Twins Paradox Thought Exp: What Happens When Reunited?

[Sophrosyne]

TL;DR Summary

A hypothetical experiment to help flesh out the confusion in the twins paradox

I have been looking through some of the threads about the twins paradox in relativity. It’s clear there’s a lot of confusion on this, and I am yet one more person very confused on this.

So I was thinking about a hypothetical experiment, and I will lay out my hypothesis of what might theoretically happen in this experiment based on my current, admittedly confused thinking, up until the part where it doesn’t make sense to me anymore. Maybe then it can be pointed out at which point I am going wrong in my thinking about this.

Let’s say the twins synchronize their watches before the first one takes off in his space ship. They both have powerful telescopes through which they can look at each other‘s watches throughout the trip and back.

As the rocket with the first twin begins to accelerate to near the speed of light, both twins will see the other’s watch slowing down. If a baby is born on the rocket at the time it has achieved constant velocity away from the first twin on earth, that baby will see the Earth twin’s watch run slow, as he is not in any accelerating frame of reference anymore. Relativistically speaking, that baby is not going to know who did the initial accelerating, the Earth or the rocket ship. And the twin on Earth is going to see the rocket ship’s clock running slow relative to him.

The same should hold true as the rocket turns around and begins coming back towards the earth. They are both going to still be in constant motion relative to each other, and whether the motion is towards or away from each other, they are still going to see the clock running slow. If another baby is born on the way back, they are still going to look at the clock on Earth and see that it is running slow relative to them. They are not going to see that watch ticking faster because they are moving towards each other rather than away. Both the baby born on the way out and the baby born on the way back are going to see a slow watch back on earth, and the guy on Earth is going to see the watch on the rocket ship running slow as well. The vector changes don’t matter, only the absolute value of the velocity relative to each other. It’s the absolute value of the velocity relative to each other that matters in how fast or slow they are seeing that clock ticking.

It seems during the whole trip and back, they are going to see each other‘s watches ticking slow.

Now if one wants to think of this in terms of Doppler effects, then the traveling twin is going to see the Earth clock slowing down during the acceleration phase away from the Earth, but the Earth twin is NOT going to see a relative slowing of the rocket twin’s clock. On the way back though, the effect should be reversed to compensate. What would the baby just born on the way back see happening on Earth? They just see a constant velocity towards the earth, and should still see a slowing of the watch back on earth.

So then when they are reunited and compare their watches, they see... what? Here is where I get lost.

 

[Nugatory]

You have overlooked the relativity of simultaneity. What time does the earthbound clock read at the same time that the baby is born, using the inertial frame in which the baby is at rest? What time does the earthbound clock read at the same time that the baby is born, using the inertial frame in which the Earth is at rest?

 

[Janus]

Summary:: A hypothetical experiment to help flesh out the confusion in the twins paradox

I have been looking through some of the threads about the twins paradox in relativity. It’s clear there’s a lot of confusion on this, and I am yet one more person very confused on this.

So I was thinking about a hypothetical experiment, and I will lay out my hypothesis of what might theoretically happen in this experiment based on my current, admittedly confused thinking, up until the part where it doesn’t make sense to me anymore. Maybe then it can be pointed out at which point I am going wrong in my thinking about this.

Let’s say the twins synchronize their watches before the first one takes off in his space ship. They both have powerful telescopes through which they can look at each other‘s watches throughout the trip and back.

As the rocket with the first twin begins to accelerate to near the speed of light, both twins will see the other’s watch slowing down. If a baby is born on the rocket at the time it has achieved constant velocity away from the first twin on earth, that baby will see the Earth twin’s watch run slow, as he is not in any accelerating frame of reference anymore. Relativistically speaking, that baby is not going to know who did the initial accelerating, the Earth or the rocket ship. And the twin on Earth is going to see the rocket ship’s clock running slow relative to him.

The same should hold true as the rocket turns around and begins coming back towards the earth. They are both going to still be in constant motion relative to each other, and whether the motion is towards or away from each other, they are still going to see the clock running slow. If another baby is born on the way back, they are still going to look at the clock on Earth and see that it is running slow relative to them. They are not going to see that watch ticking faster because they are moving towards each other rather than away.

