﻿ Twin Paradox And Sample Homework Problem - Essay for you

## Twin Paradox And Sample Homework Problem

Category: Homework

## Description

### Twin paradox question - Physics Stack Exchange

I'm trying to wrap my mind around the twin paradox, but I can't figure out this one problem from my textbook. It uses the relativistic Doppler effect to demonstrate how the paradox works. The first part of the problem is as follows:

Amelia and Casper are twins. Amelia is going on a trip to a distant planet while Casper remains on Earth. The planet is about $12$ light-years from Earth, traveling at $0.6c$. Therefore from Casper's perspective his sister's trip lasts 40 years (20 years outbound and 20 years returning). From Amelia's perspective the trip only lasts 32 years (16 years outbound and 16 years returning). Casper sends a message to Amelia once a year on his birthday. The frequency of these messages will be Doppler shifted as follows:

So Amelia will receive 8 messages on her outgoing trip ($0.5yr^<-1>*16yr)$ and 32 messages on her return trip ($2yr^<-1>*16yr$), totalling to the 40 messages that were sent by Casper.

Now the question asks me to calculate how many messages Casper would receive if Amelia sent a message to him on each of her birthdays. And my calculations are as follows: on the outgoing trip, Casper will receive $10$ messages on Amelia's outgoing trip($0.5yr^<-1>*20yr$), and $40$ messages on her return trip ($2yr^<-1>*20yr$). But that totals to $50$ messages, which is more than what Amelia sent out. Why are my calculations coming out this way?

asked Oct 13 '13 at 16:29

There are a couple of solutions based on how we interpret the sentence "Amelia sent a message to him on each of his birthday" .

1. If Amelia wanted to make sure that Casper receives the messages on HIS 40 birthdays :
She must do some calculations by using her special-and-general-relativity knowledge. First, she calculates his total age throughout the entire journey and it turns out to be 40 (as in the question). Now She does the same for her age and finds it going to be 32. She understands the symmetry is broken here because this is not special relativity anymore! At half way of her journey, she has to brake out of her inertial frame and accelerate back. OK, so she knows all these stuff and then does the ingenious task of finding out his birthdays while sitting in her own high-speed frame. This is how she does it:

• I must send messages at different rates for my away journey and return journey, otherwise Doppler shift might mess things up!

During my away journey his 1 year is my 0.8 years
$t_ = 1year\times \sqrt<1-0.36>=0.8years$
So if I sent him messages for 16 years he must get a total of 20 messages ($\frac<16><0.8>$)

But there is going to be a Doppler shift which changes my message-rate of 1msg every 0.8 years to
$\frac<1 msg><0.8 yr> \sqrt<\frac<1-0.6><1+0.6>> = \frac<0.625msg>$

This means he will get a message every 1.6 years (after his birthday! boo)
$\frac<0.625msg> = \frac<1msg><1.6 yr>$

So, I must increase my message frequency so that he gets it in time, $F_=\frac><\sqrt<\frac<1-0.6><1+0.6>>>=\frac<1><0.5>= \frac<2msg>$

Now he will get 1 msg every year, yay!
$\frac<2 msg> \sqrt<\frac<1-0.6><1+0.6>> = \frac<1msg>$
Note: She will send 32 msgs in 16 years ($16\times2$) and 8 more in the return journey. You can easily change the equations above for a blue-shift and workout the frequency. Otherwise, it might elongate the answer (you can guess it's going to be a small frequency since only 8 msgs are pending and msgs travel faster now)

2. If Amelia did the same as Casper did (How you solved the problem):
She sends messages every year (his 1.25 years). He get's them after a red shift of:
$$\frac<1 msg> \sqrt<\frac<1-0.6><1+0.6>> = \frac<0.5msg>$$
She sends 16 messages in the first 16 years (he receives only 8) and then on her way back she repeats it (but now with a blue-shift):
$$\frac<1 msg> \sqrt<\frac<1+0.6><1-0.6>> = \frac<2msg>$$
Thus, in that 20 years period (her return journey of 16 years) he gets 16 such messages along with those remaining to reach him from the away-trip. This adds up to:
$8\ delayed\ msgs + 8\ more\ delayed\ msgs + 16\ really\ fast\ msgs = 32\ total\ msgs$
Note: Interestingly, this explains that classical who aged more "paradox"! The switching of frames by Amelia is the reason why her brother aged (he did not accelerate, she did). If she was to go on forever with out stopping, both of them would think the other one is younger. But she did stop and came back, thus entering a new frame, which treated her brother as old.
Also, the first 8 messages and the next 8 ones were send to him in a red shift journey. Meanwhile she is on her return journey. 16 messages send during this blue-shift journey will reach him really fast and obviously all of them (32) will be messed up and not synchronous at all! Chances are none of these messages from his sister will actually reach him on his real birthday!

You got the wrong answer because you simply did this:
$\frac<40><32>=1.25$
$1.25\sqrt<\frac<1-0.6><1+0.6>> = 0.625$
$0.625\times16=10$
$1.25\sqrt<\frac<1+0.6><1-0.6>> = 2.5$
$2.5\times16=40$
$10+40 = 50$
1.25 is his years, not hers. Applying it in Doppler shift gives her receiving rate on the R.H.S.

answered Oct 14 '13 at 13:53

## Other articles

### Concept of the Twin Paradox in Physics

What Is the Twin Paradox? Real Time Travel

Updated September 02, 2016.

Question: What is the Twin Paradox?

The twin paradox is a thought experiment that demonstrates the curious manifestation of time dilation in modern physics, as it was introduced by Albert Einstein through the theory of relativity .

