Terrell Revisited: The Invisibility of the Lorentz Contraction

[Ken G]

That looks very nice, might I suggest you use a 1X1 aspect instead of 4X1?
(ETA: I see you are one step ahead of me!)
That would make it easier to see two interesting things:
1) it looks like its "true" 1X1 square at a time when the object is actually at its closest approach, even though of course at that time its image has not reached closest approach,
2) it looks smaller than 1X1 by just the Lorentz factor when the image is directly across from the camera, i.e., when the image is at closest approach.
The first fact stems from Terrell's proof that the image will look like the stationary image in a camera that is in the object frame as that moving camera passes our stationary camera, and the second fact stems from our analysis in the early parts of the thread. Framed in this way, we can see that Terrell's claim that length contraction is "invisible" is merely the clam that when the image appears to be at closest approach, and we know we see the "true" length contraction by our analysis, the camera that moves with the object merely sees that same degree of horizontal contraction because it sees a rotated image, given that the camera is laterally displaced from being directly opposite the object, by exactly the distance the object moves during the time it takes the light to get to us.

So I now see this whole issue as a classic example of what often happens in relativity, that two observers agree on what is being observed, but they do not agree on why it is seen that way. We might imagine ourselves looking at a distant star that has what astronomers call a radial velocity toward us, and saying the motion of that star explains why the lines are blueshifted. Just then, an alien spacecraft on its way to that distant star might zoom past Earth, with zero difference between ours and its relative line-of-sight component of velocity toward that star. Under those circumstances, it would be natural for them to say the light is blueshifted because the velocity of the spacecraft is toward the star, as that is their destination. So we can all agree that would be a mundane example of two observers using different sounding language to say the same thing. Terrell is saying that we have the same thing with the moving cameras and our camera-- we have observers in the same place and time, seeing the same things, and using very different language to describe why that's what they see-- we say we are seeing length contraction when we are directly opposite the image, the observer in the object frame says they are not directly opposite the object, and neither were we when that image was taken, so that completely describes what both images show, and "length contraction" is just how we are attributing the source of that image. That's the sense to which it is "invisible," but we agree it is not strictly so, because both observers can agree that we'll see something different in a universe that does not have length contraction. It's just the difference between a "raw image" and an inference based on a raw image.

 

[JDoolin]

Finally, I am posting the formula for the case of a ray traced image of a line of rest length L moving moving at v in the +x direction, along the line y=1, with angles measured down from the horizontal (e.g. on a approach, and angle might be -π/6, on recession -5π/6). I let c=1. I use a parmater α between 0 and 1 to reflect positions along the line in its rest frame. The sighting point is the origin. Then, to describe the range of angles seen at some time T, you simply solve (for each α):

cot(θ) = v csc(θ) + vT + αL/γ

The T corresponding to the symmetrically placed image that shows the exact same angular span (but not internal details) as a stationary ruler of length L/γ centered on the Y axis is:

T = -(L/2γ + v csc(θ_L))/v

where cot(θ_L) = -L/2γ

[Edit: It is not too hard to verify (formally) that you have stretching whenever cot(θ) < 0, and compression whenever cot(θ) > 0. ]

Ah, the sighting point is from the back of the rod to the origin. I came very close to reproducing your first equation... But I oriented the rod in the wrong direction, so there is one sign change.

If the object were passing symmetrical to the observer, what would T be? It seems to me like all the co-secants, and gammas should cancel out so you get T=0.

 

[PAllen]

Ah, the sighting point is from the back of the rod to the origin. I came very close to reproducing your first equation... But I oriented the rod in the wrong direction, so there is one sign change.

View attachment 83373If the object were passing symmetrical to the observer, what would T be? It seems to me like all the co-secants, and gammas should cancel out so you get T=0.

The reception time T where the view is symmetrical must have the rod viewing angles symmetrical about y. For your convention of starting from the front of the rod, T would be very similar to mine. It would not be zero. You have to solve for T that produces symmetry.

So you would want θ such that cotangent = L/2 γ, then T= (1/v) ( (L/ 2 γ) - v csc θ). Plug this T in your equation and you see that you get the rod half to the right of y axis, half to the left. [As always, I take c=1]

 

[JDoolin]

I spent some time, this morning, trying to answer my own question (see thumbnail) about the value of T, but I found that my premise that \csc(\theta)=T_0 didn't hold in general.

This may not be an error in PAllen's post #49, but in my post #122 I have found I assumed that the time, T0, for the front of the rod to cross the viewpoint, and the time T0, for light to reach the observer from the front of the rod were both equal. I realize now, that would not be true in general.

