Terrell Revisited: The Invisibility of the Lorentz Contraction

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JDoolin

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Terrell Revisited: "The Invisibility of the Lorentz Contraction"

I recently posted this response to a general question of how Special Relativity worked.
I'd say you are missing the vital third issue commonly known as the "relativity of simultaneity"

Imagine that you have an open-ceiling circular room ,filled with smoke (to reveal where a flash of light is), and walled with mirrors (to reflect the flash of light), and there is a bright flash of light that emits from the center, passes through the smoke in an expanding circle, bounces off walls (simultaneously), and arrives again simultaneously at the center.

From the point of view of someone hovering directly above the room, it appears that the light hits every part of the mirror simultaneously. However, to someone traveling past at 30% of the speed of light, it should appear as in the animation below.

attachment.gif


There are three main differences from the hovering viewpoint and the .3c viewpoint:
(1) The light takes longer to make its outbound and return trip. (time dilation)
(2) The room no longer appears to be circular but slightly oval. (length contraction)
(3) The lignt no longer reaches all parts of the outer circle simultaneously, but instead hits the back end first. (relativity of simultaneity.)
G H Wells Jr pointed out that
You must have some magic smoke that doesn't attenuate the primary signal significantly but still provides a secondary reflection scattered in all directions but still bright enough to be detected far, far away.
and
Isn't it obvious that eventually the angles for viewing each cycle will distort the image so that it doesn't match your animation? Once you see that, you can also see that there is distortion within each cycle right down to the very first one. That's all I'm trying to point out.
Terrell's Argument
Both of these are important issues to take into account. However, they do not change the fact that "a meter stick in motion past the observer in such a way that it is moving parallel to its length, and is momentarily seen by its observer at its point of closest approach, will surely be seen as contracted."

This last quote is directly from an article by James Terrell, in in the 1959 Physical Review "Invisibility of Lorentz Contraction" , but he argues the opposite point: He claims that "the conformality of aberration ensures that, at least over small solid angles, each [co-located observer, regardless of relative motion] will see precisely what the other sees. No Lorentz contractions will be visible, and all objects will appear normal."

Is the Lorentz contraction "invisible" as Terrell claims, or has James Terrell made a mistake which has gone unnoticed for decades?

I'll take some time to analyze Terrell's argument (or lack thereof), and check whether my own methods (analyzing intersections of world-lines and light-cones) agree with his, (transformations of angles via an aberration equation) and if they don't agree, see if I can figure out why.

Terrell finds an aberration equation from the Lorentz transformations, then uses the aberration equation to conclude that the Lorentz contraction effect "vanishes," but I find it suspicious, when by using the Lorentz transformations directly, I find that the Lorentz contraction is quite visible. and when using diagrams from the aberration equation, I still find that the Lorentz contraction is visible.

In fact, even using the aberration equation, it is plain that the objects do not "all appear normal," as Terrell claimed. They definitely have different lengths depending on their relative position, and the relative speed of the observer.

Method Using the Aberration:
I found a nice diagram of the aberration equation here: http://www.mathpages.com/rr/s2-05/2-05.htm which helps me make my point without a lot of math.

attachment.png


It appears to me, that even using the aberration equations, the Lorentz Contraction is visible, and for certain, it cannot be said that "all objects will appear normal" as Terrell claims.

In the attached diagram you can see that the length of the ruler swept out by angle A is approximately 5.5 units when in the original frame, but it is contracted to 3.6 units when the observer is going 50% of the speed of light, and to 3.1 units when the observer is going 90% of the speed of light.

The length of ruler swept out by angle B is 2.8 units, but contracted to 2.1 units when the observer is going 50% of the speed of light. However, when you go 90% of the speed of light, yes, there is ONE angle where the apparent length of the ruler is equal to the original length.

But Terrell's statements seem to indicate that he believes the "objects will appear normal" regardless of the angle viewed, which is simply not true.

It appears that this error has indeed stood up to the test of time, because there is no hint in the Wikipedia article on terrell rotation that anyone has ever bothered to correct him. http://en.wikipedia.org/wiki/Penrose-Terrell_rotation.

(My Method)
The main axiom I'm using here is that the current apparent position of an object according to an observer, is the positional component (in the observer's current rest frame) of the intersection of the objects world-line (or curve) with the [surface of] the observer's current past light cone.

