This video from Brian Greene suggests this is so: https://www.youtube.com/watch?v=2sZUNud6rRw&list=PLj6DWzIvBi4PFDXCCV1bNhVUgDLTwVbFc&index=60

It shows if you stop a pole in the barn (ignoring all the obvious engineering challenges of doing so) it will end up permanently length contracted just like the returning twin will end up permanently younger than her earth bound twin in the twin paradox. Ignoring the practical problems with infinite deceleration, she stops when she turns around and that causes her permanent age difference but does she also end up permanently flatter? Again just consider the relativistic math and not all the physical impossibilities this example entails.

Relativity allows a frame jump without deceleration, it's called a clock handoff in the twin paradox. Since a clock is used to measure length for length contraction, a clock handoff could also keep a record of both permanent age difference and permanent length contraction when the twin hands off her clock readings to a ship passing her to return to earth. There's no physical crunching of the pole in a clock handoff.

So does relativity sanction permanent length contraction along with permanent age difference in the clock handoff twin paradox?

  • 8
    $\begingroup$ By permanent length contraction, you presumably mean a change in the proper length of an object. And that's clearly possible both within relativity and outside it. For example, if I crush a soda can, it'll end up permanently shorter than before. $\endgroup$ – knzhou Oct 8 '19 at 1:17
  • 4
    $\begingroup$ In both relativity and outside it, this kind of permanent length change can occur if you squeeze an object hard, and the object doesn't bounce back. It really has nothing to do with length contraction. Length contraction is a property of a frame you use to describe an object, not the object itself, so it undoes itself perfectly as long as you don't squeeze or crush the object in the common sense way. $\endgroup$ – knzhou Oct 8 '19 at 1:18
  • 11
    $\begingroup$ If it's from a science popularizer like Brian Greene, I wouldn't really listen to it. These folks almost always put out oversimplified explanations. I rarely read a paragraph from them that doesn't have some error in it. $\endgroup$ – knzhou Oct 8 '19 at 1:39
  • 4
    $\begingroup$ There's a variant of this scenario that doesn't involve compression. A train travels through a tunnel. At the two ends, there are two enormous guillotines. When the train is completely in, they go down simultaneously and just miss the train; then immediately retract. The train passes safely. Now, in the train frame, the tunnel is shorter then the train; if the two guillotines come down at the same time, the train is destroyed. Both can't be true. But, in train frame, they don't drop at the same time, one drops and retracts, train passes, then the other does the same behind it. $\endgroup$ – Filip Milovanović Oct 8 '19 at 2:02
  • 5
    $\begingroup$ @knzhou I wouldn't describe Brian Greene as a science popularizer as much as a prominent physicist who also does science popularization. I can't remember ever reading a paragraph of his that's unambiguously wrong (as opposed to a defensible simplification that glosses over some subtleties). Greene knows his stuff. $\endgroup$ – tparker Oct 8 '19 at 22:30

Does relativity sanction permanent length contraction along with permanent age difference in the twin paradox?

No, it does not. However, given the many analogies between time and space this may seem disturbing. What makes time different from space in this context?

The issue is that a clock does something different than a ruler does: it maintains a record. A ruler merely measures the distance between its endpoints, and as a moving ruler is brought (gently) to rest that measurement agrees with a permanently resting ruler.

The device that most closely resembles a ruler for time is not a clock but rather a metronome. There is no permanent time dilation for a metronome, and as a moving metronome is brought (gently) to rest that measurement agrees with a permanently resting metronome. In this way it is symmetric with the impermanence of length contraction.

If you want a device that resembles a clock for distance that would not be a ruler, but rather an odometer. An odometer maintains a record and will register permanent length contraction in the same manner as a clock.

In this way the symmetry between time and space is recognized again. The difference was not due to differences in the physics of time and space, but rather differences in the measuring devices. We were comparing a memory-less device for space to a device with memory for time. With a proper comparison of similar devices the issue is resolved.

