What are the mechanics of time dilation and length contraction? Going beyond the mathematical equations involving light and the "speed limit of the universe", what is observed is merely a phenomenon and not a true explanation of why time dilates or length contracts. It has been proven to work out, but do we know why? Is it something that happens at a subatomic level?


10 Answers 10


It's not a mechanism so much as a misconception of the nature of space (and its relationship to time): at low velocities, everything looks linear and Euclidean so we assume it is, but in reality it is not (as can be determined by appropriate experiments). It's kind of like asking by what mechanism you can reach something to your west by traveling east: if you conceptualize the earth as flat, the ability to end up to the west by traveling east isn't going to make much sense. Once you realize the earth is a sphere, you realize that there isn't exactly a west-is-east mechanism per se; it's really that the wrong concepts were being used (though they were a good approximation locally).

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    $\begingroup$ "at low velocities, everything looks linear and Euclidean so we assume it is, but in reality it is not" What do you mean by "linear"? The Lorentz transformations are linear by the usual definition. $\endgroup$ Commented Oct 14, 2011 at 4:42
  • $\begingroup$ The analogy is great, but spacetime in special relativity is still flat. $\endgroup$
    – Siyuan Ren
    Commented Oct 14, 2011 at 10:04
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    $\begingroup$ @Mark: What I assume Rex is getting at is that Minkowski space, while flat, is not Euclidean, due to the signature. $\endgroup$ Commented Oct 14, 2011 at 13:12
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    $\begingroup$ -1 because you see it as a misconception of space rather than space-time. $\endgroup$ Commented Oct 14, 2011 at 14:33
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    $\begingroup$ @MarkEichenlaub - By linear I meant that if you travel twice as fast, it will take you (from your perspective) half as long to go from point A to B along a rigid object; the Lorentz transformations are affine in the coordinates, but other previously linear relations no longer are. $\endgroup$
    – Rex Kerr
    Commented Oct 14, 2011 at 15:08

The right way to think about this is geometry--- but the geometry mixes up space and time. I wrote some answers about this here: Einstein's postulates $\leftrightarrow$ Minkowski space for a Layman and here: Help Me Gain an Intuitive Understanding of Lorentz Contraction and if you read these first, you can easily understand the effect.

The Lorentz contraction is no more mysterious than the following everyday phenomenon: when you place a meterstick parallel to the edge of the table, it marks off a part of the edge which is one meter long. If you rotate the meterstick so that it isn't parallel to the edge anymore, and look how far along the table the stick extends, it extends less distance. You can then ask "what is the mechanism that causes the x-distance of the ruler to shrink when it is rotated into y?" And the answer, if given in terms of the mechanism of cohesion of the atoms, would be ridiculous. It is obviously a property of rotations, of space, not of the forces in the ruler.

But you can ignore this, and ask--- if I have a line of particles held by elastic forces, why does their x-separation shrink when they are tilted? The answer would then be "because the equilibrium position is given by the solution to the equation:

$$ \Delta x^2 + \Delta y^2 = a^2$$

When you restrict $\Delta y$ to be zero, you get one separation, but when you make $\Delta y$ proportional to $\Delta x$ with a different constant of proportionality, you get a different separation. If you don't believe in rotational invariance, you can consider this to be a nontrivial physical effect--- "x contraction" in response to "y-tilt"--- caused by the mysterious $x^2 + y^2$ dependence of forces inside a ruler.

If you have a ruler tilted at a slope of m, then $\Delta y= m \Delta x$, and

$$ \Delta x = {a\over \sqrt{1+m^2}}$$

This is obvious in a picture--- the tilted ruler is reduced in horizontal length by this amount.

To understand relativistic length contraction, a second geometric analogy is useful. Consider a prison-stripe fabric placed on the table, so that the stripes are along the y axis with separation a between the edges. If you rotate the fabric so that the stripes make a tilt of slope m with respect to the y axis, and you make a line parallel to the x-axis what is the distance between the intersections with the stripes?