Answer: Consider two twins, named Biff and Cliff. On their twentieth birthday, Biff decides to get in a spaceship and take off into outer space, traveling at nearly the speed of light .

He journeys around the cosmos at this speed for about 5 years, returning to the Earth when he is 25 years old.

Cliff, on the other hand, remains on the Earth. When Biff returns, it turns out that Cliff is 95 years old.

What Happened?

According to relativity, two frames of reference that move differently from each other experience time differently, a process known as time dilation. Because Biff was moving so rapidly, time was in effect moving slower for him. This can be calculated precisely using Lorentz transformations. which are a standard part of relativity.

The first twin paradox isn't really a scientific paradox, but a logical one: How old is Biff?

Biff has experienced 25 years of life, but he was also born the same moment as Cliff, which was 90 years ago. So is he 25 years old or 90 years old?

In this case, the answer is "both". depending on which way you're measuring age. According to his driver's license, which measures Earth time (and is no doubt expired), he's 90. According to his body, he's 25. Neither age is "right" or "wrong," although the social security administration might take exception if he tries to claim benefits.

The second paradox is a bit more technical, and really comes to the heart of what physicists mean when they talk about relativity. The entire scenario is based on the idea that Biff was traveling very fast, so time slowed down for him.

The problem is that in relativity, only the relative motion is involved. So what if you considered things from Biff's point of view, then he stayed stationary the whole time, and it was Cliff who was moving away at rapid speeds. Shouldn't calculations performed in this way mean that Cliff is the one who ages more slowly? Doesn't relativity imply that these situations are symmetrical?

Now, if Biff and Cliff were on spaceships traveling at constant speeds in opposite directions, this argument would be perfectly true. The rules of special relativity, which govern constant speed (inertial) frames of reference, indicate that only the relative motion between the two is what matters. In fact, if you're moving at a constant speed, there's not even an experiment that you can perform within your frame of reference which would distinguish you from being at rest. (Even if you looked outside the ship and compared yourself to some other constant frame of reference, you could only determine that one of you is moving, but not which one.)

But there's one very important distinction here: Biff is accelerating during this process. Cliff is on the Earth, which for the purposes of this is basically "at rest" (even though in reality the Earth moves, rotates, and accelerates in various ways). Biff is on a spaceship which undergoes intensive acceleration to read near lightspeed. This means, according to general relativity. that there are actually physical experiments that could be performed by Biff which would reveal to him that he's accelerating. and the same experiments would show Cliff that he's not accelerating (or at least accelerating much less than Biff is).

The key feature is that while Cliff is in one frame of reference the entire time, Biff is actually in two frames of reference - the one where he's traveling away from the Earth and the one where he's coming back to the Earth.

So Biff's situation and Cliff's situation are not actually symmetrical in our scenario. Biff is absolutely the one undergoing the more significant acceleration, and therefore he's the one who undergoes the least amount of time passage.

This paradox (in a different form) was first presented in 1911 by Paul Langevin, in which the emphasis stressed the idea that the acceleration itself was the key element that caused the distinction. In Langevin's view, acceleration therefore had an absolute meaning. In 1913, though, Max von Laue demonstrated that the two frames of reference alone are enough to explain the distinction, without having to account for the acceleration itself.

### Twin paradox - Example Problems

The twin paradox is a thought experiment in special relativity of two twin brothers, one undertaking a long space journey with a very high-speed rocket at almost the speed of light. the other remaining on Earth. When the traveler finally returns to Earth, it is observed that he is younger than the twin who stayed put. Or, as first stated by Albert Einstein (1911 ):

If we placed a living organism in a box. one could arrange that the organism, after any arbitrary lengthy flight, could be returned to its original spot in a scarcely altered condition, while corresponding organisms which had remained in their original positions had already long since given way to new generations. For the moving organism the lengthy time of the journey was a mere instant, provided the motion took place with approximately the speed of light. (in Resnick and Halliday, 1992)

Twins of different physical age can be seen as a paradox.

A paradox arises if one takes the position of the traveling twin: from his perspective, his brother on Earth is moving away quickly, and eventually comes close again. So the traveler can regard his brother on Earth to be a "moving clock" which should experience time dilation. As is commonly accepted: "Special relativity says that all observers are equivalent, and no particular frame of reference is privileged."

This is slightly different from Einsteins original text: 'Die Gesetze, nach denen sich die Zustände der physikalischen Systeme ändern, sind unabhängig davon, auf welches von zwei relativ zueinander in gleichförmiger Translationsbewegung befindlicher Koordinatensysteme diese Zustandsänderungen bezogen werden'. The physical laws are independent of coordinate systems.

This outcome is predicted by Einstein's special theory of relativity. It is due to an experimentally verified phenomenon called time dilation. in which a moving clock is found to experience a reduced amount of proper time as determined by clocks synchronized with a stationary clock. One example of this phenomenon involves muons. particles produced in the upper atmosphere which can be detected on the ground. Without time dilation, the muons would decay long before reaching the ground. Another experiment confirmed time dilation by comparing the effects of speed on two atomic clocks. one based on earth, the other aboard a supersonic plane. They were out of sync afterwards: the atomic clock on the plane was slightly behind.

Special relativity says that all observers are equivalent, and no particular frame of reference is privileged. Hence, the traveling twin, upon return to Earth, would expect to find his brother to be younger than himself, contrary to that brother's expectations. Which twin is correct?

