Terrell Revisited: The Invisibility of the Lorentz Contraction

[Ken G]

Not at all. One side gets closer to you, the other side further away.

Not in the limit of infinitesmally small images, in that limit, a rotation will contract uniformly along the horizontal direction. It has to be a linear transformation.

The angle subtended by markings closer to you will be greater than those further away. The result I got for this effect precisely matches rotation, so I find it very hard to believe this is not what Terrell is referring to. Penrose also describes this effect.

I don't understand how you can get that it exactly matches rotation, does not the scale of the effect you describe depend on the ratio of how wide the cross is, to how far it is away? But that ratio doesn't appear in the analysis, it is a limit.

I don't necessarily think their results are that limited. A small object can still have markings on it.

Yes, but all transformations on those markings must be linear, so no gradients in what happens to them.

When what you call the stationary camera coincides and sees a symmetric cross, in that camera's frame its viewing is NOT head on at that event.

This is the scenario that is not clear to me-- I can't see if there will ever be a time when the stationary camera sees a symmetric cross. But for Terrell and Baez to be right, there must be such a time, and it must be when the moving camera directly opposite the cross coincides with the stationary camera.

Aberration will have change the incoming light angle such that the image appears to still be approaching, and the light delay stretching will compensate for the length contraction such that it produces a symmetric photograph.

When the moving camera is directly opposite the cross, we can certainly agree the cross will appear symmetric. You are explaining how that is reckoned in the frame of the stationary camera, but in any event we know it must be true.

The one case I can't quite resolve is that a rapidly approaching cross still far away will have the parallel (to motion) arm greatly stretched by light delay (by much more than length contraction can compensate, when it is far away). I can't find an all stationary analog of this case.

But are you also including the rotation effect, not just length contraction? Since none of the locations we could put the moving camera will ever see a horizontal arm that is wider than the vertical arm, it must hold that the stationary camera never sees that either.

But I think your point that the horizontal arm is stretched by time delay effects is the crucial reason that there is a moment when the cross looks symmetric to the stationary camera. I believe that moment will also be when the camera directly opposite from the cross passes the stationary camera. That is, it is the moment when the cross is actually at closest approach. A moment like that would make Baez right. Note that for any orientation of the moving camera, relative to the comoving cross, there is only one moment when the stationary camera needs to see the same thing-- the moment when the two cameras are coincident. If we imagine a whole string of moving cameras, then the stationary camera will always see what the moving camera sees that is at the same place as the stationary camera-- but at no other time do they need to see the same thing.

If so, this means it is very easy to tell what shape the stationary camera will photograph-- simply ask what the moving camera would see that is at the same place when the stationary camera takes its picture, and what the moving camera sees just depends on its location relative to the comoving object. You just have to un-contract the string of moving cameras, and measure their angle to the object, and that's the angle of rotation the stationary observer will see at that moment.

 

[PAllen]

Not in the limit of infinitesmally small images, in that limit, a rotation will contract uniformly along the horizontal direction. It has to be a linear transformation.

I disagree on how restrictive Terrell/Baez conclusion is. It may only be exact in some limit, but is good 'reasonably small'.

I don't understand how you can get that it matches rotation, does not the scale of the effect you describe depend on the ratio of how wide the cross is, to how far it is away? But that ratio doesn't appear in the analysis, it is a limit.

No, I disagree. Moving a tilted ruler further away linearly scales the image, but does not change the ratio of subtended angle at one end compared to the other for e.g. centimeter markings. Per my computation, the effect does match that produced by rotation. [edit: well, as long as you are not too close. Once you are far enough that further distance is linear shrinkage, just imagine the tilted ruler against a non-tilted ruler. The whole image scales, thus preserving the ratio of subtended angle between a closer inch and a further inch, on the tilted ruler] [edit 2: OK, I see that if you allow the angular span of a tilted ruler to go to zero with distance, the ratio angles subtended by ruler lines goes to 1. However, if you fix the angular span of a tilted ruler (e.g. 2 degrees), then distance doesn't matter and the ratio front and back ruler lines remains constant. This is what I was actually modeling when comparing to rotation - all angles. I remain convinced the the rotation model is quite accurate for small, finite spans, e.g several degrees.]

Yes, but all transformations on those markings must be linear, so no gradients in what happens to them.

I disagree. I claim the result goes beyond this.

This is the scenario that is not clear to me-- I can't see if there will ever be a time when the stationary camera sees a symmetric cross. But for Terrell and Baez to be right, there must be such a time, and it must be when the moving camera directly opposite the cross coincides with the stationary camera.
When the moving camera is directly opposite the cross, we can certainly agree the cross will appear symmetric. You are explaining how that is reckoned in the frame of the stationary camera, but of course we know it must be true.
I think this is the crucial issue-- it is that stretching that allows the cross to have a moment when it looks symmetric to the stationary camera, and I believe that moment will also be when the camera directly across from the cross passes the stationary camera. That would make Baez right.

I agree, and I thought that's what I was explaining in my last few posts.