In this case, the x-axis line will intersect the rotated stripes at a longer distance, so that the stripes will change color every

$$ \Delta x = a\sqrt{1+m^2}$$

When the rotation angle approaches 90 degrees, the slope blows up, and you get an infinite distance, reflecting the fact that the stripes are now parallel to the x-axis.

Relativistic analogs

In relativity, the atoms make lines in space-time, and their equilibrium position is determined by the "minimum" relativistic distance between the lines (I put minimum in quotes, because it is a maximum, but it is analogous to the Euclidean distance between two lines, and it is only a maximum because of the minus sign in the relativistic pythagorean theorem), so that if the atoms at rest have a x-separation of a, and the force between them is relativistically invariant, when they are moving, the distance between them has to obey

$$ \Delta x^2 - \Delta t^2 = a^2 $$

where $\Delta t$ is now nonzero. The invariant distance between the lines is given by the "shortest" (actually longest) line linking them. This shortest line is the moving observer's x axis, which is tilted upward in a spacetime diagram by a slope v, just like the moving observer's t-axis is tilted by a slope of v to the right. The tilt of the axis gives that for the two space-time points at separation a, $\Delta t = v \Delta x$, and the result is

$$ \Delta x = {a\over\sqrt{1-v^2}} $$

This gives the x-distance between two endpoints on the moving ruler which are simultaneous in the ruler's frame. This distance is longer by a factor of $1\over \sqrt{1-v^2}$, just like in geometry it is shorter by $1\over \sqrt{1+m^2}$. The argument is exactly the same, except for the minus sign in the pythagorean theorem.

This thing is not usually explained in relativity books. It is the un-named phenomenon of "length dilation", and it is the direct analog of the shrinking of the x-length of a tilted ruler. This is not length contraction, which is like the prison stripe fabric.

When you have a moving ruler, you are usually not interested in the x-distance of two points which are simultaneous for somebody riding along with the ruler, but in the x-distance of two points which are simutaneous to you. To understand this case, consider a bunch of rulers end to end. These make a collection of lines parallel to the time axis which represent the endpoints in space time.

Now if all these end-to-end rules are moving, their space-time diagram is tilted to make a slope v with the time axis. You then ask how often the x axis crosses these tilted lines. The relativity formula is exactly the same as the geometry formula, except for the minus sign in the pythagorean theorem:

$$ \Delta x = a \sqrt{1-v^2}$$

so that the prison stripes (ruler ends) are closer together by $\sqrt{1-v^2}$, just as in geometry the prison stripes are further apart by $\sqrt{1+m^2}$.

In these formulas the units of length and time are chosen to make the speed of light c equal to 1. Any other choice would be as ridiculous for relativity as measuring the x coordinate in feet and the y coordinate in meters, and trying to describe a rotation.

  • $\begingroup$ This is the best answer I've read thus far! Why does this not have 700 more upvotes? The tilted ruler example is not just an analogy that's easy to understand. It literally is an example of what's happening $\endgroup$
    – Jim
    Commented Nov 24, 2014 at 16:10

Don't worry, you don't need any quantum mechanics or any knowledge about what happens at the subatomic level to understand this phenomenon. Length contraction and time dilation are purely a property of the 4 dimensional space-time continuum that we live in. It has to do with the actual measurements of length and time that can be performed by different observers that are traveling relative to each other.

The fundamental fact about our universe, that is the basis of special relativity, is the fact that all observers in inertial (non-accelerating) frames of reference always measure exactly the same value for the speed of light. This is not at all compatible with our naive intuitive understand of the way the universe works based on our experiences in our everyday lives. For example, if two cars are traveling on the freeway, one at 50MPH and one at 80MPH as measured from the ground, you would expect that as measured by the car traveling at 50MPH, the 80MPH car is only going 30MPH faster than the 50MPH car.

