ghwellsjr said:
I have looked up the papers that were cited in the wikipedia article on the Terrell Effect along with the original paper you linked to.
If you look at the footnote on the second page of Terrell's paper concerning the paper that Roger Penrose wrote at about the same time, you will see the clues that explain what's going on here. Penrose proved that a moving spherical object will always appear as a sphere rather than an ellipsoid, that is, the outline of the image will be a circle. If there are any surface features on the sphere, they can be distorted. A sphere is the one object that always looks the same no matter how it is rotated.
Terrell, on the other hand, is discussing a different situation. He is talking about the image of an object that is far away so that it "subtends a very small angle" to the eye or camera. Although he states this in his paper, the way he states some of his conclusions makes it sound like he is talking about the more general situation, that is, those that apply for a sphere.
As a result, the wikipedia article is quite misleading in that it combines the two different effects that Penrose and Terrell separately identified as if they were the same issue. It could be vastly improved if this distinction were made clear instead of leaving it up to the reader to look up the papers in its bibliography to figure this out.
I can't envision any situation where the length of the angle subtended would matter. The objects approaching will look elongated, and the objects receding will look contracted. There is only one angle, (θ'=arccos(v/c), if Terrell's paper is in any way correct) where an approaching obect would appear to be at its normal uncontracted length.
So there aren't really any mistakes in Terrell's paper and it has been extensively debated to point out the confusion surrounding it. To tell you the truth, I don't understand his paper or your analyses and I have no reason to disagree with either of you. I think you are doing some marvelous work in this area and I'm even more impressed with the animations that you are producing now than I was with your first one.
Well, Terrell's paper is just amiguous, circular, and strange. He says at one point "a linear object which was oriented in the θ=0 angle at the earlier time when light left it, will appear contracted by the rotation just to the extent of the Lorentz contraction. This does not constitute proof of the visibility of the contraction, as this relation does not hold for other orietations, angles of observation, and shapes, and since the appearance of the object is normal at all time."
How can he claim that the object "will appear contracted" but "this does not constitute a proof of the visibility of the contraction." Secondly, shouldn't he leave out the phrase "since the appearance of the object is normal at all time," being as that is what he is trying to prove?
Furthermore for some reason he picks out angles where \cos(\theta - \theta') = \sqrt{1-v^2/c^2}, and doesn't do much of an analysis anywhere else.
You asked me:
First decide whether or not you agree with me that the apparent position of object is the intersection of the observation's* past light-cone and the world-lines making up the object.
Here's what's confusing to me about this:
1) I thought we were trying to determine the apparent visual shape of an object, not it's position, unless maybe you are talking about all the postitions on an object.
The apparent visual shape of the object consists of the locus of events that happened to the object which are currently arriving at the eye.
This is the intersection of two space-time shapes: (1) the locus of events which are currently arrivng at the eye; the past light-cone, and (2) the locus of events which have and will happen happening to the object; all of its particles world-lines.
2) Isn't a past light-cone just a subset of the complete history of the light coming to the eye?
The INTERNAL portion of the light-cone is just a complete history of th light coming to the eye, but the SURFACE of the light-cone is the set of events which is arriving at the tip of the light cone at a single instant.
Wouldn't it illustrate just what a single circular pattern of optical sensors would detect in a single plane?
The TIP of the lightcone represents a SINGLE optical sensor at a single instant of time. (Even smaller really since it's a geometric point, while a single sensor is extended.) You could treat it, for instance, as the pinhole of a pinhole camera.
I'm guessing you are using the term in a broader three-dimensional sense which cannot be illustrated on a 2D picture, correct?
I am leaving out the z-component, and referring to the shape of the circle in the xy plane as the object moves by in the x direction.
I assumed that the "length" and "width" to which you were referring were in those planes. However, my argument would not change (except for further complication) if you were referring to "length" and "width" as the two directions perpendicular to your line of sight.
The object will look elongated as it approaches and contracted as it recedes, and right in the middle it will have the standard "Lorentz contracted" size.
3) It seems to me that you have to do more than just establish an intersection of an object's world line with the light cone. Don't you have to perform a complicated algorithm to detect the first occurence of an intersection (looking back from the peak of the light-cone) to establish which surface of a three-dimensional object will be the one that is viewed?
Quite right. That would be what I mean by the "surface" of the light-cone. ;)
If you are doing this just for a two-dimensional object like you originally started with then this won't matter but now you are discussing other more complicated objects.
I think there should not be too much difficulty with the x and y-axis cross-section in the z=0 plane. For the full sphere, I may have to think further on cross-sections of the sphere in the z≠0 planes.
But getting back to your original animation, I would like to see you do that again with your new technique and show what it would look like from a reasonably close distance and then from a very far distance. I am now more inclined to agree that your original animation is a close approximation to what an observer would see,
Thank you. I will see whether I can get Mathematica to work. (I have not been inspired to overcome the "password expired" message that I receive when I try to run it on my laptop these days.)