# Relativistic effects on apparent angular size

1. Aug 5, 2009

### nutgeb

I'm trying to confirm my understanding of the simple effects of SR Lorentz contraction and SR aberration on the apparent angular size of an object receding at relativistic speeds.

Consider a detector at rest in inertial frame S, which is observing light emitted uniformly from the planar surface of a large disk. The disk (the plane of which is orthogonal to the detector) is moving radially away from the detector at a relativistic velocity.

If the disc was not moving relative to frame S at a given time, its proper distance would be D. However, since the disk is moving away at a relativistic velocity, the distance between the disk and the detector is Lorentz contracted radially to a smaller distance D', as viewed from either the detector or the disk.

From the perspective of frame S, the Lorentz contraction of D' does not change the transverse diameter of the disk. Therefore (ignoring aberration for the moment), the angular size of the image of the disk as viewed at the detector is:

(a) the same angular size that would have been observed if the disk were stationary (relative to frame S) at proper distance D? or,

(b) the same angular size that would have been observed if the disk were stationary (relative to frame S) at Lorentz-contracted distance D'?

I believe that (a) is the correct answer, because otherwise radial Lorentz contraction would have the effect of causing apparent transverse stretching, which seems wrong.

In addition to the effect described above, my understanding is that relativistic aberration will cause the apparent angular size of the disk's image at the detector to be larger by a factor of (z+1) than if the disk had been stationary (relative to frame S) at distance D.

As a result, the total relativisic effect (considering both Lorentz contraction and aberration) is that the apparent angular size of the disk's image at the detector will be the same angular size that would have been observed if the disk were stationary (relative to frame S) at distance D'. Correct?

Last edited: Aug 6, 2009
2. Aug 9, 2009

### nutgeb

No takers on this question?

3. Aug 9, 2009

### A.T.

So the disc has a orthogonal ruler with the proper length D attached. And you want to know how it looks like for the observer, when the end of that ruler passes the observer?

Two things to consider here:

- You are comparing the appearance disc at two different distances (D and D'). Does this make sense?

- If you want to know how the disc looks like optically, you have to consider light signal delay. That creates a lot of weird effects: http://www.spacetimetravel.org/

4. Aug 9, 2009

### nutgeb

I suppose that's one way of setting up the problem, but what I had in mind is that the ruler is stationary in the observer's frame S, and the disk is moving radially away from the observer alongside the ruler. If the disk is stationary for an instant, the ruler measures distance D; if the disk is moving radially away at a relativistic velocity at that instant, the ruler measures a Lorentz contracted distance D'.

Actually I am comparing the apparent angular size of the disk at two different velocities (0, and relativistic). The difference in velocities causes the disk to be at different distances in the two cases, as measured by the ruler that is stationary in frame S.
The website you referenced is interesting. However, the examples there don't seem to apply to my scenario, because I have the disk moving away in a precisely radial direction. In contrast, the examples on the website have an angle between the viewer ('kamera') and the object's direction of travel.

I recognize that the flat disk might be viewed as having a slight hyperbolic distortion, but that in itself does not seem to have any effect on the apparent angular diameter measured by a viewer who is radially in line with the recession vector.

Last edited: Aug 9, 2009
5. Aug 10, 2009

### Ich

I didn't answer yet because you have a long record of not believing a single thing I say.
The measured distance is trivially the same for objects at rest or moving radially away, there's no length contraction or anything to be considered.

6. Aug 10, 2009

### A.T.

If the ruler is stationary in the observers frame then it is not contracted, and the distances are not different in the two cases in the observers frame. This is of course different from the scenario with the ruler at rest in the discs frame, I described above. It is up to you how you want to measure the distance.

7. Aug 10, 2009

### nutgeb

Thanks for the answers guys, but they don't seem correct to me. Let's do this step by step.

1. Let's have the disk begin right at the stationary detector, then have it immediately accelerate to a relativistic velocity radially away from the detector. The detector in frame S views the distance to the moving disk to be Lorentz contracted.