But if the first car is traveling at half the speed of light and the second car is replaced with a short pulse of light, both the observer on the ground and the observer traveling at half the speed of light will measure exactly the same relative speed for the pulse of light.

That is "why" the length contraction and time dilation is real for two observers moving relative to each other - so that both will measure the same value for the speed of light. Note that you, the observer on the ground, will think that the observer moving at half the speed of light has a slow clock and a shortened ruler and that the observer moving at half the speed of light will also think that your clock on the ground is slow and that your ruler is shorter. That is the "relativity" of special relativity.

This all works out such that the speed of light is exactly the same constant for all inertial observers. It is a property of the space-time continuum that we live in and has nothing to do with microscopic or subatomic physics or quantum mechanics.

  • $\begingroup$ I'm not sure that quantum mechanics (or, more importantly, its relativistic generalization) is not needed to understand the phenomenon. Remember that "observation" of time dilation and length contraction requires "clock & ruler" (for instance, atomic clock, etc.). The clock "mechanism" is not independent from this phenomenon. $\endgroup$ Commented Mar 23, 2012 at 17:49
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    $\begingroup$ All this “you, the observer on the ground, will think that the observer moving at half the speed of light has a slow clock and a shortened ruler” stuff just reiterates existence of said phenomena (assuming we know what the lab frame is) but not explains them. Not pass. $\endgroup$ Commented Oct 25, 2014 at 15:08

There are NO underlying "mechanisms" to lorentz contraction or time dilation. They are quantifiable observations stemming purely from the operational definition of "measurement." Arnold Arons presents this very nicely in chapter 36 of his long out of print textbook Development of Concepts of Physics (Addison-Wesley, 1965).


I agree very much with the previous responses, but it is important to remember that, in the rest frame, no contraction or dilation occurs.

High energy particles generated in the upper atmosphere travel at a substantial fraction of the speed of light relative to the Earth's surface. They appear to observers on the ground to decay more slowly (on average) than identical but "at rest" particles observed in the lab. There is obviously a time-dilation effect going on.

However, in the particle's reference frame, the decay happens (on average) exactly as it would for the particles "at rest" in the lab and it is you that is behaving slowly. It doesn't make any more sense to ask what is making the atmospheric particle's time dilate than to ask what is making your time dilate. Your perspective seems to be similar to that in this question. What you don't recognize is that you are being length contracted right now to all sorts of degrees depending on which reference frame you choose.


Although the time dilation and length contraction are really occurring between the two reference frames of the muon and earth, they are not occurring to the muon or earth for other reference frames.

For example, the distance from the muon's start to the earth is not contracted at all for anything moving at right-angles to the muon's direction.

The length contraction is not something mechanical happening to the earth's frame itself but is an aspect of the spatial relationship between the muon and the earth.

The length and time passing of everything we see and measure is a property not just of the thing itself but of our speed (distance x time) relative to it. To a photon, everything is as flat as a photograph and fozen in time. But we don't experience things that way! :)

  • $\begingroup$ First, Ī managed to decypher enigmatic letter X in “distance x time” as a contraction of Latin preposition “ex” as in “deus ex machina”, but still not clear enough. Second. One might say that “to a photon, everything is… f[r]ozen in time” were its proper time infinite. But photon has zero proper time (unless enters a medium with refraction). Could the author elaborate it? $\endgroup$ Commented Oct 25, 2014 at 15:43

All the undergraduate answers saying "there are no mechanisms" are incorrect because:

For any equation of physics to comply with special relativity has to force both Lorentz contraction and Time dilation upon any of its moving solutions.