So the attached diagrams show first, a one-dimensional object passing in the y=0 plane. In this plane, the distortion is always present.

attachment.png


In the second diagram, a one-dimensional object is passing in the y=d plane. In the y=d plane, there is a small region where there is no significant distortion between the "actual" length-contracted shape, and the "apparent" shape.

attachment.png

Maybe I should clarify that I am talking about the SURFACE of the light cone. I think the point you're missing is that the surface of the past light-cone with its point at (0,0,0,0) is the locus of events which can be detected at the event (0,0,0,0). This fact remains the same, regardless of the observer's reference frame.

I hope you can understand that the observer can see the locus of events on the surface of his or her past light-cone. The tip of the light-cone is the event where the information arrives. It is not a remote event, but THE local event. Naturally, it is also the one event which does not move when you perform a lorentz transformation. All of the other events move, but any event which is in the surface of the past light-cone stays in the surface of the past light-cone.

(If you have any doubt on this, think, how could it be otherwise? How could light that is arriving at an event (0,0,0,0) in one reference frame be NOT arriving at the event (0,0,0,0); the same exact event, in another reference frame? Also you can check the before and after transformation coordinates in my spreadsheet files, to check that indeed t' = -sqrt(x'^2+y'^2).)



Consider a ruler lying in the y=1 plane and the z=0 plane. Consider marks on the ruler at points (-2.0, -1.9, -1.8, -1.7, ... 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6). (You can imagine the ruler going on forever if you prefer.

Assume your position is x=0,y=0,z=0, and the time is now t=0. (This experiment will take a long time to discuss, but essentially takes zero time to perform.) Assume also that the ruler is aligned with its zero mark at x=0 (with you).

Now, you are observing several "events" on the ruler. Namely, light bounced off or emitted from the ruler sometime in the past, and you are now seeing those events which happened in the past. You can calculate when those events happened by the formula:

[tex]t=-\frac{\sqrt{x^2+y^2}}{c}[/tex]

Now we consider another observer passing through the same location and time (0,0,0,0) but traveling at a speed of 0.8c. The two of you share past light-cones, so all of the events that you are observing, the other observer is observing at the same instant.

However, to find out where he is seeing these events, we must perform a lorentz transformation on each of them.

[tex]\begin{align*} t' &= \gamma t - \beta \gamma x\\ x'&=-\beta \gamma t + \gamma x \\ \end{align*}[/tex]




When this is done, in particular, the ruler marks (-1.5, -1.4, -1.3, -1.2, -1.1, -1.0) are mapped to new positions:

-0.0963, -0.0394, 0.0202, 0.0827, 0.1488, 0.2190

We are particularly interested in the markers -1.4 and -1.2, which now appear at positions -.0394 and .0827. The uncontracted length of the ruler is [tex](-1.2) - (-1.4) = 0.2[/tex], while the apparent length is [tex].0827 - (-.0934)=.1221[/tex] The length contraction factor is [tex].1221/.2=.6105[/tex]

Which is roughly* the same as that which is expected by the lorentz contraction factor [tex]\sqrt{1-0.8^2}= 0.6[/tex]

*If you wanted more fine detail, you should make more marks on the ruler around x=-1.33

I am attaching a couple of xls files, so you can see how I calculated things. File 1: https://www.physicsforums.com/attachment.php?attachmentid=37895&d=1312984992

File 2: https://www.physicsforums.com/attachment.php?attachmentid=37896&d=1312984992

Finally, (In direct conflict with Terrell's conclusiion) the region where to look, so that the effect of lorentz contraction is exactly what would be expected (a meter stick in motion past the observer in such a way that it is moving parallel to its length, and is momentarily seen by the observer at its point of closest approach" WILL BE SEEN AS CONTRACTED:

the effect [of speed-of-light delay-times] will at a minimum at a certain angle, but would never competely go away.