Where I continue to disagree (with you, but I think agree with Terrell and Penrose) is that I think there is more to image rotation than you want to admit. Consider it from another angle, so to speak. Imagine a camera and cross stationary with respect to each other, but with the cross displaced from head on. One arm will subtend less angle than the other, and it would look (exactly) rotated relative to a colocated stationary cross turned head on to the camera. To be concrete, let us imagine the displacement is to the left. Now add a camera moving left to right, past this stationary camera. It will see a viewing angle (for the appropriate set up) of 'head on' due to aberration of viewing angle. The image seen by the stationary camera will have moved to a perpendicular viewing angle, but otherwise essentially unchanged. Thus it will see a rotated image in the head on viewing angle, with the rotation producing the contraction and explaining the distribution of ruler lines on this cross arm. Then, my final comment is that this is only one way to interpret the image. If you introduce another element that establishes rotation could not have occurred, you change your interpretation to contraction and distortion - that happens to match rotation.

 

[PAllen]

I will attempt another summary, similar to #49, that includes full understanding of Terrell/Penrose (I haven't looked as much at Baez) explaining my view that while accurate when properly understood, common statements of these results are inaccurate.

1) A common sense definition of 'seeing length contraction' means with knowledge of the object's rest characteristics. It is only relative to that there is any meaning to 'contraction'.

2) There are obvious ways to directly measure/see any changes in cross section implied by the coordinate description. Simply have the object pass very close to a sheet of film, moving along it (not towards or away) and have a bright flash from very far away so you get as close as you want to a plane wave. Then circles [and spheres] becoming ovals, and every other aspect of the coordinate description will be visible. Note that in a frame co-moving with the object, the plane wave reaching the film will be considered angled, and the exposure non-simultaneous. It is precisely because this method directly measures simultaneity across a surface in a given frame, that it directly detects the coordinate description of length contraction.

3) The impact of light delays on idealized camera image formation has nothing to do with SR. However it combines with SR in such a way that a common sense definition of 'see', length contraction is always visible (if it occurs, e.g. not for objects fully embedded in the plane perpendicular to their motion ). That is, if you establish what you would see from light delay under the assumption that the object didn't contract, and compare to what you would see given the contraction, they are different. You have thus seen (the effect of, and verified) length contraction.

4) To my mind, a correct description of the Terrell/Penrose result is that they have described a much more computationally elegant way (compared to ray tracing) to arrive at the image detected by any idealized camera, that allows one to qualitatively arrive at a result with often no computation at all.

A) Instead of ray tracing based on world tube representation of the object, simply represent the image in terms of angles for a camera at rest with respect to the object at the detection event of interest. Then apply aberration to get the angular displacement of all detected rays in a camera moving in any way at this same event. This method is completely general and exact, up to having a frame in which you can ignore the object's motion (e.g. for a swirling gas cloud where you care about the details, there is no small collection of rest frames you can use). Given the static nature of the analysis before applying aberration, this is a huge simplification.

B) For objects of smallish size (not just infinitesimal objects; size defined by subtended angle), the result of (A) is (to good approximation) to shift the stationary (with respect to object) camera image to a different viewing position (with some scaling as well). This implies apparent visual rotation in a substantive sense. Viewing a sphere with continents on it, from a moving camera, the apparent hemisphere seen will correspond to a different viewing angle the one you are sighting along. The markings on a rod will appear distorted (relative to what is expected for the viewing angle of the moving camera) as if rotated by the change in viewing angle between the stationary and moving cameras. All of these results can be had, much more laboriously, by direct ray tracing in the frame of moving camera, with the object properly represented as a world tube.

C) Summarizing A) and B) as "invisibility of length contraction is physically absurd", not just because of the logical point made in (3), but also because if additional elements are introduced into the visual scene that are stationary with respect to the camera considered moving in the 4)A) analysis, you will see that the apparent rotation of the image of the moving object is illusory, and must be replaced by an alternate interpretation of the same image - that 'actual' contraction plus light delay is the only interpretation consistent with the whole scene.

 

[Mentz114]

I will attempt another summary, similar to #49, that includes full understanding of Terrell/Penrose (I haven't looked as much at Baez) explaining my view that while accurate when properly understood, common statements of these results are inaccurate.

1) A common sense definition of 'seeing length contraction' means with knowledge of the object's rest characteristics. It is only relative to that there is any meaning to 'contraction'.
...
...
C) Summarizing A) and B) as "invisibility of length contraction is physically absurd", not just because of the logical point made in (3), but also because if additional elements are introduced into the visual scene that are stationary with respect to the camera considered moving in the 4)A) analysis, you will see that the apparent rotation of the image of the moving object is illusory, and must be replaced by an alternate interpretation of the same image - that 'actual' contraction plus light delay is the only interpretation consistent with the whole scene.

This seems well argued but I have always had a problem with 'actual contraction'. If you mean what I think you mean, I don't see how the object can have a different 'actual contraction' for different observers. I understand that different observers might 'measure' a contracted length, but it is a frame dependent measurement. In its rest frame the object does not experience contraction.

 

[Ken G]

[edit 2: OK, I see that if you allow the angular span of a tilted ruler to go to zero with distance, the ratio angles subtended by ruler lines goes to 1. However, if you fix the angular span of a tilted ruler (e.g. 2 degrees), then distance doesn't matter and the ratio front and back ruler lines remains constant. This is what I was actually modeling when comparing to rotation - all angles. I remain convinced the the rotation model is quite accurate for small, finite spans, e.g several degrees.]

This is the crux of the matter, it is what I find confusing about the language relating to "rotation." A rotation looks different at different angular sizes, because of how it makes some parts get closer, and other parts farther away. Is that being included, or just the first-order foreshortening? And what angular scales count as "sufficiently small"? Baez said:
"Well-known facts from complex analysis now tell us two things. First, circles go to circles under the pixel mapping, so a sphere will always photograph as a sphere. Second, shapes of objects are preserved in the infinitesimally small limit."