Let us for instance look at the mechanism by which the wave equation of the electromagnetic field below leads to Lorentz contraction:

\begin{equation} \frac{\partial^2 A}{\partial t^2}\ -\ c^2 \frac{\partial^2 A}{\partial x^2}\ -\ c^2 \frac{\partial^2 A}{\partial y^2}\ -\ c^2 \frac{\partial^2 A}{\partial z^2}\ =\ 0 \end{equation}

Now any arbitrary stable solution $A$ moving with $v$ automaticaly forfills the following equations, \begin{equation} \frac{\partial A}{\partial t}\ =\ -v \frac{\partial A}{\partial x} \qquad \qquad \frac{\partial^2 A}{\partial t^2}\ =\ v^2 \frac{\partial^2 A}{\partial x^2} \end{equation}

simply because of the relation $x=vt$. An example is the static field of a point charge moving with a constant velocity $v$.

If we now combine the second order equation which must hold for any stable solution moving at $x=vt$ with the wave equation of the electromagnetic field then we we can eliminate the time dependency by substitution. We get:

\begin{equation} \left(1-\frac{v^2}{c^2}\right)c^2 \frac{\partial^2 \Phi}{\partial x^2}\ +\ c^2 \frac{\partial^2 \Phi}{\partial y^2}\ +\ c^2 \frac{\partial^2 \Phi}{\partial z^2}\ =\ 0 \end{equation}

This shows that the solutions are Lorentz contracted in the direction of v by a factor $\gamma$, The first order derivatives are higher by a factor $\gamma$ and the second order by a factor $\gamma^2$. Velocities higher then c are not possible.

We see that the static field field of a point charge moving with velocity $v$ is Lorentz contracted by a factor $\gamma$ as a result of the wave equation.

From my book:

Lorentz contraction from the classical wave equation.

Time dilation from the classical wave equation.

Non-simultaneity from the classical wave equation.

Yes, even the relativity of simultaneity is the direct result of mechanisms provided by physical laws which adhere to special relativity as is demonstrated in the third chapter.


  • $\begingroup$ I'm having a hard time imagining how a solution of the first equation could move with $v\ne c$. Could you give an example of such solution? $\endgroup$
    – Ruslan
    Commented Aug 3, 2014 at 10:58
  • $\begingroup$ Hi, Ruslan. I have updated the answer with the example of the static field of a point charge moving with a constant velocity $v$. $\endgroup$ Commented Aug 5, 2014 at 20:52
  • $\begingroup$ Thanks for the update. It still seems to me that the field of a moving point charge doesn' t satisfy that equation. I get $\Box A \ne0$, namely, $\Box A=\frac{3eV}{c}\left(1-V^2/c^2 \right)\delta^3\left((x-Vt)^2+(1-V^2/c^2)(y^2+z^2)\right)$. $\endgroup$
    – Ruslan
    Commented Aug 11, 2014 at 9:24
  • $\begingroup$ In fact, it seems I now understand. You should have $\Box \vec A=\mu_0 \vec J$. In this case any moving localized charge will create Lorentz-contracted field. BTW, it's quite interesting that similar equation for scalar potential is isomorphic to an equation of membrane oscillations ("the" wave equation), so moving a source with speed close to wave speed on a membrane will also give analog of Lorentz contraction. $\endgroup$
    – Ruslan
    Commented Aug 22, 2014 at 17:20
  • $\begingroup$ @Ruslan: if you deem such analogies are interesting, then you certainly can be amazed by such phenomenon as Cherenkov radiation (if happened to miss it before). $\endgroup$ Commented Oct 25, 2014 at 15:53

Time dilation is not just in theory. It is there because certain phenomena are explained only if time dilation happens.Presence of Mu-meson on the surface of earth is such an example. We are so conditioned in such a way that our daily life experiences are all explained by classical Newtonian mechanics, so that we dont digest a concept like time dilation.

Mu-Meson 'story' is somewhat like this-(I am not giving exact figures -you can Google). Mu_mesons are produced in upper layer of atmosphere, far away from earth. and their lifespan is so short. even if they move at the speed of light, there is no possibility of them to be reaching earth withing their life time. Bu they are found on earth! Reason is that time is slower for them (time dilation as they travel closer to light speed) that they live enough time to reach earth.