You can get it as exact as you want by using events around the point t'=-1, x'=0

[tex]
\begin{pmatrix}
c t'\\ x'
\end{pmatrix}
= \Lambda ^{-1}

\begin{pmatrix}
c t\\ x
\end{pmatrix}

[/tex]

[tex]
=
\begin{pmatrix}
\gamma & \beta \gamma \\
\beta \gamma & \gamma

\end{pmatrix}

\begin{pmatrix}
-1\\ 0 \end{pmatrix} = \begin{pmatrix} -\gamma\\ -\beta \gamma
\end{pmatrix}
[/tex]

Since in this case,
[tex]\begin{matrix} \beta = 0.8 \\ \gamma = \frac{1}{\sqrt{1-.8^2}}=\frac{5}{3}\\ \beta \gamma = \frac{4}{3} \end{matrix}[/tex]

You can use points on the ruler right around x=-4/3, labeled, for instance 1.333 and -1.334, (and the corresponding t values, which you can easily calculate, using [itex]- (ct)^2 = x^2 + y^2[/itex]
) .
 

Ken G

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I'm bumping this because I still don't see any correction in the Wiki article you cite, though the current wording is quite ambiguous as to just what the "Terrell effect" actually is. I completely agree with your analysis, indeed I regard it as obvious from symmetry principles that a straight thin rod moving along its path must appear length contracted by just the Lorentz factor when it passes the point of closest approach to the observer. This is because the sightline to the front and back of the ruler are arranged symmetrically around the impact parameter of closest approach, so there can be no simultaneity problems or illusions of any kind-- what you see is what you get in that situation, and we all know that special relativity forces you to reckon the rod as being length contracted. So we can conclude that if the "Terrell effect" is the statement that "length contraction is invisible," then this is wrong. If instead the effect is "length contraction gets mixed up with visual illusions that in some situations can produce a rotated appearance that can mask the length contraction", or that "disks cannot appear length contracted" (because they are rotationally symmetric), then it's fine. Since the Wiki seems unclear about all this, I concur with you that it still needs to be fixed.
 

A.T.

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I completely agree with your analysis, indeed I regard it as obvious from symmetry principles that a straight thin rod moving along its path must appear length contracted by just the Lorentz factor when it passes the point of closest approach to the observer. This is because the sightline to the front and back of the ruler are arranged symmetrically around the impact parameter of closest approach, so there can be no simultaneity problems or illusions of any kind-- what you see is what you get in that situation, and we all know that special relativity forces you to reckon the rod as being length contracted.
Yes, also see last paragraph and the videos here:
http://www.spacetimetravel.org/bewegung/bewegung3.html

So we can conclude that if the "Terrell effect" is the statement that "length contraction is invisible," then this is wrong. If instead the effect is "length contraction gets mixed up with visual illusions that in some situations can produce a rotated appearance that can mask the length contraction", or that "disks cannot appear length contracted" (because they are rotationally symmetric), then it's fine. Since the Wiki seems unclear about all this, I concur with you that it still needs to be fixed.
There are some videos of spheres here:
http://www.spacetimetravel.org/tompkins/node1.html
http://www.spacetimetravel.org/fussball/fussball.html
Unfortunately not at the closes approach, as far I can see.
 

Ken G

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What I don't understand is what is wrong with Terrell's argument-- it's a relatively sophisticated mathematical argument, and even has Penrose's name associated with it. Is the problem simply that the Terrell effect is not the same as the conclusion of his paper that length contraction cannot be seen?
 
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Two relevant links on Penrose-Terrell rotation:

http://en.wikipedia.org/wiki/Terrell_rotation

http://math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html

I regard it as obvious from symmetry principles that a straight thin rod moving along its path must appear length contracted by just the Lorentz factor when it passes the point of closest approach to the observer. This is because the sightline to the front and back of the ruler are arranged symmetrically around the impact parameter of closest approach, so there can be no simultaneity problems
This argument is not correct, because light does not travel instantaneously along the sight lines, so the fact that the sight lines are symmetrically oriented at one instant does not mean the paths the light actually travels will be symmetric. Describing what is going on as a "simultaneity problem" isn't really correct, though; the issue is simply that the light reaching your eye at a given instant was emitted by different parts of the object at different times. The Usenet Physics FAQ article I linked to above gives a good treatment of what is going on.
 
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to find out where he is seeing these events, we must perform a lorentz transformation on each of them.
No, this is not correct. This transformation tells you what the coordinates of the events will be in the other frame, but it does not tell you how the events will actually be observed--i.e., it does not tell you how the images seen by the two observers at a given instant will be related. To calculate what is observed by the two observers at the event (0, 0, 0, 0) where they are both co-located, you need to perform a conformal transformation on the null curves forming the past light cone that they both have in common. When you do this, you will find that the image seen by one observer is a rotated version of the image seen by the other, not a contracted version. The Usenet Physics FAQ article I linked to in my previous post describes how this works.
 