I interpreted that to mean the shapes are only preserved in the infinitesmally small limit, i.e., for the Lorentz contracted cross to look like a rotated cross, it has to be infinitesmally small, so this would not include how the forward tilted arm can look longer than the backward tilted arm on a large enough angular scale. You are saying I am overinterpreting Baez here, and what's more, your own investigation shows a connection between that longer forward arm, and what Lorentzian relativity actually does. So perhaps Baez missed that, or did not mean to imply what I thought he implied.

This is what Terrell says in his abstract:
"if the apparent directions of objects are plotted as points on a sphere surrounding the observer, the Lorentz transformation corresponds to a conformal transformation on the surface of this sphere. Thus, for sufficiently small subtended solid angle, an object will appear-- optically-- the same shape to all observers."

The answer must lie in the meaning of having a conformal transformation on the sphere of apparent directions. If we use JDoolin's asterisk, instead of a cross, we can see that a rotation will foreshorten the angles of the diagonal arms, and it is clear that a conformal transformation will keep those angles fixed, so certainly Terrell is saying that the Lorentz contraction will foreshorten the angles in exactly the same way. But what about the contrast in the apparent lengths of the arms tilted toward us and the arms tilted away, is that contrast also preserved in the conformal transformation? You are saying that it does, and that seems to be the key issue. We do have one more clue from Terrell's abstract:
"Observers photographing the meter stick simultaneously from the same position will obtain precisely the same picture, except for a change in scale given by the Doppler shift ratio"
So the word "precisely" says a lot, but what is meant by this change in scale, and is that change in scale uniform or only locally determined? You are saying that it looks precisely like a rotation, including the contrast between the fore and aft distortions, not just the first-order foreshortening effect.

A sphere with continents on it might be a good case to answer this. We all agree the sphere still looks like a sphere, and in some sense it looks rotated because we see different continents than we might have expected. But the key question that remains open is, do the continents in the apparent forward regions of the sphere appear larger than the continents in the most distant parts of the sphere, or is that element not preserved in the conformal transformation between the moving and stationary cameras? I agree Terrell's key result is essentially that it is easier to predict what you will see for small shapes by using the comoving camera at the same place and time as the stationary camera, but what we are wondering about is over what angular scale, and what types of detail, we should expect the two photos to agree on. The mapping between the two cameras is conformal, but it is not the identity mapping, so can we conclude the continents will look the same size in both photos? Certainly distortions on the surfaces of large spheres should look different between the two photos, but even large spheres will still look like spheres.

 

[PAllen]

This seems well argued but I have always had a problem with 'actual contraction'. If you mean what I think you mean, I don't see how the object can have a different 'actual contraction' for different observers. I understand that different observers might 'measure' a contracted length, but it is a frame dependent measurement. In its rest frame the object does not experience contraction.

Substitute "actual contraction per some camera's frame of reference", if you prefer. The contraction is actual up to inclusion of interaction fields, e.g. an EM model of an object moving in some frame will have the EM field represented such that equilibrium distances of moving charges will be closer than modeled in a frame where the charges are not moving.

 

[JDoolin]

Could it be the difference between first-order-small effects that don't show any difference, and larger-solid-angle pictures where you start to see the distortions? Terrell only ever claimed that small shapes appeared the same, larger images require some type of cobbling together that might involve bringing in "non-literal" information, analogous to how all local frames in GR are Minkowski but the equivalence principle breaks down on larger scales.

Actually, I'm starting to think maybe the game designers rendered some of the objects in the game with the full aberration effect, and other objects in the game without it.

Here are four screen-captures from the promotional video at http://gamelab.mit.edu/games/a-slower-speed-of-light/

This is a very short part of the promotional video but it captures several things. For instance the distance between the two poles increases when the observer moves to the right, and it shrinks when the observer moves to the left (so long as the poles are on the right-side of the observer's view.

Looking at the warping of this one structure in the game, it seems like they attempted to get the shapes right. The circles on the ground don't look quite circular, but rather they look flattened ovals, as I think they should.

 

[PAllen]

This is the crux of the matter, it is what I find confusing about the language relating to "rotation." A rotation looks different at different angular sizes, because of how it makes some parts get closer, and other parts farther away. Is that being included, or just the first-order foreshortening? And what angular scales count as "sufficiently small"? Baez said:
"Well-known facts from complex analysis now tell us two things. First, circles go to circles under the pixel mapping, so a sphere will always photograph as a sphere. Second, shapes of objects are preserved in the infinitesimally small limit."

In interpreted that to mean the shapes are only preserved in the infinitesmally small limit, i.e., for the Lorentz contracted cross to look like a rotated cross, it has to be infinitesmally small, so this would not include how the forward tilted arm can look longer than the backward tilted arm on a large enough angular scale. You are saying I am overinterpreting Baez here, and what's more, your own investigation shows a connection between that longer forward arm, and what Lorentzian relativity actually does. So perhaps Baez missed that, or did not mean to imply what I thought he implied.

This is what Terrell says in his abstract:
"if the apparent directions of objects are plotted as points on a sphere surrounding the observer, the Lorentz transformation corresponds to a conformal transformation on the surface of this sphere. Thus, for sufficiently small subtended solid angle, an object will appear-- optically-- the same shape to all observers."

