Special relativity and accelerating twins

Question

Imagine two twins synchronise their clocks and then twin A quickly accelerates to velocity v. After a time T twin B quickly accelerates to 2v and catches up with twin A. Which one will be younger? How do you pick a "preferred " frame when both had acceleration at some point.

Answer 1

For any observer there is a property called the proper time that is equal to the time shown by the clock they are holding. So to find out how the elapsed times of the twins compare we just have to calculated the elapsed proper times of their travels.

Despite the fearsome reputation that relativity enjoys, the equation to calculate the proper time is surprisingly simple. Suppose we observe someone to travel a distance $\Delta x$ in a time $\Delta t$ then the proper time for the journey $\Delta\tau$ is given by:

$$ c^2 \Delta\tau^2 = c^2\Delta t^2 - \Delta x^2 $$

This equation is known as the Minkowski metric, and it's the fundamental equation that underlies all of special relativity.

Anyhow, let's see how this works for the twins. $A$ sets off at time $t = 0$ and position $x = 0$ with some velocity $v$, and at time $T$ has reached a distance $vT$. So the proper time for A's journey is:

$$ c^2 \tau_a{}^2 = c^2 T^2 - (vT)^2 $$

Giving:

$$ \tau_a = T \sqrt{1 - \frac{v^2}{c^2}} \tag{1}$$

This is just the usual equation for time dilation:

$$ \tau = \frac{t}{\gamma} $$

where $\gamma$ is the Lorentz factor, and I'm sure you have seen this before.

Now let's do the same calculation for B's journey. B does this in two stages. In the first stage B just sits motionless at the starting point for a time $\tfrac12T$, so for this stage B moves from $t=0, x=0$ to $t=\tfrac12T, x = 0$. So the proper time for this step is:

$$ c^2\Delta\tau_{b1}{}^2 = c^2\tfrac14T^2 - 0 $$

and we just get $\tau_{b1} = \tfrac12T$. Not very exciting so far. In the second stage B accelerates to a speed of $2v$ and travels at the speed for a further time $\tfrac12T$ so they reach a distance of $vT$ at time $T$ i.e. just in time to meet A. For this stage $\Delta t = \tfrac12T$ but now $\Delta x = vT$ so the proper time for this step is:

$$ c^2\tau_{b2}^2 = c^2\tfrac14T^2 - v^2T^2 $$

Giving:

$$ \tau_{b2} = \tfrac12 T \sqrt{1 - \frac{4v^2}{c^2}} $$

So the total proper time for B's journey, $\tau_{b1}+\tau_{b2}$ is:

$$\begin{align} \tau_b &= \tfrac12 T + \tfrac12 T\sqrt{1 - \frac{4v^2}{c^2}} \\ &= T \left(\tfrac12 + \tfrac12 \sqrt{1 - \frac{4v^2}{c^2}}\right) \tag{2} \end{align}$$

It's a little hard to compare the equations (1) and (2) to see which twin has the shorter proper time, so I've graphed the ratio $\tau/T$ for both twins for a range of speeds up to $v = 0.5c$ - $v$ can't be greater than this otherwise twin B would be moving faster than light:

You can see immediately from the graph that twin B has a smaller proper time than twin A for all speeds greater than zero, so twin B has the greater overall time dilation. What is happening is that B has no time dilation for the first half of the experiment, but for the second half B's time dilation is more than twice A's time dilation. The end result is that B ends up with a greater total time dilation.

I've done this calculation using the frame of the observer at rest at the starting point, so all the times and positions are measured by this stationary observer. However the beauty of using the proper time calculation is it works for any reference frame. I could have used the rest frame of twin A or B and arrived at the same end result, though the calculation would have been harder. When calculating the proper time we can use whatever frame is most convenient, and we can be sure that we'll get the correct answer regardless of the frame.

Answer 2

UPDATE BELOW.

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Among twins, the inertial one is the "preferred" one.
Between two fixed events, the inertial twin visiting both events ages the most.
However, in general, one has to compute the elapsed proper time along each worldline, e.g. using a calculation or possibly graphically, as below.

Think of "age" as "elapsed wristwatch-time [proper-time] along a worldline",
akin to "distance-traveled" as "odometer reading along a path in space".

from my answer to How Does the Multiple Traveler Situation or Paradox Relate To The Twin Paradox.

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UPDATE:

To more directly address the scenario you described, here's a spacetime diagram (drawn on "rotated graph paper") of your scenario (with more convenient relative velocities).

Since nothing is sacred about using twice the first twin's velocity, I have chosen velocities $v$ from 0c, (3/5)c, and (4/5)c. These have corresponding doppler factors $k=\sqrt{\frac{1+v}{1-v}}$ of 1, 2, and 3 [which are rational, and are thus associated with Pythagorean triples].

We have drawn the ticks of traveler-proper-time (as traced out by a light-clock carried along a piecewise-inertial worldline-segment).

• The doppler-factor helps us draw these "light-clock diamonds" by stretching a grid diamond [a light-clock diamond for the diagram's rest frame] by $k$ in the forward-future direction and shrinking by $k$ in the backward-future direction (which preserves the slope of the edges (speed of light) and the area of the diamond (since a boost has unit determinant)). [Areas of causal diamonds are equal to the square-interval of the diagonal.]
  
  So, we have 3 inertial-frames depicted on this diagram used by these 2 piecewise-inertial travelers.
  [In the earlier diagram, there are 5 inertial-frames used by 5 piecewise-inertial travelers (one of which is inertial).]

So, inertial-trip OZ logs 16 ticks, whereas piecewise-inertial OTZ logs (5+9=)14 ticks. This is essentially the clock-effect with unequal outgoing and incoming relative speeds.
In this portion, "OZ is preferred" because OZ is inertial. A boost can make OZ vertical on a spacetime diagram that faithfully represents all events in spacetime.
This can't happen for OTZ. (Why? See my answer https://physics.stackexchange.com/a/507592/148184
to What is the proper way to explain the twin paradox? )

So, by counting,
traveler SOZE logs 20 ticks, whereas the late-but-will-catchup SOTZE logs 18 ticks.
Neither traveler is "preferred"... one just has to (in general) add up the proper-times along the various segments.

Of course, you can certainly use formulas and handle more general situations.

But, to develop some intuition for relativity and spacetime physics, you can use this graphical method (which I call "relativity on rotated graph paper") to make up all sorts of situations ... and the mathematics needed to carry out the calculations is probably more accessible. (For more info, see my blog contribution: https://www.physicsforums.com/insights/relativity-rotated-graph-paper/ )

Answer 3

This is the usual twin paradox scenario, with A being the "stay at home" twin and B being the "traveling" twin, but viewed from the inertial reference frame co-moving with B on the outbound portion of their journey. And hence, as usual, B will be younger than A when they meet again.

The principle of relativity says that all inertial reference frames are equivalent, and you can analyze an experiment in any of them and get the same outcome. The twin "paradox" is only a problem if you forget that B is not inertial for the whole experiment.