Twins paradox, Christmas photo album edition

Question

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.

Answer 1

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.

Answer 2

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"?

Answer 3

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.