The answer must lie in the meaning of having a conformal transformation on the sphere of apparent directions. If we use JDoolin's asterisk, instead of a cross, we can see that a rotation will foreshorten the angles of the diagonal arms, and it is clear that a conformal transformation will keep those angles fixed, so certainly Terrell is saying that the Lorentz contraction will foreshorten the angles in exactly the same way. But what about the contrast in the apparent lengths of the arms tilted toward us and the arms tilted away, is that contrast also preserved in the conformal transformation? You are saying that it does, and that seems to be the key issue. We do have one more clue from Terrell's abstract:
"Observers photographing the meter stick simultaneously from the same position will obtain precisely the same picture, except for a change in scale given by the Doppler shift ratio"
So the word "precisely" says a lot, but what is meant by this change in scale, and is that change in scale uniform or only locally determined? You are saying that it looks precisely like a rotation, including the contrast between the fore and aft distortions, not just the first-order foreshortening effect.

Focusing on Terrell's statement above, and on my description of the exact method of 4)A) in post #93, the key is that the the conformal transform is applied to the image from a different viewing angle - thus it preserves the angular distortion produced by the overall shift in viewing angle of the smallish object. What is conformally mapped is the image from the camera stationary with respect to the object. But this image, even to first order, is rotated by the overall change in viewing angle for the moving camera, compared to what the moving camera would expect at its apparent viewing angle.

 

[Ken G]

Focusing on Terrell's statement above, and on my description of the exact method of 4)A) in post #93, the key is that the the conformal transform is applied to the image from a different viewing angle - thus it preserves the angular distortion produced by the overall shift in viewing angle of the smallish object. What is conformally mapped is the image from the camera stationary with respect to the object. But this image, even to first order, is rotated by the overall change in viewing angle for the moving camera, compared to what the moving camera would expect at its apparent viewing angle.

Yes, the globe with continents will show different continents from what would be expected if the globe was not in relative motion. The question is, will the continents look larger in the forward parts, as a static image would give, or will their relative sizes be distorted from that? In other words, the conformal transformation maps spheres to spheres, but it need not be the identity mapping on the surfaces of those spheres, so distortions can appear between the two photographs if the spheres are not small. It is not clear that aspect of what a "rotation" does is intended to be taken literally in Terrell-Penrose rotation, it might just be the fact that you see the different continents and no more than that can be relied on in general. It seems to me what is crucial is that a cross seen by a comoving camera directly across from it will look symmetric, so a stationary camera that sees the cross as moving must at the appropriate moment also see the cross as symmetric, that's the first-order "invisibility" of the length contraction. We are wondering if there is also a higher-order effect, where you can take contrasts in the fore and aft parts of the rotated object as part of that "invisibility" as well.

 

[PAllen]

Yes, the globe with continents will show different continents from what would be expected if the globe was not in relative motion. The question is, will the continents look larger in the forward parts, as a static image would give, or will their relative sizes be distorted from that? In other words, the conformal transformation maps spheres to spheres, but it need not be the identity mapping on the surfaces of those spheres, so distortions can appear between the two photographs if the spheres are not small. It is not clear that aspect of what a "rotation" does is intended to be taken literally in Terrell-Penrose rotation, it might just be the fact that you see the different continents and no more.

I am not sure how to convince you. The aberration is applied to the rays forming an image at viewing angle x. To first order, for modest subtended angle, they rotate all the rays by the change in viewing angle. This produces a distortion in the positions of ruler lines that I independently verify with direct ray tracing computation. Perhaps I overstated precise - my computational comparison was numerical to 4 significant digits, for a two degree subtended ruler.

Consider, for example, in the camera stationary with respect to a ruler viewed off to the left. Suppose the angle between 1 cm markings is .02 degrees on one side and .01 degrees on the other side. If all of these rays are rotated by the overall aberration change in viewing angle, these angles are preserved.

 

[Ken G]

Looking at the warping of this one structure in the game, it seems like they attempted to get the shapes right. The circles on the ground don't look quite circular, but rather they look flattened ovals, as I think they should.

Yes, flat disks are supposed to look rotated. I think when Terrell says this makes length contraction "invisible", he only mean the observer who does not know relativity will not see anything that sets off an alarm in just the nature of that flattened disk. The observer cannot say "hey, wait a minute, that's a length contracted disk," they can just say, "hey, who rotated that disk?" Of course, we agree that "seeing" is allowed to invoke additional information, like, there's no one there to rotate that disk.

 

[Ken G]

I am not sure how to convince you. The aberration is applied to the rays forming an image at viewing angle x. To first order, for modest subtended angle, they rotate all the rays by the change in viewing angle. This produces a distortion in the positions of ruler lines that I independently verify with direct ray tracing computation. Perhaps I overstated precise - my computational comparison was numerical to 4 significant digits, for a two degree subtended ruler.

What would convince me is a calculation of the angular size of the forward tilted arm, contrasted with a calculation of the angular size of the backward tilted arm, where that contrast is the same in the ray-tracing calculation as in a simple rotation. That would mean that the conformal mapping of the globe does not just map the rotated image into what the aberrated picture looks like in terms of seeing all the same continents, but also that all distortions that a rotation produces in the relative apparent sizes of the continents is also reproduced in the aberrated photo. That's an issue that does not come up to first order, because to first order, there is no difference in the apparent size of a continent that is "closer" to us.