I'm going to stop you right there. They will see the Earth clock running fast in their telescope. What they see in their telescope is due to two factors, time dilation, and Doppler shift. The second factor is due to the fact that since the distance between ship and Earth is decreasing, and thus takes less and less time to travel between the two. ( the opposite happens when it is moving away from the Earth)
The Earth, will also see the ship time speed up in their telescope. But here is the difference:
The ship will see Earth time go from running slow to running fast the moment they reverse course (because this is a result of something they themselves are doing.
The Earth however doesn't see this switch until the light leaving the ship when it turns around arrive at the Earth . So for example, if the ship is 10 ly away when it does this, the Earth doesn't see this happen until 10 yrs later (by which time the ship has traveled some part of the distance back to Earth.)
So if the ship travels for 10 years by its clock( seeing the Earth clock run 1/2 as fast), then turns around, and travels another 10 yrs back to Earth( seeing the Earrth clock run twice as fast), then on the outbound leg it sees the Earth age 5 yrs, and on the return leg 20 yrs, for a total of 25 yrs to the 20 that passes on its own clock.
The Earth on the other hand, will see the ship aging half as fast for 20 year, and aging twice as fast for 5 years, accumulating 20 years of time in 25 years of Earth time.

 

[Ibix]

In terms of what they actually literally see, they see the other's clocks tick slow when they see them moving away and fast when they see them moving closer - the Doppler effect dominates. However, the time at which they see the other start to get closer again is different. The twin who turns around sees the other start to get closer when they turn around, but the stay-at-home doesn't see the traveller turn around until much later when the light from the turnaround event reaches Earth. So the two twins see fast and slow ticks for different lengths of time, so the fact that there are different total numbers of ticks for each twin shouldn't be too surprising.

In terms of calculated clock rates, both twins calculate that the other's clock is ticking slowly (once they've corrected for the varying light travel time). However, if the twin who turns around wishes to use this calculation, they need to account correctly for switching inertial reference frames. The "inbound" and "outbound" rest frames have different definitions of what "at the moment of turn around, on Earth" means, and that difference is the extra time that makes the Earth twin older. Failing to do this is a bit like printing a map of Dover and a map of Calais, placing them next to each other, and wondering why anybody bothers with a ferry when they're next to each other on your map. The problem is that there's a chunk of space missed out, as there's a chunk of spacetime missed out of the naive approach to the twin paradox.

 

[etotheipi]

As my French teacher assured me, it's called the English Channel because when the French cat raced the English cat the un deux trois cat sank. But I'd be more worried about running into one of these:
https://www.mcsuk.org/news/five-sharks-found-uk-waters/

 

[russ_watters]

The same should hold true as the rocket turns around and begins coming back towards the earth. They are both going to still be in constant motion relative to each other, and whether the motion is towards or away from each other, they are still going to see the clock running slow.

Turning around and coming back is acceleration, not "constant [velocity] motion".

 

[Dale]

The babies are irrelevant, throw them out. (Out of the problem statement, not the airlock!)

They both have powerful telescopes through which they can look at each other‘s watches throughout the trip and back.

Here it seems that you are interested in what the twins actually visually see through the telescopes.

It seems during the whole trip and back, they are going to see each other‘s watches ticking slow.

But this statement is not about what they will visually see through the telescopes. This statement refers to what they calculate after calculating and correcting their visual observations for the light travel delay.

Can you please be explicit? Are you interested in what they actually visually see through the telescopes or in what they calculate after correcting for the light travel delay.

 

[PeterDonis]

I have been looking through some of the threads about the twins paradox in relativity.

I would strongly recommend also reading the Usenet Physics FAQ entry on the twin paradox:

https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

It does a good job of clearing up a lot of common misconceptions.

 

[Sophrosyne]

I would strongly recommend also reading the Usenet Physics FAQ entry on the twin paradox:

https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

It does a good job of clearing up a lot of common misconceptions.

This link looks quite good. I may need a day or so to read through it and try to digest it. Thanks!

 

[Sophrosyne]

The babies are irrelevant, throw them out. (Out of the problem statement, not the airlock!)

Here it seems that you are interested in what the twins actually visually see through the telescopes.

But this statement is not about what they will visually see through the telescopes. This statement refers to what they calculate after calculating and correcting their visual observations for the light travel delay.

Can you please be explicit? Are you interested in what they actually visually see through the telescopes or in what they calculate after correcting for the light travel delay.

OK, but maybe after I read PeterDonis' link. That looks like it answers my question. I'll respond after I finish reading and digesting it. As always, he seems to cut to the chase pretty quickly.