This is the same as in the Milne cosmology model, where all fundamental comovers begin at the origin, and some of them move radially away at almost c. The distance from the origin to each such fast-moving comover is Lorentz contracted, as viewed in the origin's frame.

2. The ruler is stationary in the detector's frame, so S is its rest frame also. If there are stationary observers positioned at intervals along the ruler, S is their rest frame too, and they each will view the radial distance from themself to the moving disk (which is different for each of them) to be Lorentz contracted by the same proportion.

The mere fact that a stationary ruler lies along the moving disk's radial path away from the detector cannot somehow negate the Lorentz contraction of the distance to the moving disk, as viewed by the detector. For example, removing the ruler has no effect on the Lorentz contraction of the distance to the moving disk, as viewed by the detector.

3. If instead the ruler is attached to the disk, and therefore is comoving with the disk, the detector would view the ruler itself to be Lorentz contracted, as well as the distance to the moving disk. Using a Lorentz contracted ruler to measure a Lorentz contracted distance would result in the marks on the ruler showing that the distance to the disk is NOT contracted. This does not mean that there is no Lorentz contraction of the distance, as viewed by the detector. It's just a case of using a shortened ruler to measure a distance that is shortened by the same proportion. So that scenario isn't very helpful. In order to measure distance from the perspective of frame S, we don't want to use a contracted ruler, we want to use a full-size ruler.

Last edited: Aug 10, 2009
8. Aug 10, 2009

### A.T.

This depends entirely on how you define "distance".

Let's say the ruler is stationary and the detector measures the angular size when it sees the disc pass the end of the ruler. The angular size will be the same as with the disc at rest at the end of the ruler. This is trivial, just imagine both cases in one experiment: The detector sees the moving disc pass an identical stationary disc at the end of the ruler. They cannot have different angular sizes at that moment.

Your entire idea, that the distance to an object that moves away from the observer is Lorentz contracted doesn't make sense. Apply it to photons that move away from you at c. How would they ever get away, when the distance to them is contracted to zero? It seem you have a misconception about length contraction, similar to the one that due to time dilation the moving object must slow down again.

Last edited: Aug 10, 2009
9. Aug 10, 2009

### nutgeb

Well yes, that's what I said in answer (a) in my OP. But I'm not sure your principle is correct. Remember that there will be a failure of simultaneity as to when the moving disk passes the stationary disk, as viewed from the detector (frame S).
The fact that relativistic speed causes distances (and not just the lengths of objects) to become Lorentz contracted is common knowledge -- not some misconception on my part. As one among dozens of good sources, see chapter 4.7 of http://books.google.com/books?id=PD...epage&q=distance lorentz contraction&f=false". Treat the rocket traveler (travelling at relativistic velocity relative to earth) as being stationary in his own rest frame, with earth approaching him at relativistic speed. Clearly the rocket traveler measures the distance to earth to be Lorentz contracted.

It would make no sense if the lengths of objects were Lorentz contracted but distances were not. Imagine that the disk departs the detector at a relativistic radial velocity, trailing a very long string behind it. (The string is longer than the distance to the detector when we make measurements). So the string is comoving with the disk. Of course the length of the string will be Lorentz contracted as viewed in the detector's frame. If the string is Lorentz contracted along its length, then the distance between the detector and the disk must be equally Lorentz contracted.

The whole concept of the Milne cosmology model is that distances from the origin to comovers is Lorentz contracted (in the origin's frame), depending on each comover's recession velocity relative to the origin. Please explain how Milne would work without Lorentz contraction of radial distances (?)

Last edited by a moderator: Apr 24, 2017
10. Aug 10, 2009

### sylas

This ought to be easy. The angular size of an object, in special relativity, depends on the distance from where light is emitted to where it is received, as given by the frame of the observer.

You can derive the same result for angular size in the frame of the observer, or any other inertial frame by considering Lorentz contraction on a pinhole camera of a moving viewer.