Consider, for example, in the camera stationary with respect to a ruler viewed off to the left. Suppose the angle between 1 cm markings is .02 degrees on one side and .01 degrees on the other side. If all of these rays are rotated by the overall aberration change in viewing angle, these angles are preserved.

Yes, but a conformal mapping preserves the angles on the sphere being mapped, it doesn't preserve angles from points on the sphere to the center of the sphere, which is the angles you are talking about. Still, I can see that if the rod is very rotated, it could have a significant length yet still be confined to a very small angle, so perhaps in that case the conformal mapping does have to preserve the distortions you are talking about.

 

[PAllen]

What would convince me is a calculation of the angular size of the forward tilted arm, contrasted with a calculation of the angular size of the backward tilted arm, where that contrast is the same in the ray-tracing calculation as in a simple rotation. That would mean that the conformal mapping of the globe does not just map the rotated image into what the aberrated picture looks like in terms of seeing all the same continents, but also that all distortions that a rotation produces in the relative apparent sizes of the continents is also reproduced in the aberrated photo. That's an issue that does not come up to first order, because to first order, there is no difference in the apparent size of a continent that is "closer" to us.
Yes, but a conformal mapping preserves the angles on the sphere being mapped, it doesn't preserve angles from points on the sphere to the center of the sphere, which is the angles you are talking about. Can you really get a factor of 2 contrast in those angles while still enforcing a small angle between those two sides?

The conformal mapping applies to the image, directly from its derivation via aberration. It does not apply the object being imaged.

Yes, you can get large front to back distortion for a small subtended angle ruler. Holding subtended angle of ruler constant, while increasing velocity, increases front to back distortion. It can be made very large for a visually small ruler, for v close to c. Think of this as a very long ruler, rotated near head on, at a distance such as to preserve the subtended angle of the ruler as a whole.

Again, to first order you are shifting all the image rays by the same amount. Thus, if two pairs of rays in the image differ by a factor of 2 in subtended angle, they still will do so after shifting all the rays the same way.

 

[Ken G]

The conformal mapping applies to the image, directly from its derivation via aberration. It does not apply the object being imaged.

Sure, it is a 2D transformation on the sphere of apparent angles.

Yes, you can get large front to back distortion for a small subtended angle ruler. Holding subtended angle of ruler constant, while increasing velocity, increases front to back distortion. It can be made very larger for a visually small ruler, for v close to c.

Again, to first order you are shifting all the image rays by the same amount. Thus, if two pairs of rays in the image differ by a factor of 2, they still will do so after shifting all the rays the same way.

Yes, I now see that you can pack a long rod into a small angle by rotating it toward you, achieving a lot of fore/aft distortion over a narrow angle, perhaps narrow enough to allow us to apply Terrell's argument. So it seems you are right that not only the foreshortening, but also the fore/aft distortions over small scales, look just like a rotation.

Ironically, that strengthens Terrell's case for claiming the length contraction is not literally visible, because it produces even more similarity between the photo of the moving and nonmoving camera. I think what Terrell is really saying is he is imagining two coincident cameras asking "do objects look like they are length contracted", and saying "let's compare our photos to find out-- oh, we can't see any difference." In other words, the implied comparison is not between a moving object and a stationary object in some general sense, it is between the observation of two observers in relative motion made at the same place and time. In that instant, one observer regards the object as stationary, and the other as moving, and in that instant, they cannot see any length contraction. It is only if they use more than just that instant, but tell a compete story of the situation (including things like the finite speed of light), that they can infer the length contraction, so it comes down to what contextual information can be included in the act of "seeing." So it's an issue of language, but at least if we can get all the physics ironed out, that's what matters, so thank you for clarifying these points.

Above all, I agree with you and JDoolin that the term "invisible" makes over-extended implications, and that the real value of what Terrell did is showing the much simpler way to figure out what a movie of an object in motion will look like, by imagining a string of cameras comoving with the object, and just borrowing the appropriate snapshots as those cameras coincide with the movie camera.

 

[JDoolin]

Are you sure it's not the other way around? The back (still approaching) should look stretched, and the front (already receding) should look compressed. See Fig.1 here:
http://www.spacetimetravel.org/bewegung/bewegung3.html

You are right. I did that math a while ago, re-did it this morning. I had remembered it backwards.

You guys have gone into really subtle details that I hope to get into sometime soon. But I just wanted to come back to this simple statement about flat lines. I noticed that the link A.T. posted to actually shows videos of straight lines passing
• left-to-right in the distance (Lorentz Contraction at "small angle")
• right-to-left in the distance (Lorentz Contraction at "small angle")
• back-to-front underfoot (Lorentz + Moving Away)
• front-to-back underfoot (Lorentz + Moving Closer)

Now, what it doesn't show, is the "large angle", that is, if the camera panned to "the feet" as the lines passed underneath the observer. If you could imagine "strafing" a fence, you should expect that to one side, the fence should be obviously stretched (in the direction you're moving toward), and to the other side, the fence should be obviously contracted (in the direction you're moving away from), and directly in front of you, the fence should be as it is in the "Lorentz Contraction at 'small angle'" examples.

We're all agreed on this point, right?

