I understand that if an object is moving at relativistic speeds in space, it will noticeably contract in the direction that it was traveling, but I'm not sure "towards" which point the contraction occurs (i.e. which point is fixed). e.g. Suppose you had the following object:


and you observed it to accelerate to relativistic speeds. Then, in comparison to its position if length contraction did not occur (but say stuff like time dilation and whatnot did, not sure what this would affect, let me know if this assumption is wherein lies the issue), then these all seem to be possible candidates for the contracted object:




My issue with this actually goes a bit further: whatever the answer to this question is, I claim something annoying happens when you split the object into components and then measure contraction on each part: you'll get disjoint parts, which are seemingly connected by... nothing? For instance,


becomes (for example):

  <--->        <--->        <--->

One argument I could see being made is that the space within also contracts. But, if you had a Lorenz factor of say 1/2, the space between can only be contracted to 1/2 the amount; the components will still remain separated by a non-zero amount of space. But this leads to two (in fact, infinite) contradictory views of what a length contracted object would become: either one contiguous segment or multiple disjoint segments.

  • $\begingroup$ Doesn’t contraction happen relative to a frame and isn’t there one frame for every pairing of the object in question with every other object in the universe, so wouldn’t contraction simultaneously happen differently in each of these frames or am I understanding relativistic contraction incorrectly? $\endgroup$
    – bob
    Commented May 5 at 19:22

8 Answers 8


Consider your object,


If each of the segments accelerates at the same rate at the same time (in your frame), then the object will indeed become

  <--->        <--->        <--->

This is Bell's spaceship paradox. In order for an accelerating object to remain rigid, the back needs to accelerate faster than the front. The result would then be


Within the frame of the object, the necessity for the back to accelerate faster than the front can be understood as a consequence of gravitational time dilation. Acceleration is equivalent to a uniform gravitational field, and the back of the object is lower in the associated gravitational potential, so less time passes for it. Thus, it needs a larger acceleration to keep up.

  • $\begingroup$ Du Sten! det var da godt. -Niels Jakob $\endgroup$ Commented May 3 at 1:57
  • $\begingroup$ can you do that without referencing gravitational time dilation? $\endgroup$
    – JEB
    Commented May 3 at 3:37
  • 4
    $\begingroup$ @JEB Relativity of simultaneity is another way. Relative to an inertial observer, the object's "now" surface continuously rotates (in spacetime) as it accelerates, so that forward in the object frame points more and more into the future of the observer frame. Or just analyze Rindler coordinates, in which the behavior arises explicitly. $\endgroup$
    – Sten
    Commented May 3 at 4:54
  • 1
    $\begingroup$ @Sten exactly. I also use Bell's Space Ship paradox to explain the electrons in youtube.com/watch?v=1TKSfAkWWN0 $\endgroup$
    – JEB
    Commented May 3 at 6:27

To accelerate a rod, you have to push or pull on it. Whether it contracts or expands (or for that matter whether it breaks), and the exact locations of the ends, depend entirely on how you push or pull.

Whatever its length and location in your own frame, it will always be longer in its own.

For example, one of many possible scenarios is that you accelerate the rod uniformly along its length, so that each individual point on the rod accelerates by the same amount at every instant (all as measured in your own frame of course). In that case, the length of the rod is unchanged, and it's easy to calculate the new location of the rod after any given amount of time. But in the frame of the now-accelerated rod, the acceleration at the front started before the acceleration at the back, so the rod has been stretched (as long as it hasn't broken).

The rod has two endpoints. To calculate the locations of those endpoints in your own frame post-acceleration, you do not need to know anything about relativity; you just need to do some elementary calculus, starting with the acceleration profile and integrating a couple of times to get the new locations of the endpoints. If you want to know the locations of those endpoints in the rod's own new frame, that's where relativity comes in.

And of course because you can accelerate a rod any way you want to, there is no general answer to the question of where its endpoints are located after the fact.