It turns out that the traveling twin's expectation is mistaken: special relativity does not say that all observers are equivalent, only that all observers in inertial reference frames are equivalent. But the traveling twin jumps frames (accelerates) when he does a U-turn. The twin on Earth rests in the same inertial frame for the whole duration of the flight (no accelerating or decelerating forces apply to him) and he is therefore able to distinguish himself from the traveling twin.

Note: it is wrong to think twin paradoxes are simply due to acceleration effects. One needs no acceleration to achieve a twin paradox in flat spacetime (cf. Brans and Stewart). Thus, a twin paradox situation does not always imply acceleration.

There are not two but three relevant inertial frames: the one in which the stay-at-home twin remains at rest, the one in which the traveling twin is at rest on his outward trip, and the one in which he is at rest on his way home. It is during the acceleration and deceleration of the departure and arrival to Earth and similar accelerations at the U-turn when the traveling twin switches frames. That's when he must adjust the calculated age of the twin at rest. This is a purely artificial effect caused by the change in the definition of simultaneity when changing frames. Here's why.

In special relativity there is no concept of absolute present. A present is defined as a set of events that are simultaneous from the point of view of a given observer. The notion of simultaneity depends on the frame of reference, so switching between frames requires an adjustment in the definition of the present. If one imagines a present as a (three-dimensional) simultaneity plane in Minkowski space. then switching frames results in changing the inclination of the plane.

In the spacetime diagram on the right, the first twin's lifeline coincides with the vertical axis (his position is constant in space, moving only in time). On the first leg of the trip, the second twin moves to the right (black sloped line); and on the second leg, back to the left. Blue lines show the planes of simultaneity for the traveling twin during the first leg of the journey; red lines, during the second leg. During the U-turn the plane of simultaneity jumps from blue to red and very quickly sweeps a large segment of the lifeline of the resting twin. Suddenly the resting twin "ages" very fast in the reckoning of the traveling twin.

It is sometimes claimed that the twin paradox cannot be resolved without the use of general relativity. since one of the twins must undergo acceleration during the U-turn. This is false, for two reasons. First, most simply, the acceleration can easily be made to be a negligible part of the trip by making the inertial legs long enough. Second, it is no problem, in principle, to describe the effects of acceleration in special relativity as long as one does so using the laws of physics formulated in an inertial frame of reference — general relativity is only needed to make the laws of physics in the accelerated frame the same as in an inertial frame with a gravitational field. As Hermann Bondi once quipped on this question (French, 1968), "it is obvious that no theory denying the observability of acceleration could survive a car trip on a bumpy road," and special relativity certainly does not deny acceleration. This can actually be accomplished through the use of the Frenet-Serret_formulas.

Consider a space ship going from Earth to the nearest star system; 4.45 light years away; at 0.866 c. The above image shows the ship with its 0.5 length contraction. To an observer on Earth the trip will take 5.14 years, producing a round trip time of 10.28 years.

However, to the ship's crew the stars and the distance between them will be shortened in the direction of motion to 0.5 of what is observed on Earth, resulting in a travel distance of only 2.23 light years and a corresponding 2.57 year travel time. This produces a round trip time of only 5.14 years. Thus the ship's crew would experience less time than those on Earth.

If one of the astronauts on the ship had a twin that stayed on Earth, he would return home to find his brother about 5 years older than himself.

Now, this takes care of the physical effect of time dilation but it does not reconcile how each twin would observe the other during the trip. The solution to this observational problem can be found in the relativistic Doppler effect. It includes the effects of both time dilation and distance, hence we can calculate just how fast each twin will observe time to flow in the other's reference frame.

Left: Earth to ship.
Right:Ship to Earth.

• The black line is the Earth's path through space time.
• The dark blue line is the ship's outward path through space time.
• The purple line is its return path through space time.
• The green dots are equally spaced points in time within their respective frame of reference.
• The red lines are the path of light between Earth and the ship during the outward trip.
• Light blue dots indicate arrival point of light.

On the outward trip the relativistic Doppler effect shows each twin would see 1 second pass for the other twin for every 3.7 seconds of his own time. The twin on the ship would see this effect for 2.57 years; as such he would see his twin age 0.69 years. However, since the light showing the ship's arrival would not arrive at Earth for 4.45 years, the twin on Earth would see this effect for 9.59 years, as such he would see his twin age 2.57 years.

Left: Earth to ship.
Right:Ship to Earth.

• Blue lines are the path of light between Earth and the ship during the return trip.

On the inward trip, the relativistic Doppler effect shows each twin would see 3.7 seconds pass for the other twin for every second of his own time. The twin on the ship would see this effect for 2.57 years; as such he would see his twin age 9.59 years, for a total of 10.28 years. However since the light showing the ships departure would not arrive at Earth for 4.45 years, the twin on Earth would see this effect for only 0.69 years since the ship returns after 10.28 years; as such he would see his twin age 2.57 years, for a total of 5.14 years.

Left: Earth to ship.
Right:Ship to Earth.

• This diagram shows the light path's for both legs of the flight.
• Note that there is more blue Earth-ship and more red Ship-Earth. That is why the ship's crew sees the Earth age more than they do.

Thus the twin paradox is resolved both in terms of the physical effect of time dilation and the observation of both twins, since both the physical affect and observation agree on how much each twin would age.

References
• A. P. French, Special Relativity (W. W. Norton: New York, 1968).
• Robert Resnick and David Halliday, Basic Concepts in Relativity (Macmillan: New York, 1992).
• Tipler, Paul; Llewellyn, Ralph (2002). Modern Physics (4th ed.). W. H. Freeman. ISBN 0716743450.
• C. B. Brans and D. R. Stewart, Unaccelerated-Returning-Twin Paradox in Flat Space-Time, Phys. Rev. D 8, 1662-1666 (1973).