  • 14
    $\begingroup$ @Albert I believe what is being said is that the ruler measures the distance between two points in space just as the metronome measures the “distance” between points in time. If you set off a metronome at one tick per second and then move it at close to C then the ticking with slow down. Just as the ruler will contract. But when you bring it back to relative resting it will still be ticking at 1 tick per second. Whereas the clock in the paradox is measuring the time interval from leaving to returning. Just as the odometer is measuring the perceived distance travelled from leaving to returning. $\endgroup$ – Fogmeister Oct 8 '19 at 12:12
  • 6
    $\begingroup$ @Albert Because the clock tells you the time. If you took two clocks that both read 12:00 and sent one on a relativistic trip before bringing it back to rest, you could end up with, say, 12:05 and 12:10. This 5 minute difference will never vanish. The clocks maintain a permanent record of their dilation. Metronomes have no capacity for this. $\endgroup$ – HTNW Oct 8 '19 at 13:09
  • 9
    $\begingroup$ The point is that the twin who travelled and then came back does not spend the rest of their life moving in slow motion, or any flatter. They now move at ordinary speed (although are younger than they should be). In a similar way they are ordinary in shape (no space compression) but the total distance they have travelled (according to a machine they brought with them on their space voyage) is a lot less than the total distance that you would have thought it was to their destination and back. $\endgroup$ – Dast Oct 8 '19 at 13:10
  • 9
    $\begingroup$ @Albert a metronome is not a clock. If you put a counter on it then it becomes a clock and is no longer a metronome. $\endgroup$ – OrangeDog Oct 8 '19 at 14:16
  • 3
    $\begingroup$ @Albert a metronome is a frequency standard. It measures temporal intervals, $\Delta t$. A clock is a metronome plus a counter. It measures elapsed time $t=n \Delta t$. Similarly a ruler or rod measures spatial intervals, $\Delta x$. An odometer is a ruler plus a counter. It measures elapsed distance $x=n\Delta x$. Hence the similarity. Time dilation affects metronomes the same way that length contraction affects rulers. Time dilation affects clocks the same way that length contraction affects odometers. $\endgroup$ – Dale Oct 8 '19 at 17:35

No, length contraction is not permanent. The travelling twin ends up permanently younger, and the spatial analogue is that she has permanently travelled a longer distance than her twin. The temporal analogue of being permanently length contracted would be permanently aging more slowly, but neither of these things happen.

Also, the pole in the barn doesn't end up permanently length contracted either. The true answer to the pole in barn paradox is that when you try to bring the pole to a halt inside the barn, it will collide violently with the barn, causing one or both of them to break. The pole is only permanently length contracted in the sense of "shattered into multiple pieces". (According to special relativity, infinitely strong materials are not only practically impossible, but theoretically impossible.)


First: Despite appearances, this question has absolutely nothing to do with relativity, because it asks about the length of a traveler at various stages in the journey with all measurements made in a single frame (namely the earthbound frame). So the entire question comes down to: Does decelerating change your length?

The answer entirely depends on how the traveler twin decelerates. Suppose he's heading toward earth head-first, and he abruptly stops moving. If (in the earth-frame) his head stops before his feet do, he's going to contract. If his head and feet stop at the same moment, he's going to remain the same length he was while traveling. If his feet stops before his head does, he's going to stretch out.

Of course all the same things are true in the traveling frame. It's perfectly possible, for example, that his head and feet stop simultaneously in the earth frame but non-simultaneously in the traveling frame, so his length stays fixed in the earth frame and not in the traveling frame. Or vice versa. That's where relativity comes in, but it has nothing to do with the question that was asked.