If you have the opportunity, PAllen, could you post a more rigorous explanation of your premises, definitions, and equations from post #49? I thought I had correctly understood what all the variables represented, but it looks like I was in error.

 

[JDoolin]

Added a purple square in front and a blue square in back.

 

[JDoolin]

Had a bit of success I think today. The workhorse of this animation is the equation

Solve[\frac{x-x0}{v}==t_{now}-\sqrt{x^2-y^2-z^2},x]

This should have read:

Solve[\frac{x-x0}{v}==t_{now}-\sqrt{x^2+y^2+z^2},x]

...if anybody was wondering.

 

[PAllen]

I spent some time, this morning, trying to answer my own question (see thumbnail) about the value of T, but I found that my premise that \csc(\theta)=T_0 didn't hold in general.
View attachment 83429

This may not be an error in PAllen's post #49, but in my post #122 I have found I assumed that the time, T0, for the front of the rod to cross the viewpoint, and the time T0, for light to reach the observer from the front of the rod were both equal. I realize now, that would not be true in general.

If you have the opportunity, PAllen, could you post a more rigorous explanation of your premises, definitions, and equations from post #49? I thought I had correctly understood what all the variables represented, but it looks like I was in error.

I'm not sure when I'll have a chance to post the derivation, but I am not sure what is unclear about the variables. T, for example, is just a reception time - a moment a 'picture' is taken by a camera at the origin. The source signals were emitted at all different (earlier) times, but that doesn't show up in my equations because it was not of interest. The angles you compute for alpha from 0 to 1 are just the viewing angle you would see for the corresponding point of the ruler, with alpha being defined per the ruler rest frame. I don't compute anything about apparent distance (e.g. parallax), just image angles present at T for e.g. a pinhole camera (or single eye).

Note that the equation can be cast as quadratic equation in cotangent, since csc = √ (1+cot²), for non-numeric solution. You have to be careful which root to pick, of course.

 

[PAllen]

Here is an outline of the derivation of the main formula in #49:

The congruence of world lines describing the moving ruler is:

x = vt + αL/γ
y=1

with α defining a particular 'element' along the ruler. The the set of all events reached by light emitted from any event in the ruler congruence is simply:

x = vt + αL/γ + (T -t) cos(θ)
y= 1 + (T-t) sin(θ)

where T is a possible detection time obviously > t. We simply want to know, for a chosen T, what are all the possible ways, from all earlier events on the ruler, that can reach (0,0). By my convention, sighting angle and the emission angle (the θ in the formula above) are both measured relative to the x axis, they are the same. The sighting angle is simply measured up from the x axis, while the emission angle is measured down. So, we want to solve for all possible angles for a chosen T that satisfy:

0 = vt + αL/γ + (T -t) cos(θ)
and
0 = 1 + (T-t) sin(θ)

Solving the second equation for t allows its elimination from the first, and then algebra leads to the equation I gave in #49.

To arrive at the statement I made about compression versus stretching, note that for a stationary ruler, the deriviative of cotangent by α is L/γ. A change in contangent at a greater rate than this would be perceived as stretching, while a lesser rate would be seen as compression. Then, evaluating these derivatives for the moving case, you can demonstrate that whenever the cotangent < 0, the derivative by α is > L/γ, and whenever it is > 0, the derivative is < L/γ. Thus, stretching on approach, compression on recession.

 

[JDoolin]

0 = vt + αL/γ + (T -t) cos(θ)
and
0 = 1 + (T-t) sin(θ)

Solving the second equation for t allows its elimination from the first, and then algebra leads to the equation I gave in #49.

cot(θ) = v csc(θ) + vT + αL/γ

Thanks, that is much more clear, but I think you may have left out a detail, or I'm missing something. I solved the first equation for

\cos \theta = \frac{-v t - \frac{\alpha L}{\gamma} }{T-t}

and the second equation for

\sin \theta = -\frac{1 }{T-t}

I divided them and I get

\cot \theta = v t + \frac{\alpha L}{\gamma}

So I'm missing the v \csc \theta term.

 

[PAllen]

You're missing that t is useless to have as the variable. You need the angles corresponding to a given T (reception = where the light is). You have to solve for T. You have emission events of all different t arriving at an eye or camera at some given T (and you have no interest in what those varying t values are for describing the image).