  • 1
    $\begingroup$ "But in the frame of the now-accelerated rod, the acceleration at the front started before the acceleration at the back" This is not true. At the time when the acceleration starts, there is no velocity difference between the frames of reference, so the stationary inertial observer and the rod will agree that the entire rod starts accelerating simultaneously. On the other hand, the front end will accelerate faster than the back end, according to the rod itself. $\endgroup$
    – Arthur
    Commented May 3 at 10:31
  • 1
    $\begingroup$ @Arthur: "the frame of the now-accelerated rod" means the inertial frame of the rod after the rod has accelerated. In thag frame, the front accelerated first.9 $\endgroup$
    – WillO
    Commented May 3 at 13:15
  • 1
    $\begingroup$ Right, I was thinking about the (non-inertial) reference frame of the rod itself. In the inertial frame that the rod eventually catches up to, you are right, of course, the front started accelerating first. $\endgroup$
    – Arthur
    Commented May 3 at 13:31
  • $\begingroup$ Good answer, but I think it is clear from the OP's question that he means a type of acceleration that does not induce physical stresses in the rod. That is, the resulting length contraction is purely due to relativistic effects, and not due to physical stresses. $\endgroup$
    – fishinear
    Commented May 5 at 16:12
  • $\begingroup$ @fishinear: Nothing changes length for any reason other than physical stresses. Of course those stresses (and the resulting length change) are frame dependent and relativity tells you how to compare them across frames. But the analysis in a single frame (as the OP is asking for) is strictly about the effects of pushing and pulling. $\endgroup$
    – WillO
    Commented May 31 at 2:47

Length contraction happens because time in one frame is out of synch with time in another. When you consider the length of a passing object, you are effectively asking where the two ends of the object are 'now', where 'now' means a given time in your frame. In the frame of the object, your 'now' at the front of the object is earlier than your 'now' at the rear, so from the object's perspective you are locating its trailing end after its leading end, which means the trailing end moves forward in the meantime and hence the object seems shorter to you. As the other answers have said, the physics becomes complicated if your lengthy object accelerates, so instead, assume it stays at rest and you accelerate instead. The object will appear to shrink in such a way that the positions of the parts furthest from you change more than the positions of the parts closest to you. So if you were accelerating from the right hand end of the object, it would appear to shrink from the left hand end initially. By the time you had reached the middle, it will appear to have shrunk equally either side of you.


Your question seems a bit confused, as it implies that a moving object contracts in its own frame. Moreover, relying on acceleration or internal compression from g-loading to remove the discrepancy misses the point.

Consider a relativistic train. You can analyze that such that an engine is pushing from behind, or it starts pulling from the front, and come to conclusions about the finite speed of forces within materials (Born Rigidity).

But then I can design a train with a motor on every wheel and tell you the whole thing starts and accelerates simultaneously, and then all your hard work is out the window.

It's actually a good problem, but it has been solved with an engine and a caboose, except they are both rocket ships and the train cars are replaced with a string. (Bell's Spaceship Paradox).

If you understand that problem, then you'll know the answer to your question, and that has been solved elsewhere on this site.

But the real problem is "accelerating to relativistic speeds". No mater how fast the object goes, it is still at rest in its own frame. When you place the burden of contraction on the moving object, you are implicitly saying there is an absolute rest-frame, because there are frames where it is contracted and frames where it is not. Such frames do not exist.

The other way to get the relativistic train to high speed, for the train station observer, is for him to get on rocket ship going the other way.

Now in this case, the Lorentz contraction is no different from case where the train left the station and the observer did not.

If the relative velocity is $v$, then the observer sees the train contracted by:

$$ L = L_0/\gamma $$

So now we have a problem. All the explanations that talked about forces, gravity and Born-Rigidity have a problem: The train never moved, but it is shorter. But obviously a stationary train doesn't get shorter because SOMEONE ELSE IS MOVING. It makes no sense.

Lorentz contraction is a property of the observer's coordinates, not the "contracted" object. Obviously, for an accelerating train, the two ends of the train have to have different acceleration profiles if the train changes length, but that is explained by the relativity of simultaneity, since clock bias depends on distance from the observer's origin, and the 2 ends of the train have different distances from the origin.