  • $\begingroup$ I think that this is by far the best answer to this question, but I can't get behind your first sentence, because that it seems to imply that special relativity is only concerned with changes in Lorentz frame, and that nothing "relativistic" ever happens that can be described within a single Lorentz frame. But phenomena like Lorentz contraction and time dilation certainly "have to do with relativity" even though "all measurements are made within a single frame." $\endgroup$ – tparker Oct 9 '19 at 4:22
  • $\begingroup$ @tparker: Thank you for your kind words, but I want to defend my first sentence. An object is headed toward earth. It decelerates. What happens to its length, as measured from earth? Answer: It depends on the details of how it decelerates. That is the question at hand, and the correct answer is exactly the same as the answer you'd give if you'd never heard of relativity. If knowing about relativity doesn't change the answer then (by a reasonable definition) the question has nothing to do with relativity. $\endgroup$ – WillO Oct 9 '19 at 4:44
  • $\begingroup$ @tparker: (Of course if you then go on to ask how things look in some other frame, then you're doing relatiivity.....) $\endgroup$ – WillO Oct 9 '19 at 4:44
  • $\begingroup$ Okay, I'm starting to come around to your first sentence. Taken in isolation, it seems misleading, because you could imagine a different problem where Lorentz contraction really is conceptually crucial, in which cases this same sentence would be incorrect. But in the context of this particular question and the rest of the answer, I think it's okay. $\endgroup$ – tparker Oct 9 '19 at 12:44
  • 1
    $\begingroup$ @tparker : If you're assuming that decleration is instanteous and uniform (in the earth frame) then it's natural to assume the same about the initial acceleration at the beginning of the journey. With that assumption, the traveler never contracted in the first place. $\endgroup$ – WillO Oct 9 '19 at 16:49

No, length contraction occurs only while there is relative motion, so assuming the travelling twin comes to rest in the frame of the stationary twin, they will once again agree that they are the same length.

The age difference is not a permanent time dilation- the clocks of the twins will run at the same speed once they are back in the same frame.

You should also remember that the effects are entirely symmetric, so that the stationary twin appears shorter to the travelling twin. Ironically that means that even if the length contraction were permanent (which it isn't) there would be no way to tell, as each twin would believe the other was permanently shortened by the same amount.

The scenario Brian Greene is asking you to imagine is one in which the people in the barn apply forces to the pole to bring it to a halt. They think they are bringing all parts of the pole to a halt simultaneously, so that the front of the pole is halted at exactly the same time as the rear. They believe that because they are applying the forces all along the length of the pole at the same instant they are not changing the length of the pole but are simply capturing it as is.

However, from the perspective of the pole the forces are being applied out of phase, so when the first grab is applied at the front, there is no restraining grab on the rest of the pole so it continues forward ploughing into the front and thus compressing itself. His example would have been clearer had there been just two people grabbing in the barn, one at the front and one at the rear.

Either way, the shortening of the pole has been made permanent by compressive forces being applied by the people in the barn. If the grabbing had not occurred the pole would not have been physically shortened in its own frame of reference.

  • $\begingroup$ This is the right answer. If the twin kept on moving once she reached her end destination the contraction will stay. But since she moves at a significant fraction of the speed of light this would make it hard to compare the two twins. If the twin decelerates she will expand to her rest length again. $\endgroup$ – AccidentalTaylorExpansion Oct 8 '19 at 23:10
  • $\begingroup$ @user3502079: "If the twin decelerates she will expand to her rest length again." This is equally likely to be either true or false, depending on the details of the deceleration. See my answer. $\endgroup$ – WillO Oct 8 '19 at 23:23
  • 1
    $\begingroup$ @WillO Realistically, no. A person is made up of many atoms which don't like to be compressed. If you replace the twin by a spring you will get that the spring is length contracted when it is moving but as soon as it stops it will take on its rest length. To stop the spring you have to apply a force at some point on the spring. As this force propagates through the spring it starts to expand because its velocity changes. A person is basically a very complex spring. $\endgroup$ – AccidentalTaylorExpansion Oct 9 '19 at 11:20

Ok, someone has explained this to me on another forum. When a frame jump occurs, time and space effects are no longer reciprocal. Only one participant ages less but the space imbalance is in the distance travelled, not in a permanent flattening of the returning ship. Hence if a non-time based odometer could be made, it would record the ship has travelled a contracted distance but you could not expect the return of a flattened ship because that would make the space effect reciprocal which it no longer is due to the frame jump.

