Why or why not?

This is closely related to another question I posited here, Does it make sense to say that objects moving at relativistic velocities appear in space-time cross-section?

What I mean is the following: Saying that moving objects appear in space-time cross-section along a space-like hypersurface (the observed 3D space view at a given moment) in the geometry of SR implies that they must undergo length contraction.

Motivation: I used a similar statement in a couple of videos I made on length contraction and the length contraction paradox and got mixed responses. My intent was in part an unorthodox approach that highlights the strong connection between length contraction and relativity of simultaneity beyond the common examples. Although the videos use strictly standard special relativity, somebody commented to me that such statements sound unprofessional and likely leave the impression of crackpottery. I posted a similar question on Reddit/r/Physics and got mostly the same reaction, which to me is a big surprise.

Are there really any conceptual objections to this that I overlooked? Or is it just a matter of the audience being unprepared for it?

Note: The original version posted on Reddit was "Moving objects appear to undergo length contraction because they are perceived in space-time cross-section". A major objection over there was that using words like "appear" and "perceive" suggest I consider length contraction some sort of optical illusion dependent on the observer. To the contrary. I never intended to debate that it is a real, measurable effect. Only meant to emphasize that length contraction appears differently in different frames, as demonstrated by measured coordinates. Rephrased here to eliminate this sort of misunderstanding.

Let me refine and clarify please:

Let us limit the question to special relativity only, for convenience and clarity. Let us also leave aside optical effects in SR, including doppler, beaming, Terrell-Penrose, etc. In other words, the question does not refer to the optical observation of length contraction by a human observer or a camera. Let us discuss only in terms of measured coordinates.

Let object A move at relativistic velocity relative to an inertial frame O. In 4D space-time (Minkowski diagram) the space view of O at any given moment of its own time is a space-like hyperplane. What O observes of A at any time is the cross-section of A's world-tube by O's corresponding constant-time hyperplane. In this cross-section A undergoes length contraction. Geometrically speaking, the length contraction is a consequence of the fact that O's constant-time hyperplanes "slice" A's world-tube at a different angle than do A's own hyperplanes of constant proper-time (corresponding to the 3D space view in its rest frame).

In view of this 4D representation, is it ok to say that, as seen in O, A undergoes length contraction because it appears in a different space-time cross-section than it does in its rest frame?

  • 2
    $\begingroup$ I have no idea what you are talking about! What do you mean by "space-time cross-section"? Special relativity is a very well established theory and Lorentz contraction is a "real, measurable effect". $\endgroup$ Commented Aug 7, 2015 at 16:07
  • 2
    $\begingroup$ Then you either don't really know SR or your critical reading score is really low. As I said, above "I never intended to debate that it is a real, measurable effect". $\endgroup$
    – udrv
    Commented Aug 7, 2015 at 16:27
  • $\begingroup$ @udrv : I'm not clear on your question either. Maybe it needs a bit of a polish? Meanwhile, please do note that if you accelerate in your gedanken spaceship, some nearby star doesn't obligingly flatten into a pancake version of its former self. It doesn't change one iota. Its length contraction is a measureable effect, but it's only real to you. Because you change. The star doesn't. $\endgroup$ Commented Aug 7, 2015 at 16:32
  • $\begingroup$ What spaceship are you talking about? What star? The statement of the problem in the other post is as follows: Let object A move at relativistic velocity relative to a frame O. In 4D space-time (Minkowski diagram) the space view of O at any given moment of its own time is a space-like hypersurface (hyperplane, if O is inertial). What O observes of A at any time is the cross-section of A's world-tube by O's corresponding constant-time hypersurface. If so, are there any formal objections to saying that "moving objects are observed in space-time cross-section"? Never mention any spaceship, star. $\endgroup$
    – udrv
    Commented Aug 7, 2015 at 16:35
  • $\begingroup$ OK, now I understand I know nothing about special relativity! $\endgroup$ Commented Aug 7, 2015 at 16:56

1 Answer 1


It's the same question but now we've specialized on length contraction. Right on.

