0.866c$ so we should see the same time dilation and therefore the clocks should show the same time. But what we will find is that in my frame events 2 and 3 are not simultaneous. You have introduced an asymmetry by using your definition of simultaneous and this is observer dependent.

So let's see how things look in my frame. To do this we use the Lorentz transformations:

$$\begin{align} t' &= \gamma \left( t - \frac{vx}{c^2} \right ) \\ x' &= \gamma \left( x - vt \right) \end{align}$$

I won't go through all the details, but in my frame the events are:

  1. $t=0$, $x = 0$ i.e. the same as you

  2. $t$ = ¹⁄₂ year, $x = 0$

  3. $t = 2$ years, $x = -\sqrt{3}$ light years

So in my frame I take my photograph when my time $t'$ = ¹⁄₂ year, that is your photo is of a clock showing one year and my photo is of a clock showing half a year. But notice that event 3 in my frame happens when my clock shows two years i.e. in my frame events 2 and 3 are not simultaneous. This is the key to understanding the asymmetry. You defined what was meant by simultaneous and that is not the same as my definition of simultaneous, and that introduced the asymmetry.

Incidentally note that in my frame your photograph recorded your clock showing one year when my clock showed two years, so both of us observed the other person's clock to be running at half speed i.e. the time dilation is the same for us both.

John Rennie
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  • Thanks for the detailed answer but you already start out assuming that you are the one travelling in a rocket. How about if we take the Earth out and we are both in spaceships in deep space? We sync clocks and I tell you that my buddy is trailing me 0.866 light years behind travelling at the same speed and you should photograph your clock when you meet him. You tell me you have a similar buddy in the opposite direction and I should photograph my clock when I meet her. We have no reason to say which pair of us is travelling and both of us would do the calculations symmetrically. – Jay Mar 10 '23 at 08:45
  • But in my frame your buddy is only 0.433 light years behind you (i.e. ¹⁄₂ of the distance you measure), and in your frame my buddy is only 0.433 light years behind me. Our meeting up with each others buddies won't be simultaneous for the both of us. Again we get a breakdown of simultaneity. – John Rennie Mar 10 '23 at 08:52
  • @Jay the Lorentz transformations tell us how things look in different frames. In your frame your buddy is 0.866 light years behind you. When we use the LTs to find how things look in my frame we find I observe your buddy to be only 0.433 lyrs behind you. If you're asking why we get this result from the LTs then that's a complicated answer because it is due to a symmetry called Lorentz covariance. If you're interested I explain how this works here. – John Rennie Mar 10 '23 at 09:24
  • But to understand this is going to require you to learn SR properly. There simply isn't any way to explain this to your grandmother without some hand waving. – John Rennie Mar 10 '23 at 09:25
  • OK, thanks for your time. I will keep researching... – Jay Mar 10 '23 at 09:28
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    @Jay It is tempting to think there must be an intuitive way to understand this, but down that route lies the slippery slope to crackpottery. "Intuition" just means "what we are used to" and we are not used to SR because we never encounter it in everyday life. If you learn SR you will find it becomes intuitive, but the hard work of learning it comes first and intuition follows. To put intuition first and say "this can't be true because my intuition tells me otherwise" is to make the same mistake as generations of relativity deniers, and you do not want to join their ranks. – John Rennie Mar 10 '23 at 09:40