Twin Paradox: Spacetime Physics Ch.4 & 202 Year Difference

[Office_Shredder]

TL;DR Summary

The twin paradox is resolved by claiming if you instantaneously switch frames, your new frame's measured earth clock instantaneously changes. Try to compute that

Spacetime physics chapter 4 describes this wonderfully well chosen set of speeds/distances for the twin paradox.

A traveler departs from Earth at a speed of 99/101 (1=speed of light), traveling to a star that is 99light years away. From Earth's perspective, the traveler takes 101 years to go to the star, and 101 years to come back. Because of time dilation, Earth thinks the travelers' clock advances by $101\sqrt{1-(99/101)^2} = 20$ years outbound, and another 20 inbound. From the rocket's perspective, because of length contraction it thinks the Earth and the star are $99\sqrt{1-(99/101)^2}\approx 19.6$ light years away, and since the earth/star are moving at a speed of 99/101 times the speed of light, the time between the Earth passing and the star passsing takes $19.*101/99=20$ years each way on the rocket traveler's clock. Thanks to time contraction, from the rocket traveler's perspective, the Earth's clock only appears to advance $20\sqrt{1-(99/101)^2}\approx 3.96$ years each trip.

So the twin paradox is worse than expected. The traveler goes out to the star and back, and when they arrive, they should only expect the Earth to have aged about 8 years, when actually it has aged 202. The book explains the discrepancy by noting that for the traveler to go to the star and back, it must switch frames from outward to inward. Therefore, all is consistent as long as that frame switch instantaneously moves the traveler's perspective of the clock time on Earth by about 194 years.

I wanted to try to compute this (no hint as to how you would derive this other than 'it must be so for consistency!' is given). My thought was to imagine a third rocket traveler, who starts on the far side of the star from earth, and travels towards Earth at a speed of 99/101. Then the inbound rocket traveler can compute how much time it thinks has passed on Earth between the two events of the outbound rocket leaving, and the the outbound rocket reaching the star. I guess the claim here is that the always-inbound rocket traveler should measure the Earth clock as advancing by should advance by about 198 years.
So I guess here's my weak attempt at it. From the always-inbound rocket perspective, the Earth is moving at a speed of 99/101. The outbound rocket is moving at a speed of 99/101 in the Earth frame, so the velocity addition formula says that from the perspective of the always-inbound rocket, the outbound rocket moves at a speed of $\frac{2 \frac{99}{101}}{1+(\frac{99}{101})^2}\approx 0.9998$. That's pretty fast! From here, I get pretty confused. Because of length contraction, the always-inbound rocket should think the outbound rocket needs to travel 19.6 light years at a speed of 0.9998, which takes, well, about 19.6 years to do. Because of time contraction, it should think the Earth clock only moves $19.6\sqrt{1-(99/101)^2} \approx 3.88$ years between these two events. This seems to be moving in the wrong direction, since 3.88 is smaller than 3.96, and it's certainly not anywhere close to 198 years.

I'm not sure if I'm missing the point of the claim that the apparent Earth clock jumps instantaneously when the frame switches, or if I'm just doing a bad job of attempting to compute the Earth clock from the point of view of the inbound rocket. Any advice or insight into this is appreciated.

 

[Orodruin]

You are confusing yourself by blindly trying to apply length contraction and time dilation. This is the reason that the twin paradox exists in the first place. Instead, write down the coordinates of particular events in one system and Lorentz transform between systems.

In the end, the point is that the event on Earth that is simultaneous with the turnaround event is not the same in the out/inbound frames.

This may also help https://www.physicsforums.com/insights/geometrical-view-time-dilation-twin-paradox/

 

[PeterDonis]

Summary:: The twin paradox is resolved by claiming if you instantaneously switch frames, your new frame's measured Earth clock instantaneously changes.

Saying that the twin paradox is "resolved" by this claim is a bit much. In fact, this "resolution" is only one of a number of different ways of viewing the twin paradox scenario. See the Usenet Physics FAQ article on the twin paradox:

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

Your proposal is discussed on the "Time Gap Objection" page, which, you will note, is one of a number of pages, each describing a different viewpoint. To me, the most useful viewpoints are the Doppler Shift and Spacetime Diagram viewpoints; the latter, as the FAQ article notes, is really the most general one since it can be used to construct all the others.