  • 1
    $\begingroup$ "but the space imbalance is in the distance travelled, not in a permanent flattening of the returning ship" This is where you are getting confused. Aging is the "time travelled". If the twins come to relative rest at the end of the experiment (rateher than the travelling twin just flying by) then they find that their clocks are once again ticking at the same rate. Their time measuring devices are no more pemenantly affected than their rulers. It is only the accumulated time that differs. But then, so do their odometer readings.. $\endgroup$ – dmckee --- ex-moderator kitten Oct 10 '19 at 19:12
  • $\begingroup$ Their clocks tick at the same rate and their accumulated time differs like you said so their measuring devices are permanently affected. One's body clock will have aged less permanently. $\endgroup$ – ralfcis Oct 10 '19 at 19:17
  • $\begingroup$ No more (or less) than their rulers are permanently affected simply because they clocked up different distances. Their clocks are once again measuring the same time intervals and their rulers are once again measuring the same space intervals. $\endgroup$ – dmckee --- ex-moderator kitten Oct 10 '19 at 19:25
  • 1
    $\begingroup$ The non-time-based odometer is "reciprocal" to the clock; the ruler isn't. $\endgroup$ – wizzwizz4 Oct 11 '19 at 6:25
  • $\begingroup$ Your comment is interesting. Are you saying time dilation is reciprocal to length contraction and reciprocal time dilation or reciprocal length contraction are not what's important? $\endgroup$ – ralfcis Oct 12 '19 at 3:54

To my understanding, it does neither. Special relativity says that if two entities are traveling at a high relative speed to one another, each will observe the other as changed in time, length, mass. This precludes the goofy idea that one of them will be younger than the other when the velocities reverse and then stop (become non-moving relative to each other.

Furthermore, if I leave earth and return, I am not at a constant relative velocity, so does special relativity even apply? So, who’s wrong, me, or dozens of sci-fi writers?

  • $\begingroup$ You`re wrong. Here's a video for you: youtube.com/…. Here's another one for all those on here who believe acceleration has magical powers to create this phenomenon:youtube.com/… $\endgroup$ – ralfcis Oct 10 '19 at 12:11
  • $\begingroup$ Neither video disproves my point. What I have repeatedly read in science FICTION is that when Gracies comes back, George IS older than her. Special relativity does NOT predict that. Neither video has Gracie coming back. First video says that Gracie will expect George to be younger and George will expect Gracie to be younger. Find me a video that convinces me that if Gracie does return, she will find her prediction wrong and George's correct, as the stories claim that I complain about. $\endgroup$ – WGroleau Oct 10 '19 at 17:19
  • $\begingroup$ I don't think you saw the 1st video to it's conclusion. It tears down that initial assumption. Technically it proves Gracie reunites younger than George, she ages slower. Yes relativity does predict that as you will see in the video. Here's another at 11:50 youtube.com/… $\endgroup$ – ralfcis Oct 10 '19 at 17:24

I slept on it and came up with a partial answer. As I said I spoke with Don Lincoln years ago on a forum and he impressed into my brain that space and time are two sides of the same coin; whatever problem you can solve using the time phenomena of relativity you can also solve using its symmetrical counterpart space. So I decided Greene's example was the space equivalent of the muon example.

A spacetime path must begin and end with the participants co-located. The muon's path does not begin co-located with earth, so the clocks are not sync'd, and there is no frame jump (only constant velocity) in that path. Hence, even though the muon ends up co-located with earth, there is no permanent twin paradox age difference. In fact any clocks that co-locate must have the same clock reading unless a frame jump has occurred.

So in the classic twin paradox example of Alice going and returning at .6c 3 ly out, Bob on earth will age 10 yrs and Alice only 8 when they re-unite. That age difference will be seen in a spacetime diagram. In the muon example, since the clocks did not start co-located and no frame jump occurred, the two clocks will not indicate that the muon aged less a la twin paradox but that its time dilated from an unknown start time. The spacetime diagram should have no difference between the clocks at co-location. Please don't confuse permanent age difference of 1 participant resulting from a frame jump with the reciprocal time dilation of both participants due to constant relative velocity.

