Relativistic effects on apparent angular size

[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?

I'm not asking about Luminosity Distance or flux in this question.

 

[nutgeb]

No takers on this question?

 

[A.T.]

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.

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/

 

[nutgeb]

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?

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'.

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

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.

- 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/

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.

 

[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.

 

[A.T.]

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.

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.

 

[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.

 

[A.T.]

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 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.

 

[nutgeb]

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.

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).

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.

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 (?)

 

[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

 

[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?

 

[sylas]

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?

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

 

[A.T.]

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.

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).

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.

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.

Clearly the rocket traveler measures the distance to Earth to be Lorentz contracted.

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.

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.

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.

 

[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?

 

[A.T.]

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?

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.

 

[nutgeb]

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.

Agreed. In this example I don't care about an observer who is at rest in Earth's frame. There is no contradiction.

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.

Agreed. That's how I set up my scenario.

 

[nutgeb]

You need to use coordinates consistently, not apply some relativistic concepts to situations where they don't fit.

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.

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?

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.

 

[A.T.]

That's how I set up my scenario.

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

 

[nutgeb]

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

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!

 

[Ich]

You need to use coordinates consistently, not apply some relativistic concepts to situations where they don't fit.

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 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.

I am now changing my favored answer in my OP to (b).

That's not a good idea.

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.

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.

 

[A.T.]

The distance between the detector and the moving disk is Lorentz contracted as viewed in BOTH the detector and disk frames.

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?

Lorentz contraction depends only on relative velocity, and neither frame is 'preferred'.

"Neither frame is preferred" means that the situation is symmetrical, not identical. Different frames measure different values for distance between the same two points.

Therefore the angle also is bigger in both frames.

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.

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)

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.

And of course it's contracted only as viewed from the disk frame, the opposite of what you say.

This is not opposite of what I say. I mention the contracted detector only in the description of the disc's frame.

No object is ever contracted in its own rest 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?

 

[nutgeb]

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?

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.

"Neither frame is preferred" means that the situation is symmetrical, not identical. Different frames measure different values for distance between the same two points.

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.

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.

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 occurring 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).

This is not opposite of what I say. I mention the contracted detector only in the description of the disc's 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.

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?

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.

 

[A.T.]

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.

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.

 

[nutgeb]

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.

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.

 

[nutgeb]

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.

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.

 

[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.

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.

Ahhhh... so the Earth and moon observers, which are at rest to each other, disagree on simultaneity. Amazing.

 

[sylas]

Ahhhh... so the Earth and moon observers, which are at rest to each other, disagree on simultaneity. Amazing.

I take it this is sarcasm? I'm sometimes tone deaf to various nuances.

I agree with Ich , [post=2305919]msg #20[/post].

Nutgeb, you appear to like working from secondary considerations like contraction and dilation, rather than working with the underlying transformations directly. That's not a good idea.

This statement also makes no sense:

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.

Lorentz contraction refers to a ruler, or something with an associated distance. For a photon, the wavelength qualifies... and it IS affected by Lorentz contraction. The trajectory of something (photon, particle, anything you like) is not a ruler. A ruler has TWO endpoints, and hence two trajectories. You can have a photon going from one end of a ruler to another, and the distance between the two events is Lorentz contracted depending on the observer, of course.

Nutgeb... you badly need to go back to basic transformations. When someone tries to think in terms of Lorentz contraction and dilation and so on, it's very easy to go wrong. Go back to the basic Lorentz transformation, and Lorentz contraction or time dilation and so on will fall out correctly. Go the other way around, and its too easy to make mistakes.

Cheers -- sylas

 

[nutgeb]

Nutgeb, you appear to like working from secondary considerations like contraction and dilation, rather than working with the underlying transformations directly. That's not a good idea.

Sylas, it's easy to critisize without making any effort to actually answer the question. If you're as good at this as you like to pretend, demonstrate the transformation for all of us that proves that the definitive answer is either (a) or (b).

You can have a photon going from one end of a ruler to another, and the distance between the two events is Lorentz contracted depending on the observer, of course.

Of course. But in your trivial scenario, the stationary observer (at the near end of the ruler) does not measure the distance to the photon to be Lorentz contracted by virtue of the the photon's own recession velocity. That doesn't happen with a null geodesic.

