Terrell Revisited: The Invisibility of the Lorentz Contraction

[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.

 

[Ken G]

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

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

 

[JDoolin]

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

cot(θ) = v csc(θ) + vT + αL/γ

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

T = -(L/2γ + v csc(θ_L))/v

where cot(θ_L) = -L/2γ

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

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

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

 

[PAllen]

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

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

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

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

 

[JDoolin]

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

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

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

 

[JDoolin]

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

 

[JDoolin]

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

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

This should have read:

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

...if anybody was wondering.

 

[PAllen]

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

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

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

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

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

 

[PAllen]

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

The congruence of world lines describing the moving ruler is:

x = vt + αL/γ
y=1

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

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

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

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

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

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

 

[JDoolin]

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

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

cot(θ) = v csc(θ) + vT + αL/γ

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

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

and the second equation for

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

I divided them and I get

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

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

 

[PAllen]

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

 

[JDoolin]

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

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

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

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

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

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

 

[PAllen]

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

Answering specific questions:

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

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

 

[Nick666]

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

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

 

[PAllen]

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

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

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

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

 

[jartsa]

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

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

When velocity is 0.9 c, the spaceships stop accelerating.

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

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

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

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

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

 

[PAllen]

Jartsa,

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

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

 

[PAllen]

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

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

 

[JDoolin]

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

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

and the second equation for

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

I divided them and I get

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

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

Answer: t=T+csc θ

I was using the wrong "Tee"

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

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

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

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

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

 

[PAllen]

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

 

[jartsa]

Jartsa,

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

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

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

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