# Terrell Revisited: The Invisibility of the Lorentz Contraction

1. Aug 11, 2011

### JDoolin

Terrell Revisited: "The Invisibility of the Lorentz Contraction"

I recently posted this response to a general question of how Special Relativity worked.
G H Wells Jr pointed out that
and
Terrell's Argument
Both of these are important issues to take into account. However, they do not change the fact that "a meter stick in motion past the observer in such a way that it is moving parallel to its length, and is momentarily seen by its observer at its point of closest approach, will surely be seen as contracted."

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

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

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

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

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

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

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

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

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

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

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

(My Method)

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

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

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

$$t=-\frac{\sqrt{x^2+y^2}}{c}$$

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

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

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

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

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

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

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

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

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

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

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

2. Apr 28, 2015

### Ken G

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

3. Apr 28, 2015

### A.T.

Yes, also see last paragraph and the videos here:
http://www.spacetimetravel.org/bewegung/bewegung3.html

There are some videos of spheres here:
http://www.spacetimetravel.org/tompkins/node1.html
http://www.spacetimetravel.org/fussball/fussball.html
Unfortunately not at the closes approach, as far I can see.

4. Apr 28, 2015

### Ken G

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

5. Apr 28, 2015

### Staff: Mentor

Two relevant links on Penrose-Terrell rotation:

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

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

This argument is not correct, because light does not travel instantaneously along the sight lines, so the fact that the sight lines are symmetrically oriented at one instant does not mean the paths the light actually travels will be symmetric. Describing what is going on as a "simultaneity problem" isn't really correct, though; the issue is simply that the light reaching your eye at a given instant was emitted by different parts of the object at different times. The Usenet Physics FAQ article I linked to above gives a good treatment of what is going on.

6. Apr 28, 2015

### Staff: Mentor

No, this is not correct. This transformation tells you what the coordinates of the events will be in the other frame, but it does not tell you how the events will actually be observed--i.e., it does not tell you how the images seen by the two observers at a given instant will be related. To calculate what is observed by the two observers at the event (0, 0, 0, 0) where they are both co-located, you need to perform a conformal transformation on the null curves forming the past light cone that they both have in common. When you do this, you will find that the image seen by one observer is a rotated version of the image seen by the other, not a contracted version. The Usenet Physics FAQ article I linked to in my previous post describes how this works.

7. Apr 28, 2015

### m4r35n357

I have up till now understood that Terrell "rotation" (is this the term you were referring to?) is the visual effect of seeing the back of a structure as you go past it, but because it is "projected" in front of you by aberration, it gives the visual illusion that it is turning inside out in front of you. There is an example of it at about 1 min 25 secs into this video. The video was actually produced by massaging the equations from the mathpages link from the OP in a raytracer.

BTW I find the quote "each [co-located observer, regardless of relative motion] will see precisely what the other sees" astonishing, surely Terrell didn't really say that?

8. Apr 28, 2015

### Staff: Mentor

In the idealized case described in the Usenet Physics FAQ link, every pixel in one observer's image has a corresponding pixel in the other observer's image; the only difference is the location of the pixels on the respective images--one is a distorted version of the other. So the same light rays are present in both images, just not in the same places.

9. Apr 28, 2015

### Ken G

I'll give that a look, but I still cannot see what could be wrong with my argument. Imagine the rod slides along a string. As we watch the rod, there must be some instant where we see a rod at its closest approach to us, with the point of closest approach bisecting our image of the rod at that instant. That must occur, correct? The rod cannot appear to be rotated, because its ends would not appear along the string-- we could not see the string enter the rod ends. So we must be looking at a rod along that string, and we must reckon the length of the rod to be its usual Lorentz-contracted length, because we will always reckon that length once we correct for the finite speed of light. By definition, we are seeing light, at that instant, from the front and back ends of the rod at the same moment in our frame, and we are reckoning the time of flight of those light beams as the same (by symmetry), so we must see an image that looks like a length-contracted rod or else we would not be reckoning its length to be Lorentz contracted. In other words, the image of the rod we see at close approach must be exactly the length of the rod that we would reckon, given that image.