 

[PeterDonis]

I wanted to try to compute this

Draw the spacetime diagram. For an instantaneous turnaround, you have a triangle in spacetime formed by the stay-at-home twin's segment and the two segments of the traveling twin's trip (outbound and inbound). Then draw the two lines of simultaneity from the turnaround point to the stay-at-home twin's segment, one for the outbound and one for the inbound traveling twin frames. Then just compute the length along the stay-at-home twin's segment between the two intersection points. By drawing the triangles for different traveling twin speeds, it should be easy to see that the length of the "in between" segment, as a fraction of the stay-at-home twin's total segment, increases as the traveling twin speed increases.

 

[FactChecker]

In the end, the point is that the event on Earth that is simultaneous with the turnaround event is not the same in the out/inbound frames.

This puzzles me. I am used to the discussion of the simultaneity of two, physically separated events. Clearly, one event is the "hand-off" between the out/inbound frames at the point of the turn-around. What is the second event on Earth?

 

[PeterDonis]

Draw the spacetime diagram.

Or, alternatively, compute everything in the stay-at-home twin's frame; it's easy to write down the equations for the two lines of simultaneity in that frame, since you know both a point on each line (the turnaround point) and their slope (it's just the complementary slope to that of the corresponding traveling twin segment). Then you can just compute where those lines intersect the line $x = 0$, which is where the stay-at-home twin is.

 

[Nugatory]

Clearly, one event is the "hand-off" between the out/inbound frames at the point of the turn-around. What is the second event on Earth?

There are three relevant events on Earth (more precisely, on earth’s worldline):
1) that event which happens at the same time as the turnaround, when using the frame in which the Earth is at rest
2) that event which happens at the same time as the turnaround, when using the frame in which the traveling twin is at rest on the outbound leg
3) that event which happens at the same time as the turnaround, when using the frame in which the traveling twin is at rest on the return leg

If you are want to use time dilation and length contraction to understand the twin paradox (although I would like to discourage you - this is the most complicated and least illuminating way of approaching the problem) you must also consider relatively of simultaneity and especially @Orodruin’s point that events 2 and 3 are different events,

 

[FactChecker]

There are three relevant events on Earth (more precisely, on earth’s worldline):
1) that event which happens at the same time as the turnaround, when using the frame in which the Earth is at rest
2) that event which happens at the same time as the turnaround, when using the frame in which the traveling twin is at rest on the outbound leg
3) that event which happens at the same time as the turnaround, when using the frame in which the traveling twin is at rest on the return leg

I agree. It seems to me that this approach to explaining the Twin's Paradox depends on instantly switching from one of these events to another and yet considering them the same event.

 

[robphy]

I agree. It seems to me that this approach to explaining the Twin's Paradox depends on instantly switching from one of these events to another and yet considering them the same event.

Two distinct events that are simultaneous to an inertial observer need not be simultaneous according to another inertial observer.

Reiterating what @Nugatory said…

The stay at home observer is inertial from separation to reunion… and determines an event on his worldline that simultaneous with the (distant) turnaround event.

The outbound leg is inertial… and experiences the turnaround event, but determines a different event on the stay-at-home worldline to be simultaneous with the turnaround event.

The inbound leg is inertial … and experiences the turnaround event, but determines a different event (from the other two) to be simultaneous with the turnaround event.

All three inertial observers agree on the turnaround event, but each disagree with what event on the stay-at-home worldline is simultaneous with the turnaround event.

If you consider the Euclidean analogue… construct a “perpendicular from each inertial worldline” that meets the turnaround vertex. Those perpendiculars determine three distinct points on the side opposite the vertex.

(Of course, the traveler is non-inertial, in spite of being piecewise inertial… a ball at rest on a frictionless table in the traveler frame will move at the turnaround event.)

 

[jartsa]

Summary:: The twin paradox is resolved by claiming if you instantaneously switch frames, your new frame's measured Earth clock instantaneously changes. Try to compute that

Thoughts of accelerating dude before the acceleration: I think now that the Earth is now distance x away. The Earth thinks now that I'm the same distance x away from it.