Greene's example is very similar to the muon. The pole starts out unsync'd to the barn clocks so there can be no discernible permanent length difference when it's stopped in the barn. Just like the muon, there is a crash at the end but that doesn't constitute a frame jump when the clocks are co-located. A frame jump has to occur at an appreciable distance for there to be permanent effects on either the moving frame's time or space relative to the stationary frame which is chosen to be the barn in this case.

Greene's analysis may be wrong but what if he had properly done it for a true twin paradox example with a valid spacetime path. He'd have to start the pole and barn together, the pole then goes out and returns to the barn for a valid spacetime path. The pole would have to be permanently length contracted at the turn around point, not at the barn. One thing he did get right was the length contraction is not caused by a physical crushing of the pole but by the incremental relative simultaneity of the clocks as the pole is being stopped by subsequent hands.

Like I said, the clock handoff scenario of the twin paradox involves no crushing of the pole only the effects of relative simultaneity on measuring the pole's length. In the clock handoff, Alice going out meets Charlie returning the pole's measurements (not a physical pole) for her. Neither Alice or Charlie are affected by a frame jump but the data passing between them is.

Charlie coming in from deep space towards earth has no sync'd clock to earth's clock. He is the muon example incarnate. He sync's his clock to earth's when he gets Alice's clock data. Since neither Alice nor Charlie experience a frame jump, they do not experience a permanent age difference to earth, they just experience reciprocal time dilation relative to earth's clock. What does experience the frame jump is the data. So the data doesn't really represent that either Charlie or Alice have aged less during their journeys. All Charlie has really done is drag a distant time reading into the co-located present with earth when he reaches it. He has aged normally and earth can't really tell he has aged slower, only that his inherited clock data from Alice has aged slower.

So the question is has time and space itself been warped to cause the clock to take real measurements of time and length or has the info of time and space been warped by the delay of distance and the effect velocity has on rate of returning information. For example if a clock face were receding from you, the info from the clock face would have a rate of delay which you could easily misinterpret as time itself being slowed. If only the info is getting distorted, is there any real permanent age difference and permanent length contraction that comes out of a frame jump? Those who answer only permanent age difference is real are forgetting that there's a symmetry between time and space and their position breaks that symmetry. Relativistic effects can only be calculated for either space or time. The muon doesn't cross the atmosphere length contracted AND in a dilated time, it's one or the other. So if permanent age difference does exist according to the twin paradox, then the twin can't also return concurrently flatter as well, it's one or the other. That's my answer interpreting relativity but I'm now personally leaning to the belief that neither occur just as neither really occurs in the clock handoff example.

PS. No I'm wrong in my last paragraph. The muon example proves that it's not just the information of its clock that velocity affects, it's time itself that the muon's clock measures. Otherwise the muon would not be able to really make it to earth.


No, because the length difference is essentially an optical illusion. It happens because of red/blue shift as an object moves toward or away from your perspective. It is only dependent on the direction of travel relative to you, and only happens in that dimension. An object moving toward you appears thinner, it does not also appear taller or wider. That's why you only get frequency (red/blue) shift and not amplitude shift.

The key is that time, the 4th dimension, changes on us constantly. It's always getting larger (more nanoseconds since the big bang, never the same amount, never less). We progress along the timeline. A clock measures the velocity of this progression (in my timeline, I read my clock as progressing at 1 second per second; if I watch my twin's timeline, I read their clock as progressing at 1.2 of their seconds per 1 of my seconds). Thus, the clocks are a record that they once traveled at different rates.

E.G., two cars pop into existence with 0 miles on the odometer, going 30mph. One accelerates to 100mph, then they both drive around for 10 minutes, then they both stop at the same spot. They are now both going 0mph. But one car has more miles on its odometer, proving that car traveled faster than the other one at some point in the past.

The fist 3 dimensions of X,Y,Z (length, width, height) are static. The X axis does not continually grow or shrink on us. We do not progress along a size-line. Thus, there can be no record that the twins progressed on that line at different rates in the past. If we did progress on a size-line, then yes, it would be obvious that they had moved at relativistic speeds in the past. One would be much longer, or wider, or taller, or just overall bigger (larger in all 3 axes) than the other. Their rulers would be different sizes.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.