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I am learning Terrell-Penrose's effect and it is often argued that the effect allows an observer to see the back of a cube. Many references even explicitly show the "back sides" of the moving object to stress the magic of Terrell-Penrose effect. I have difficulty to understand this.

This reference provides details of the calculation and I agree most of its derivations https://andrewyork.net/Math/TerrellRotation_York.html In the reference it was shown that the two dimensional square and three dimensional cube are distorted and rotated by Terrell-Penrose effect.

In the case that the object is a sphere, people quote 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"

My question is that, how come a light coming out of any point "from the back of the object" has any chance to reach the observer? Surely the object is moving, so the lights reached the observer simultaneously came from different times and locations in space. However, the speed of the object cannot be higher than the speed of light, so even if any point on the back of the object can emit a light ray toward the observer, the light ray would be block by the (moving) object. In other words, the object has no chance to move out of the way of light rays.

In the case of 2D square from the linked reference. Taking the example from the reference. I agree that the square (the square with vertices PQSR looks like not a square because of Lorentz contraction in action) would look like a parallelogram if lights coming out of every point of the square toward the observer at point m can be seen. Basically, the calculation in the reference shows that lights coming out of points on the boundary of parallelogram at different times reach the observer at the same time.

However, my theory is that the lights coming from the blue boundaries (from edge PS and edge SR at various times) cannot reach the observer because the light rays are always blocked by the square itself. The square blocked lights from edge PS and edge SR (at various times) towards the observer at point m because light is faster than the object.

Is my understanding correct? if no, what did I miss?

Criteria to not be blocked: I guess this is what most of references missed if I am not mistaking. Suppose the square or object moves toward the right (direction of positive x-axis), and a light ray emits from a distance $D$ above the observer (i.e. their y-coordinates have difference $D$). Suppose the distance between the "light source" and the observer is $L$. In order for the light ray to not be blocked by the object, the distance D must satisfy $D/L \geqslant \sqrt{1 - v^2/c^2}$. Or, $$ \theta \leqslant \arccos(\sqrt{1 - v^2/c^2}), $$ where $\theta$ is the angle deviated from the vertical line to the negative x-axis. If the angle is too large, the light ray is actually blocked by the object and cannot be seen by the observer.

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  • $\begingroup$ No. If you do all the calculations you will find that when you see the "back" of the cube it is actually behind you! It only looks like it is in front because of aberration. en.wikipedia.org/wiki/Relativistic_aberration $\endgroup$
    – m4r35n357
    Commented Jun 2 at 11:45
  • $\begingroup$ Actually if the observer is far enough transversely, the object can indeed "get out of the way of light ray" in time so that the observer sees the back of the cube even when the cube is still approaching the observer. $\endgroup$
    – chichi
    Commented Jun 2 at 11:48
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    $\begingroup$ Seems to me you are more proficient in "calculus" then me. It looks to me there is a formula for PS to start to be seen at some point on approach to the observer. As for SR, I think it is never seen. So it is "from edge SR and edge PS at various times", not the other way around, maybe you just mistyped. $\endgroup$ Commented Jun 2 at 15:33
  • $\begingroup$ @Martian2020, I agree that PS starts to be seen at some point on approach to the observer. It is still unclear for me if SR would be completely blocked. $\endgroup$
    – chichi
    Commented Jun 2 at 16:21
  • $\begingroup$ @chichi you see the part "you would have seen later". You never ever see the "part you would never have anyway seen later". That's all it is. $\endgroup$
    – Fattie
    Commented Jun 3 at 18:17

3 Answers 3

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multiexposure render of (top) a die proceeding at relativistic speed in the direction of its "three" face, with the viewer on the "one" face side and (bottom) a die proceeding at nonrelativistic speed in the same way; the "four" face opposite the "three" face is visible in the relativistic case even before passing the viewer because of the relativistic distortion

The answer depends on what you call the "back of the cube".

In the above image, the side opposite the side with 1 pip has 6 pips, and that face remains out of sight. If you were standing beside a railway track and imagined that the cube represents a train going past, then the driver of train would be where the 3 is, and the the rear of the train would be where the 4 is. If that is what you consider to be the back of the train, then you can really see it. You would also see it in the none-relativistic context once the train has gone past your position beside the track.

Another way to convince yourself that you will never see the 6 side is to consider the reference frame where the cube is considered stationary, and the track-side observer is moving. (This does not change what the observer sees.) In this point of view, the cube obviously cannot get out of the way of rays coming from the 6 side (because the cube is not moving). Only rays coming from the 1, 3 & 4 faces can reach the observer without passing through the cube. (Ignoring the top and bottom 5 and 2 faces.)

