What is happening when you take into account distances with the twin paradox?

Question

The twin paradox is based on special relativity alone. Bob is the reference (immobile), and measure the speed of Alice: 0.99c (Lorentz factor =7), they both set their clock at zero. After one year of Alice (Alice and Bob are measuring 1 year using Alice's clock), Alice should do her U-turn.

Bob has been watching Alice's clock and noticed the time-flow dilation of a factor 7. When Alice has aged by 1 years, Bob has aged by 7 years. That is the paradox (or half of it, as the paradox includes the return and that return with the U-turn is used to solve the paradox).

The paradox never treats distances (see Wikipedia).

What is the distance separating Alice and Bob at the U-turn? For Alice, it is 1 year at 0.99c so 0.99ly? For Bob it is 7 years at 0.99c so 6.93ly.

One suggestion was to add the contraction of Alice's rod observed by Bob, mathematically it is a good idea as we would be back at 0.99ly but Bob uses his rod to measure the speed, not Alice's rod. I don't think this can help.

Answer 1

Bob is watching Alice and notices the time-flow dilation of a factor 7. When Alice should do her U-turn, she has aged by 1 years and Bob by 7 years. That is the paradox (or half of it, as the paradox includes the return).

No, this is not part of the paradox; in order for the scenario to count as the "twin paradox", it has to include the U-turn. Two observers moving relative to each other with disagreement on time intervals between two events isn't the twin paradox scenario.

You are just questioning how time dilation can be symmetric, which is covered very well in the PSE post How can time dilation be symmetric? You need to be more specific on what you mean by "After one year"; is this time measured in Alice's frame or in Bob's frame? From there, you can apply time dilation to figure out the time the other twin would measure between the same events that determine what "one year is" in the frame you are mentioning.

If you want to keep track of distances, you also need to specify the rest frame that is calculating those distances, since for one of the twins that distance will be length-contracted.

A main concept to nail down with relativity is that you can't just say a distance or a time without specifying the frame in which it was measured.

Answer 2

My question was just special relativity without U-turn (or before U-turn).

Unfortunately, you did not make this clear in your question. There is no clear definition of who is older when there is no turnaround. For example, we can find an observer who considers the Earth and Bob to go one way and Alice going away in the opposite direction at the same speed as Bob. From this new observer's point of view, they are both the same age; Bob considers himself to be older than Alice, and Alice considers herself to be older than Bob. No one in the universe can say with absolute certainty who is older in this scenario. If Bob decided he was missing Alice and chased after her, it would turn out that Bob would be younger than Alice when he caught up with Alice. (This is the opposite of the result in the classic twin paradox.) In your proposed scenario where everyone is always moving inertially, everyone will have different opinions about the ages. They will also have different opinions about the separation distances, which appears to be what the rest of your question is about, and I address that below:

When Alice has aged by $1 \ \text{year}$, Bob has aged by $7 \ \text{years}$.

This is as measured by Bob.

For a gamma factor of $7$, the velocity of Alice measured by Bob is $$\frac{v}{c} = \sqrt{1-1/7^2} \approx 0.98974332.$$ After $7 \ \text{years}$ by Bob's clock, Bob will calculate that Alice has travelled a distance of $0.9874332 \times 7 = 6.9282 \ \text{ly}$.

After Alice has accelerated away from Earth and reached a constant velocity, she will measure the proper distance to the distant galaxy as $6.9282/7 = 0.989743332 \ \text{ly}$ in her new reference frame. When Alice arrives at the distant galaxy, she notes that $1 \ \text{year}$ of her proper time has elapsed, so she calculates her velocity for the trip to be $d/t = 0.989743332c$, which is the same as the speed Bob measured, so no contradictions there. It will take Alice another year of her proper time to return to Earth, so when she gets back home, she will have aged by $2 \ \text{years}$ during her round trip, and she will notice that Bob has aged by $14 \ \text{years}$ during her absence.

What is the distance separating Alice and Bob at the U-turn? For Alice, it is $1 \ \text{year}$ at $0.99c$, so $0.99 \ \text{ly}$? For Bob, it is $7 \ \text{years}$ at $0.99c$, so $6.93 \ \text{ly}$.

This is essentially correct, and our results agree when we allow for rounding. In a nutshell, when Alice is travelling, she measures the distance to the galaxy as $1/7$, the distance that Bob measures to the galaxy. To Alice, a rod stretching from the Earth to the galaxy is moving relative to her and is length contracted by a factor of $7$.

