54
$\begingroup$

It's my understanding that when something is going near the speed of light in reference to an observer, time dilation occurs and time goes slower for that fast-moving object. However, when that object goes back to "rest", it has genuinely aged compared to the observer. It's not like time goes slow for a while, and then speeds back to "normal," so that the age of the observer once again matches the object. The time dilation is permanent. Why wouldn't the same thing happen with length contraction? Since the two are so related, you'd think if one is permanent, the other would be also. And from everything I've read so far, length contraction is not permanent. An object will be at rest touching an observer, go far away near light speed, return to touching the observer, and be the same length it was at the beginning. It shortens, and then grows long again, as if its shrinkage was an illusion the whole time. Did I just not read the right things or what? Were my facts gathered incorrectly?

$\endgroup$
  • 13
    $\begingroup$ The relative movement of time (which is what we call time dilation) is just as observer dependent as the relative distances in space. When the velocity between the observed object and the observer goes to zero, both effects disappear equally. The actual time dilation between two clocks in the same reference frame is zero, but because time itself is cumulative, we retain a memory of the relative state of motion in the eigentime of the object, i.e. in the total amount of time that has passed according to a clock in the object's frame of reference. $\endgroup$ – CuriousOne Dec 19 '14 at 2:32
  • 4
    $\begingroup$ Time dilation is not permanent. If you bring a moving clock to rest then it is no longer time dilated. $\endgroup$ – Hraish kumar Dec 19 '14 at 3:20
  • $\begingroup$ An object will be at rest with a clock moving at the same speed as an observer, go far away near light speed, return to the observer, and the clocks will still be moving at the same speed. $\endgroup$ – immibis Dec 19 '14 at 5:25
  • 6
    $\begingroup$ The length of space passed is in fact permanently contracted. Without loss of generality, consider a particle moving at C. For it the space passed contracted to 0 and the time passed was 0. If the space travelled were not permanently contracted, the particle would not be permanently at the new location after 0 time. Note that this does not imply that the distance can be traversed again in its contracted state. $\endgroup$ – dotancohen Dec 19 '14 at 12:42
  • 11
    $\begingroup$ Or, to summarize things in an easy to remember way: The effect of length contraction has the same kind of permanency on your odometer as time dilation does on your clock. $\endgroup$ – Random832 Dec 19 '14 at 14:56

11 Answers 11

36
$\begingroup$

Time dilation is a comparison of rates. When an object is moving fast with respect to you, it's clock rate is slow, and when it comes to rest with respect to you its clock rate returns to normal. The time difference between the two clocks at this time is due to the accumulation due to these different time rates. That is the leftover effect of the time dilation but not the time dilation itself.

Length contraction, like time dilation, exists when there is relative motion and goes away when there is no relative motion, but there isn't any "accumulation" with length contraction, so there is nothing to be "left over".

$\endgroup$
15
$\begingroup$

The way I see it, time dilation is the real effect here.

Length contraction (in SR) is just a consequence of the fact that the "length" of a rod is the distance between simultaneous positions of the rod's endpoints. But two observers with different velocities will have different ideas about what simultaneous is, and this means they measure different lengths.

The best paradox to think about here is the "ladder" or "train" paradox. I think if you have got your head around that you understand length contraction.

$\endgroup$
  • $\begingroup$ +1 Best answer. Time can be measured at a single location using a clock, but length is a measurement between two separate locations, which depends on simultaneity. Welcome to StackExchange! $\endgroup$ – Pulsar Dec 20 '14 at 8:19
13
$\begingroup$

It's my understanding that when something is going near the speed of light in reference to an observer, time dilation occurs and time goes slower for that fast-moving object.

According to the 'something', it is the observer's clock that runs slower and it is the observer's rulers that are contracted. That is to say, the time dilation and length contraction are symmetrical. Neither clock can be objectively said to be running slow and neither ruler can objectively said to be contracted.

However, when that object goes back to "rest",

Now the symmetry is lost; the object's accelerometer registered non-zero acceleration for some time while the observer remained inertial. This means that there is now an objective difference between the object (non-inertial) and observer (inertial) and, thus, an objective difference in elapsed times.

Why wouldn't the same thing happen with length contraction?

In fact, in SR, acceleration of an extended object must be handled with great care. If an object is to not to stretch or compress during acceleration, different parts of the object must have different (proper) acceleration.

See, for example, this question for additional information and links.

$\endgroup$
8
$\begingroup$

Length contraction effects may be permanent in the same way as time dilation! You just have to choose the right example.