 

[PAllen]

You guys have gone into really subtle details that I hope to get into sometime soon. But I just wanted to come back to this simple statement about flat lines. I noticed that the link A.T. posted to actually shows videos of straight lines passing
• left-to-right in the distance (Lorentz Contraction at "small angle")
• right-to-left in the distance (Lorentz Contraction at "small angle")
• back-to-front underfoot (Lorentz + Moving Away)
• front-to-back underfoot (Lorentz + Moving Closer)

Now, what it doesn't show, is the "large angle", that is, if the camera panned to "the feet" as the lines passed underneath the observer. If you could imagine "strafing" a fence, you should expect that to one side, the fence should be obviously stretched (in the direction you're moving toward), and to the other side, the fence should be obviously contracted (in the direction you're moving away from), and directly in front of you, the fence should be as it is in the "Lorentz Contraction at 'small angle'" examples.

We're all agreed on this point, right?

Yes, that is right, visually. Of course, the time for the fence to go by completely will be straight Lorentz contraction, being yet another way to directly measure it.

 

[JDoolin]

There's a video here of oncoming dice.

http://www.spacetimetravel.org/tompkins/tompkins.html

The claim seems to be that the forward "3" face of the die rotates away from the viewer, and the back "4" face rotates toward the viewer.

What do you guys think? Does this animation successfully demonstrate Penrose/Terrell Rotation? And is this rotation what actually should happen?

My intuition says that vector from the back of the die to the front of the die should maintain the same heading, while the rest of the die may be skewed forward--more of a hyperbolic rotation of the faces of the die than a spherical rotation of the die. Of course I realize the frailties of intuition, but I have a pretty good idea what I would do to try to model it, and test my hypothesis via simulation.

Based on my own arguments in the video in Post 71 I can see that it is possible for all forward observers to see the back of the die. However, I don't see how they could fail to see the front of the die at the same time.

I think the shape of the die would be skewed forward instead of rotated.

Now, having stated that argument as clearly as possible, I'm already beginning to have some doubts. For instance, the ray of light coming off the "3" face ranges from the normal, to tangent to the surface. That beam along the tangent surface, though, is going to miss the observer, before the observer passes the face.

 

[PAllen]

It looks reasonable to me. I assume this accurately done. It is based on one of the author's doctoral dissertation, so it got heavily evaluated, I presume. It claims to use ray tracing, the most universal method (the same method I used to derive the formulas I posted in #49 for the trivial case of moving 1-d rod). [I didn't worry about light sources and reflections. I just considered the rod luminous. I didn't care about color or brightness.]

 

[A.T.]

I can see that it is possible for all forward observers to see the back of the die.

Explained in more detail here:
http://www.spacetimetravel.org/bewegung/bewegung5.html

However, I don't see how they could fail to see the front of the die at the same time.

Basically the same reason as above: To get from the front to the camera at c, the forward component of the photons velocity would have to be less than v, so it couldn't outrun the dice.

Now, having stated that argument as clearly as possible, I'm already beginning to have some doubts. For instance, the ray of light coming off the "3" face ranges from the normal, to tangent to the surface. That beam along the tangent surface, though, is going to miss the observer, before the observer passes the face.

Yes, that is another way to put it.

 

[JDoolin]

I'm going to be comparing two animations of special relativity from spacetimetravel.org. The first is a 2-dimensional simulation of lines moving toward and away from the observer.

The second will be a three-dimensional simulation of dice moving toward the observer.

Now, both of these animations show a set of oncoming objects as they travel along at relativistic speeds. The rods over on the right are receding or approching at 70% of the speed of light, whereas the dice over on the left are approaching at 90% of the speed of light.

Now, my eye seems to detect the motion of the line on the right as entirely straight-line motion, whereas on the left, mye eye detects the motion of the edges of the dice as a rotationg motion.When I play the straight-line motion, the oncoming lines seem to be stretched by the same amount as the distance between them.

Another phenomenon that I'm seeing in the dice is that while the distance between individual dice seems to be stretched, the actual dice themselved don't appear to be stretched.

The other is that the original organization of the dice had around two or three dice spaces between them, so it shouldn't be a surprise that they are further apart than the lines in the diagram on the right.

Now, how can I say for sure that the oncoming lines are stretched by the same amount as the distance between them? Let's pause these videos at an opportune time.

One thing I can say for sure about the blue lines on the right. They definitely appear to maintain linearity.. They are lined up with the checker grid in the picture, and stay lined up with the checker-grid in the picture.

On the other hand, my eye tells me that the edges of the dice are not lined up properly. But is that real, or optical illusion?

What I'm testing is the edges on the bottom side and top side of the one-face of the die. To see if they are actually aligned, even though it looks like they are not.

So I have two observations to make here. One is that it does indeed appear that the top and bottom sides of the 1-face actually do appear to keep to the straight line paths.

One other point I probably ought to make is that this checkerboard pattern shouldn't quite remain straight as it passes underfoot.

I decided to look up panoramic views in the google to see if I couldn't make this point clearly. Here's a panoramic view of a fence near St. Bartholomae, I found on wikimedia commons.

You can see that this fence seems to be angled upwards to the right, then it is flat in the middle, then it goes downward on the top. Even though that is a straight fence, it does not appear from the perspective of my eye to be a straight line.

If I could do a similar panoramic view with the die face, I should expect when the 1-face becomes perpendicular to the observer, If the angles from the observer were equal. Then this would be the Lorentz-Contracted length of the 1-face.