 

[Sophrosyne]

I'm going to stop you right there. They will see the Earth clock running fast in their telescope. What they see in their telescope is due to two factors, time dilation, and Doppler shift. The second factor is due to the fact that since the distance between ship and Earth is decreasing, and thus takes less and less time to travel between the two. ( the opposite happens when it is moving away from the Earth)
The Earth, will also see the ship time speed up in their telescope. But here is the difference:
The ship will see Earth time go from running slow to running fast the moment they reverse course (because this is a result of something they themselves are doing.
The Earth however doesn't see this switch until the light leaving the ship when it turns around arrive at the Earth . So for example, if the ship is 10 ly away when it does this, the Earth doesn't see this happen until 10 yrs later (by which time the ship has traveled some part of the distance back to Earth.)
So if the ship travels for 10 years by its clock( seeing the Earth clock run 1/2 as fast), then turns around, and travels another 10 yrs back to Earth( seeing the Earrth clock run twice as fast), then on the outbound leg it sees the Earth age 5 yrs, and on the return leg 20 yrs, for a total of 25 yrs to the 20 that passes on its own clock.
The Earth on the other hand, will see the ship aging half as fast for 20 year, and aging twice as fast for 5 years, accumulating 20 years of time in 25 years of Earth time.

OK, I have to sit here and think about this for a minute, but it seems to make sense. Thanks. I didn't realize that BOTH factors were so important here: the relativistic effects, as well as the doppler effects.

 

[PeterDonis]

I didn't realize that BOTH factors were so important here: the relativistic effects, as well as the doppler effects.

The "Doppler Shift Analysis" page in the FAQ I linked to discusses this. IMO it is one of the key things to understand.

 

[Sophrosyne]

So as

The "Doppler Shift Analysis" page in the FAQ I linked to discusses this. IMO it is one of the key things to understand.

Again, thanks for that link. I am still working on reading and digesting it.

But as a related question: if two spaceships are approaching each other at near the speed of light at constant velocity, relativistically they would see each other’s watches running slow. But by Doppler effects they would be seeing them go faster, right? Which effect wins out? It seems they would sort of cancel each other, no?

Of course, once they pass each other, both effects would be contributing to seeing each other‘s clocks go slower.

 

[PeterDonis]

if two spaceships are approaching each other at near the speed of light at constant velocity, relativistically they would see each other’s watches running slow.

No, they would not see each other's watches running slow. They would see each other's watches running fast, by the Doppler factor. They would calculate that, by the simultaneity convention of their respective inertial frames (which are different) that each other's watches are running slow, since that calculation would correct for the decreasing light travel time. But that's a calculation, not something anyone sees.

by Doppler effects they would be seeing them go faster, right?

Yes, and that is all they would see. See above.

Which effect wins out?

There is no such thing as one "effect winning out" over the other. One is a direct observation, the other is a calculation. But the calculation starts from the direct observation; each one sees the other's watch running fast, then calculates a correction based on the decreasing light travel time since they are moving towards each other. So it's not a matter of two effects competing. They aren't even the same kind of thing.

Of course, once they pass each other, both effects would be contributing to seeing each other‘s clocks go slower.

No. Same comment as above: there aren't two effects combining with each other. There is what is directly observed--which now is that each one sees the other's watch running slow by the Doppler factor--and then there is what each one calculates to be the other's "rate of time flow" given the simultaneity convention of their inertial frame.

Note, btw, that in this case, the calculation actually results in each one calculating the other's watch to run less slow than it is seen to run directly. So the two "effects" are still in "opposite directions" in this case. Do the math and see.

 

[robphy]

Time dilation depends on relative speed… and relates spacelike-related events.

Doppler effect depends on the relative speed and the sign of the relative velocity and relates lightlike-related events.

If there is any competition of effects, it’s indirect.

 

[Ibix]

If you are traveling at speed $v$ away from me and emit flashes with frequency $f_0$, relativity says I receive them with frequency$$f=f_0\sqrt{\frac{c-v}{c+v}}$$If $v$ is negative you are coming towards me, but the same formula works. Time dilation is already factored into this. You can see that the received frequency is lower if you are moving away from me and higher if you are moving towards me.

 

[PeroK]

But as a related question: if two spaceships are approaching each other at near the speed of light at constant velocity, relativistically they would see each other’s watches running slow. But by Doppler effects they would be seeing them go faster, right? Which effect wins out? It seems they would sort of cancel each other, no?

You could, of course, always calculate that for yourself, as it only requires basic kinematics.

 

[stevendaryl]

Something about Doppler shifts and relativity that might not be common knowledge:

Assume that light travels at speed c in one “rest” frame, and that there are two observers sending light signals to one another, at a rate of$f$ per second, one observer at rest, and one moving away from the first at speed v. Ignoring time dilation, you can find:

1. If the receiver is at rest, then the Doppler shift will be $\dfrac{1}{1+\frac{v}{c}}$. That is, he will receive the signals at a rate $f’= f \dfrac{1}{1+\frac{v}{c}}$.
   