 

[JDoolin]

The varying values of emission event times, t, arriving at an eye or camera at some given T are

t(x,y,z) =T - \sqrt{x^2+y^2+z^2}

You can also solve x = v t + \frac{\alpha L}{\gamma} for t.

Set the two values of t equal and solve for x, which would give you the location of the object, the emission event, and the image observed at time T, if you already know y and z.

I was thinking that you were setting the value of t as a constant, and attempting to evaluate a range of values for T. Were you actually using a single value of T and finding a range of possible t's and θ's?

I'm trying to see if the two different methods are compatible.

 

[PAllen]

I was trying to get a single equation that showed a complete image of the ruler given reception time at origin. That is pick a reception time, solve for theta for each alpha, and you have the complete image of the ruler, including stretching and compression. I have no interest in when the light I see at a given viewing angle was emitted. I wanted and got emission time to disappear from the equation. I could reproduce all the effects for a rod in that website's visualization, as well as discovering that the distribution pattern of viewing angles for alpha was consistent with rotation.

Answering specific questions:

1) Your formula for emission time in terms of T and (x,y,z) is, of course fine.

2) I was picking a particular T and solving for θ for each α. I had no interest in t whatsoever. I wanted it gone, because a camera doesn't know or care when the light it receives was emitted. It only cares about angle of reception (= angle of emission by my conventions).

 

[Nick666]

I'm trying to wrap my head around this... so the shadow of a disk or sphere moving at relativistic speeds is or isn't a circle ?

In other words, if we have a timer, on a railroad line, that records for how long it doesn't detect sunlight photons, and a 299792458 meters long train moving at almost c, the timer will record for a fraction of a second ?

 

[PAllen]

I'm trying to wrap my head around this... so the shadow of a disk or sphere moving at relativistic speeds is or isn't a circle ?

In other words, if we have a timer, on a railroad line, that records for how long it doesn't detect sunlight photons, and a 299792458 meters long train moving at almost c, the timer will record for a fraction of a second ?

A shadow, formed by plane wave pulse, will show an oval for a moving sphere. Its largest diameter will match its rest diameter, while its shortest will shorter by the factor gamma. A disk moving parallel to the plane wave front will have the same shadow as a corresponding sphere. Your experiment will also detect expected length contraction.

However, a visual image formed by an eye or a camera will always show a sphere to be of normal size and shape. However, if will appear rotated in the sense that what features you see on the sphere will correspond to what you expect for a different angle than your momentary light of sight to the sphere. A visual image of a (relatively small) disk will appear rotated, and this will be consistent with both shape change and pattern observed on the disk. Specifically, when it visually appears to be at closest approach angle, you will see an oval contracted cross section (with the smallest diameter consistent with gamma), but the the patterning on the disk will be consistent with rotation as the cause of this oval cross section.

 

[jartsa]

Let's say we have a long static ruler, and one static spaceship at ruler position 0 km, and another static spaceship at ruler position 1 km.

Now both spaceships start to accelerate along the ruler, in such way that their proper distance stays constant. As we know, the trailing spaceship must accelerate harder. Both spaceships will send a radio message every time they pass a mark on the ruler.

When velocity is 0.9 c, the spaceships stop accelerating.

An observer standing at ruler position 1000000 km is listening to the radio messages. He notes that the trailing spaceship sends messages at faster rate during the acceleration phase. Apparently the trailing spaceship is closing on the leading spaceship. If the observer used his eyes, he would see the positions of the spaceships to be in agreement with the radio messages.

(I ignored the fact that the leading spaceship appears to start accelerating earlier than the trailing spaceship, because it also appears to stop accelerating earlier.)

What happens when the leading spaceship passes the observer? The rate of messages from the leading spaceship decreases, because the messages become red shifted. So the apparent distance between the ships shrinks even more.

What happens if the observer is 10 km away from the ruler? When the spaceships are far away, the 10 km does not matter. When the spaceships are far away and approaching, the apparent distance between the spaceships is shrunken. When the spaceships are far away and receding, the apparent distance between the spaceships is shrunken even more.

And when the spaceships are closest to the observer, the apparent distance between the spaceships is changing from shrunken to even more shrunken, that seem logical to me.

 

[PAllen]

Jartsa,

I don't have time to untangle your whole post, but: if two spaceships start accelerating from rest in a given frame such that their mutual distance is constant per each spaceship (per normal conventions), then, per the starting frame:

- they start accelerating at the same time in the starting frame, it is just that the leading one will have lower rate of acceleration
- the leading ship stops accelerating later - it continues until reaching the stopping speed for the trailing ship. At this point, the distance between them will stop shrinking

 

[PAllen]

A shadow, formed by plane wave pulse, will show an oval for a moving sphere. Its largest diameter will match its rest diameter, while its shortest will shorter by the factor gamma. A disk moving parallel to the plane wave front will have the same shadow as a corresponding sphere. Your experiment will also detect expected length contraction.