For Alice, during her round trip, the distance between Earth and the galaxy really was just under a light year, and since she can consider herself to be stationary, from her point of view, the galaxy was travelling towards her at $0.989743332c$. It took the galaxy a year of its time to arrive at her location. When she briefly stopped at the galaxy to turn around, she would have noticed the distance to Earth briefly appeared to be $7 \ \text{lightyears}$ again.

To an observer which is permanently moving at $0.98974332c$ relative to the Earth, the distance from Earth to the galaxy is permanently just under a light year, and it takes $2 \ \text{years}$ for a light signal to travel from Earth to the galaxy and back by his measurements and it takes about $2 \ \text{years}$ by his measurements for a rocket travelling at close to the speed of light to travel from Earth to the galaxy and back.

One point of view is that Alice has travelled nearly $14 \ \text{lightyears}$ in $2 \ \text{years}$ of her proper time, and this appears to exceed the speed of light. However, this is not the case. This is not the normal way to measure speed. Speed is coordinate distance divided by coordinate time in a given reference frame. Using time measured in one reference frame and distance measured in another gives odd results. This way of measuring speed is called proper velocity or celerity. There is no suggestion that Alice has exceeded the speed of light. If a light signal is sent from Earth to the galaxy when Alice sets off to the galaxy, the light signal arrives at the galaxy before Alice arrives, and all observers will agree on this observation.

Answer 3

To amplify the answer by @KDP,
here's a spacetime diagram drawn on "rotated graph paper" so we can more easily visualize the tickmarks for arithmetically-nice velocities like $v=\pm(4/5)c$, whose Doppler factor is rational $k=\sqrt{\frac{1+(v/c)}{1-(v/c)}}=3^{\pm 1}$ (that is, $3$ and $\frac{1}{3}$, which are the stretching factors of the light-clock diamonds).

(click to magnify)

• The main point of the "Clock effect" is that the proper time elapsed (the wristwatch time) from event O to event Z depends on the worldline path taken from O to Z, and that the longest wristwatch time occurs for the inertial observer from O to Z. So, here we see that inertial observer OTZ ages 20 years but non-inertial observer [with piecewise-inertial legs] OQZ ages 12 years (for outgoing and incoming speed (4/5)c).

• The main point of the "Twin Paradox" is the mistaken belief that the we can analyze the problem from "the non-inertial observer's OQZ frame" where the non-inertial observer is "at rest" and obtain the contradictory result that it is OTZ who should be younger (and thus argue that that there is no clock effect so that OQZ and OTZ have the same wristwatch time).
  The misconception is the mistaken belief that "being-at-rest is the equivalent to being-inertial".
  So, the "Twin Paradox" should really be called the "Twin Misconception".

• Why is distance not treated on the assumption that we have already addressed the clock readings?
  It's because we learn nothing new.
  However, we can use it as check of consistency [as @KDP did].

So, here is the analysis.

With the turn around event Q,

• (RED) Bob says that T is simultaneous with Q.
  $OT=10$ and $TQ=8$, with $OT \perp_M TQ$ (Minkowski-perpendicular).
  So, (BLUE) $v_{AliceOUT}=\displaystyle\frac{OPP}{ADJ}=\frac{TQ}{OT}=\left( \frac{8}{10} \right)$
  and $\gamma_{AliceOUT}=\displaystyle\frac{ADJ}{HYP}=\frac{OT}{OQ}=\frac{10}{6}=\left( \frac{5}{3} \right)$.
  As a check $(OT)^2-(TQ)^2=(10)^2-(8)^2=(6)^2=(OQ)^2$.
  
  Similarly,
  $TZ=10$ and $TQ=8$, with $TZ \perp_M TQ$
  So, (GREEN) $v_{AliceIN}=\displaystyle\frac{OPP}{ADJ}=\frac{TQ}{ZT}=\frac{8}{-10}= \left( -\frac{8}{10}\right)$
  and $\gamma_{AliceIN}=\displaystyle\frac{ADJ}{HYP}=\frac{ZT}{ZQ}=\frac{-10}{-6}=\left( \frac{5}{3} \right)$.
  As a check $(ZT)^2-(TQ)^2=(-10)^2-(8)^2=(6)^2=(ZQ)^2$.
  