We need some additional structure in the dice video to represent some additional structure that is conveyed by the

Additional structure of straight power lines, or a straight fence would provide the extra detail to the environment to see the Lorentz Contraction at the point of the dice-face's nearest approach to the observer.

Let's look at one other detail, by pausing these two videos at an appropriate time.

What I want to see is the apparent elongation of the parallel lines. Here, comparing the back-to-front length of the oncoming dash to the back-to-front length of the stationary dash.

In the dice video, these two lines (planes) simply don't feel parallel,

Okay. I really can't tell at all, but what I would want is to have the same kind of stationary structure in the background in the dice-video so we can easily tell what path the edges of the 1-face are taking.

I think if you brought the 1-face around until top and bottom edge of the 1-face of the die were parallel, and the velocity vector was perpendicular to our point-of-view vector, you could measure Lorentz Contraction across the face of the die.

But you'd need that stationary structure in place--a fence, or power-lines to help identify the more familiar panoramic distortions that are easy to recognize, but maybe a bit hard to account for.

http://www.spacetimetravel.org/bewegung/bewegung3.html

http://www.spacetimetravel.org/tompkins/tompkins.html

http://commons.wikimedia.org/wiki/File:St_Bartholomae_panoramic_view.jpg

 

[A.T.]

the distance between individual dice seems to be stretched

From the description here:

http://www.spacetimetravel.org/tompkins/node3.html

it's not clear whether the proper distances or the distances in the camera frame are equal between the two rows of dice. It looks like it's the later, because otherwise the length contraction and the signal delay would cancel like they do for the dice themselves, so the gaps would be about the same for both rows.

Here more on the distortion:

http://www.spacetimetravel.org/tompkins/node4.html

 

[JDoolin]

In the dice video, these two lines (planes) simply don't feel parallel,

Okay. I really can't tell at all, but what I would want is to have the same kind of stationary structure in the background in the dice-video so we can easily tell what path the edges of the 1-face are taking.

I think if you brought the 1-face around until top and bottom edge of the 1-face of the die were parallel, and the velocity vector was perpendicular to our point-of-view vector, you could measure Lorentz Contraction across the face of the die.

But you'd need that stationary structure in place--a fence, or power-lines to help identify the more familiar panoramic distortions that are easy to recognize, but maybe a bit hard to account for.

What follows is actually a "guess" about what was modeled in the simulation.

I thought for a time that it "looked like" the dice above were actually behind the dice in the front.

What I've realized is that I cannot tell, from four non-parallel lines projected onto a two-dimensional surface, whether they share the same plane or not!

I realized I could just as well put the 1-faces in the same plane, and that, most likely, is the way it was rendered. So I added some "stationary structure" lines to the diagram to indicate that perspective.

You can see in this picture, that the apparent parallel edges of the moving dice appear very elongated in the background, and only barely elongated in the foreground.

At 90% of the speed of light, we should expect that when the 1-face comes around perpendicular to our point-of-view, we should see it contracted to
\sqrt{1-.9^2}=.436 times it's un-contracted length, but the demo doesn't to pan the camera angle to that perspective to see it.

 

[Ken G]

At 90% of the speed of light, we should expect that when the 1-face comes around perpendicular to our point-of-view, we should see it contracted to
\sqrt{1-.9^2}=.436 times it's un-contracted length, but the demo doesn't to pan the camera angle to that perspective to see it.

Right, and as I understand the situation, we will be able to attribute that shrinking to an apparent rotation. That's because what we will see is a distorted image of what a camera moving with the die would see, except displaced laterally from the die so as to see the die as rotated. That's Terrell's meaning of "invisible", merely that the stationary camera and the moving one at the same place and time see basically the same image, but when the stationary one sees a die that appears to be directly across from it and length contracted, the moving one sees a die that is shifted from across from it and thus rotated, just as that camera has always seen it. The individual images can't tell the difference without more information (information that we have agreed can be viewed as part of "seeing").

 

[JDoolin]

Right, and as I understand the situation, we will be able to attribute that shrinking to an apparent rotation. That's because what we will see is a distorted image of what a camera moving with the die would see, except displaced laterally from the die so as to see the die as rotated. That's Terrell's meaning of "invisible", merely that the stationary camera and the moving one at the same place and time see basically the same image, but when the stationary one sees a die that appears to be directly across from it and length contracted, the moving one sees a die that is shifted from across from it and thus rotated, just as that camera has always seen it. The individual images can't tell the difference without more information (information that we have agreed can be viewed as part of "seeing").

I don't think you would attribute that shrinking to an apparent rotation for long. If the animation continued, and the camera panned, the 1-face would stop rotating "away" from the camera. If you rotate the 1-face of the die away from you, the angle between the lines making up the top and bottom of the 1-face would continue to spread apart.

But that won't happen with the relativistic motion. It will come to a point where the lines stop diverging, and start coming back together again.

But the top and bottom lines on the "1-face" of the die would be parallel as it passed by, relativistically will be parallel to each other. So it shouldn't look like rotation at all when it gets there.

 

[Ken G]

I don't think you would attribute that shrinking to an apparent rotation for long. If the animation continued, and the camera panned, the 1-face would stop rotating "away" from the camera. If you rotate the 1-face of the die away from you, the angle between the lines making up the top and bottom of the 1-face would continue to spread apart.

I'm also not clear on how Terrell attributes the distortions in the full image, both he and Baez referred to ideas like "sufficiently small" images.