### Wolfram Demonstrations Project

The much-cited twin paradox consists of three stages: (1) the traveling twin takes off; (2) he turns around; and (3) he arrives back home. Those three events are connected by the green line segments. In this simulation, imagine that the traveling twin is moving between two walls or space stations, with the path marked in blue. The presence of these two space stations suggests three more observable events, namely, what can be seen happening at the opposite space station by the traveling twin when he is at: (4) takeoff; (5) turnaround; and (6) arrival home. You can adjust the light cone parameter to the viewpoint of each of these six events, and adjust the "beta" parameter, to see the space-time diagram in different reference frames.

THINGS TO TRY

The scale of the and axes is not given, but you may choose light-seconds and seconds or light-years and years, and so on. To see the space-time diagram for the right-bound trip, make the lower (green) leg of the trip vertical. Then for the left-bound trip, adjust the upper (green) leg of the trip to be vertical. The vertical distance between events represents the time experienced by each observer. You can see the time experienced by the stay-at-home twin by making the blue line vertical. When both legs of the twin's journey (green) are successively made vertical, you can see that the total time elapsed for the traveling twin is less than that for the stay-at-home twin. Of particular interest is the perceived distance to the opposite space station, which depends on the velocity of the twin. This phenomenon is commonly called "aberration" and is described as a shrinking of the image in the direction of acceleration. However, the actual phenomenon is not a mere shrinking of an image, but an actual increase in the distance to the visible event. As objects approach the speed of light, the image appears to move faster than the speed of light, because the light from the object arrives only shortly before the object itself. This phenomenon is called "superluminal motion".

### What is the twin paradox and how does it work?

Okay, this is long, but in 5-minutes you should understand the underlying physics of the Twin Paradox.

The Twin Paradox is not "caused" by anything

Our age, the time over which we age is our proper time. Nothing affects this: not speed nor gravity. Our age is a measure of the total distance we have traveled through spacetime.

This distance, our age, can be made mathematically rigorous which shows that our age $\Delta \tau$ is only and explicitly a consequence of the path we travel in spacetime.

Like two ships that sail out of a harbor, or two hikers embarking on a trip in the mountains, they may both leave together and meet up together again at some later time - with each having traveled a different distance. It relativity the idea is the same: It's the geometry and path taken that results in the Twin Paradox (there really is no paradox of course!).

The Twin Paradox was first articulated in 1911 by Paul Langevin which featured the classic set up of a stay-at-home observer and their traveling twin, sometimes called the Langevin observer. It is true that in this scenario the accelerated observer comes home younger, yet this is still just a geometric consequence of the principle of extremal aging.

**yes, in order for the principle of relativity to avoid being vacuous it is necessary (if not somewhat circular) to postulate a preferred class of worldlines, ie, inertial. The significant point to realize that the degree of acceleration can be changed arbitrarily or eliminated altogether, as in the Three Brothers version of the paradox, without altering the proper time lapse between twins. Here is a calculation where the accelerated twin ends up being the one who's older: Page on arxiv.org

To make the point clear for readers with little background in relativity: Imagine 2 hikers walking along a path. The path forks; the easy path takes a shortcut along a level mountain pass, the other goes over the top of the mountain. When they reach the fork in the road, one hiker pushes and shoves the other hiker onto the longer path. The point here is that the "pushing" the hiker did not make the mountain tall, and the total distance over the mountaintop had nothing to do with how hard the hiker was initially pushed. In spacetime, pushing on something doesn't drive it into younger-hood, it just moves it onto a different path, a non-geodesic path, through spacetime and the proper time lapse then depends upon the details of the geometry.

Anyway, in the years since 1911 there have numerous publications demonstrating the wide variety of Twin Paradox cases. There are examples on different geometries and topologies (cylindrical and toroidal spacetimes), and basically any combination of inertial and non-inertial observers that are uncorrelated with which twin is older or younger, save this one reality - the greater the proper time distance, the older the twin. Here are two interesting examples to consider:

Orbiting Twins. Consider twins in orbiting spacecraft, one in a low circular orbit and the other in a large elliptical orbit. From time to time there orbital paths coincide (without crashing!) and their lapse times can be compared. Both twins equally have ZERO acceleration: it there any time lapse between twins?

Twin Prisoners. A pair of twins is imprisoned, one in an empty jail cell on Earth, the other, in an identical jail cell on board a spaceship. Both twins receive the same 1g accelerations from the floor of their jail cell. Both twins experience identical NON-ZERO accelerations.

In both cases it is the naive expectation that the Twin Paradox effect disappears because the accelerations are all equal. However in both cases the twins age at different rates, despite having equal acceleration.

There is nothing remarkable about different path lengths. and the reason it seems so is just the conditioning of our intuition.

From a young age we are exposed to the nice squares of Cartesian graph paper. This is reinforced over the years, with just a small percentage of students learning about, say, log-log graph paper. We likely evolved on the African Savannah and flatland is our inherited sense. We are both natively and trained Euclidean thinkers and the peculiarities of spacetime geometry are initially uncomfortable and take repeated exposure before they feel intuitive.

Here are a pair of suggestions to help build a feel for the underlying physics of the paradox:

Hypersurfaces of Simultaneity

A surface of simultaneity is a surface that at an event or spacetime point, slices spacetime into the past and future and on it, all clocks can be synchronized. While there are many methods of generating correct answers for the TP, understanding simultaneity and how clocks de-synchronize (leading clocks lag) is the best, in my experience, Special Relativistic description that gives some sense of what is happening.