  • 1
    $\begingroup$ Your "other way to get the relativistic train to high speed" is marvellous, & underlines the importance of identifying which frame of reference makes the problem easiest to (a) understand and/or (b) do the maths on. It's like the Earth vs the Universe: there's simplicity in taking the Earth as the stationary frame with the entire Universe revolving around it, but the physics quickly gets awkward when you start bringing other bodies (e.g. planets) into it. The geocentric frame isn't wrong, but it certainly has its limitations. $\endgroup$ Commented May 10 at 1:11

I will use Epstein's diagrams (see appendix), which are more illustrative because they are treated in Euclidean geometry, avoiding cumbersome hyperbolic geometry. see https://youtu.be/wTwYbkyENTE. In Fig 1 we see the diagram of a train that moves at a relativistic speed, the speed is v= C cos @, the train has length Lo in its own reference system, in the earth system its length is seen to be: L= Lo cos @ . Note that the clocks on the train red line t' =- xtan@ = - gamma (xv) (we use C=1.) enter image description here

What happens when we accelerate the train? We can accelerate the train in a way that varies the angle @, see fig 2. enter image description here

We have aligned the graphs on the original coordinate origin for clarity, Note that: 1.- the value of time in the train t' is greater in the back of the train than in the front. This difference is the reason why the train looks shorter from the ground, since seen from the train, to measure the lenght of the train the rear point of the train is first taken (shorter time, past the train) and then approaching towards the front, at longer time .

2.-increasing @, the speed increases and the length seen from the ground decreases. This acceleration is not the same at all points of the train, but the distance AB seen from the earth is not limited to always remaining the same, so there is no stress on the train. In his paradox Bell gives an example in which the distance AB measured from the ground has to remain constant, but to do so the train must elongate forward as illustrated in the blue line in fig 2.


While Lorentz contraction is not a bit different from perspective contraction of coordinates in Euclidean space by an oblique point of view, an accelerated object with spatial extension is not a a rigid object of Lorentz geometry but a collection of world lines of points whose passage of a plane of constant coordinate time marks the 3-geometry by any field of observers in their time synchronized geometry of Euclidean 3-spaces.

As a paradigm of classical relativistic electrodynamics, a point mass can be accelerated with a constant $\frac{d}{d\tau} x_1(\tau)$, if it has an electric charge $q$ and starts from $$x_1(\tau=0) = 1, \dot x_1(\tau=0), \dot x_0(\tau=0) $$ in a constant electric field.

$$\ddot x_0(\tau) = q F_{01}\dot x_1(\tau), \quad \ddot x_1(\tau) = q F_{10} \dot x_0(\tau) $$

with a solution of a Lorentz hyperbola. By translation invariance start point $x(0)$ and velocity point $\dot x(0)$ are free.

$$F = E_1 dx dt = F_{01} dt\wedge dx $$ is the electric field and transforms along the hyperbola to the local tangent rest system of the accelerated particle by

$$ F_{01} dt \wedge dx \to F_{01} (dt \cosh u + dx \sinh u ) \wedge (dt \sinh u + dx \cosh u ) = F_{01}\ (\cosh^2 u - \sinh^2 u) dt\wedge dx $$

This central theorem of electrodynamics makes it possible to speak about constant charge and accelerating longitudinal component of the electric field being constant for the observers, with a felt constant acceleration by his accompanying (tripod-clock), measured gravitationally by rigid body elastics using a spring balance on the floor of the rocket.

While electric accelerated point charges are understood mechanically, the coupling to their radiation reaction field is a pure quantum topic; indeed it was the source of the evolution from Maxwell-Lorentz dynamics to Einstein-Dirac-Feynman QED.

If a macroscopic rigid body is accelerated at a constant rate by proper time rate, the constant force is acting at a part of the volume, lets say a charge at the tip. Then as known from a starting train, the parts of the train begin to oscillate beyond limits of there is not a damping element parallel to the coupling springs.