  

• (BLUE) AliceOUT says that Qs1 is simultaneous with Q.
  $OQ=6$ and $QQs1=-4.8$, with $OQ \perp_M QQs1$.
  So, (RED) $v_{BobOUT}=\displaystyle\frac{OPP}{ADJ}=\frac{QQs1}{OQ}= \frac{-4.8}{6} =\left( -0.8 \right)$ and $\gamma_{BobOUT}=\displaystyle\frac{ADJ}{HYP}=\frac{OQ}{OQs1}=\frac{6} {3.6}=\left( \frac{5}{3} \right)$.
  As a check $(OQ)^2-(QQs1)^2=(6)^2-(-4.8)^2=(3.6)^2=(OQs1)^2$.
  
  

• (GREEN) AliceIN says that Qs2 is simultaneous with Q.
  $QZ=6$ and $QQs2=-4.8$, with $ZQ \perp_M QQs2$.
  So, (RED) $v_{BobIN}=\displaystyle\frac{OPP}{ADJ}=\frac{QQs2}{ZQ}= \frac{-4.8}{-6} =\left( +0.8 \right)$.
  and $\gamma_{BobIN}=\displaystyle\frac{ADJ}{HYP}=\frac{ZQ}{ZQs2}=\frac{-6} {-3.6}=\left( \frac{5}{3} \right)$.
  
  As a check $(ZQ)^2-(QQs2)^2=(-6)^2-(-4.8)^2=(3.6)^2=(Qs2Z)^2$.

So, the distances don't really add anything new to explaining the Clock Effect/Twin Paradox, but they do help check (for example) that the relative-speeds between a pair of inertial observers are equal, as required by the Relativity Principle.

You can use the OP's $v=0.99$ (more precisely by @KDP, $v=0.9897$ for $\gamma=7$) in the above to get the results of @KDP.

[Note $(v/c) = \tanh\left(\operatorname{arccosh}\left(\gamma\right)\right)= \tanh\left(\operatorname{arccosh}\left(7\right)\right)\approx 0.9897$ and $k=\exp\left(\operatorname{arccosh}\left(\gamma\right)\right)= \exp\left(\operatorname{arccosh}\left(7\right)\right)\approx 13.92$]

Answer 4

The key principle that you have to learn about SR is that 'now' is frame-dependent. If you are moving inertially, then for you 'now' is a flat slice through spacetime at right angles to your time axis. If I am moving relative to you, then 'now' for me is a slice through spacetime that is at some angle to your 'now'. When we are in the same place, we agree on a common 'now', but when we are far apart we disagree on what time it is 'now' at each other's location. Time dilation and length contraction are both consequences of that fact.

When you say that Alice travels for a year, during which time Bob ages by seven years, what is really happening is that the extra six years accrued by Bob arises because he is using a different 'now' as the baseline by which to determine what is happening. To make it specific, let's suppose it is 2024 when Alice leaves Bob, and that a third person, Carol, is stationary relative to Bob seven light years away in Bob's frame. According to Bob, 'now' in Carol's location is 2024, when she is seven light years away. However, in Alice's frame, Carol's calendar 'now' already shows 2030. In Alice's frame, 'now' at the start of her journey corresponds to a much later point along Carol's world-line where Carol is only a light year away. And that is how length contraction happens. Carol is 'now' at one point in spacetime in Bob's frame, while for Alice 'now' means a much later time when Carol has already advanced through spacetime to be much closer.

Answer 5

What is the distance separating Alice and Bob at the U-turn?

Instead of call Bob's frame as "immobile", it is more precise to take its frame as the reference for coordinate time. That means: there are an array of synchronized clocks at many points in the Bob's frame, that is: with fixed distances to Bob's position.

Alice is travelling at a relative velocity of $0.99 c$ with respect to Bob's frame. After 1 year in Alice's clock, (her proper time), if she looks at the clock of the array just in front of her, it will show: $t_B = \gamma * 1 = \frac{1}{\sqrt{1 - 0.99^2}} = 7.09$ years.

The distance from the initial point is: $0,99 * 7.09 = 7,02$ ly in Bob's frame (proper distance), and $0,99 * 1 = 0.99$ ly in Alice's frame.

The advantage of that approach is that we can also suppose that Bob (with all its synchronized clocks) moves at $0.99$ c, while Alice is stationary. The result doesn't change.