Twin paradox problem with acceleration, turning issues removed

Question

Twin Anna leaves Chicago in a supersonic jet, accelerates to Mach 3 then immediately decelerates and lands in New York. Then she climbs into a rowboat and rows slowly to England, landing 3 months later. Twin Bob waits for 3 months then leaves Chicago in a jet, accelerates to Mach 3 and flies (just above the water) all the way to an aircraft carrier near England, then climbs into a rowboat and rows the last 100 metres. Which twin is younger?

UPDATE: I have created the spacetime diagram to show how each moves relative to the other. The basis of ANY twin paradox question is not how they move relative to other observers (like people in Chicago or New York) but only how they move relative to each other. As you can see in the diagram, from Anna's reference point, Bob moves westward first fast for a few minutes, then slowly for 3 months, then suddenly moves eastward fast for an hour to rejoin her. From Bob's reference point, Anna moves eastward first fast for a few minutes, then slowly for 3 months, then moves westward fast for an hour to rejoin him.

Twin paradox questions are not concerned with how outside observers see the situation. The paradox is, when they rejoin Bob would say that Anna is younger while Anna would say that Bob is younger.

We can all agree that Bob is younger because we, as outside observers, can tell that Bob travelled faster than Anna. And this is the problem with interpretations of special relativity that say that "all motion is relative".

I have read the Physics Forums Explanation but it does not seem to cover this situation.

Answer 1

I have read the Physics Forums Explanation but it does not seem to cover this situation

The spacetime geometry explanation is completely general. It works for all possible paths through all possible spacetimes. That includes this scenario.

Unfortunately, the Physics Forums link does not show the math. The formula is $$\Delta \tau =\int d\tau = \int \frac{1}{c}\sqrt{-g_{\mu\nu}dx^\mu dx^\nu}$$ In this formula $\tau$ is the proper time measured by one twin, i.e. how much they age. $g$ is the spacetime metric in the frame of interest and $x$ is the coordinates of the twin in the frame of interest. The integral is taken over the worldline of the twin.

Applying this one formula to any situation always gets the correct prediction, including this scenario. You simply calculate the above formula for each twin and compare. The smaller value is the younger twin at the reunion. This formula is manifestly invariant, so all reference frames will agree on the $\tau$ calculated for each twin

Answer 2

The point of your question appears to be this- how can one twin age less than the other when time dilation is symmetrical and you have designed your thought experiment to include the same degree of acceleration in the case of each twin?

The answer is that your thought experiment is not symmetrical, and its asymmetry allows one twin to age less than the other.

Time dilation is symmetrical between two inertial frames, so that when the twins are coasting, each is time dilated to the same extent in the frame of the other. However, when the twins switch reference frames by changing speed, you have to take the relativity of simultaneity into account. The key point here is that the effect depends not just on the change of speed but also on how far apart the twins are when one of them accelerates. You can readily see that if you draw a spacetime diagram. When a twin accelerates, their plane of simultaneity tilts, and the effect of the tilt is to advance the time that corresponds to 'now' at the location of the other twin.

The effect of the tilting plane of simultaneity increases with the distance between the twins. When they are close together, the effect is negligible. In your set up, the essential asymmetry arises because Anna undergoes deceleration when she is roughly 700 miles from Bob (ie the distance from Chicago to New York), whereas Bob accelerates when he is about 4,000 miles from Anna (the distance from Chicago to London). The effect of a plane of simultaneity sweeping upwards in the other twin's frame is greatest in Bob's case, and therefore it will be Bob who will age less.

You can get another insight into the nature of the asymmetry if you consider the Doppler effect. Let's suppose the twins are far apart and are sending light pulses to each other at microsecond intervals. If one of the twins accelerates toward the other, they will immediately see an increase in the timing of the pulses from the other twin, whereas there will be a time lag before the other twin sees any effect on the pulses received from the accelerating twin. If you dwell on that point, and perhaps sketch it in a diagram, you will see that it is consistent with the idea that when one twin accelerates toward the other, their tilting plane of simultaneity results in a jump forward in the time that corresponds to their 'now' at the location of the remote twin.

Answer 3

If Alice travels at speed $v$ for time $t$ (according to those of us who stay home), she ages by $t/\sqrt{1-v^2}$ (again, according to us). Carefully write down the duration and velocity that you're assuming for each leg of her trip, calculate how much she ages on that leg, and add up over all the legs. Do the same for Bob.

What is stopping you from doing this?

Answer 4

A problem with this question is the completely non-relativistic set up. Chicago to London is milliseconds, and speeds are slow. SR is not a factor. The 3 cities aren't collinear, which complicate a quantitative analysis.

There is also the problem that it's on Earth, which will invoke failures to abstract, such as rotating reference frames and gravitation time dilations.

To fix that, imagine 3 collinear deep space stations called CHI, NYC and LON, all with relative velocities zero.