Translated to quantum thermodynamic field theory: There is nothing like an accelerating steady state of rigid body. Radiation and internal friction is a central element of such a theory, that would be - if a working model can be constructed - a building block of the inclusion of gravity in a QFT model.


enter image description here

Consider the above diagram where two spaceships each have a thruster at their centre (the dashed curves) and each accelerates with constant proper acceleration. To an inertial observer, the acceleration of the centres of the rockets remains equal to each other at any instant, but the coordinate acceleration is not constant but slows down over time. It can be seen that due to length contraction, each rocket length contracts towards its individual centre. It can also be noted that the distance between the centres of the two spaceships remains constant in the initial inertial reference frame while to the observers onboard the spaceships, the distance between the centres of the two spaceships is expanding over time. The length contraction of the green spaceship subtracts from the acceleration of its front, while the length contraction of the blue rocket adds to the acceleration of its rear part and the end result is that the gap between the rockets increases. If there was a string connecting the two spaceships, it would be progressively stretched and eventually break (even if the string connected the two centres), because its proper length is increasing. This is essentially the explanation of Bell's rocket paradox. There is a slight simplification here. Due to inertia the back of each spaceship would lag slightly behind and the front of each spaceship would be compressed towards its centre if there was only one thruster at the centre of each spaceship. In order to perfectly accelerate the rockets so that that in their own reference frames they see no change in their proper length, we would need a thruster for each atom of the spaceship or an external force field accelerating the spaceships. This is called Born rigid acceleration.


If we had three rods as depicted above, and had a thruster at the centre of each rod and simultaneously accelerated the rods with equal acceleration at any instant, we would indeed end up with 3 spatially separated rods as depicted below:

  <--->        <--->        <--->

However, if we connected the rods with links (o) and had a single thruster at the centre of the central rod, we would end up with:


where the 3 rods length contract towards their mutual centre. The links to the left would be under tension and the link to the right would be under compression and a comoving observer would notice slight changes in the combined proper length of the rods because this is not ideal Born rigid acceleration. However, if the acceleration at the back was 1/k and the acceleration of each part of each rocket had acceleration 1/(k+x), where x is the distance from the back, then it would be ideal Born rigid acceleration and the rods would remain together even without the links and the proper lengths of the rods would remain constant from the point of view of a comoving accelerating observer.

One way we can apply equal proper acceleration to each atom of a long series of connected rods without requiring a thruster for each atom, is to arrange the rods in a loop around the perimeter of a cylinder and spin up the cylinder around its axis of rotational symmetry. The resulting length contraction of the individual rods would cause tension in the connecting links which would eventually break the links (assuming the links are weaker than the rods themselves). This phenomena is at the heart of the Ehrenfest paradox and the resolution is that it is basically impossible to spin up a solid cylinder in a Born rigid manner that does not cause tension in the perimeter. You can see my more detailed explanation of the related Ehrenfest paradox in this old answer, where I borrowed the top diagram from.

In summary, what happens depends on the interconnections of the particles and how they are accelerated relative to each other.

Bell's rocket paradox and the Ehrenfest paradox demonstrate that length contraction is not just a perception and can have real physical consequences such as objects being torn apart.


TL;DR Lorentz contraction is towards the moving origin.

I think your confusion is best clarified in the simpler situation of inertial frames$^{[0]}$, which I consider first. (For the case of accelerated frames, I'll append later)

The entire space$^{[1]}$ of a moving reference frame appears Lorentz contracted towards the origin of that reference frame, to stationary observers. In other words, the origin of the moving reference frame is the fixed point under Lorentz contractions (LC). There is nothing special about relativity in this regard - whenever a scaling transform is applied to a vector space, its origin remains unchanged.

fig1: A marked ruler as it appears in its rest frame $S'$ (top) and as it appears to observers with respect to whom it is moving with uniform constant velocity (frame $S$, bottom). The appearances have been aligned so that the origins coincide. Clearly, the ruler appears contracted towards the origin ($\gamma=1.2$).

