Terrell-Penrose effect and surface reflectance

Question

All explanations of Terrell-Penrose effect seem imply that the effect makes some of the back-facing walls of a passing-by object visible. You can see some examples of those in many different references. However, from my understanding many of those sources assume that the light reflect or emitted from the surface travels in all directions, which is not at all how surface reflectance works.

Even for a perfect Lambertian reflector the outgoing light will travel only in the direction of the hemisphere around the surface normal – none of the light is reflected "into" the wall. How would it be possible to see any light reflected from a back-facing wall if that light was never reflected in the observer's direction in the first place?

I can convince myself of being able to see a wall that is facing at most 90° away from me, based on the assumption that I'd see photons emitted from it in the direction parallel to the surface. How could I ever observer photons from a wall that's facing more than 90° away? Some of the references clearly show walls that were originally facing away from the observer as visible, like the wall with 4 dots on the moving dices:

Are those mistaken?

Answer 1

Concentrate on the nearest top die on the right. The reason you can see the "four" face is that you have already passed the "one" face of the die! It is just that aberration has distorted the image to place it in front of you. There is nothing more to "Terrell Rotation" than this simple fact.

You can find a more complete discussion here.

Answer 2

If you want to know what an inertially moving nonrotating rigid object will look like when you're at any spacetime location and moving with any relative speed, the easiest way to do it is:

1. Work out what it would look like from that spacetime location if you were at relative rest. (Once you boost to the frame in which everything is at rest, this is just an ordinary 3D rendering problem.)

2. Transform that picture according to the rules of relativistic aberration and Doppler shift.

Aberration and Doppler shift only depend on the 2D rendered picture, not on any other property of the object you were looking at, such as its distance or what's on the faces that OpenGL didn't draw because their surface normals pointed away from the camera. As long as you're at that spacetime location, regardless of your motion, you'll see the same sides of the cube, the same specular highlights, and so on. You're seeing the same light: it just hits your retina or camera film at different points depending on the motion (and angle, obviously) of the eye/camera.

The Terrell-Penrose rotation is basically a result of not thinking carefully about what you're seeing.

Suppose the die is at rest and you do a flyby while shooting pictures. But, oddly, you accelerate to a stop relative to the die before taking each picture, then accelerate back to your former velocity to move to the next location. Your path is such that the earlier photos show the 1 and 3 faces, the middle photos show the 1 face alone, and the later photos show the 1 and 4 faces. They look normal and undistorted since you're at relative rest.

Now you do the same flyby, taking pictures at the same locations, but without stopping when you take them. The pictures you take on the second flyby will be like the pictures on the first flyby but distorted by aberration and Doppler shift. Doppler shift is irrelevant here so we'll ignore it. The effect of aberration is to shift the location of the cube on the film toward the direction of your motion (the so-called headlight effect). It also distorts the straight edges of the cube into arcs of circles, but that's not important.

Now – here's the key part – you decide to only look at the photo from each flyby where it appears the cube is located at a right angle to your direction of motion. In the first flyby, that's the middle photo, and it shows only the 1 face. But in the second flyby, it's an earlier photo, because aberration moved the cube closer to the forward direction in all of the photos. The earlier photo shows the 1 and 3 faces.

Your motion doesn't make any face visible that wasn't visible when you were at rest. You'll never see the 6 face (opposite 1) while moving on this path because you'd never see it in the photos taken at rest on that path. You only see different things in the two selected photos because you used a somewhat silly criterion for selecting them.

Answer 3

Sure, you cannot see the “back” wall of the dice and Terrell – Penrose effect does not imply that you can.

As a rule this effect is considered in a frame of “stationary” camera; the camera opens its shutter to catch that image, which was emitted some time ago, when the object was at points of closest approach with the camera. When the camera made a “click” the moving dice was not at point of closest approach with the camera, but was far away already, because it takes some time for the rays to come to the aperture.

Please also note, that light pulse which came to the stationary observer at right angle (in this observer's frame) due to aberration of light was once emitted "backward" in the frame of the moving dice.

It is easier to understand the effect if you change frames and ascribe motion to the camera and state of rest to the object. In this case the object was permanently “shining” and the camera was “sunbathing” in reflected or emitted rays of light.

Let's consider pictures as taken by “regular, non-relativistic” camera first. If the camera was “moving” and the picture had been taken Terrell-way it turns out that it takes a picture of a “stationary” object at some angle, that’s why the object “rotates”.

For example your friend asked you to take a picture of his face; but you took the picture not from that point which was directly opposite but walked aside. In this case your friend’s whole one ear would be seen on the picture. You can show him this picture and say that the “ Terrell's rotation” affected it.

The higher the relative speed was, the further from the dice would be that point from which the picture was taken (according to Terrell's method). The more the dice would appear "rotated".

Sure, you can take a picture of the dice at that instant, when the object is at points of closest approach. In this case you would only see its "face" or "the nearest" wall, which is oriented towards aperture. Due to relativistic Lorentz contraction the image would be stretched. The greater relative velocity was, the more the image would be stretched.

Relativistic effects (aberration and Lorentz contraction) distort an image in a specific way, but do not change the essence of the matter.

Answer 4

From mentioned in Wikipedia paper by R.Penrose:

the light from the trailing part reaches the observer from behind the sphere, which it can do since the sphere is continuously moving out of its way

As far as I know that effect has not been demonstrated experimentally, but if we try to imagine light travel in discreet time intervals, we can see sphere (and the dice I think too) has time to move out of light's way (object is also length contracted in observers frame of reference, which afaik is not essential to the effect, but allow us to see even more "behind", at least for some shapes). Note: the image below I have drawn shows length contraction of about 50% which is at speed of ~0.85c, not 0.5c.

...The rest maybe better posted as comments, but my rating is low and the site does not allow me write them at the time of writing this answer (in my frame of reference of cause ;-) The aberration effect mentioned by previous answers has nothing to do with seeing face 4. The relativistic aberration of light, and aberration show that object is displaced relative to its "true" position, that is position in moment of simultaneity in observers frame of reference.

During the time it takes the light beam to reach the observer the light source moves in the observer's frame, and the 'true position' of the light source is displaced relative to the apparent position the observer sees, as explained by light-time correction. Finally, the beam in the observer's frame at the moment of observation is tilted compared to the beam in source's frame, which can be understood as an aberrational effect. Thus, a person in the light source's frame would describe the apparent tilting of the beam in terms of aberration, while a person in the observer's frame would describe it as a light-time effect.

From observer view, "light-time effect" means we will see dice at right angle to us as it was some time ago at that (at points of closest approach) position, not as we could see it in previous position as other answers try to convince. One answer correctly states that light in dice frame is emitted "backwards", but does not point it as main reason for the effect.

P.S.The light travels straight lines in all inertial frames of references, so we cannot imagine camera is moving instead of object and assume result is completely the same. When reading Understanding the Headlight/Beaming effect in Special Relativity I imagined eye moved and after light is went through eye's lenses and by the time it hit retina retina displaced little. It would explain change in apparent position, but then I realized it is incorrect idea, light will travel in moving frame of the eye straight and won't curve in it; aberration is fully explained by time it takes light to reach the eye, as stated in wiki citation above.