Simplified twin paradox

Question

I'm struggling to understand the idea that if two observers are moving relative to each other, they both see the other's clock as running more slowly. This is a well-known paradox in special relativity, and is supposed to be dealt with by the classic 'twin paradox', but I don't feel the way the twin paradox is resolved really answers this question. It relies on the asymmetry of a two-way journey, but what if we eliminated the asymmetry?

Here is a simplified version of the twin paradox I thought of: a spaceship leaves Earth for a planet 10 light years away travelling at 0.9c. Assume the time taken to accelerate to and decelerate from 0.9c is negligible. When it reaches the planet, the crew beams the date recorded on their clock to Earth. (This eliminates the asymmetry, since the message is sent before the return journey. Maybe the crew don't even make a return journey.) What date will be recorded and what will the date be on Earth when the message reaches Earth?

Answer 1

Here's a spacetime diagram of your proposed experiment. The coordinates $x$ and $ct$ are the coordinates in the reference frame of the observer on Earth. The grid spacing is $2$ lightyears.

Worldline 2 is the trajectory of the spaceship, moving with velocity $0.9 c$. It's not difficult to see that the ship reaches its destination at $t = \frac{10\ \mathrm{ Ly}}{0.9 c} \approx 11.11$ years.

For the passengers on the ship, how much time has passed? The spacetime interval is $$c^2\Delta \tau^2 = c^2 \Delta t^2 - \Delta x^2 \approx 23.45\ \mathrm{ Ly}^2$$ $$c\Delta \tau \approx 4.843 \ \mathrm{ Ly} \iff \Delta \tau \approx 4.843\ \mathrm{ y}$$

They immediately send a signal back towards Earth, shown by worldline 3. The message is received at $t \approx 21.11$ years, which is the proper time along worldline 1 (since for those observers, $\Delta x=0$).

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So from the Earth perspective, the ship's clock was ticking more slowly. But from the ship's perspective, the Earth clock was ticking more quickly.

This is not the case. For example, imagine that prior to the ship's departure, both observers agreed that the Earth observer would send a signal after 1.111 years, so the signal arrives at the ship at the moment the ship arrives at the planet. I denote the signal from Earth with a red squiggly line in the diagram below.

Now, from the ship's perspective, Earth has been moving away at a speed of $0.9c$. The ship receives the signal after $4.843$ years, so at what time $T$ (by the ship's reckoning) was the signal sent?

Well, the Earthly observer waited a time $T$, after which time the light signal had to traverse a distance $D = 0.9cT$. Therefore, we must have $$T + \frac{D}{c} = T + 0.9 T = 1.9T = 4.843\ \mathrm y$$ $$\implies T \approx 2.549 \ \mathrm y$$

According to the observer on Earth, they waited $1.11$ years to send the signal - but according to the observers on the ship, the signal was sent after $2.549$ years. From the ship's perspective the clocks on Earth are running slow by a factor of $\frac{2.55}{1.1}\approx 2.294$.

But then again, according to the observer on the ship it took $4.84$ years to reach the planet - but according to the observers on Earth, they arrived after $11.1$ years. So from the Earth's perspective the clocks on the ship are running slow by a factor of $\frac{11.11}{4.843}\approx 2.294$.

So we see that each observer sees the other's clock running slow by a factor of $1/\sqrt{1-(0.9)^2} \approx 2.294$, precisely as one would expect from the time-dilation formula.

Answer 2

Here's the situation for the arithmetically-simpler $v=3/5$ case.
I provide the equivalent values for $v=0.9$ that @J.Murray had determined earlier.

With this choice of velocity (leading to a Doppler factor $k=\sqrt{\frac{1+v}{1-v}}=2$) and the use of the rotated graph paper, it is easy to read off many results. The light-clock diamonds are traced-out by the signals in an observer's light-clock. (The areas of all clock diamonds are equal by Lorentz invariance.)

Some results:

First, the velocity of the ship according to the earth is $\displaystyle v_{SE}=\frac{TQ}{OT}=\frac{3}{5}$,
where $T$ and $Q$ are simultaneous according to the earth

and the velocity of the earth according to the ship is $\displaystyle v_{ES}=\frac{T'Q'}{OT'}=\frac{-3}{5}$,
where $T'$ and $Q'$ are simultaneous according to the ship.