Why we can say what we're saying

You can work this out from Lorentz transforms if you want, but Lorentz invariants let you simply write it down: a particle has an invariant 4-velocity $\gamma~[c, \vec v]$ and we can form the inner product of this with the local 4-positions $[ct, \vec r]$ to find the equation for that other particle's spacelike hyperplane in my coordinates, as in their coordinates this 4-velocity is $[c, 0]$ and picks out only the time component $c^2 ~ t'$ in the local coordinates. Their "present moments" therefore have the form $$c t - v_x ~ x - v_y ~ y - v_z ~ z = C = c^2 ~ t' / \gamma.$$ Therefore, if you have a spacelike hyperplane $a_\mu x^\mu = A$, the parameters $u^i = -c~a_i / a_0$ tell you the 3D components of $\vec v$ directly. Since you have $\vec v$ you have everything you need to derive the Lorentz transform that makes that hyperplane constant-time. However, if the hyperplane is too tilted, you will presumably get a $\vec v$ with magnitude greater than $c$, which among other things will be unphysical.

Interpretation nuances

While we have everything we need to determine the Lorentz boost, we have to be really careful with how we interpret the geometry here.

One example: the light cone $c t - x - y - z = 0$ must look like a light cone in all other reference frames due to the mathematics of Lorentz transforms. But if you remember your conic sections, the relevant conic section is an ellipse, a circle that's been stretched along one axis. What we call light cones are actually uniformly expanding bubbles at speed $c$, but the point still holds: the shape that is seen in the hyperplane is not a sphere, but is linearly extended in some direction. This direction, indeed, is $\vec v$, and the extension of the sphere in our version of the hyperplane is corrected for exactly by the Lorentz contraction of the hyperplane when you make the Lorentz transform into the local coordinates.

In other words, it is too simplistic to say "A undergoes length contraction because it appears in a different space-time cross-section than it does in its rest frame." In fact we see that any "vertical tube" (i.e. a sphere of constant radius staying in one place over time) has a lengthened cross-section in the hyperplane, which must be corrected for by the relativistic $\gamma$ to make it length-contract. Instead we have to play a game where this timelike hyperplane intersect a light cone which intersects all of the "perpendicular" (i.e. to $\vec v$) spatial dimensions at the same time, but intersects the other dimension $x$ at different points in time. The light cone's conic section is bigger, and when we scale it down to a circle, then the tube has contracted in the $x$ direction.

  • $\begingroup$ Thanks again Chris. I understand what you say and it would make sense, except one thing: you are referring to the wrong hyperplane. From frame O's point of view, the hyperplane that determines the cross-section in which A appears length contracted is always one of O's constant-time hyperplanes. It is the "horizontal" hyperplane, not one of object A's tilted constant-time hyperplanes. $\endgroup$
    – udrv
    Commented Aug 8, 2015 at 5:00
  • $\begingroup$ So: 1) it and the cross-section it generates are space-like by definition and can never be "too tilted"; 2) its intersection with the light cone is always going to generate a spherical cross-section because it is normal to the cone's axis, nothing wrong here; 3) we don't need to play any game to retrieve length contraction from the intersection of this "horizontal" hyperplane and A's tilted hyperplanes of constant-time, although you are right to say that A does appear elongated by a factor of γ in the latter. $\endgroup$
    – udrv
    Commented Aug 8, 2015 at 5:02
  • $\begingroup$ On a second thought, the apparent elongation factor of A in its own tilted constant-time hyperplane is not γ. Blunder, sorry. $\endgroup$
    – udrv
    Commented Aug 8, 2015 at 5:45
  • $\begingroup$ Funny thing to me, the video I did considers this exact problem and a way it could be tested in the real world. Watch here if you like: youtu.be/byIfbzcyjAc. Should be patient with longish intro for laymen, but the 2nd part is all about the Minkowski diagram. Sorry I can't accept the answer though. Note: posted wrong link the 1st time around. $\endgroup$
    – udrv
    Commented Aug 8, 2015 at 16:24
  • $\begingroup$ @udrv: Its x-coordinate is elongated by a factor of $\gamma$; the Euclidean distance should be $\gamma\sqrt{1 + \beta^2}.$ Basically, I'm saying "you're right, but we have to be really careful about how we interpret the geometry, especially if the object is deforming or changing color: because the changing object will actually appear rotated-and-stretched-larger by the Lorentz boost in the spacetime diagram, and it's only after you take a non-perpendicular cross section (seeing different colors across the object) that you overcorrect this from $\gamma$ down to $1/\gamma$. $\endgroup$
    – CR Drost
    Commented Aug 10, 2015 at 14:22

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