Thoughts of accelerating dude after the acceleration: I think now that the Earth is now distance y away. The Earth thinks now that I'm the same distance y away from it.

Then the accelerated dude ponders something like this: So the time on the Earth changed from that time when the Earth said the distance is x, to that time when the Earth said the distance is y. And from that I can calculate how much time passed on earth.

 

[PeterDonis]

Thoughts of accelerating dude before the acceleration: I think now that the Earth is now distance x away. The Earth thinks now that I'm the same distance x away from it.

Thoughts of accelerating dude after the acceleration: I think now that the Earth is now distance y away. The Earth thinks now that I'm the same distance y away from it.

If by "before" and "after" the acceleration, you mean immediately before and after, then x = y. What changes instantaneously at the turnaround, on the "time gap" viewpoint, is not the distance to Earth but the time that Earth's clock reads "now".

 

[jartsa]

If by "before" and "after" the acceleration, you mean immediately before and after, then x = y. What changes instantaneously at the turnaround, on the "time gap" viewpoint, is not the distance to Earth but the time that Earth's clock reads "now".

Hmm ... I was thinking of a simpler case where the accelerated person accelerates from the rest fame of the Earth towards the earth. Then it should work, unless I got something wrong.

Or is it wrong for the accelereted guy to say that: "at this moment on the Earth the distance to me is the same as my distance to the Earth according to myself"?

 

[PeterDonis]

I was thinking of a simpler case where the accelerated person accelerates from the rest fame of the Earth towards the earth

How is that case relevant to this thread?

is it wrong for the accelereted guy to say that: "at this moment on the Earth the distance to me is the same as my distance to the Earth according to myself"?

If you mean during the acceleration, the accelerated observer would have to define what "my distance to the Earth according to myself" means. There is no unique way for the accelerated observer to construct his "rest frame" during the acceleration the way inertial frames can be constructed for non-accelerated observers.

Before and after the acceleration, the observer is at rest in an inertial frame and can use that inertial frame's notion of "distance to the earth". Before the acceleration, the observer's rest frame is the same as the earth's, so his notion of "distance to the earth" will match the earth's. After the acceleration, the observer's rest frame is different from the earth's, so his notion of "distance to the earth" will not match the earth's.

However, as noted above, the case you describe, where the observer starts out at rest relative to the Earth and ends up moving towards it, is not relevant to the twin paradox scenario being discussed in this thread.

 

[Ibix]

Or is it wrong for the accelereted guy to say that: "at this moment on the Earth the distance to me is the same as my distance to the Earth according to myself"?

Depends what frame the Earth is using to make its measurement and what frame you are using to define "at this momentc on Earth.

 

[PeroK]

So the twin paradox is worse than expected. The traveler goes out to the star and back, and when they arrive, they should only expect the Earth to have aged about 8 years, when actually it has aged 202. The book explains the discrepancy by noting that for the traveler to go to the star and back, it must switch frames from outward to inward. Therefore, all is consistent as long as that frame switch instantaneously moves the traveler's perspective of the clock time on Earth by about 194 years.

The fundamental issue here is understanding that "now" at all points in space is not universal. In the original 1905 paper, Einstein elaborates at considerable length on why we must have some formal mechanism for assigning a time to a distant location. We cannot just casually say "now" on Earth and "now" on Mars and expect those events to be simultaneous in all inertial reference frames.

Your reluctance to change the time of "now" back on Earth when you change reference frames shows that, deep down, you are still wedded to universal time and universal simultaneity. You now have a veneer of understanding of SR, but deep down "now" everywhere is still "now" everywhere".

The problem is seeing too much physical significance in what is happening "now" at a distant location. Take take two objects at rest relative to each other, separated by a proper distance of $D$, with a synchronised clock at each location (using Einstein clock synchronisation). In a frame where the objects are moving at speed $v$, the clocks are out of sync by $$\Delta t = \frac {vD}{c^2}$$
If we take the upper limit for this speed ($v \approx c$), then the upper limit for the loss of simultaneity is $$\Delta t = \frac D c$$ We can see that "now" on Earth and "now" at a location 200 light years away (in the shared rest frame) must vary by up to $\pm 200$ years when we use a coordinate system of a near light-speed rocket.