The the image below is spacetime drawn up in what are called radar coordinates, an elegant representation of the surfaces of simultaneity. What I hope is a little bit intuitive is that different paths connecting common events will not necessarily be of equal length.

For those needing a more gentle and thorough explanation could try here:
Spacetime, Tachyon, Twins.

There is a really beautiful way of understanding the basic Twin Paradox using General Relativity. It does not necessarily involve differential geometry (unless you want to) but relies on just gravitational time dilation and the equivalence principle.

I couldn't find a good picture or reference, but I did find a nice video:

Ultimately, understanding the Twin Paradox involves taking the time to work through examples and convince oneself of the details. While there is nothing magical happening it is still worth the effort to understand how the effect takes place.

2.2k Views · View Upvotes · Not for Reproduction

There are two aspects to the Twin Paradox. The first, which I like to call the Clock Paradox to keep it separate, is that in the first part of the trip, when the twins are moving apart, if they use the frames they're stationary in, they each claim the other is time dilated.

The second, the Twin Pparadox proper, is how this symmetry gets broken when one of the twins turns around.

Both of these are ultimately due to the fact that it's not actually the business of clocks in relativity to measure the time coordinate of any particular measurement frame. Rather, they're spacetime odometers and they measure a distance-like quantity called Spacetime interval. This can be calculated in terms of the time and space coordinates of any frame, but that gets the causality backwards: it's the spacetime interval that's physically significant and the space and time coordinates are constructed after the fact to make calculating it easier.

To see this, consider the closely related Odometer Paradox. If you walk an odometer wheel straight down a football field, it will tick over 100 yards. If you walk it diagonally, it will tick over more than 100 yards between the 0 yard line and the 100 yard line. Call this "odometer overrun", and yes, it's the opposite of what you see with clocks - more on that presently. If due to a snafu there are line markings for a second football field on the same ground but at a slight angle, you can walk the odometer down Field 1 and see odometer overrun relative to Field 2, but you can walk the odometer down Field 2 and see odometer overrun relative to Field 1. Paradoxically symmetric odometer overrun! Nonsense! Cartesian geometry is a lie fomented by cheese-eating surrender monkeys! ( Descartes was French.) But, seriously, that's not a big mystery, that's just how 2D geometry works. An odometer just isn't naturally a football yardage measuring device. If you do want a football yardage measuring device, you need something more elaborate than an odometer, e.g. an odometer constrained to run on a track, with a transverse sight to look across the field.

Time dilation then is very similar, except with key signs reversed. The quantity that an odometer measures is $\Delta l=\sqrt<\Delta x^2+\Delta y^2>$. which is basically just Pythagorus' theorem applied to small increments of x (yardage) and y (transverse distance). By contrast, clocks measure $\Delta s=\sqrt<\Delta t^2-\Delta x^2/c^2>$. A straight line is the longest distance between two points, but an inertial trajectory (coasting without external forces) is the longest path (in terms of clock reading) between two events. Einstein synchronisation is the equivalent of the yard lines. See Mark Barton's answer to How do you resolve this paradox in time dilation? for a diagram of how this plays out.

Against this background, the Twin Paradox proper is, on one level, trivial. Of course if you take an odometer around two sides of a triangle it's going to have a different reading than if you take it straight along the third side. It's just easy to get lost in the details if you're trying to relate it to football yardage lines as you go along.

Odometers and football fields are also key to the Twin Paradox proper. In the two-football-field case above, if you use the markings for either field consistently you can use Pythagorus' theorem to get the odometer length of a path that zigzags. But if you mix and match you're going to get nonsense. In particular, if you've been marching up Field 1 and you pivot to march up Field 2, everything at a distance is now on a different yard line, and that's going to bite you in the bum in terms of telling a coherent story.

In the case of the actual Twin Paradox, pop-science accounts of relativity unhelpfully encourage you to focus on the actual human observer, when in fact everything is specified with respect to some particular choice of (t, x) coordinate system that's conceptually been painted on all of history. If you unwisely insist that the traveler changes frames at the turnaround, that's tantamount to redefining "now" for the traveler through the entire universe. In particular it has the effect of throwing away a bunch of earth history. It's not that the earth history didn't happen, it's just that the traveler's account is then broken and doesn't include it. See the diagram at Twin paradox.

386 Views · View Upvotes · Not for Reproduction · Answer requested by Damodar Ghimire

The twin paradox arises in the special theory of relativity (SR ). This theory predicts that if one of a pair of twins were to fly away from the Earth all the way to Alpha Centauri (for instance) and then fly back again, when he got home he’d find he was younger than his twin brother.

The faster the travelling twin travels during the journey, the bigger the age difference when he returns.

This happens because, according to SR, moving clocks run slow, and moving objects contract along their direction of motion. Or to put it another way, a moving observer sees distances as being shorter than a stationary observer.

An important point is that in SR, speed is only relative. There is no absolute standard of “moving” or “stationary”. Observers who are in relative motion measure time and distance differently. They are both right in their own frames of reference.

This is an effect of perspective, but is “real” — in the same way that if I turn a book sideways and look at its spine, it looks very narrow — but it is “really” narrow, because I can then fit it through a letter box — as long as the letter box hasn’t also been rotated.

A person who flies past the Earth in a spaceship at a very high speed will think clocks on the Earth are running slow, perhaps marking only half an hour when they should mark two hours. But here on the Earth we will think clocks on the spaceship are running slow by the same factor.