2. If the sender is at rest, and the receiver is moving away, then the receiver will experience a frequency Doppler shift of $1-\frac{v}{c}$. That is, he will receive the signals at a rate $f’= f (1-\frac{v}{c})$.

When $v\ll c$, these two will be approximately the same, but a careful comparison could tell you which was moving, in violation of the relativity principle.

However, if you assume that the “moving” observer’s clock is running slow by a factor of $\gamma$ (as measured by the observer at “rest”), then his light signals will be sent at a lower frequency, $f/\gamma$. This will combine with the nonrelativistic Doppler shift to mean that the receiver at rest will measure a Doppler shift of $\dfrac{1}{\gamma(1+\frac{v}{c})}$

When the moving observer is the receiver, he will experience the light signals to be sped up by factor of $\gamma$, compared to the situation without introducing clock slowing. So he will measure a Doppler shift of $\gamma (1-\frac{v}{c})$

If $\gamma$ has the magical value of $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$, then the two Doppler shifts will be the same, meaning that you can’t tell who is “moving”, in keeping with the relativity principle.

 

[Grasshopper]

If you are traveling at speed $v$ away from me and emit flashes with frequency $f_0$, relativity says I receive them with frequency$$f=f_0\sqrt{\frac{c-v}{c+v}}$$If $v$ is negative you are coming towards me, but the same formula works. Time dilation is already factored into this. You can see that the received frequency is lower if you are moving away from me and higher if you are moving towards me.

Does this indicate that clocks moving toward the observer will always appear to be ticking fast to the observer? Because in this case, unless I’m having a senior moment, the radical should always be greater than 1.

 

[PeroK]

Does this indicate that clocks moving toward the observer will always appear to be ticking fast to the observer? Because in this case, unless I’m having a senior moment, the radical should always be greater than 1.

I'm not sure I understand your question. That's the standard relativistic Doppler shift.

 

[Ibix]

Does this indicate that clocks moving toward the observer will always appear to be ticking fast to the observer? Because in this case, unless I’m having a senior moment, the radical should always be greater than 1.

$v$ is signed. If the source is coming towards me at half light speed, $v=-0.5c$ and the root is $\sqrt 3$. If the source is moving away, $v=0.5c$ and it is $1/\sqrt 3$.

Does that answer your question?

 

[Grasshopper]

$v$ is signed. If the source is coming towards me at half light speed, $v=-0.5c$ and the root is $\sqrt 3$. If the source is moving away, $v=0.5c$ and it is $1/\sqrt 3$.

Does that answer your question?

It confirms what I thought of the math, but is the physical implication that when the radical is greater than 1, the clock will appear (visually) to be ticking fast?

 

[Ibix]

It confirms what I thought of the math, but is the physical implication that when the radical is greater than 1, the clock will appear (visually) to be ticking fast?

Yes. However, if you subtract out the effect of changing distance you will find time dilation remains and clocks tick slow by a factor of $\gamma$ independent of direction. Many sources are sloppy and say you will see moving clocks tick slowly, but taking "see" literally this is false.

 

[PeroK]

It confirms what I thought of the math, but is the physical implication that when the radical is greater than 1, the clock will appear (visually) to be ticking fast?

It's called the Doppler effect!

 

[stevendaryl]

It confirms what I thought of the math, but is the physical implication that when the radical is greater than 1, the clock will appear (visually) to be ticking fast?

Yes. Let $t$ be the time in your reference frame, and let $T$ be the time in the clock's reference frame.

Suppose that a clock is traveling toward you at 60% of the speed of light. That means that its time-dilation factor is 0.8 (that is, $\sqrt{1 - \frac{v^2}{c^2}}$, with $v = 0.6 c$). Suppose that initially, the clock is a distance of 6 light-seconds away. Suppose that initially, the clock shows time $T=0$.

So some numbers:

• Time for the light showing $T=0$ to reach you: 6 seconds
• Time for the clock to reach you: 10 seconds
• Time showing on the clock when it reaches you: 8 seconds (because the clock is slowed by a factor of 0.8)

That means that you will see the clock showing $T=0$ at 6 seconds, your time. Then 4 seconds later, your time, you will see the clock showing $T=8\ \text{seconds}$. So the image of the clock as seen by you will appear to advance 8 seconds, from $T=0$ to $T=8$, while your clock only advances from $t=6\ \text{seconds}$ to $t=10\ \text{seconds}$. So $\dfrac{\Delta T}{\Delta t} = \frac{8}{4} = 2$. The image of the clock seems to be advancing at twice the rate of your own clock.