I think the point of "relativistic effect" is that we can see 4 even when it is approaching. How do we understand this in the other frame where the cube is stationary and the observer is moving? I find it difficult to imagine how the cube is distorted in this frame too. – chichi

In that frame the observer is moving and is subject to the aberration effect (demonstrated in the above animation), which is like running in the rain. Vertically falling raindrops appear to be coming down diagonally from a different location to the running observer. In the cube example, the moving observer is really seeing light that came from the rear of the cube when it had already passed her, but due to the aberration, it appears like the cube is still approaching her. You could still see an effective rotation (without length contraction) with Newtonian physics alone (assuming finite light speed).

animated overhead view of the same scenario in a reference frame where the die is stationary and the viewer is moving at relativistic speed; light rays from the die's edges and their apparent positions from the viewer perspective are also shown

The above animation is in the rest frame of the die, which appears to be moving left-to-right in the camera frame. The moving oblate shape represents a spherical pinhole camera that is length contracted to an elliptical cross section. The dashed rays represent where the light rays appear to be coming from to the camera, and hopefully the aberration effect due to the motion of the camera can be seen as an apparent change in direction of the light rays. Some of the photons have to emitted at an earlier time in order to arrive at the aperture more or less simultaneously, in order to pass through in the short time the pinhole is open. The ray from the purple vertex at the top right has to pass through the cube so it cannot arrive at the camera, but it is just there to show where that vertex would appear to be, if the cube was transparent, so that the apparent profile of the cube's cross section can be represented.

For completeness, here is one more animation in the rest frame of the observer that sees the cube as moving.

enter image description here

It can be seen that the back of the dice (face 4 - green) can indeed be seen because the light from that face does not pass through the cube (unlike faces 6 and 3). The cube can get out of the way of light coming from the rear face, despite moving at less than the speed of light, because the horizontal velocity component of the diagonally moving light ray is less than the velocity of the cube. Note that for ease of construction and for clarity the all the light does not arrive at the observer simultaneously in this construction, but the general principle and the ray tracing remains valid. For the sake of demonstration, imagine the faces are covered in coloured LED's that are all switched on briefly and simultaneously in the reference frame of the camera - a form of flash photography.

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    $\begingroup$ I think for the purposes of this discussion the back is the "four" side. $\endgroup$
    – m4r35n357
    Commented Jun 2 at 19:05
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    $\begingroup$ I think the point of "relativistic effect" is that we can see 4 even when it is approaching. How do we understand this in the other frame where the cube is stationary and the observer is moving? I find it difficult to imagine how the cube is distorted in this frame too. $\endgroup$
    – chichi
    Commented Jun 3 at 0:55
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    $\begingroup$ @chichi "I think the point of "relativistic effect" is that we can see 4 even when it is approaching. " I agree. "How do we understand this in the other frame where the cube is stationary and the observer is moving?" In that frame the observer is moving and is subject to the aberration effect, which is like running in the rain. Vertically falling raindrops appear to be coming down diagonally from a different location to the running observer. In the cube example, the moving observer is really seeing light that came from the rear of the cube when it had already passed her, but .... $\endgroup$
    – KDP
    Commented Jun 3 at 3:11
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    $\begingroup$ ... due to the aberration, it appears like the cube is still approaching her. You could still see an effective rotation with Newtonian physics alone (assuming finite light speed). Does that clarify things? $\endgroup$
    – KDP
    Commented Jun 3 at 3:12
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    $\begingroup$ @chcichi I have added an animation to my answer to demonstrate the aberration effect. I realise you get that that part, but I can't help myself. Animations are my hobby :-) $\endgroup$
    – KDP
    Commented Jun 3 at 5:21
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It is actually incorrect to describe this effect as a rotation. As is clearly evident from the animated figures 8 and 9 in the very reference you linked to, the square is not rotated but merely sheared. Yes, you can see the backside (left side) of the square, but only if it is a sufficiently short distance away and if you are not on the exact line of motion. For the configuration you pictured in your post, the square is just not close enough for the bottom blue side to be visible from the position of the observer at $m$. If you display this a bit later (I just did a screenshot of the animation at the right moment), it looks as follows

enter image description here

and here this side is visible from $m$ even though it still has not quite reached it yet

This effect has as such nothing to do with Relativity but is a consequence of the fact that any finite signal propagation speed results in a distorted appearance of the object due to different signal travel times from its different parts. It would be applicable to sound waves for instance as well. The only factor that Relativity adds here is the additional length contraction of the object when it passes you.