The closest things to actual direct tests of this are things like the Hafele–Keating experiment - Wikipedia and the observed fact that muons formed in the upper atmosphere can make it to the ground, even though their lifespans are so short that it seems like they generally shouldn’t. But note that the Hafele-Keating experiment involves gravity and a circular path, which requires general relativity (GR ) for a proper treatment.

Strictly speaking, we can handle accelerations using only special relativity, but it’s annoyingly complicated. A circular path in classical physics is considered to involve constant acceleration towards the centre of the path, so circular motion counts as acceleration in SR.

SR normally only deals with inertial frames of reference — observers who are not accelerating, in other words. If a set of observers are not in relative motion to each other, we say they all share the same “frame of reference”.

The apparent paradox arises from the following chain of reasoning. In SR, all inertial (non-accelerating) motion is relative. Therefore, the travelling twin could argue that in fact he never went anywhere. What happened was, he stayed still while the Earth jetted off into space, and Alpha Centauri hurtled towards him. Then, when Alpha Centauri reached him, both astronomical bodies reversed their directions and the Earth sped towards him, and he eventually landed back on it.

If “moving clocks run slow”, the traveller should always have seen the Earth’s clocks as moving slow, therefore he should end up older than his twin who stayed on Earth. The Earth twin might have timed twenty years when the travelling twin timed only ten years.

Since in fact SR is only really ideal for dealing with inertial motion but the travelling twin had to accelerate and decelerate several times, it’s tempting to think we simply cannot apply SR here. Something must happen during these accelerations which can only be explained by GR.

In fact we can apply SR perfectly well here, and we’ll see that there is no paradox. But to do so without a lot of complication, we have to get rid of those accelerations somehow. We either have to just imagine that changes in speed can happen instantaneously, or we can reconstruct the problem by adding an extra traveller into the picture.

Reframing the Problem to Make It Easier to Apply Special Relativity

We can reconstruct the problem using only inertial frames. We’ll involve three observers to do that.

Let’s say the “Earth observer” stays on the Earth, and times the whole sequence of events.

An “outbound traveller/observer” flies past the Earth, moving in the direction of Alpha Centauri at speed “v”.

When he passes the Earth, he synchronises his clock with the Earth observer’s clock. For simplicity, let’s say they both set their clocks to zero at that time.

An “inbound traveller/observer” is also travelling at speed v, but in the opposite direction to the outbound traveller, flying in from the space beyond Alpha Centauri.

When the outbound traveller reaches Alpha Centauri, the inbound traveller also reaches Alpha Centauri. As they pass each other at Alpha Centauri, the inbound traveller synchronises his clock with whatever the outbound traveller’s clock says.

Sometime later, the inbound traveller reaches Earth, exactly when the travelling twin would have arrived in the original form of the twin paradox.

Applying SR, we find that when the inbound traveller reaches the Earth, his clock reads less than the Earth observer’s clock.

In fact, if the outbound traveller had taken a cat with him and handed it over to the inbound traveller as they passed at Alpha Centauri, that cat would be younger than its twin cat when it reached the Earth.

Detailed Explanation of What Happens and Who Measures What

As far as I know, everything I’m going to say here is correct and would be agreed to by all “mainstream” physicists, but since I’m no Einstein, someone please correct me if I err.

Some people dislike some aspects of the language that is often used in connection with relativity, like the word “disagree”. Of course if all observers understand relativity, they will not really “disagree”, but will only make different measurements, which they can explain via their relative motion.

The following things really are all true (assuming SR is true and the Earth and Alpha Centauri are not in motion relative to each other!):

1. The Earth observer thinks the distance to Alpha Centauri is more than the inbound and outbound travellers think it is. The two travellers both agree with each other on how far Alpha Centauri is from Earth, and that’s less than the Earth observer’s measurements.
2. Both travellers think the Earth observer’s clock runs slow (by a factor known as the gamma factor or Lorentz factor). The Earth observer thinks both travellers’ clocks run slow, both by the same factor again.
3. The inbound traveller’s clock nevertheless reads less than the Earth observer’s clock when he passes the Earth at the end of the experiment.
4. No-one is ever “wrong” about their measurements. Their measurements are always correct in their own inertial frames of reference, and correspond to “reality” in every sense.
5. All observers can argue with equal legitimacy that they are “stationary”, since they all measure light to travel at the same speed and the laws of physics are the same for all of them.

Measurements made by the Earth observer are easy to calculate, using ordinary notions of time, speed and distance.

It’s fairly easy to calculate the measurements made by the outbound observer, using the Lorentz transformation - Wikipedia. or the time dilation and length contraction formulas, which involve only the Lorentz factor - Wikipedia .

We find the outbound traveller thinks he arrives at Alpha Centauri earlier than the Earth observer thinks the outbound traveller arrives there. The outbound traveller might say it took him twenty years to get there when the Earth observer thinks the outbound traveller arrives there after thirty years.

But if the outbound traveller uses a powerful (!) telescope to observe the Earth clock, he finds the Earth clock measured even less than his own clock when he reached Alpha Centauri.

Since light from the Earth takes 4.4 years to reach Alpha Centauri, the outbound traveller is not going to actually see that until he’s long since passed Alpha Centauri and headed off into space. But of course he knows that light travels at a finite speed and that the Earth is also moving away from him, so he can allow for that and calculate what the Earth clock said when he was at Alpha Centauri.

The trouble is, the two of them just don’t agree about “when” the outbound traveller reaches Alpha Centauri.