ETA: I've looked at the Baez description, and I find this statement telling: " Its apparent size will also change, but not its shape (to a first approximation)." That sure sounds to me like the rod will still look like a rod-- but will have its apparent size shortened. That means my argument could be consistent with what Baez is saying, and also means that length contraction is not "invisible." If my interpretation is correct, it means we should instead say "length contraction will not distort the images of the shapes of things."

Last edited: Apr 28, 2015
10. Apr 28, 2015

### Staff: Mentor

No. That would only be true if light traveled instantaneously. Try actually assigning coordinates to events and calculating the null worldlines that the light rays follow. This is one of those cases where intuition is too easily misled; there's really no substitute for doing the math.

11. Apr 28, 2015

### m4r35n357

OK, that makes perfect sense, I interpreted the comment very differently ;)

12. Apr 28, 2015

### PAllen

the web site AT linked to, based on computer implementation of the math, clearly shows the a rod moving by perpendicular to your light of sight appears visually contracted; its shortest visual size, as it goes by, is as expected from the LT. If the has thickness, it looks more twisted than rotated.

13. Apr 28, 2015

### Ken G

Yet this logical argument must be addressed:
1) we will always see a rod along the string. (This follows because the string is not moving, so cannot be distorted, and we must see it enter the ends of the rod.)
2) the rod will at some moment have the centroid of its image at the point of closest approach (this follows because at first the centroid of its image will not yet be at the point of closest approach, and later it will have its centroid past the point of closest approach, so there must be an instant where it is at the point of closest approach).
3) at that moment, we are seeing an image of a rod that straddles the point of closest approach. We can easily reckon the length of the rod by accounting for the time of flight of light from the symmetrically separated endpoints of the rod, and the time of flight from both endpoints is the same. Let's call that time of flight t. Then we can certainly say, in our coordinates, that a time t ago, the rod endpoints were at those two points, at the same time. Hence we can reckon the length of the rod a time t ago, and it must be the Lorentz contracted length (because we'll always get that after correcting for time of flight, but here that correction doesn't do anything).
4) hence, we must be seeing a rod that looks length contracted.

14. Apr 28, 2015

### Staff: Mentor

Meaning, the centroid of the image will be in the direction of the point of closest approach at some instant. Also, by "centroid" I assume you mean the point in the image that appears to be equidistant from both edges. Yes, this is true.

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

No, it is not. As above, the light ray in the centroid of the image at this instant does not come from the physical center of the rod. Once again, try actually assigning coordinates to events and calculating the null worldlines that the light rays follow.

15. Apr 28, 2015

### Ken G

Correct, the ray to the center of the image is the ray to the point of closest approach.
Yes.
Actually, I don't need to know if it was emitted by the center of the rod or not, because I am only finding the distance between the endpoints, at a moment when the endpoints straddle the point of closest approach. That is the only thing the logical train I just gave used, so you must defeat that logical argument based only on the things it used.

16. Apr 28, 2015

### Staff: Mentor

Ok, but then you can no longer assume that the image received by the observer one light-travel-time later will only contain light coming from between those two points in space. I don't think that will be true.

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

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

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

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

17. Apr 28, 2015

### Ken G

You are right I am simply assuming that, but I can change the problem to avoid needing to. Imagine having a red light at one end, and a blue light at the other. Then imagine not even having a rod between them-- we can just have those two light sources, moving at constant speed, and reckon the distance between them. If that distance appears to length contract in the scenario I describe, then we cannot rightly say that "length contraction is invisible."

18. Apr 28, 2015

### PAllen

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

Last edited: Apr 29, 2015
19. Apr 28, 2015

### JDoolin

Thanks for bumping this.

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

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

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

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

Last edited: Apr 28, 2015
20. Apr 28, 2015

### Ken G

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