An analogy is like people in place:X lives only for 10 days and he can travel at 100km per day. so, Max distance they can reach out is 10x100=1000km in their life span. but how will you explain if one such guy is found alive at a place 2000km away from place:X? unexplainable! but as he travels at 100km/day if time goe slower for im, then it is probable that he can reach farther than 1000 kms. at normal sppeds it does not happen considerably. But it really happens at speeds closerto that of light.

  • $\begingroup$ Yet another Mu-Meson 'story' ☹ $\endgroup$ Commented Oct 25, 2014 at 15:57
  • $\begingroup$ They haven't been called "mu-mesons" in decades mostly because they are not mesons (strongly interacting states of a quark and an anti-quark) but rather are leptons (the same category as electrons). The modern parlance is "muon". $\endgroup$ Commented Sep 15, 2015 at 15:00

It depends on what you mean by mechanics. The mechanics of how electromagnetic systems undergo contraction and time dilation was worked out by Lorentz and Larmour. A good review of this was done by J. S. Bell in "How to Teach Special Relativity" (which unfortunately is not on the internet - you can either buy Bell's "Speakable and Unspeakable" book or get a copy through interlibrary loan). This approach is called "constructive relativity" and there is a lot of disagreement among physicists as to whether it is valid.

The quantum mechanics of how systems undergo contraction and time dilation is unaddressed. It is one of my research goals to publish something on this, but it is difficult to get rid of the circularities.

Many physicists have concluded that the underlying "cause" is Minkowski spacetime geometry, and some answers here have pointed that out. Swann in the 1940s-60s wrote several papers pointing out that only quantum systems automatically adjust their clocks and lengths to conform to relativistic predictions, and that macroscopic systems of clocks and so forth do not. This suggests that there is no fundamental reality to Minkowski space, that instead it emerges from quantum mechanics. Harvey R. Brown has written a book titled "Physical Relativity" which takes this point of view.

If we only have two reference frame systems, then the relativistic changes are mutual and it is impossible to decide whether one is more fundamental than another, but . . .

If we consider a large number of reference frame systems, and ONE of them accelerates, then all but that one will agree that only that one changed. The one which accelerates will see ALL others change. So really, it is pretty easy to determine that some physical change occurred in the one which accelerated. However, it is impossible (so far) to determine the nature of that change, because no one can identify a preferred frame of motion.

Light can be captured in fiber optics and dragged around at will, and the light will only move relative to the fiber. Similarly, light can be captured by a gravitational field in a photon orbit or deep hyperbola, and if the object is dragged around (a little difficult since it must be massive, but it will be dragged around by other massive objects) the light will go with it. Therefore from outside the gravitational field of the object, it is easy to detect a preferred frame of motion for the captured light.

However, when asking about the motion of light in "free space" or "the background" as it is called in General Relativity, it is impossible to step outside the background (that would be outside the universe) and make a similar determination of the preferred frame of motion. Some people suggest the CMB frame, but this is only a guess and cannot be verified.

Most people do not realize that de Broglie waves (matter waves) arise because of the distortion of simultaneity due to relativistic motion. In the rest frame, the de Broglie wave is everywhere in phase and has infinite wavelength. It automatically synchronizes its wave phase clocks, so to speak, into an Einstein reference frame. This is direct evidence of a quantum system automatically adjusting itself to its rest frame, but does not entirely explain why. For more information about this puzzle see the paper "Leading Clocks Lag" (intended for students) on my website http://mc1soft.com/papers/