If we consider two events here: 1) the event when the Earth observer reads thirty years on his clock and 2) the event when the outbound traveller reaches Alpha Centauri, the two observers do not both consider these two events to be simultaneous. Only the Earth observer thinks they happen at the same time.

If these two had stayed in touch via radio signals for the whole trip, the Earth observer would have said the outbound traveller’s clock was always running slow, while the outbound traveller would have said the Earth’s clock was always running slow.

Both would have had to correct for the fact that the distance between them is increasing and light travels at a finite speed. But that’s OK — they can both do that, and no relativity is needed.

There’s a kind of Doppler effect here on the radio or light signals, like the effect that makes the siren of an ambulance sound higher-pitched than it “really” is when it’s coming towards you. If the Earth observer had sent a signal every ten seconds, but the distance between the Earth observer and the outbound observer is always increasing, the outbound observer will receive the signal perhaps only every fifteen seconds. This follows just from almost common sense, and has nothing to do with relativity.

When the outbound traveller looks at the Earth observer’s clock with his telescope, it seems to run very slow. But he realises that most of that is due to Doppler effect. But even after correcting for that, the Earth clock still looks like it’s running slow.

When the outbound observer reaches Alpha Centauri, the inbound observer passes him. For simplicity (although it makes no difference to the reality of what happens), the inbound observer can synchronise his clock with the outbound observer’s clock at that time.

So the two travellers’ agree that the clock time is twenty years when they pass at Alpha Centauri.

We then follow the inbound traveller as he moves towards the Earth. If he looks at the Earth observer’s clock with a telescope, the clock seems to be going much too quickly. But he knows most of this is due to Doppler effect. He’s heading towards and intercepting the light that is emitted or reflected from the Earth observer’s clock. Or the Earth is hurtling towards him, from his point of view.

He knows that and of course corrects for it, since he sees the Earth is hurtling towards him.

After the correction, he finds the Earth observer’s clock is in fact running slow.

If the outbound observer thought he passed Alpha Centauri twenty years after passing the Earth, the inbound observer will believe he passes the Earth twenty years after Alpha Centauri passed by him.

That means that when the inbound observer reaches the Earth (or when the Earth reaches him, in his view), his clock reads forty years. He’s added his own measurement together with the outbound traveller’s measurement.

The clock on the Earth timed thirty years for the outbound observer to reach Alpha Centauri, and another thirty years for the inbound observer to reach the Earth after passing Alpha Centauri. So the whole thing took sixty years according to people on Earth.

The simplest way to state the resolution of the paradox is just to say that, in the usual form of the twin paradox, the situation is not symmetrical between the twin who stays at home and the twin who goes on a trip.

The twin who goes on a trip has to shift from one inertial frame to another when he “turns around and goes back”.

It’s not the fact of turning that’s important (he can reverse all the way back if he likes!) and it’s not that the acceleration is the key, as such. It’s the fact that he has to move in the opposite direction going back to when he was going out.

He will certainly realise this, because he’ll feel his deceleration and then acceleration in the opposite direction.

If we consider the version of the paradox reformulated to make it easier to work with in SR, how is it that both travellers always thought the Earth clock was running slow, and yet when the inbound traveller passes Earth, his clock reads less than the Earth’s clock? If anything it looks like the Earth’s clock was running fast, not slow.

The problem is that the inbound traveller accepted the outbound traveller’s assertion that the Earth clock had always ran slow. In this frame of reference (the outbound traveller’s), that was true. But from the perspective of the inbound traveller, the outbound traveller’s own clock is running very slow! After all, they are passing each other at a relative speed of 2v.

The inbound traveller feels that the outbound traveller “should” have measured the Earth’s clock to be going too fast, in fact, if he takes the outbound traveller’s measurement of time and factors in the horrible slowness of the outbound traveller’s clock, even though he himself also sees the Earth’s clock as running slow.

There isn’t actually any objective fact about whether the Earth’s clock is running slow or fast at any point. It’s always only a matter of perspective, and we can’t compare measurements made in different inertial frames without using SR.

If a single traveller had made this trip, he would always have observed the Earth’s clock to have been slow — except at one point. When he turned around at Alpha Centauri for the return journey, the Earth observer’s clock would have appeared to suddenly gain rapidly from his perspective (even after correcting for Doppler effect), settling down only when the travelling twin reaches his cruising speed towards the Earth and stops accelerating. His twin on Earth would have appeared to age rapidly during the deceleration/acceleration phase.

After all, consider what happens if the twin just “stops” at Alpha Centauri instead of turning around and going back. Once he has “stopped”, he is in the same inertial frames as the Earth. He will feel that the Earth observer’s clock is going at the same rate at his own clock. But his own clock will read less time than the Earth’s, as if it had been set wrongly.

He can now consider himself to have previously underestimated the distance to Alpha Centauri, since it now appears further away than it did when he was in motion relative to it. So of course, he can consider himself to have previously timed the voyage incorrectly. But this is all a matter of perspective; his previous measurements were equally valid, and were correct in his previous inertial frame, while travelling from the Earth to Alpha Centauri.

785 Views · View Upvotes · Not for Reproduction

Start with this idea: time passes at the same rate for all people from their own point of view, no matter where they go.

The rate of all physical processes we can possibly use to measure time doesn't appear to ever change for clocks that move along with us: clocks just seem to tick at the same rate, no matter where you go or how you move.

But the thing is: the amount of time that passes for you, relative to the amount that passes for someone else does depend on the two paths that both of you follow through space and time.