The above, is incorrect, as to a sphere. For reasons wholly unrelated to the imaging effects covered by the Terrell/Penrose analysis, a rapidly moving sphere wil not cast an oval shadow, irrespective of whether the light source is continuous lasers or a plane wave pulse. The problem is that any light path reaching, e.g. film at some time T from the trailing edge of the sphere would have had to be inside the sphere a moment before. I have not analyzed what the shadow shape would be in detail, but it would certainly be larger than the contracted sphere diameter. Ken G. made this point much earlier, and I erroneously disputed it.

 

[JDoolin]

Thanks, that is much more clear, but I think you may have left out a detail, or I'm missing something. I solved the first equation for

\cos \theta = \frac{-v t - \frac{\alpha L}{\gamma} }{T-t}

and the second equation for

\sin \theta = -\frac{1 }{T-t}

I divided them and I get

\cot \theta = v t + \frac{\alpha L}{\gamma}

So I'm missing the v \csc \theta term.

Answer: t=T+csc θ

I was using the wrong "Tee"

\cot \theta = v t + \frac{\alpha L}{\gamma}

\cot \theta =v(T+\csc \theta)+\frac{\alpha L}{\gamma}Both equations are equally valid, but the second equation is more "useful" because you can be given observation time T, eliminate t, and find the relationship between θ, and α.
I think there's something weird going on with the angles--if T is necessarily greater than t, then you're using angles between (-180,0) or between (180,360)

Here's something I notice, trying to make the two compatible

If I draw a right triangle with angle theta, Adjacent side, x and Opposite side, 1, then \sin \theta = \frac{1}{\sqrt{x^2+1}}, and so

t=T-\csc(\theta)=T-\sqrt{x^2+1^2+0^2}

 

[PAllen]

I'm using angles between 0 and -pi (-180). That was a deliberate choice on my part. More precisely, the other angles reflect light emitted upwards, which simply don't figure in the solution for light getting from (x,y)=(x,1) to (0,0).

 

[jartsa]

Jartsa,

I don't have time to untangle your whole post, but: if two spaceships start accelerating from rest in a given frame such that their mutual distance is constant per each spaceship (per normal conventions), then, per the starting frame:

- they start accelerating at the same time in the starting frame, it is just that the leading one will have lower rate of acceleration
- the leading ship stops accelerating later - it continues until reaching the stopping speed for the trailing ship. At this point, the distance between them will stop shrinking

Well those things I know. I was talking about apparent this and apparent that, by which I meant what the observer sees with his eyes.

But I noticed that the general opinion here seems to be that an approaching rod appears to be contracted, and when the rod starts to recede, it contracts more. So my heuristic argument is not needed.

 

[PAllen]

Well those things I know. I was talking about apparent this and apparent that, by which I meant what the observer sees.

But I noticed that the general opinion here seems to be that an approaching rod appears to be contracted, and when the rod starts to recede, it contracts more. So my heuristic argument is not needed.

Visually, an approaching rod appears expanded.

 

[Ken G]

We know a couple things from the Terrell analysis, for rods that take up a small solid angle (and can appear distorted in the way PAllen described above, where one side looks closer to us so has expanded looking tickmarks which I erroneously disputed, but here I'm just talking about the appearance of the full length of the rod). The key point is that the observer who sees the rod in motion can never see anything in an instantaneous image that some observer moving with the rod cannot also see in such an image, the only question is which moving observer is the right one to use. So PAllen's analysis has quantified which observer to use, but it turns out to be the one that is in the same place and time as the stationary observer! But more to the point here, the stationary observer will see a series of images that correspond to comoving observers' views that start out being ahead of the rod, then are straight across from the rod, then are behind the rod. The one thing I'm not clear about is that Baez says there can be distortions in the overall scale when you compare those images, so I'm not clear if that will mess up the argument, but assuming there is indeed a comoving observer that sees the same thing, then we can tell the stationary observer will see a rod getting both closer and less foreshortened at first, will see the maximum length when it looks like an unforeshortened rod at its closest distance (but that will come before the image reaches closest approach), with the image at closest approach being foreshortened by the Lorentz factor.