. . . .enter image description here

$$To \enspace begin,\qquad L^2=L'^2+b^2\qquad \qquad and\qquad \qquad c^2=a^2+v^2 \qquad $$ $$Thus\qquad L'^2/L^2+b^2/L'^2=1\qquad and\qquad a^2/c^2+v^2/c^2=1$$ $$ \enspace \qquad \qquad And\enspace \qquad \qquad x^o=y^o, \qquad \qquad Thus\enspace \quad L'^2/L^2+v^2/c^2=1 \qquad \qquad $$ $$\rightarrow \quad L'^2/L^2=1-v^2/c^2\quad \rightarrow \quad L'^2=L^2(1-v^2/c^2)$$ $$\rightarrow \quad L'=L\sqrt{1-v^2/c^2}$$ $$To \enspace begin,\qquad a^2/c^2=1-v^2/c^2\qquad and\qquad t'^2/t^2=a^2/c^2 \qquad $$ $$Thus\qquad t'^2/t^2=1-v^2/c^2\quad \rightarrow \quad t'^2=t^2(1-v^2/c^2)$$ $$\rightarrow \quad t'=t\sqrt{1-v^2/c^2}$$

As mentioned by physicist Brian Greene in his book The Elegant Universe, all objects are constantly on the move within Space-Time, and they do so at the speed of light. Thus if at rest in space, such as in Twin Space Ship A as shown above, yes the Space Ship will extend across space, but it will be moving across the dimension of time at the speed of light .

By stacking both constant motion vectors, object lengths, and depths across Space-Time, it is extremely easy to use simple geometry to create all of the SR equations in mere minutes.

All of the equations are valid between Twin Space Ship A, which is at rest in space, and Twin Space Ship B which is in motion across space at v = 260,000 kps.

However, if Twin Space Ship A started to move across space, the clocks would slow down, the rulers contract, and the clocks located at the (R)ear and (F)ront are no longer in sync, thus the measurement instruments are not in the same condition that they were when at rest in space. Its measurement instruments have changed accordingly. Thus even under these changing conditions, the equations are still valid between Twin Space Ship A, and Twin Space Ship B.

Thus the equations are valid between a body at rest and a body in motion, and they are also valid between body at motion and another body in motion. Thus there is no way possible to tell if either of the 2 bodies are truly at rest in space, or not. Thus relativity is born.

However, if Twin Space Ship A used its measurement instruments and measured the spatial length of Twin Space Ship B, they would see that Twin Space Ship B is at half of its rest length and that the clocks onboard are ticking at half speed. But what's even more interesting, is that if Twin Space Ship B used its measurement instruments and measured the spatial length of Twin Space Ship A, they would see that Twin Space Ship A seems to be at half of its rest length and that the clocks onboard are ticking at half speed, even though this is not the actual case. All of this has occurred due to the change of Twin Space Ship B's measurement instruments.

Thus in one case, Twin Space Ship B, the appearance of length contraction is due to actual rotation within Space-Time, thus in turn causing partial extension across the dimension of time and contraction across space. And, the appearance of Time dilation is due to the Twin Space Ship B no longer being in motion across time only, via now being in motion across space as well.

In the other case, Twin Space Ship A, the appearance of its length contraction and Time dilation as seen from Twin Space Ship B, is entirely due to the change of Twin Space Ship B's measurement instruments.

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    $\begingroup$ This doesn't address the question. The question asks for a mechanism, not a derivation. $\endgroup$
    – user4552
    Commented Sep 1, 2014 at 23:13
  • $\begingroup$ The mechanism is made clear. Objects are constantly in motion at the speed of light within Space-Time. A change in the direction of this constant motion leads to time slowing down, and also this change in the direction of travel causes rotation, which then leads to spatial length contraction due to having partial extension across the dimension of time. The derivation is merely a verification. $\endgroup$
    – Sean
    Commented Sep 2, 2014 at 0:17
  • $\begingroup$ Also, see ( i.sstatic.net/7hCUh.png ) to view physicist Brian Greene's equations he used to conclude constant "c" motion of all objects existing within Space-Time. See ( goo.gl/fz4R0I ) to view 9 mini YouTube videos for a geometric/mechanistic analysis of the "c" motion of all objects. $\endgroup$
    – Sean
    Commented Sep 20, 2015 at 19:40

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