I've described it, with some diagrams, in [post=2217788]msg #21[/post] of "most basic of thought experiments in special relativity".

I'm not going to argue the point here; but if people have genuine questions expressed clearly and concisely, I'll be happy to go into it further. But if you know SR, this shouldn't be necessary.

Cheers -- sylas

11. Aug 10, 2009

### nutgeb

Sylas, I want to be sure I understand what you're saying.

Your msg #21 gives a good picture of what goes on inside the camera after the light passes through the pinhole. But it doesn't say anything about whether the distance from the camera to the subject is Lorentz contracted in the camera's frame when the camera is receding from the subject at a relativistic velocity.

In that scenario the distance from the camera pinhole to the subject is Lorentz contracted, so that could be interpreted to mean that the angle of the light as it approaches the pinhole is changed -- to a wider angle than if the camera were not moving relative to the subject. Is that what you mean?

12. Aug 10, 2009

### sylas

Just pick your events; light at the pinhole, light at the screen, light at the emitter. Do Lorentz transformations. You know when things are Lorentz contracted. Come on! You don't need me for this.

Cheers -- sylas

13. Aug 11, 2009

### A.T.

Simultaneity is not a problem here since there is no spatial separation between the discs when they pass each other. Let's say one is a ring and the disc fits exactly trough. Every observer will at some point see that happen, with both at the same size fitting exactly.
Yes in his frame. Which is completely irrelevant for the observer on earth who measures the angular size of the rocket, at a certain distance as measured in the earth's frame.
As I said: This depends entirely on how you define "distance". But usually distance is measured by a ruler at rest in the observers frame.

Last edited: Aug 11, 2009
14. Aug 11, 2009

### Ich

nutgeb,

your application of Lorentz contraction seems quite strange to me, same with your use of aberration. You need to use coordinates consistently, not apply some relativistic concepts to situations where they don't fit.
I'll give you a conundrum; to solve it, use coordinates and the Lorentz transformations. That should clear things up.

Aberration - contrary to Wikipedia's claim - is not a function of the relative velocity of emitter and observer. It is not dependent on the emitter's velocity; it depends solely on the observer's velocity.
How do you reconcile that with the principle of relativity?

15. Aug 11, 2009

### A.T.

Quick shot, don't quote me on this: It is just semantics.
- In the frame of the emitter it is called aberration due to receiver movement (what you describe above)
- In the frame of the receiver you say that due to finite signal velocity the signal comes from an old emitter position, and not from where the emitter is now.

16. Aug 11, 2009

### nutgeb

Agreed. In this example I don't care about an observer who is at rest in earth's frame. There is no contradiction.
Agreed. That's how I set up my scenario.

17. Aug 11, 2009

### nutgeb

That kind of confusion results when I have to respond to 3 different posters who all want to take the thread in different directions. I've tried to keep my own posts clear and specific, and note any differences between scenarios in a step by step process.
By Jove, Ich, you've hit the nail on the head! Aberration simply is not observed if the detector is "stationary" or moving at a constant radial velocity, and the source is moving. This is discussed in the Wikipedia article "Aberration of Light" and the referenced sources. E.g., this abstract from Edward Eisner's article in AmJPhys 1967:

"It is widely believed that aberration, like the Doppler effect, depends on the relative velocity of source and observer. It is here shown that, if this were true, binary stars would mostly look widely separated and rapidly rotating. Not only is this not observed, but it would appear to conflict with Kepler's third law if it were. It is argued that aberration does not depend on the relative velocities of source and observer: it depends only on the change in velocity of the observer between the times when the two measurements from which the aberration is deduced are made. The misconception is due to a faulty customary interpretation of the correct standard treatment."

This understanding of how aberration works has been studied by numerous authors who conclude that it is perfectly consistent with relativity, but in a non-intuitive way. (Again, for more explanation see the papers referenced in the Wikipedia article.) The relativistic aberration equation is applied only to the observer's change in velocity between two measurements. The obvious example is earth's velocity as it circles the sun, which is constantly changing in direction (during the orbital period) relative to a distant star.