When a non-photon object, such as a disk, is moving radially away from the observer at a relativistic speed, the observer does measure the distance to the disk to be Lorentz contracted. Please make yourself clear, do you agree or disagree with this statement? A simple yes or no will suffice.

 

[sylas]

When a non-photon object, such as a disk, is moving radially away from the observer at a relativistic speed, the observer does measure the distance to the disk to be Lorentz contracted. Please make yourself clear, do you agree or disagree with this statement? A simple yes or no will suffice.

No.

What is contracted is the thickness of the disk; that is the distance along the direction of motion between one end and the other. The distance to the disk is independent of the disks motion.

I am not pretending anything. That's unnecessarily rude.

Cheers -- sylas

 

[nutgeb]

No.

What is contracted is the thickness of the disk; that is the distance along the direction of motion between one end and the other. The distance to the disk is independent of the disks motion.

Thanks for the straight answer. Unfortunately it's wrong. Please read section 4.7 of the Taylor & Wheeler's textbook which I linked to in an earlier post.

I am not pretending anything. That's unnecessarily rude.

OK I apologize. I would appreciate if instead of telling me to "go work the problem", you show me how you work it to obtain a different result from mine -- or point me to a reliable source that shows it. Especially when the question relates to the heuristics rather than to the math itself. It's easy to generate an answer that is mathematically correct but does not correctly reflect the workings of a particular scenario.

I will offer one other thought about my scenario. It is a basic tenant of relativity that clocks which are synchronized in one frame (say, in frame S) must be unsynchronized from the perspective of a different frame (say, S') which is moving at a constant velocity relative to the first frame. The line of simultaneity is horizontal in S, but it is angled in frame S'.

Therefore, in AT's scenario, if the rocket passes next to the moon and then passes next to the earth, the rocket will perceive the Earth and moon clocks to be unsynchronized from each other, despite the fact that those two clocks are synchronized in the earth-moon inertial frame. Thus in the rocket's frame, when it sees itself just passing the moon, the rocket observer will calculate that the Earth clock will record a different time than the moon clock. There is no agreement between any two of the three clocks as to when the rocket passed the moon. This happens despite the fact that two of the clocks are in the same inertial frame; they are in different spacetime locations.

 

[A.T.]

Therefore, in AT's scenario, if the rocket passes next to the moon and then passes next to the earth, the rocket will perceive the Earth and moon clocks to be unsynchronized from each other, despite the fact that those two clocks are synchronized in the earth-moon inertial frame. Thus in the rocket's frame,...

Why do you keep talking about the rocket's frame and rocket passing the earth, while completely ignoring my very simple question about the observer in the Earth's frame? Let me reiterate it:

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?

Note that "rocket passing by the moon" is a single event in spacetime. And I'm interested in the length measured in a single frame of reference: the Earth's frame. So please stop bringing up "failure of simultaneity".

 

[sylas]

Thanks for the straight answer. Unfortunately it's wrong. Please read section 4.7 of the Taylor & Wheeler's textbook which I linked to in an earlier post.

I think you do not understand that basic text.

Let me make the situation of your original post precise, using actual numbers. We have a disk, at a distance of 6 light years from an observer. It is at rest with respect to the observer. The disk is 1 meter thick, and 10 meters in radius, in the rest frame of the observer, and face on to an observer.

We accelerate the disk, in a very short period of time (1 nanosecond), up to 60% of light speed, directly away from the observer.

In the frame of the observer, the disk is now 80 centimeters thick. It is Lorentz contracted, along the direction of motion. During the acceleration, it can have moved no more than 18 centimeters, which is 60% of a light-nanosecond. Hence it is still 6 light years distant, and it is still 10 meters in radius.

Of course, it takes six years from the acceleration for the observer see the disk undergo that acceleration.

There is no Lorentz contraction of the distance to the disk. There is a Lorentz contraction of the disk itself.

The angular size of the disk depends on its radius, not its thickness. That is unaltered.

We know that the distance to the disk is unaltered by the velocity of the disk, because a twin disk which was not accelerated is also seen six years later. The light travel time from the disk to the observer, as measured by the observer, is independent of the velocity of the disk.