Elaborating further, when we talk of the LC of a moving rod, what we really mean is the length between the simultaneous measurements of the positions of its ends (by stationary observers). The fact that this length is smaller (wrt. its value in its rest frame) doesn't imply that one end is contracting towards the other as the other end remains fixed. As you point out, this is ambiguous. Instead distances of both ends measured from the origin, contract towards the origin, by the same factor, and so does their difference. Indeed, every point of the rod, nay, the entire space$^{[1]}$ contracts towards the origin. Since the contraction is uniform, there is no fracture.

Besides the last line, the lack of fracture is expected for a more foundational reason: simply by changing your frame of reference, an object can't suddenly be made to appear fractured - after all, in its rest frame there are no forces acting on it, and so it sits unchanged for all time; this physical reality of the rod is identically observed by all observers - inertial or not.

Is there anything special about the origin of the moving frame? Absolutely not. The fixed point is determined by the initial alignment of the origins of the two frames that are moving relative to each other, and the initial sync of their clocks. Indeed, with an appropriate choice, the origin can be arbitrarily shifted. This is why associating the quality of 'towardness' to scaling isn't really useful.


This skippable section puts the above explanation on a firmer footing. let $S'$ be a reference frame ('rod frame') moving with constant uniform velocity $\beta\hat x$$^{[2]}$ wrt. the frame $S$ ('lab/observer frame'). Further, by choice, let their origins coincide and axes align when their clocks are synchronized. Let some arbitrary event $E$ be described by the coordinates $(t',x')$ in $S'$ and $(t,x)$ in $S^{[3]}$. These are connected by a Lorentz transformation:

$$ \begin{align} x'&=\gamma(x-\beta t)\\ t'&=\gamma(t-\beta x) \end{align}\tag{1} $$ where $\gamma=(1-\beta^2)^{-1/2}$.

Consider the two events as described by observers in $S$ (for some $L>0$).

$$ \begin{align} E_0&:(t,\beta t)\\ E_1&:(t,\beta t\pm L)\tag{in $S;\>2$} \end{align} $$

As you may have guessed, $E_0$ is the spacetime coordinate of the origin of the moving frame, and $E_1$ an arbitrary point at distance $L$ (wrt. observers in $S$) from it. Note the same time coordinate: the events must be simultaneous for observers in $S$ to constitute a length measurement.

Observers in frame $S'$, on the other hand, describe the same events with coordinates

$$ \begin{align} E_0&:(t/\gamma,0)\\ E_1&:(t/\gamma\mp\beta\gamma L,\pm \gamma L)\tag{in $S';\>3$} \end{align}$$ Ignore the time coordinate (it's complicated because of time dilation). The space coordinate of $E_0$ is always $0$, which it must be since to observers in $S'$ it corresponds to their origin. Note the correct contraction: observers in $S'$ conclude that observers of $S$ are measuring a contracted $L$ between the two events instead of the 'correct' $\pm \gamma L$. Notice how the $\pm$ denotes the contraction towards the origin.

But $E_1$ could have referred to any point. So the length of all points measured from the point $\beta t$ at time $t$, appears contracted by $\gamma$ to the observers in $S$, in the opinion of observers in $S'$. Indeed, as seen in eqns. $(1)$ for the Lorentz transformation, the contraction happens on the length $x-\beta t$ i.e. length measured (by observers in $S$) from the origin of the moving ref. frame.


"Then, in comparison to its position if length contraction did not occur (but say stuff like time dilation and whatnot did, not sure what this would affect, let me know if this assumption is wherein lies the issue)"

If "length contraction did not occur" neither can time dilation. By definition, the non-contracted length is the length of the object as measured in its rest frame; wrt. this all other frames measure smaller lengths. Same goes for the time intervals. A time interval, by definition, is undilated in the rest frame of the object, wrt. which all other frames measure it to be larger.


$^0$the rod's not accelerating but moving uniformly w.r.t. the observer.

$^1$ Components of vectors along the direction of relative motion. Orthogonal components aren't contracted.


$^3$ coordinates $y,z$ don't change and are suppressed


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