Now pick reasonable velocities, one slow, and one fast:

$$ v_s = \frac{13}{85}$$ $$ v_f = \frac{84}{85}$$

with corresponding Lorentz factors:

$$ \gamma_s = \frac{85}{84}$$ $$ \gamma_f = \frac{85}{13} $$

With CHI defining the origin of the $S$ frame , the world line of CHI is:

$$ W_{CHI}(t) = (t, 0) $$

we'll place the other stations at:

$$ W_{NYC}(t) = (t, 84) $$ $$ W_{LON}(t) = (t, 97) $$

Here $(t, x)$ refers to years and light-years in the $S$ frame.

It's very important to label relevant events when setting up and SR problem. Firing off long sentences describing them is not clear.

So everyone starts at:

$$E_0 = (0,0)$$

At $t=0$, Alice (it's always Alice, not Anna) leaves for NYC, traveling in the $S'$ frame at $v_f$, arriving at:

$$E_1 = (\frac{D_{C, N}}{v_f}, D_{C, N})=(85, 84) $$

and immediately slowing to $v_s$, traveling in the $S''$ frame to arrive at LON:

$$E_2 = (170, 97) $$

At that point, Bob leaves for a direct flight to LON at $v_f$:

$$E_3 = (170, 0) $$

traveling in $S'''$, arriving at LON at:

$$ E_4 = \big(85(2+\frac{97}{85}), 97\big) $$

where he slows down to $v_s$ for $\epsilon$, in frame $S''''$, and then stopping, returning to $S$ and reuniting with Alice at:

$$ E_5 \approx E_4 $$

So now we have 6 events and 5 frames to deal with, instead of the standard 3 events and 3 frames in the Twin Paradox.

To figure out the elapsed proper times, we just add the proper time of the straight legs. Nominally, this requires transforming into all the frames, but that is not necessary here.

From $E_0$ to $E_1$, Alice experiences:

$$ \tau^A_{0,1} = \frac{D_{C, N}}{v_f\gamma_f}=13\,{\rm years} $$

From $E_1$ to $E_2$:

$$ \tau^A_{1,2} = 195.15\,{\rm years} $$

Finally, from $E_2$ to $E_5$:

$$ \tau^A_{2,5} = \frac{97\cdot 85}{84}=98.15 \,{\rm years} $$

Her total elapsed time is:

$$ \tau^A_{0, 5} = 307.3\,{\rm years}$$

Meanwhile, Bob is simpler:

$$ \tau_{0,3}^B = 170\,{\rm years} $$ $$ \tau_{3,5}^B = 15.0\,{\rm years} $$

His total elapsed time is:

$$ \tau^A_{0, 5} = 185\,{\rm years}$$

Much less than Alice.

So where's the paradox? A paradox is an apparent contradiction, and there are no contradiction in SR: it is self-consistent. Since I analyzed the whole thing in Minkowski Space, there are no apparent contradiction, thus: no paradox.

I have to breakdown some 3+1 frames to find a possible paradox. That means: how much time did Alice see Bob's clock tick?

From $E_0$ to $E_1$:

$$ t^{A, B}_{(0, 1)} = \frac{D_{C, N} }{ v_f \gamma^2_f} =1.99\,{\rm years} $$

From $E_1$ to $E_2$:

$$ t^{A, B}_{(1, 2)} = 12.7\,{\rm years} $$

Finally $E_2$ to $E_5$:

$$ t^{A, B}_{(2, 5)} = 15.0\,{\rm years} $$

for a total time of:

$$ t^{A, B}_{(0, 5)} = 29.7\,{\rm years} $$

which appears to contradict his proper time of $185\,{\rm years}$.

This is an apparent contradiction, aka: a paradox. The resolution involves the relativity of simultaneity. When Alice changes frames, her definition of "now" back on Earth changes, skipping much of Bob's existence sitting around on Earth aging.

Note that this is not time-dilation, this is the relativity of simultaneity. All observers in SR thought experiments have an infinite lattice of synchronized clocks and rulers, and when Alice goes from $S'$ to $S''$ (and from $S''$ to $S'''$), her lattice of clocks goes out of synch, and she needs to pick new ones.

If you work out the linear Lorentz Transformations:

$$ t' = mt + b $$

you will find the missing time. The slope, $m$, is time dilations, while the intercept, $b$, is clock synchronization.

Note that in the problem here, it is much more tedious than just doing the standard Twin Paradox, where the time on Earth jumps forward at space twin's turn around. Since I provided labeled events: just do the Lorentz Transformations.

This can be analyzed in general relativity as a gravitational time dilation, as GR contains SR, and GR is self-consistent away from singularities, that is a nice exercise when studying GR. It is not necessary. One problem is that the time on Earth can go forward and backward, depending on the direction of the velocity change. (See: Rietdijk–Putnam argument).