That's exactly the prediction of the relativistic Doppler shift:

$\sqrt{\dfrac{1 - \frac{v}{c}}{1 + \frac{v}{c}}}$

With $\frac{v}{c} = - 0.6$ (negative, since it is coming toward you), this becomes:
$\sqrt{\dfrac{1 - (-0.6)}{1 + (-0.6)}} = \sqrt{\frac{1.6}{0.4}} = \sqrt{4} = 2$

 

[robphy]

It's called the Doppler effect!

to add ... "red shift", "blue shift" (akin to the siren of a car receding or approaching).

 

[meekerdb]

First, start thinking in terms of the triplet paradox version, so that you don't confuse yourself by thinking acceleration has something to do with it. It's pure geometry; the acceleration, if any, is incidental. Here's a diagram of the triplet version. A stays on Earth. B passes by Earth at high 0.5c going to the right. He syncs his clock with A as he passes. Five light-years out (in A's reference frame) he passes C who is inbound toward Earth at 0.5c. As C passes he syncs his clock to B's. The pips along the lines are evenly spaced ticks of the identical clocks measuring proper time along the paths. So you can count the ticks and see that

A's clock ticked off 20yrs. B's clock ticked off 8 and a fraction to his passing C, so C set his clock to 8.7yr. And then C's clock ticked another 8.7yrs until he passes A on Earth. So the B+C path measured 17.7yr while A measured 20yrs. That's the paradox. You could also imagine a twin who traveled with B and jumped to C as they passed, to get the classical twin version, but the triplet version makes it clear that it's not a question of stress due to acceleration. It's just a matter of a straight path is longer (in Mikowski metric) than a un-straight path. Now just do a Lorentz transform to B's reference frame:

Note that all the counts of ticks are the same. And note that A, by symmetry (or radar), sees that B and C met at the 10yr mark of A's time. But not according to B, who records it happening at 8.66yrs. It's the relativity of simultaneity. In A's frame I showed C starting toward A at the same time B left A. But in B's frame you see that C must have started much earlier. For completeness we should look at C's frame too.

Part of your confusion comes from thinking about what A, B, and C see, instead of what they measure. What they see depends on the finite speed of light which produces Doppler shifts. Here's a diagram in A's frame in which B is the red rocket and C is the blue rocket and B and C send evenly spaced pulses to A (the cyan ones with slope -1) and A sends evenly space pulses to the right (the green ones). So you can see the big Doppler effect.

At the passing point B and C, whose clocks read 2011 see A's clock as reading only late 2008. A doesn't see the passing point until early 2015 and sees all 8.7 ticks of C over just the last 2 years.

 

[robphy]

The arithmetic is easier with v=(3/5)c or v=(4/5)c.

 

[Ibix]

Nice diagrams. One point - it's conventional to use scales for the time and spatial axes such that light rays travel on 45° lines. I was briefly confused by your diagrams, wondering how your ships were supposed to be traveling at $c$, before I noticed the scales.

 

[Grasshopper]

It's called the Doppler effect!

It seems the misconception is people get the idea that the time dilation effect and Doppler effect are competing with each other for approaching clocks. But the equation says what it says fairly unambiguously.

 

[PeroK]

It seems the misconception is people get the idea that the time dilation effect and Doppler effect are competing with each other for approaching clocks. But the equation says what it says fairly unambiguously.

To be precise, that's the relativistic Doppler effect, which involves an element of time dilation. There is, for example, the transverse Doppler effect, which is essentially all due to time dilation.

In general, whether there is a redshift or blueshift also depends on the angle between the source and receiver. When the angle is zero there is always a blueshift (increased frequency).

 

[Ibix]

It seems the misconception is people get the idea that the time dilation effect and Doppler effect are competing with each other for approaching clocks. But the equation says what it says fairly unambiguously.

You can see the relativistic Doppler effect as the product of the Lorentz gamma factor and the naive distance-change-only Doppler factor of non-relativistic physics. In that sense, yes, the naive Doppler effect dominates the relativistic time dilation effect to get a net speed-up for sources approaching you.

But I'd say it's more sensible to call relativistic Doppler "Doppler" and refer to the distance-change-only effect by another term (distance-change-only, for example). Relativistic Doppler is the measurable thing - the other is a frame-dependent effect, so less important.