In case of further interest, I have published a paper in European Journal of Physics a couple of years ago that illustrates the effect of a finite signal propagation on the shape and density distribution of various moving and/or rotating objects. Full text version is available here.

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  • $\begingroup$ Have you published the code too? Under what's license/where? TIA $\endgroup$ Commented Jun 3 at 0:11
  • $\begingroup$ Agree. I guess people call it relativistic effect rather than finite speed effect (if we forget about length contraction) because relativity sets the condition that signal is faster than object so that "the back side can be seen as approaching" is possible. $\endgroup$
    – chichi
    Commented Jun 3 at 1:00
  • $\begingroup$ @Martian2020 There is just a bit of Mathematica code that I used to produce the plots shown in my paper, based on the transformation equations derived analytically there. This is not published, but I could make it available if you want. $\endgroup$
    – Thomas
    Commented Jun 3 at 17:41
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Two people at the same place at the same time will always see the same thing, even if they're in relative motion, because they're seeing the same light. If they're in relative motion, the scene will be distorted in different ways by aberration and Doppler shift, but those effects act on the light at one's location, which encodes a 2D projected image of the world. They don't act on the 4D reality. The two viewers may see differently colored and shaped images of a die, but they will count the same number of faces and the same number of pips on each.

So the easiest way to figure out which faces of a moving die will be visible to you from a spacetime point is to transform to the rest frame of the die and render it using ordinary 3D graphics techniques with the camera at the spatial part of your spacetime location. Those are the faces you'll see.

The reason people call it a "rotation" is that aberration changes the apparent position of the object in the sky. If, in the rest frame of the die, it is "southeast" of your observation point, and you are moving "north" relative to that frame, then for a certain choice of your northward speed, the die will appear to be exactly east of you because of aberration, but you will see the same faces as if it were at rest southeast of you. Hence it's "rotated".

I recommend against trying to work out light travel times from different faces of a moving cube. It's much harder, and it's unnecessary. A surprising number of people seem to believe that special relativity requires you to use your own inertial rest frame to figure out what you see. What the principle of relativity actually says is that you can use any inertial frame to solve any problem, because they're all equivalent. This is a great example of a problem in which using the viewer's rest frame is a bad idea. Furthermore solving the problem in different frames for relatively moving viewers misses the essential fact that they are seeing the same light. There's no sense in explaining the origin of the same light in two different ways.

If you really want to do it, I think your mistake is this:

The square blocked lights [...] towards the observer at point m because light is faster than the object.

The light may be slower than the object in the direction the object is moving. The light's velocity has components parallel and perpendicular to the object's velocity, and only the vector sum of those has to have length $c$, not the parallel component.

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  • $\begingroup$ In the frame where object is moving and observer is stationary, the light ray from "back side" of cube has to go in the direction of "attempting to penetrate the cube" in order to reach the observer on approaching. But the need for such "penetrating light ray" is no longer needed in the frame where object is stationary and observer is moving. How to understand this asymmetry between frames? Is this requirement of light ray direction an optical effect too? (such requirement may be translated as a physical requirement of the need of scattering light or emitting light) $\endgroup$
    – chichi
    Commented Jun 3 at 4:56
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    $\begingroup$ @chichi You're not thinking fourth-dimensionally. If you slice spacetime into 3D planes according to an inertial frame's $t$ coordinate, and imagine the light ray moves first along a slice and then through time to the next slice, then for some choices of frame the ray will enter the cube's world-tube in the first half of that zigzag ("tries to penetrate the cube") and leave it in the second half ("the cube moves out of the way"). For other choices of frame it won't. In reality, the light's path is a straight diagonal in spacetime that never intersects the world-tube, independent of frame. $\endgroup$
    – benrg
    Commented Jun 3 at 6:13
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    $\begingroup$ @chichi The whole point here is that the observer is not exactly on the line of motion but offset from it. So strictly speaking there is actually no backside. An observer exactly on the line of motion would never see anything but the front side until the cube has passed him. $\endgroup$
    – Thomas
    Commented Jun 3 at 18:02
  • $\begingroup$ I don't see how an observer at rest with regard to the cube should ever be able to see one of its remote sides, whatever the distance between the two. The whole point here is that the sides of the cube must appear to be sheared to a rhombus, which only happens if there is some relative motion, the signal propagation speed is finite, and the rhombus is close enough so that you can 'look around the corner'. $\endgroup$
    – Thomas
    Commented Jun 3 at 19:59

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