This says something deep about the nature of time, space, relative motion and relative simultaneity. Note that nothing physical happens to the clock on Earth in your scenario. It's the concept of a distant "now" that changes.

 

[Orodruin]

I agree. It seems to me that this approach to explaining the Twin's Paradox depends on instantly switching from one of these events to another and yet considering them the same event.

No. The approach to consider these events the same is what leads to the twin paradox in the first place. Thus, realising that they are indeed not the same event resolves the paradox.

 

[Mister T]

I wanted to try to compute this (no hint as to how you would derive this other than 'it must be so for consistency!' is given).

It's explained later in the book as being due to the relativity of simultaneity.

 

[Office_Shredder]

Draw the spacetime diagram. For an instantaneous turnaround, you have a triangle in spacetime formed by the stay-at-home twin's segment and the two segments of the traveling twin's trip (outbound and inbound). Then draw the two lines of simultaneity from the turnaround point to the stay-at-home twin's segment, one for the outbound and one for the inbound traveling twin frames. Then just compute the length along the stay-at-home twin's segment between the two intersection points. By drawing the triangles for different traveling twin speeds, it should be easy to see that the length of the "in between" segment, as a fraction of the stay-at-home twin's total segment, increases as the traveling twin speed increases.

Thanks. Spacetime diagrams are the next chapter, so I will be sure to revisit this when I read it.

The stay at home observer is inertial from separation to reunion… and determines an event on his worldline that simultaneous with the (distant) turnaround event.

The outbound leg is inertial… and experiences the turnaround event, but determines a different event on the stay-at-home worldline to be simultaneous with the turnaround event.

The inbound leg is inertial … and experiences the turnaround event, but determines a different event (from the other two) to be simultaneous with the turnaround event.

This really clicked for me. The point is that if you have a frame with two points distance d apart that release flashes of light simultaneously, then in a frame where these points are moving for the flashes of lights to reach the middle at the same time, the observer has to think the point that's at the rear of the motion released its light first. So imagine Earth and the star release flashes of light that intersect at the midpoint of the Earth and the star. In the outbound frame, the star needs to emit the light first since the Earth is at the front of the earth/star motion from the outbound perspective. So from the outbound perspective, the events of "earth clock hits 101" and "rocket reaches star" are not simultaneous, and the rocket reaches star one happened first. From the inbound rocket perspective, the Earth event happened first, by a symmetrical amount of time.

Thanks!

 

[vela]

So I guess here's my weak attempt at it. From the always-inbound rocket perspective, the Earth is moving at a speed of 99/101. The outbound rocket is moving at a speed of 99/101 in the Earth frame, so the velocity addition formula says that from the perspective of the always-inbound rocket, the outbound rocket moves at a speed of $\frac{2 \frac{99}{101}}{1+(\frac{99}{101})^2}\approx 0.9998$. That's pretty fast! From here, I get pretty confused. Because of length contraction, the always-inbound rocket should think the outbound rocket needs to travel 19.6 light years at a speed of 0.9998, which takes, well, about 19.6 years to do.

The thing you overlooked here was that according to the always-inbound observer, the star is moving at a speed of 99/101, so the relative velocity of the outbound rocket and the star is about 0.01960. So according to the always-inbound observer, it takes the outbound rocket 1000.1 years to reach the star.

 

[robphy]

Any advice or insight into this is appreciated.

Make life easier by choosing an arithmetically simpler velocity, like (3/5)c or (4/5)c. The resulting calculations are generally simple fractions.

 

[Sagittarius A-Star]

Two frames perspective of the twin paradox
...
I'm not sure if I'm missing the point of the claim that the apparent Earth clock jumps instantaneously when the frame switches, or if I'm just doing a bad job of attempting to compute the Earth clock from the point of view of the inbound rocket. Any advice or insight into this is appreciated.

The reason for "apparent Earth clock jumps instantaneously" is, that the turnaround is idealize to be an event.

In reality, the turnaround needs a finite time and the rocket accelerates towards the earth, until the velocity has changed from $v$ to $-v$.