 

[JDoolin]

Here's an animation I made of "what a person would see" of three square elements of dots moving at 0.866c.

It uses 25 red+20 blue +20 purple worldlines of objects following x=.866 t + \frac{x_0}{\gamma}

The object's "comoving" length is half the distance between the green fence-posts.

To me, it looks very clear in the animation that

(1) Lorentz Contraction is clearly visible when the view passes the middle sphere, as it's apparent length is about 1/4 of the distance between the fence-posts
(2) the red plane of dots appears stretched greatly as it is oncoming. In the first few frames of the animation the red dots cover a distance of several fence-posts. It is hard to tell exactly how many, but it's a lot.
(3) The view of the three planes does NOT appear like it would if the cube were simply rotated. Rather, the front and back face are approximately the shape that one would expect to see from a stationary cube, but always tilted back in a hyperbolic shape.I feel confident that PAllen's equation t = T + csc \theta and my equation: t=T-\sqrt{x^2+y^2+z^2} are derived from exactly the same ideas. So we're not disagreeing on the math.

So when Ken says:

The key point is that the observer who sees the rod in motion can never see anything in an instantaneous image that some observer moving with the rod cannot also see in such an image, the only question is which moving observer is the right one to use.

I'm trying to figure out how I can agree with him, because my first impression is that there is practically NOTHING in that animation that looks "the same" as it would in the perspective of a comoving viewer. I would say instead "the observer who sees the rod in motion can never see any event in an instantaneous image that some observer moving with the rod cannot also see in such an image"

However, there is major disagreement on the positions and time where and when those events occurred. The differences in positions make it so that the shape appears warped. And that apparent difference could NOT be mistaken for ordinary rotation--at least not in the case of a cubic object.

As far as spheres and disks go, though, they have particular symmetries that may come into play. I look forward to making another set of animations with the dots in spherical and disk patterns to see. I don't know the mathematical elegance with which Roger Penrose showed that the sphere stayed looking like a sphere, but I will brute force it with software.

I would have to do quite a lot more work to figure out how a shadow would appear--I think you'd have to take into account the position the velocity of the light-source would have... But it would create a more complicated problem than I want to think about right now. I'd rather focus on the light that is coming directly off the moving body toward the observer.

 

[JDoolin]

And that apparent difference could NOT be mistaken for ordinary rotation--at least not in the case of a cubic object.

Well, I shouldn't say it couldn't be mistaken for ordinary rotation... It's going by pretty quick, and you might not have the presence of mind to measure the length of the parallel side, and watching it shrink against the fence-posts as it goes by... It could be mistaken for ordinary rotation, but I don't understand why you would stress the fact that you COULD mistake it for ordinary rotation when you should be stressing the DIFFERENCES which clearly indicate that it is different from ordinary rotation.

 

[PAllen]

I think there is something interesting and useful about the Terrell/Penrose rotation idea, as long as it is not mis-represented.

1) It can provide a computational shortcut such that you never have to explicitly worry about light delays. Compute a stationary image, and apply SR aberration to each of its image angles. This is an exact procedure, without limit to size or shape, and will also handle surface features. To me, this result is computationally and conceptually interesting.

2) It immediately follows from the above that what might be called the 'stationary basis' image for a given viewing angle of a moving body corresponds to a different viewing angle. This gives rise to a first order visual effect that looks like rotation.

3) But since the rotation analog is only first order accurate, the larger the object the more distortion there is from pure rotation.

4) I think we all have consensus (including Penrose) that introducing other reference points in the scene (fence posts, tracks, etc.) makes rotation rather than contraction an untenable visual interpretation, but if all you saw was a movie of an isolated moving object without any knowledge of context, and it subtended small angle at closest approach, it would appear very close to an image of the rest object that was rotating as it moved.

 

[Ken G]

Yes, I see that as a good summary of the situation. What's ironic is that if we take the speed of light to infinity, or slow down the object, what we see becomes more familiar, and fits better with Galilean relativity. But in that limit, Galilean relativity and Lorentzian relativity don't make different predictions, and both would look weird at speeds approaching that of light (JDoolin-- feel like doing a contrasting picture under Galilean relativity?) so we really have no right to say that one is more familiar than the other. What's more, even if we take c to infinity so both look "normal", we still see something that looks like a rotation-- it's just that the right relativistic answer looks like the rotation is happening in a strange way that seems impossible. But wouldn't Galilean relativity also create a rotation that looked impossible, just due to light travel-time effects? So I'm not sure we can claim that weird looking things tell us we are seeing relativistic effects.