I learned something new today! Previously I have been misled by some papers on this subject (and by prior threads on this forum.)

I am now changing my favored answer in my OP to (b). The Lorentz contraction of the distance between the moving (receding) disk and the stationary detector causes the apparent angular size of the disk to increase, such that the apparent angular size is the same that would be observed if the disk were stationary (relative to the detector) at the Lorentz contracted distance D'. Moreover, aberration has no effect on the apparent angular size of the disk.

Last edited: Aug 11, 2009
18. Aug 11, 2009

### A.T.

Just stick with one frame:

Detector frame:
- distance is not contracted
- same measured angular size as at rest

Disc frame:
- distance is contracted
- angle is bigger
- but this bigger angle is measured by a contracted detector, yielding the same measured angular size as at rest

19. Aug 11, 2009

### nutgeb

Sorry A.T., you have it wrong. The distance between the detector and the moving disk is Lorentz contracted as viewed in BOTH the detector and disk frames. Lorentz contraction depends only on relative velocity, and neither frame is 'preferred'.

Therefore the angle also is bigger in both frames.

It is probably meaningless to say that the detector is 'contracted', since it's contracted only in the radial dimension (not in the transverse dimension), which may be irrelevant (e.g., if it's a flat CCD detector or the like). And of course it's contracted only as viewed from the disk frame, the opposite of what you say. No object is ever contracted in its own rest frame!

Last edited: Aug 11, 2009
20. Aug 12, 2009

### Ich

I was referring to your OP, and my advice is still the same: Do not use those derived concepts (LC, TD, Aberration...), and don't rely on the literature to find out how they work. Check if out for yourself by using the generic coordinate approach, understand it, and then derive the concepts on your own.
That's not a good idea.

The only non-intuitive thing about aberration is the name itself, as it implies a deviation from some natural state.
If you observe a photon (event O) which has been emitted at an event E, its direction for the observer is simply the spatial part of (O-E) in the observer's frame. Nothing more to it. No length contraction, no aberration.
You have to stop reading and start calculating to see how everything fits together.

21. Aug 12, 2009

### A.T.

Using your definition of the word "distance": What is the distance between an emitter and an emitted photon in the frame of the emitter, 1 second after emission measured by a clock at rest in the emitters frame?
"Neither frame is preferred" means that the situation is symmetrical, not identical. Different frames measure different values for distance between the same two points.
When is it bigger? Definitely not when the moving disc passes the identical disc at rest in the detector's frame. All observers will see the discs at the same size at that moment.
How would you measure the apparent angular size with such a flat detector? You cannot. Every measurement device will have a certain radial size. And in the moving disc's frame every such device will be contracted in the same way as the distance, and the view cone formed by the device and the disc. Therefore the observer in the disc's frame will still conclude the the device will measure the same angle, as the one concluded by the observer in the device's frame.
This is not opposite of what I say. I mention the contracted detector only in the description of the disc's frame.
And that's why you claim that distance measurements by a ruler in his own rest frame are affected by the movement of some stuff relative to the ruler?

Last edited: Aug 12, 2009
22. Aug 12, 2009

### nutgeb

That's an interesting question AT, but not the subject of this thread. I believe that neither photons themselves, nor the distance between an emitter and an in-flight photon, are ever affected by Lorentz contraction. Light travels along null geodesics which are not Lorentz contracted in any observer's frame.

Not true, with respect to the frames of the two objects in motion relative to each other (which are the only two frames relevant to my scenario). If the relative speed as between object A and object B is .9c, each object (from the perspective of its own rest frame) views the distance between them to equally Lorentz contracted. Without regard to whether one of them is "stationary" and the other is "moving." This is a basic aspect of relativity.