Example: An astronaut is traveling at v=0,99 c to an exoplanet, according to Earth frame he is traveling 198 light years in 200 years. According to his frame (reciprocal gamma = 0,141) he is traveling 27,9 light years in 28,2 years. After his arrival on the exoplanet, he is permanently younger than (and outliving) his twin brother on Earth, and he is permanently at a distance of 198 light years from Earth, a distance which he could never have traveled without length contraction.

$\endgroup$
  • $\begingroup$ "A distance which he could never have traveled without length contraction" -- But there is a perfectly consistent analysis of this situation in the Earth/exoplanet rest frame which doesn't involve length contraction at all, just time dilation which causes him to age 28.2 years over the 200 years of coordinate time it took him to cross 198 light years of coordinate distance. Whereas with proper time, all frames agree that the the accelerated twin in the twin paradox aged less than the inertial one. $\endgroup$ – Hypnosifl Dec 19 '14 at 15:52
  • $\begingroup$ @Hypnosifl - No, time dilation and length contraction are separate effects, each of them increasing the autonomy of the astronaut. You may not mix up Earth frame and astronaut's frame because the result would be a velocity faster than speed of light (198 light years in 28,2 years), t=28,2 years does not belong to the Earth frame. $\endgroup$ – Moonraker Dec 19 '14 at 16:37
  • 1
    $\begingroup$ What statement of mine are you saying "no" to? I didn't claim they weren't separate effects, nor did I mix up different frames. I just said that the age of the astronaut on arriving at the exoplanet can be accounted for in different ways depending on what frame you use. If you use only the Earth/exoplanet frame, the coordinate distance is 198 l.y. and the coordinate time is 200 y, but the astronaut's clock only elapses 28.2 y due to time dilation. If you use only the astronaut frame, the coordinate distance is 27.9 l.y. due to length contraction, so he takes 28.2 y to get there. $\endgroup$ – Hypnosifl Dec 19 '14 at 16:41
8
$\begingroup$

There is actually an equivalent to "total elapsed proper time" along time-like curves in spacetime (which can represent the worldlines of particles moving slower than light), and that is the "proper distance" along a space-like curve (which cannot be any real particle's worldline). See the spacetime wikipedia article for more on time-like vs. space-like, particular the basic concepts section dealing with different types of intervals in special relativity, and the spacetime in general relativity generalizing that discussion.

The simplest physical interpretation of proper time on a time-like curve is just the total elapsed time on an ideal clock that has that curve as its worldline. But just as an arbitrary curve can be approximated as a polygonal shape consisting of a series of straight segments connected at their endpoints, so an arbitrary time-like curve can be approximated as a series of short inertial segments, which could represent bits of the worldlines of a bunch of different inertial clocks which cross paths with one another at the point the segments join. Then if you add the time elapsed by each inertial clock on each segment, this is approximately the proper time on the whole curve. Analogously, an arbitrary space-like curve can be approximated by a series of space-like segments, and the endpoints of each segment can be events at either end of a short inertial ruler which is moving at just the right velocity so that its plane of simultaneity is parallel to the the segment. Then the total proper distance is just the sum of the proper length of the rulers for all the segments. But this will probably only make sense if you already have a decent familiarity with spacetime diagrams in special relativity.

To give a mathematical example, suppose we are dealing with curves in SR which can be described in the coordinates of some inertial frame, and suppose the curves only vary their position along the x-axis so we can ignore the y and z space coordinates, and just describe the curves by some x(t) function. Then a time-like curve is one where $\frac{dx}{dt} < c$ everywhere, and a space-like curve is one where $\frac{dx}{dt} > c$ everywhere. If the time-like curve is approximated by a "polygonal" path made up of a series of inertial segments that each have a constant velocity $v$ over a time-interval $\Delta t$ in the inertial frame, then the elapsed proper time on each segment is $\sqrt{1 - v^2/c^2} \Delta t$ (this is just the time dilation equation), and the total proper time along the whole polygonal path is the sum or the proper time for each each segment. In the limit as the time-intervals become infinitesimal, this sum becomes an integral, and in this limit the error in the polygonal approximation goes to zero, so the actual proper time along the curve is $\int \sqrt{1 - v(t)^2/c^2} \, dt$.

Similarly, the space-like curve can be approximated by a polygonal path made up of a series of space-like segments whose endpoints have a spatial interval of $\Delta x$ between them, and with each segment having a constant value of $v^{\prime} = \frac{dx}{dt}$, where $v^{\prime} > c$. Each segment will be parallel to the simultaneity plane of a ruler moving at a slower-than-light speed $v = \frac{c^2}{v^{\prime}}$, and if the ruler's ends line up with the endpoints of the spacelike segment, that means the ruler has a contracted length of $\Delta x$ in the inertial frame we're using, which means the ruler's proper length is $\frac{1}{\sqrt{1 - v^2/c^2}} \Delta x$ (this is just the length contraction equation). So the total proper distance along the polygonal path is just the sum of the proper length for each ruler, and in the limit as the rulers' proper lengths become infinitesimal the sum becomes an integral and the error goes to zero, so the actual proper distance along the curve is $\int \frac{1}{\sqrt{1 - v(t)^2/c^2}} \, dx$.