When the ship reaches the planet at event $Q$, 4 ticks have elapsed for the ship, and ship's clock-face display has been broadcast to the earth. The earth receives that image of "4 on the ship clock" at event $R$ 8 ticks after separation at event O.
This ratio $\displaystyle \frac{OR}{OQ}= \frac{8}{4}=2$ is the Doppler factor $\displaystyle k=\sqrt{\frac{1+v}{1-v}}$. (Note that $QR$ is a light-signal.)
(For $v=9/10$, we have $k=\sqrt{19}\approx 4.3588$, which is consistent with @J.Murray's $\displaystyle\frac{21.11}{4.843}$.)
Said another way... if the ship broadcasted 1-hour tv shows, the earth views those shows running at half-speed (taking the earth 2 hours to view the content of the ship's 1-hour show).

Note that if the earth broadcasted 1-hr tv shows, the ship would also view them at half-speed. In fact, the ship receives the image of "4 on the earth clock" (broadcast at event $Q'$) at event $R'$ 8 ticks after separation at event O---in accordance with the Principle of Relativity for these two inertial observers.

Events connected by light-signals are associated with the Doppler effect.

Contrast this with....

Events connected by lines-of-simultaneity are associated with the Time-Dilation effect.

The earth says that distant event $Q$ (when the ship reaches the planet, when the ship clock reads 4 ticks) is simultaneous with local event $T$ on the earth worldline. ($TQ$ is parallel to the spacelike-diagonal of the earth's light-clock diamonds.)
The ratio $\displaystyle\frac{OT}{OQ}=\frac{5}{4}$ is the time-dilation factor $\displaystyle\gamma=\frac{1}{\sqrt{1-v^2}}$. ($TQ$ is on the earth's line-of-simultaneity.)
(For $v=9/10$, we have $\gamma=\sqrt{\frac{100}{19}}\approx 2.294$, which is consistent with @J.Murray's $\displaystyle\frac{11.11}{4.843}$, as stated there.)

Operationally, the local event $T$ is determined by a radar experiment by the earth to measure distant event $Q$. The earth sends a light-signal that reaches $Q$ and awaits its echo. The mid-time event between transmission (at 2 ticks) and reception (at 8 ticks) determines that the event at $\frac{1}{2}(8+2)=5$ ticks is simultaneous with $Q$.

Since the earth says $T$ and $Q$ are simultaneous, signals emitted (from firecrackers exploding at $T$ and $Q$) are received at a common event that the earth says is midway, at $\frac{1}{2}TQ=\frac{1}{2}3=\frac{3}{2}$, at a time $\frac{3}{2}$ later (the event at $(t,x)=(6.5,1.5)$ according to the earth). (This event is the future-intersection of the light-cone of event $T$ and the light-cone of event $Q$.)

The earth can actually use this radar information to determine the elapsed time $OQ$ on the remote segment of the ship's worldline: $OQ=\sqrt{(8)(2)}=4$. (This follows from $\Delta t= (t_{rec}+t_{send})/2$ and $\Delta x= (t_{rec}-t_{send})/2$. Then $t_{rec}t_{send}=\Delta t^2 -\Delta x^2$. By the way, the ship would say $OQ=\sqrt{(4)(4)}=4$ using a very short radar experiment to measure $Q$.)

(For $v=9/10$, the earth would send at 1.11111 ticks and receive at 21.11111 ticks. Then $\sqrt{(21.11111)(1.11111)}=4.843$, as @J.Murray calculated.)

In accordance with the principle of relativity, I think you can see that the ship says that distant event $Q'$ (when the earth clock reads 4) is simultaneous with local event $T'$ (at 5 ticks) on the ship worldline. ($T'Q'$ is parallel to the spacelike-diagonal of the ships's light-clock diamonds.) The ship would obtain all of the results measuring the earth that the earth did measuring the ship.

...all completely symmetric for these inertial observers, in accordance with the principle of relativity.

Hopefully, this visualization (motivated by the light-clock and other relativistic concepts) encodes and summarizes these and other results in a numerically accurate and physically interpretable way.

For more aspects of this symmetry, consult my answer https://physics.stackexchange.com/a/383363/148184 and other answers from How can time dilation be symmetric?

Finally, as mentioned in the comments to the OP, this problem is really about the symmetry of time-dilation---not the clock-effect/twin-paradox involving a non-inertial twin that reunites with the first twin.