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I have up till now understood that Terrell "rotation" (is this the term you were referring to?) is the visual effect of seeing the back of a structure as you go past it, but because it is "projected" in front of you by aberration, it gives the visual illusion that it is turning inside out in front of you. There is an example of it at about 1 min 25 secs into this video. The video was actually produced by massaging the equations from the mathpages link from the OP in a raytracer.

BTW I find the quote "each [co-located observer, regardless of relative motion] will see precisely what the other sees" astonishing, surely Terrell didn't really say that?
 
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I find the quote "each [co-located observer, regardless of relative motion] will see precisely what the other sees" astonishing, surely Terrell didn't really say that?
In the idealized case described in the Usenet Physics FAQ link, every pixel in one observer's image has a corresponding pixel in the other observer's image; the only difference is the location of the pixels on the respective images--one is a distorted version of the other. So the same light rays are present in both images, just not in the same places.
 

Ken G

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This argument is not correct, because light does not travel instantaneously along the sight lines, so the fact that the sight lines are symmetrically oriented at one instant does not mean the paths the light actually travels will be symmetric. Describing what is going on as a "simultaneity problem" isn't really correct, though; the issue is simply that the light reaching your eye at a given instant was emitted by different parts of the object at different times. The Usenet Physics FAQ article I linked to above gives a good treatment of what is going on.
I'll give that a look, but I still cannot see what could be wrong with my argument. Imagine the rod slides along a string. As we watch the rod, there must be some instant where we see a rod at its closest approach to us, with the point of closest approach bisecting our image of the rod at that instant. That must occur, correct? The rod cannot appear to be rotated, because its ends would not appear along the string-- we could not see the string enter the rod ends. So we must be looking at a rod along that string, and we must reckon the length of the rod to be its usual Lorentz-contracted length, because we will always reckon that length once we correct for the finite speed of light. By definition, we are seeing light, at that instant, from the front and back ends of the rod at the same moment in our frame, and we are reckoning the time of flight of those light beams as the same (by symmetry), so we must see an image that looks like a length-contracted rod or else we would not be reckoning its length to be Lorentz contracted. In other words, the image of the rod we see at close approach must be exactly the length of the rod that we would reckon, given that image.

ETA: I've looked at the Baez description, and I find this statement telling: " Its apparent size will also change, but not its shape (to a first approximation)." That sure sounds to me like the rod will still look like a rod-- but will have its apparent size shortened. That means my argument could be consistent with what Baez is saying, and also means that length contraction is not "invisible." If my interpretation is correct, it means we should instead say "length contraction will not distort the images of the shapes of things."
 
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As we watch the rod, there must be some instant where we see a rod at its closest approach to us, with the point of closest approach bisecting our image of the rod at that instant. That must occur, correct?
No. That would only be true if light traveled instantaneously. Try actually assigning coordinates to events and calculating the null worldlines that the light rays follow. This is one of those cases where intuition is too easily misled; there's really no substitute for doing the math.
 
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In the idealized case described in the Usenet Physics FAQ link, every pixel in one observer's image has a corresponding pixel in the other observer's image; the only difference is the location of the pixels on the respective images--one is a distorted version of the other. So the same light rays are present in both images, just not in the same places.
OK, that makes perfect sense, I interpreted the comment very differently ;)
 

PAllen

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No. That would only be true if light traveled instantaneously. Try actually assigning coordinates to events and calculating the null worldlines that the light rays follow. This is one of those cases where intuition is too easily misled; there's really no substitute for doing the math.
the web site AT linked to, based on computer implementation of the math, clearly shows the a rod moving by perpendicular to your light of sight appears visually contracted; its shortest visual size, as it goes by, is as expected from the LT. If the has thickness, it looks more twisted than rotated.
 