So, you can see that in the first integral for proper time the factor in the integral is the same one that appears in the time dilation equation $dt_{proper} = \sqrt{1 - v^2/c^2} \, dt$, and in the second integral for proper distance the factor in the integral is the same one that appears in the length contraction equation $dx_{proper} = \frac{1}{\sqrt{1 - v^2/c^2}} \, dx$.

$\endgroup$
6
$\begingroup$

There is an asymmetry between space and time, and that is the reason why time dilation can be permanent and length contraction can not. The reason is that you can travel back and forth in space but not in time. In turn, this is related with an asymmetry between time and space in relativity, not in the laws of motion, but in the theory itself: we cannot go continuously from moving below the speed of light to moving faster than the speed of light, thus any physical reference frame is always moving either forward or backward in time, but cannot change from one to another. If they could, then length contraction could be made permanent.

Why could you make length contraction permanent if you could move backwards in time?
To see why you need to travel backwards in time remember the twin paradox:

a thought experiment in special relativity involving identical twins, one of whom makes a journey into space in a high-speed rocket and returns home to find that the twin who remained on Earth has aged more. This result appears puzzling because each twin sees the other twin as moving, and so, according to an incorrect naive application of time dilation and the principle of relativity, each should paradoxically find the other to have aged more slowly

You can explain why the twin paradox breaks the symmetry, that is, make the time difference permanent for one of them, by watching the figure below. There the twin traveler changes of inertial reference frame in the middle of the trip to initiate its return. In the graph the change in speed is instantaneous, thus the acceleration infinite and the brake in the asymmetry instantaneous: one point for the traveler maps into a segment for the stationary twin. If the change in speed were not instantaneous, then we would see that a small time segment (that is, during acceleration) in the traveling twin maps into a much larger time segment in the stationary tween, so the stationary twin will be correct.

enter image description here

Now relabel the graph $ct$ to $x$ and $x$ to $ct$. You get exactly the same problem, but with $t$ and $x$ reversed. Thus, to make length contraction permanent you need, instead of a traveler in space, a traveler in time. It should work like this: in a stationary reference frame, two meters, meter one and meter two have the same length. But now one of the meters (meter two) moves with traveler two. In the travelers reference frame meter two will be larger than meter one.The same will think the stationary meter, meter one, who will think he is longer. Same when he is in a different reference frame when returning. No symmetry break for now. But if the traveler is allowed to move back in time when returning, a similar asymmetry than in the twin paradox will occur. The final result being that the stationary meter was correct, and the returning meter will be shorter than the original.

$\endgroup$
  • 3
    $\begingroup$ This seems very confusing and not very well explained. $\endgroup$ – M.Herzkamp Dec 19 '14 at 10:53
  • $\begingroup$ Traveling back and forth in time would correspond to having a spacelike worldline, no? I don't see what that has to do with the thermodynamic arrow of time, even if the universe were at maximum entropy so there was no thermodynamic arrow, all particle worldlines would either be timelike or lightlike, assuming there are no tachyons. Also, does your comment imply length contraction could be permanent in some sense if particles could travel back and forth in time, i.e. if there were spacelike worldlines? How would that work? Maybe we could imagine tachyon "clocks" that measure proper distance? $\endgroup$ – Hypnosifl Dec 19 '14 at 16:57
  • $\begingroup$ @Hypnosifl, yes, but then, where does the asymmetry of the fact we cannot travel back in time, but we can in space, comes from? I might be wrong, is it from the absence of tachyons? $\endgroup$ – Wolphram jonny Dec 19 '14 at 17:10
  • $\begingroup$ Yes, I would say it's due to the absence of tachyons, since a space-like trajectory is always going backwards in time relative to some inertial frames, and you can come up with paths that are space-like everywhere and loop around to cross themselves, or visit any arbitrary sequence of out-of-order time coordinates (like a path that passes through an event in 2007, the further along the path it passes through an event in 1832, then further along the path it passes through an event in 3018, etc.) $\endgroup$ – Hypnosifl Dec 19 '14 at 17:15
  • $\begingroup$ @Hypnosifl you I am not talking about time travel because of a reference frame faster than light, but actually without going over the speed of light. I am not sure now that that idea makes sense. Any thoughts are welcomed! (I edited the answer to try to explain the process in more detail) $\endgroup$ – Wolphram jonny Dec 19 '14 at 17:27
2
$\begingroup$

Putting CuriousOne's comment into an answer,

In the theory of relativity, time dilation is an actual difference of elapsed time between two events as measured by observers either moving relative to each other or differently situated from gravitational masses. Wikipedia

I see that such a definition might be misleading as it talks about time dilation in an "elapsed time" sense. Although I can't say it is technically wrong, perhaps a better way to understand it would be in terms of speed that time goes at for observers moving relative to each other.