But that won't happen with the relativistic motion. It will come to a point where the lines stop diverging, and start coming back together again.

I'm not sure if these are entering, but there are optical illusions associated with a visual field that is not small. For example, some people say that an infinitely long power line on a flat Earth would look like it curves as we track it from our closest point to the horizon. But a straight line angled away from us photographed by a pinhole camera onto a flat film does not curve. So it seems that vision is like the pinhole camera exposing a spherical film, as in Baez' example-- if we used a flat film with the pinhole camera, the photograph would look distorted, simply for not including the distortions we are accustomed to from that spherical film! That's why bringing in the larger angular scales makes it confusing as to what is a true physical distortion and what is an illusion. It seems the Terrell argument is not intended to apply to those scales, because the act of seeing (when binocular vision is of limited help) already involves a mapping from the 3D space of locations to a 2D sphere of directions, and that mapping is not conformal, so in some sense introduces worse distortions than the Lorentz transformation does!

But the top and bottom lines on the "1-face" of the die could be parallel as it passed by, relativistically will be parallel to each other. So it shouldn't look like rotation at all when it gets there.

On small enough scales, every image has to look just like a rotation, though not necessarily a rotation at constant angular velocity if you are watching a movie instead of looking at a single photo. Also, there can be cues that the rotation is impossible (like not lining up with some straight track the die is sliding along), but these kinds of cues involve additional information about the setup. Terrell should really just have said, if you want to know what a small image will look like, imagine a camera moving with the object that is at the same place and time as your image is taken, and borrow that image. Doing this over and over for lots of tiny images will allow you to reconstruct the full image, but you will need to know where to put the tiny images into your full image, and how to isotropically rescale their size, as per the things conformal mappings do. Perhaps the way you would need to do that to get agreement with your picture is where you will find evidence for length contraction, but not in the tiny images themselves.

 

[JDoolin]

Well, I'm still working on it... Here's a "Low Velocity" animation of the motion of a vertical stick

I want to make a high velocity animation of a cube, though.

Let's see if the rest of this project takes me a week, a month, or a year. Haha!

Happy Mothers Day!

 

[Ken G]

Looking good so far-- can you put a horizontal segment on it? That might be easier than a cube!

 

[PAllen]

I thought I would mention how the Terrell-Penrose method could be use to determine the visual appearance of an object if one assumed there is no length contraction (more generally, frame dependence of shape as represented in standard coordinates). The direct way is simply to represent the rest frame description of an object in a chosen frame, and just assume that is the spatial cross section (per that frame) of the moving object world tube. Then do ray tracing accounting for light delay.

However, Terrell-Penrose allows one to sidestep much of this computational effort given a description of the object in its rest frame (you just do static imaging, then use relativistic aberration of the angles in this image). You can put together a complete movie by computing a sequence of static images, without worry about light delay, and then transform each. So the question is how to get this computational simplification for the case of assuming the object's shape in some frame is not affected by its motion in that frame?

You certainly can't use Bradley aberration, since that is based on light speed being affected by the motion of the emitter. It is close in its predictions for stellar aberration to SR aberration only because the motion of the Earth is slow compared to c, so v^2/c^2 corrections are not significant. You certainly cannot use Galilean transform, since that produces isotropy of light speed only in one frame.

What you can do, is perform what I will call an "anit-boost". Given a rest frame description of an object, which by fiat we want to say holds in a frame in which it is moving, compute what rest frame description would be required per SR such that after a boost, you end up with a coordinate description matching this desired description. Roughly this would be a length expansion, but for irregular objects, there would be additional shape distortions. This is one time computation of no greater complexity than a Lorentz transform. Then, you can use Terrell-Penrose to produce the imaging of this anti-boost description. Voila, you have imaging under the assumption of no contraction without having to worry about light delays. Further, this shows why length contraction really is visible - each frame, compared to what you would really see, would differ by being an image derived from the anti-boosted rest description compared to being based on the actual rest description.

 

[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]

where y and z are figured from the distance to the fence, and z is the height on the fence, and x0 is the place where each object is at t=0, and v is the velocity of the dots as a proportion of the speed of light. The only unknown is x, so it figures out where the image is.

It figures out where the path of the object intersects with the past light-cone.

However, I just assumed that the object's x0 path (e-g, the x-coordinate where the object is at t=0), would be the Lorentz contracted x0, rather than the proper x0 of its own frame.

Edit: THE ORIGINAL DIMENSION of the dot pattern is 2 fence-lengths long by half a fence-length tall!

So it's a 4x1 rectangle lorentz-contracted to nearly square in the middle.

And it's 90% of the speed of light.

If you want to see some other animations of the same thing, at different distances from the fence, check out

http://www.spoonfedrelativity.com/web_images/ViewFollowing02.gif

http://www.spoonfedrelativity.com/web_images/ViewFollowing04.gif

and

http://www.spoonfedrelativity.com/web_images/ViewFollowing05.gif

I think it shouldn't be too simulate another plane in there, like the front or top of the cube.

 

[JDoolin]

Here. I decided it would be worth it to make one more animation tonight.

I set the parameters so that the "uncontracted" square of red dots was half a fence-length wide and half-a fence-length tall.

Also, I set the velocity to 0.866c. So Lorentz Contraction should be almost exactly 1/2.

Now there's a bit less confusion, since the uncontracted image should be square.