If you go deep into a very, very strong gravity field, and your twin doesn't, but simply hovers far away from the object producing the gravity, and then you return to meet your twin, then much less time will have passed for you than for your twin.

That is the twin paradox in a nutshell, though it is usually stated in the context of special relativity instead, where it can be shown that the twin who leaves on a long journey, travels very fast, and then returns experiences much less time than the twin who stays put.

It seems paradoxical but it isn't: because there is no symmetry between the two twins. They move along different paths. The so-called proper time, the clock time along their two paths, is actually different.

Gravity doesn't change this at all, since general relativity contains special relativity as a limit.

So if the twin paradox made sense to you in special relativity, then you should still be happy with it when there is gravity!

Here on the Earth these effects are tiny - people living at the altitude of Denver, Colorado experience 100 years, while those living at sea level experience 100 years less about two seconds.

So it takes atomic clocks to detect the effect, but it has been done.

898 Views · View Upvotes · Not for Reproduction

The so-called twin “paradox” isn’t a real paradox at all, it’s just a counter-intuitive result.

What it says is that if two identical clocks are in the same location and synchronised, and one is then taken away at high speed and then brought back, the clock that was accelerated will register less time elapsed than the one that didn’t.

The reason it’s called the “twin” paradox is that it’s often told in the form of a story. There are two twins, and one stays at home while the other gets in a rocket and travels to a distant star. When she returns she has aged only ten years, while her twin has aged sixty. However, it hasn’t been tested in this form. It has, though, been tested, by having two synchronised atomic clocks and leaving one in the lab while the other was mounted in a plane and flown at high speed for a long time. The moving clock went slower than the static one.

This has nothing to do with any mysterious effect of acceleration as such, and you can’t escape it by having higher or lower acceleration (though the precise amount of the time difference will be affected).

Probably the best way of thinking of it is in terms of the well known geometrical fact known as the Triangle Inequality. This says that the length of any side of a triangle will be less than the total length of the other two sides. This is a straightforward example of a straight line being the shortest distance between two points - any path between two points that takes a more roundabout route than the straight line will, unavoidably, be longer.

The path of an object through four-dimensional spacetime is called its worldline. An object that doesn’t undergo any acceleration during a time interval will have a worldline (from its location at the beginning of the period, to its location at the end) that follows a straight line. In the case of an object that starts at x=0 and just stays there, that straight line lies along the t (time) axis. The length of that line is referred to as the “proper time” - that is, the time experienced by someone following that path.

Now, this is probably the moment to mention a peculiarity of spacetime. Unlike in the geometry we’re familiar with, the version of Pythagoras’ Theorem that applies to a worldline says that the length of the worldline between two events is the square root, not of the sum of the squares on the two sides (i.e. the time elapsed and the distance travelled) but of their difference. This means that a straight line is the longest path between two events. You have to measure things in compatible units, by the way - years and light years, for example.

So the spacetime version of the Triangle Inequality is that the length of a path with a kink in it is shorter than the length of a straight path between the same two points. And as that length corresponds to time elapsed, the stay-at-home clock registers more time elapsed than the travelling clock.

From this you should be able to work out the so-called “symmetric” twin paradox - if two of a set of triplets head off in opposite directions while one stays at home, the worldlines of the two travellers will look like a kite shape or a diamond shape in the spacetime diagram, while the homebody’s worldline will just be the vertical straight line joining the bottom vertex to the top vertex. The travellers will age by the same amount as each other, but less than their home-loving sibling.

1.4k Views · View Upvotes · Not for Reproduction

The paradox arises out of ignorance, specifically, ignorance as a result of improper teaching of relativity theory.

To understand the twin "paradox" in arbitrary situations, it is best to think about world lines and proper times. To begin, it is useful to understand how points in four dimensional spacetime correspond to the notion of events, and how the "life" of a particle is represented by a line in spacetime, namely its world line.

There is the notion of the length of a world line, but counterintuitively, in the pseudo -Euclidean geometry of spacetime, the "length" is longest for a straight line between two points. This "length" is called proper time. on account that this is exactly what a clock would measure that is traveling along that world line. (That is to say, your watch or your body's biological clock both "measure" proper time along the world line that your body traces in spacetime.)

Now take two events, the first and the second meeting of the twins. These two events are connected by the two (distinct) world lines of the twins. Say, one twin does no accelerating (straight world line), the other accelerating away and back. The twin who did no accelerating will always measure more time between the two events than the twin who accelerates. If both accelerate, then the twin who accelerates more "wins" (i.e. stays younger than his sibling.)

601 Views · View Upvotes · Not for Reproduction · Answer requested by Damodar Ghimire

• Can you explain the twin paradox theory in a simple way?

• What is the principle involved in twin paradox?

• How do you explain the twin paradox in special relativity?

• What is the last date to apply for UG program in IISc?

• Does your new book go into detail on the various twin Paradox setting?

• Can we choose only one subject for IISC UG?

• What is the life of an IISc UG student like?

• Does twin paradox occur in reality?

• What is the Gibbs Paradox?

• What is an explanation of the solution of the twin paradox where we use the concept of throwing light pulses between the two twins?

• From the Twin Paradox, you age slower the faster you move. But within your body are you not at rest from the perspective of biological processes?

• How does twin paradox exactly work and how does metabolism depend upon time?

• What is wrong with the twins paradox concept?

• How do you use the Minkowski inequality to deduce the twin paradox?

• Why are many Quorans obsessed about time? Time travel, twin paradoxes et al. Is it a prerequisite to appear smart or this is just genuine interest?

• What is the solution to the symmetric Twin Paradox?