Ken G

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No. That would only be true if light traveled instantaneously. Try actually assigning coordinates to events and calculating the null worldlines that the light rays follow. This is one of those cases where intuition is too easily misled; there's really no substitute for doing the math.
Yet this logical argument must be addressed:
1) we will always see a rod along the string. (This follows because the string is not moving, so cannot be distorted, and we must see it enter the ends of the rod.)
2) the rod will at some moment have the centroid of its image at the point of closest approach (this follows because at first the centroid of its image will not yet be at the point of closest approach, and later it will have its centroid past the point of closest approach, so there must be an instant where it is at the point of closest approach).
3) at that moment, we are seeing an image of a rod that straddles the point of closest approach. We can easily reckon the length of the rod by accounting for the time of flight of light from the symmetrically separated endpoints of the rod, and the time of flight from both endpoints is the same. Let's call that time of flight t. Then we can certainly say, in our coordinates, that a time t ago, the rod endpoints were at those two points, at the same time. Hence we can reckon the length of the rod a time t ago, and it must be the Lorentz contracted length (because we'll always get that after correcting for time of flight, but here that correction doesn't do anything).
4) hence, we must be seeing a rod that looks length contracted.
 
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the rod will at some moment have the centroid of its image at the point of closest approach
Meaning, the centroid of the image will be in the direction of the point of closest approach at some instant. Also, by "centroid" I assume you mean the point in the image that appears to be equidistant from both edges. Yes, this is true.

However, you are assuming that the light seen in the direction of the centroid of the image at this instant, was emitted from the point of the rod that is actually in the physical center of the rod. That is not true. Work backwards from the event of that light ray's reception to see why.

We can easily reckon the length of the rod by accounting for the time of flight of light from the symmetrically separated endpoints of the rod, and the time of flight from both endpoints is the same.
No, it is not. As above, the light ray in the centroid of the image at this instant does not come from the physical center of the rod. Once again, try actually assigning coordinates to events and calculating the null worldlines that the light rays follow.
 

Ken G

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Meaning, the centroid of the image will be in the direction of the point of closest approach at some instant.
Correct, the ray to the center of the image is the ray to the point of closest approach.
Also, by "centroid" I assume you mean the point in the image that appears to be equidistant from both edges.
Yes.
However, you are assuming that the light seen in the direction of the centroid of the image at this instant, was emitted from the point of the rod that is actually in the physical center of the rod.
Actually, I don't need to know if it was emitted by the center of the rod or not, because I am only finding the distance between the endpoints, at a moment when the endpoints straddle the point of closest approach. That is the only thing the logical train I just gave used, so you must defeat that logical argument based only on the things it used.
 
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I am only finding the distance between the endpoints, at a moment when the endpoints straddle the point of closest approach.
Ok, but then you can no longer assume that the image received by the observer one light-travel-time later will only contain light coming from between those two points in space. I don't think that will be true.

Call the two spatial locations at which the endpoints emit light at the instant in question (when the endpoint straddle the point of closest approach) F and R, for front and rear. Call the instant at which the light is emitted T0. The light travel time from both F and R to the observer will be the same; call that time T. What other light is arriving at the observer at time T1 = T0 + T, and what direction is it coming from?

It is true that no light can be arriving at time T1 from a point further in the "front" direction than point F, because the rod has not yet occupied any of those points in space at time T0, and the light travel time from any of those points will be longer than T, so any light emitted after time T0 by the front end of the rod can't have arrived at the observer by time T1.

However, I think there will be light arriving at time T1 from points further in the "rear" direction than point R, because the rod did occupy those points in space prior to time T0, and the light travel time is, again, longer than T, so light emitted prior to time T0 from points further to the "rear" than point R will still be arriving at the observer at time T1.

At this point I need to take my own advice and work out the math explicitly. :wink:
 

Ken G

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Ok, but then you can no longer assume that the image received by the observer one light-travel-time later will only contain light coming from between those two points in space. I don't think that will be true.
You are right I am simply assuming that, but I can change the problem to avoid needing to. Imagine having a red light at one end, and a blue light at the other. Then imagine not even having a rod between them-- we can just have those two light sources, moving at constant speed, and reckon the distance between them. If that distance appears to length contract in the scenario I describe, then we cannot rightly say that "length contraction is invisible."
 