I an accelerated frame, the tick-rate of a clock depends not only on it's velocity, but also on it's x-cooddinate. Therefore, in the accelerated frame, the clock on Earth ticks faster than the clock in the rocket (while the turnaround). The "travelling" twin experiences an artificial "gravitational" field (in flat spacetime), that comes along with a gravitational time-dilation between different x-coordinates.

 

[PeroK]

The reason for "apparent Earth clock jumps instantaneously" is, that the turnaround is idealize to be an event.

In reality, the turnaround needs a finite time and the rocket accelerates towards the earth, until the velocity has changed from $v$ to $-v$.

I an accelerated frame, the tick-rate of a clock depends not only on it's velocity, but also on it's x-cooddinate. Therefore, in the accelerated frame, the clock on Earth ticks faster than the clock in the rocket (while the turnaround). The "travelling" twin experiences an artificial "gravitational" field (in flat spacetime), that comes along with a gravitational time-dilation between different x-coordinates.

It can be explained more simply by the relativity of simultaneity and doesn't require introducing the complexities of a pseudo-gravitational field.

 

[Sagittarius A-Star]

It can be explained more simply by the relativity of simultaneity and doesn't require introducing the complexities of a pseudo-gravitational field.

That's correct. On the other hand, the OP asked in the headline for the "Two frames perspective" (not for example 3 frames).

 

[vanhees71]

And even more simple is the explanation using invariants, i.e., the proper times of each twin from their departure to their meeting again. As any physically interpretible quantity this are invariant, i.e., frame and coordinate independent quanities (in this case scalars).

 

[Sagittarius A-Star]

Summary:: The twin paradox is resolved by claiming if you instantaneously switch frames, your new frame's measured Earth clock instantaneously changes. Try to compute that
...
Therefore, all is consistent as long as that frame switch instantaneously moves the traveler's perspective of the clock time on Earth by about 194 years.

I wanted to try to compute this (no hint as to how you would derive this other than 'it must be so for consistency!' is given).

Event $E_3$ shall be "Depart Canopus" (3rd event in Table 4-1 on page 130).
Event $C_{3ETP}$ shall be "Clock on Earth shows departure time in (returning) traveler's perspective".

Between these events is valid (rocket frame = primed frame, $c := 1$):
$\Delta t' = 0$
$\Delta x = 99 \ years\ \ \ \ \$ (=distance Earth-Canopus).

$v = \frac{\Delta t}{\Delta x}\ \ \ \ \$ (* Reason: see below quote)
$\Delta t = v \Delta x = \frac {99}{101} * 99 \ years = 97.04 \ years$
$t_{C3ETP} = 101 \ years + \Delta t$
In an analog way, you can compute $t_{C2ETP} = 101 \ years - \Delta t$
The clock time on Earth (traveler's perspective) "jumps" by $2 * \Delta t = 194.08 \ years$.

* Reason for used velocity formula:

$v=\frac{\Delta x}{\Delta t}$ if in the other frame $\Delta x' = 0\ \ \$ (relativity of "same location")
$v=\frac{\Delta t}{\Delta x}$ if in the other frame $\Delta t' = 0\ \ \ \ \$(relativity of simultaneity)

 

[stevendaryl]

It can be explained more simply by the relativity of simultaneity and doesn't require introducing the complexities of a pseudo-gravitational field.

It’s also the case that explaining the twin paradox using the time dilation of a pseudo-gravitational field is a bit circular. The way you (typically) derive acceleration-dependent time dilation is by understanding how things work in an inertial frame and transforming to get a description in an accelerated frame.

 

[PeroK]

It’s also the case that explaining the twin paradox using the time dilation of a pseudo-gravitational field is a bit circular. The way you (typically) derive acceleration-dependent time dilation is by understanding how things work in an inertial frame and transforming to get a description in an accelerated frame.

I prefer the simple geometry that the twin paradox in Minkowski spacetime is analogous to the triangle inequality in Euclidean space.

 

[vanhees71]

It's clear that the most simple way is simply to calculate the proper time of the two twins between departing from the same place and meeting again at (maybe another) same place. This physical meaningful quantities are comletely independent of the choice of the reference frame you calculate them, because they are scalars. They only depend on the world lines of the twins and, in the GR case, the spacetime metric determined by the distribution of energy and momentum ("matter") and the corresponding gravitational field.