I think I disagree with that but I'm not sure. Consider the OP scenario but with 2 additional stationary disks, one placed at proper distance D radially away from the detector, and the other placed at (the stationary equivalent of) the shorter distance D', all as measured in the detector's frame (using a ruler that is stationary in the detector's frame). Then the 3rd disk, the one moving radially away from the detector, begins at distance D and immediately reaches a relativistic radial velocity. We will choose to set the shorter distance D' equal to what the Lorentz contracted distance to the moving disk is, relative to the original uncontracted distance D, all as perceived in the detector's frame. For example, if the Lorentz contraction gamma factor of the moving disk is 2, distance D' will be half of distance D, as measured by the detector.

When the photons arrive at the detector, which were the first photons emitted from the moving disk after it began moving at distance D, will the apparent angular size of the moving disk be equal to the angular size of the stationary disk located (a) at distance D? or (b) at distance D'? The simplest answer is that the moving disk is adjacent to the stationary disk at D' at that instant, as viewed in the detector's frame. But one could also argue that it is adjacent to the stationary disk at D at the same instant, as viewed in the moving disk's frame.

These paradoxical results suggest to me that a failure of simultaneity is occuring here. Note that the stationary disk at D' does not ever view the moving disk to be adjacent to itself; at the first instant of movement, this stationary disk views the moving disk to be located halfway between D' and D. At the same time, the stationary disk at D views the moving disk to be directly adjacent to itself. This occurs even though the stationary disks at D' and D are both at rest in the same extended inertial frame. (The detector also is at rest in that same inertial frame).
OK, I misread what you said. The detector may be radially contracted in the disk's frame, but as Sylas pointed out, because the disk is moving away from the detector, the angular size viewed through a pinhole camera will be increased, not reduced, due to the increase in light-travel distance inside the camera, as seen from the moving disk's rest frame. If you disagree, I'll leave that discussion to you and Sylas, because what happens inside telescopes or cameras is tangential to the particular subject I'm interested in.
AT, you are the only writer I can recall who claims that the distance is not Lorentz contracted as between two objects which have relativistic radial velocities relative to each other. Instead of rolling your eyes, I suggest you read up and cite a reliable source you can reference to us. Either you misunderstand relativity, or you are misinterpreting the scenario. Making vague statements like "it depends on what you mean by distance" add no clarity at all to your arguments. Please state what you mean by the distance between two objects in relative radial motion.

Last edited: Aug 12, 2009
23. Aug 12, 2009

### A.T.

So when a rocket moving at 0.9c relative to earth passes by the moon, an observer on earth views the distance to moon & rocket Lorentz contracted? I can only hope that when the rocket is gone, the moon is back in it's place.

24. Aug 12, 2009

### nutgeb

If the moon and earth are both at rest in the same inertial frame, the distance between them will not be Lorentz contracted, as observed from either the moon or the earth.

However, at the instant when the moon observer sees the rocket passing by, the earth observer sees the rocket as being somewhere between the earth and the moon, not at the moon. If the gamma factor is 2, the earth observer will see the rocket to be at the midpoint between the earth and moon. The earth observer perceives the rocket to pass by the moon at a time when the moon observer perceives the rocket to still be some distance beyond the moon.

I believe that the earth and moon observers disagree with each other about the location of the rocket because of a failure of simultaneity in their respective relationships to the rocket.

25. Aug 12, 2009

### nutgeb

Ich I agree there is no aberration, as stated in my prior post.

Your statement about "no length contraction" is not clear. By definition, length contraction cannot affect the "direction" from which a single photon is observed. All received photons are received in a direction that is purely radial from the point of emission.

The situation is more complex when the emitter has a large transverse size. Then we consider two photons emitted simultaneously from opposite edges of the emitter. The respective radial approach paths of these "edge photons" will form an angle at the detector. (The transverse size of the detector is many orders of magnitude smaller than the transverse size of the emitter.)

If the emitter is moving radially away from the detector at a relativistic speed, then we need to determine whether the angle at which the 'edge photons' arrive at the detector is a function of distance D, or rather of distance D'. Distance D is the un-contracted distance to the emitter, and distance D' is the Lorentz contracted distance.