PAllen

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However, I think there will be light arriving at time T1 from points further in the "rear" direction than point R, because the rod did occupy those points in space prior to time T0, and the light travel time is, again, longer than T, so light emitted prior to time T0 from points further to the "rear" than point R will still be arriving at the observer at time T1.
Consider each 'atom' of the rod to have a world line and 'be' a visible clock. Then, in SR, with no weird lensing, any observer will see a strictly sequential history for each 'atom clock'. No one will see multiple clock times for a given rod atom. Further, again without GR lensing, you see one clock time for every atom and they form a contiguous sequence between the R and F, at any given observational moment. I find, as suggested, that at the point when R and F are seen equidistant around the perpendicular between me and the line of motion, that the image center didn't come from the rod center. However, that simply visually produces distortion of the ruler lines. It has no bearing on the fact that the length that point shows full Lorentz contraction and the rod forms a continuous horizontal image (with linear stretching of the ruler marks - compressed toward the rear, and stretched toward the front).
 
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JDoolin

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Thanks for bumping this.

I've never worked out a closed-form solution mathematically, but I did use Mathematica some time back to produce a good animation. PeterDonis is right. You can't just do a Lorentz Transform of a couple of events, but what I had in mind was a LT of the whole set of events... and finding their intersection with the observers' past light-cone.

The animation on this page makes it show up better than in my diagram above.

http://www.spoonfedrelativity.com/pages/Is-Lorentz-Contraction-Invisible.php

The simplifying idea that I would offer is that you should take the intersection of the world-lines making up the passing object, and find where they intersect with the observers' past light cone.
 
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Ken G

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What's odd is that the Terrell paper does clearly assert that two cameras at the same point photographing the rod, even at the point of closest approach, would take the same picture, even if one camera is moving with the rod, and the other stationary. So there's something subtle going on here. I just don't see what could be wrong with my argument that we have two photons arriving from two directions, symmetrically straddling the closest approach, and they can be used to recreate an image of their two source points that must be length contracted.
 

JDoolin

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By the way, there's a question at the bottom of that page... ". At any given moment, draw two lines from the tip of the cone to both visible edges of the oval shape. Will the angle subtended by those two lines equal the height of the sphere? "

I never really answered that question for myself, in detail, (I eyed it as I watched the animation, but I never worked through the math) but it might be the subtlety that Ken G is looking for.
 

PAllen

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What's odd is that the Terrell paper does clearly assert that two cameras at the same point photographing the rod, even at the point of closest approach, would take the same picture, even if one camera is moving with the rod, and the other stationary. So there's something subtle going on here. I just don't see what could be wrong with my argument that we have two photons arriving from two directions, symmetrically straddling the closest approach, and they can be used to recreate an image of their two source points that must be length contracted.
I also basically agree with your argument. I think what is wrong with the argument you just quoted from Terrell is that two coinciding cameras in relative motion would, indeed, take the same picture as interpreted by some observer. That is any one observer would claim the two cameras captured the same angular width per that observer. What is not realized is that each camera in relative motion would then disagree, per themselves, what that angle was, for the simple reason that angles change under Lorentz transform.

To try to be more clear, two coinciding cameras certainly must capture the same light rays. This does not mean that each camera agrees on the angular span of those light rays.

If that is what Terrell actually claimed, it is simply wrong.

[edit: I think there is even more wrong with the statement attributed to Terrell (I have not checked what he actually said, as I don't know where to see his original paper). For two cameras in relative motion near c, the behavior of the identical waves (per any give observer) will be radically different through the optics, and paths to the film. There are many amusing analyses of how optical media in relative motion react very differently to the same waves.]
 
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A.T.

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so the fact that the sight lines are symmetrically oriented at one instant does not mean the paths the light actually travels will be symmetric.
I don't get this. At closest approach (of the rod's center) both ends of the rod have the same distance to the observer. So light emitted by them will reach the observer simultaneously, and give the visual impression of a contracted rod.
 

JDoolin

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This does not mean that each camera agrees on the angular span of those light rays.

If that is what Terrell actually claimed, it is simply wrong.
Agreed.

I think he is usually interpreted as saying "Lorentz Contraction is Invisible", because that is the verbal conclusion he comes to, and that is the title of the paper.

Since I'm sure that's a false conclusion, I figure, there is either a flaw in the math, or there is a flaw in claiming that the math leads to that conclusion.
 
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TThe animation on this page makes it show up better than in my diagram above.

http://www.spoonfedrelativity.com/pages/Is-Lorentz-Contraction-Invisible.php

The simplifying idea that I would offer is that you should take the intersection of the world-lines making up the passing object, and find where they intersect with the observers' past light cone.
Just for the record, here is a hard to find Wikipedia page with similar animations, and here is the page it was linked from.
 

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