Just like length contraction, time dilation, interpreted as difference of speed of time flow also disappears as the observers again come at rest relative to each other. But, the elapsed time is a cumulative quantity. That difference can't be restored. Total time or any such concept may not be covered by General Relativity or any such present theory, as far as my limited knowledge goes.

$\endgroup$
  • $\begingroup$ Would this mean that time and space have different properties since time has a cumulative property that space apparently lacks? If so, why is there so much literature uniting them as space-time? I'm not trying to win an argument or anything, I'm just thoroughly confused. So far the only thing I've seen justifying space-time as a united thing is a four-dimensional coordinate system. $\endgroup$ – theboombody Dec 19 '14 at 3:09
  • $\begingroup$ Total time is more tied with entropy rather than space. Its flow however can be united with space into the 4-vector. $\endgroup$ – Cheeku Dec 19 '14 at 3:12
2
$\begingroup$

No, your facts weren't gathered incorrectly, your reasoning is just incorrect. It doesn't even take knowledge of physics to answer the question, just logical reasoning. (Don't take my language as a personal insult, I'm just trying to be clear.)

"It's not like time goes slow for a while, and then speeds back to "normal," so that the age of the observer once again matches the object."

Well, actually, yes, time does go back to "normal" when the moving clock comes back to rest. (All relative to the observer, of course). Once the clock comes to rest, it will once again tick at its normal rate, which is faster than the rate it ticked at when it was moving.

The clock will be behind, however, because it spent some time being a slowpoke. Your idea that once the clock begins to tick at its normal rate it will somehow "catch up" with the other clock is incorrect.

That's like saying if one marathon runner spends an hour walking, while his competitor runs, once he starts running again, he'll immediately catch up to the other guy. No, he'll be behind because of the time he spent walking while the other guy was running. Same idea with the clocks.

$\endgroup$
2
$\begingroup$

Time dilation does disappear as relative velocity approaches zero. The things experienced during the time experienced do not disappear; cells which have died remain dead and second hands which have ticked ahead do not reverse direction. To undo those things would require time reversal.

Sine we as humans only perceive time in one direction, time reversal is irrelevant: if an object is travelling in direction a at 1m/-s we would perceive and record it as travelling in direction -a at 1m/s, or direction a at -1m/s. We always record and perceive time as forward moving, but it can as easily be seen in the other direction.

$\endgroup$
2
$\begingroup$

As you say, time dilation and length contraction occur when two frames of reference (observer and observed) travel at two different speeds. Both of these effects "go away" if the two frames of reference subsequently travel at the same speed; that is, time will pass at the same rate and two yardsticks will have the same length.

But the EFFECTS of these relativistic effects are permanent in BOTH cases. For time dilation it's easy to imagine (i.e. the "old twin" scenario that you mentioned). So here's an example for length contraction:

Imagine that there's an immense opaque disc between you (on Earth) and a big stellar nebula. The disc is big, but not so big that it completely obscures the nebula. Some of the photons coming from the nebula are blocked by the disc.

Now imagine that the same scenario, but the disc is travelling very fast tangentially to you and the nebula. In other words, it's moving across your field of vision. Now, at such immense distances, you won't be able to easily see the disc moving, but it WILL be length-contracted. So fewer photons from the nebula will be blocked by the disc, allowing you to see more of the nebula.

This is a permanent effect! Those extra photons that slipped by the foreshortened disc are now streaming out into the universe, interacting with things and hitting retinas (maybe yours) long after the disc slows down (relative to you).

Nothing that happens in the universe every really "goes away". I didn't even mention the increased mass of the disc, which will distort the paths of those photons and everything else around it. Every distortion is "permanent" in that respect.

$\endgroup$
  • $\begingroup$ maybe it would be better to say, "Every distortion leaves permanent effects?" $\endgroup$ – Anthony Dec 20 '14 at 19:54
  • $\begingroup$ Good recommendation. I clarified my answer. $\endgroup$ – TheGerm Dec 22 '14 at 20:22
-2
$\begingroup$

Any change in time at all is only "permanent" because of the second law of thermodynamics and the resulting arrow of time.

$\endgroup$

protected by Qmechanic Dec 20 '14 at 16:49

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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