 

[PAllen]

Yes, I see that as a good summary of the situation. What's ironic is that if we take the speed of light to infinity, or slow down the object, what we see becomes more familiar, and fits better with Galilean relativity. But in that limit, Galilean relativity and Lorentzian relativity don't make different predictions, and both would look weird at speeds approaching that of light (JDoolin-- feel like doing a contrasting picture under Galilean relativity?) so we really have no right to say that one is more familiar than the other. What's more, even if we take c to infinity so both look "normal", we still see something that looks like a rotation-- it's just that the right relativistic answer looks like the rotation is happening in a strange way that seems impossible. But wouldn't Galilean relativity also create a rotation that looked impossible, just due to light travel-time effects? So I'm not sure we can claim that weird looking things tell us we are seeing relativistic effects.

I think my post #118 is of interest here.

 

[Ken G]

I didn't quite follow the anti-boost idea, it sounds like you are trying to take advantage of the conformal properties of the Lorentz mapping between observers to generalize what would be seen under non-conformal conditions like the Galilean transformation. What seems like a possibility is that Galilean relativity would look even weirder at speeds close to that of light, putting us in the ironic position of being able to tell that our universe length-contracts from how much less distorted that makes fast-moving objects appear.

 

[PAllen]

I think you are missing some of my points. As I see it, Lorentz versus Galilean transform transform is irrelevant for analyzing imaging in one frame accounting for light speed. All that matters is coordinates description of the moving object. Where a transform would matter is if you go from a frame where light speed is assumed isotropic to another frame. With Galilean, you would have to accept anisotropy in such other frame. What I am doing, though, is simply assuming we are in the preferred frame where light speed is isotropic and c, by fiat. Then, per Galilean relativity, we assume that the coordinate description of a body never changes from its rest description. Then ask, how would such an object look? This is indistinguishable from all the direct signal display computations we have been discussing except we unilaterally use the rest description of the object when it is moving. Irrespective of this, the Terrell-Penrose work establishes how to compute images of an object in motion from images of the object at rest assuming the respective coordinate descriptions are related by Lorentz transform. So given, by fiat, the rest frame object description being used for the moving object, to apply Terrell-Penrose, all we need is a rest frame description that would Lorentz transform into that. That is what I am calling an anti-boost. Please try to think about this more. I am certain my method is correct for my stated assumptions, and would answer precisely the question of how an object would look if there were no length contraction (but all else held the same, e.g. isotropic light speed).

 

[Ken G]

It sounds like you are looking for a simpler way to get what it would look like in a Galilean universe than just doing the time-of-flight ray tracing. Either way, we say the stationary camera is in the isotropic-c frame, and then the time-of-flight effects are the same in either the Galilean or Lorentzian universe. The sole difference will then be the absence of length contraction. So if G is what it looks like in the Galilean universe, and L is what it looks like in the Lorentian universe, and A is the anti-boost that removes length contraction, then you are saying G = AL, where L is easy because we can use a comoving camera. That all sounds correct, though since JDoolin already has a ray-tracing calculation, he can probably just go in and get the Galilen result G by removing the appropriate expressions that trace to the Lorentz transformation instead of Galilean. But you are saying that we can picture what it's going to end up looking like by taking what he has now, and apply the anti-boost, which is just a constant stretching in the longitudinal direction, prior to rotation.

That sounds right, so it means that instead of just seeing a rotating cube, we will see a rotating stretched-cube. If so, I think we can argue that would indeed look more strange-- so we have the odd result that we can tell we are in a Lorentzian universe by the fact that the speedy cube looks less weird than it would in a Galilean universe. Penrose's point that the wheels wouldn't look right against the rails is also true in the Galilean case, so you could tell it's not actually a rotation, but the only reason you would even be tempted to imagine it was a rotation is that it otherwise does look like a rotation in the Lorentzian case, where it is more obviously not a rotation in the Galilean case, so you wouldn't even have to ask the question in a universe like that. Non-contraction in a Galilean universe would be more obvious than length contraction is in a Lorentzian universe!

(By the way, note that wheels would be especially weird, because the point where the wheel touches the rail is presumably not moving instantaneously, so it would not be rotated-- actual wheels would look tortuously twisted, but the point where they meet the rail would indeed line up properly. Relativistic wheels not only look weird, they would be experiencing significant internal stresses due to the acceleration. It sounds like a Born-rigid wheel is an interesting problem in its own right, but that's fodder for a different thread.)