Why you think that the act of accelerating the disk results in a different distance, I am unsure. You apparently think Taylor and Wheeler imply such a thing. But they don’t. It is true that distances to things can be Lorentz contracted… for example, if it was the observer that was accelerated rather than the disk.

But in the case you describe, there is no significant effect on the distance to the disk over the brief interval of acceleration, because the disk hasn’t had time to move to a new location. There’s no Lorentz contraction involved.

OK I apologize. I would appreciate if instead of telling me to "go work the problem", you show me how you work it to obtain a different result from mine -- or point me to a reliable source that shows it. Especially when the question relates to the heuristics rather than to the math itself. It's easy to generate an answer that is mathematically correct but does not correctly reflect the workings of a particular scenario.

As I have said before, I find it very hard indeed to follow your reasoning. It might just be me; there are other folks who seem to follow rather better. So I don't know where you are going wrong, and I often find it hard to tell what you are saying. In the initial posts of this thread, however, the situation seems clear enough, and corresponds to what I have tried to express above in numbers. If this is what you did intend to describe, then you are making inferences that do not follow from Taylor and Wheeler.

My advice -- for whatever it is worth -- remains precisely the same as given previously. Specify clearly the events you mean, in some inertial frame. Use Lorentz transformations to get the time and distance in another inertial frame.

The Lorentz transformations are independent of the velocity of objects being observed. They just depend on the frame of the observer, and let you map co-ordinates of an event between different frames.

Cheers -- sylas

 

[DrGreg]

nutgeb

Lorentz contraction applies when different inertial observers each measure the distance between the same two objects that are a constant distance apart. Because each measures the distance to be constant, it doesn't matter exactly when each observer makes their measurement.

When two objects are moving relative to each other, it's not so simple. As each observer now measures a varying distance, it matters how the two observers' measurements are synchronised to each other. There is no simple yes/no answer as to whether length contraction occurs, it depends on the details of a specific problem and when each observer makes their measurement.

So simply claiming that Lorentz contraction occurs is meaningless unless you spell out exactly which two measurements you are comparing and exactly when each measurement is taken. The advice that others have given you, to forget Lorentz contraction and apply the full Lorentz transformation, is good advice.

Another misconception that seems to arise on this forum from time to time is that Lorentz contraction applies to two different measurements made by the same observer. There's no automatic reason for that to be true, unless there are additional reasons such that the conditions I spelled out in my opening sentence can be applied.

Now, considering the specific example in this thread of the rocket flying past the moon. Both the rocket and the moon are (momentarily) at the same place at the same time. This is something all observers must agree on. Let's say the rocket actually crashed into the moon with an explosion. Any observer watching this would see the moon and the rocket approach each other and the explosion occur just as they met. And any observer is forced to agree that the distance to the moon, the distance to the rocket and the distance to the explosion are all equal at that event (and very nearly equal just before the explosion). It would be nonsense to say that the distance to the rocket was half the distance to the moon milliseconds before the explosion (both distances being measured by the same observer).

 

[Austin0]

#1 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.

#2 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".

Cheers -- sylas

#1 Would seem to mean that angular distortion doesn't occur because the distance as measured in the observer frame is by definition, not contracted, and the image must arrive from exactly where it was located .

#2 The pinhole camera illustration is very clear .
Is the interpretation of this, that angular size distortion as observed from a moving frame is purely an optical effect occurring in a camera or or human eye?
In light of #1 this would seem a necessary conclusion.

Thanks

 

[sylas]

#1 Would seem to mean that angular distortion doesn't occur because the distance as measured in the observer frame is by definition, not contracted, and the image must arrive from exactly where it was located .

I'm not sure what you mean.

The angular size of an object of a known size can be calculated directly in the frame where the observer is at rest. In this frame, you simply need to know the distance from the object being viewed to the observer. Put together with the actual size of the object, and you can calculate the subtended angle.

There's no "contraction" involved in this; it's the usual way of calculating subtended angles.

#2 The pinhole camera illustration is very clear .
Is the interpretation of this, that angular size distortion as observed from a moving frame is purely an optical effect occurring in a camera or or human eye?
In light of #1 this would seem a necessary conclusion.

The second calculation is in a frame where you infer the angular size that is apparent to an observer, represented as the camera. This is calculated using numbers for a different frame. The camera is "Lorentz contracted", and the distance to the event of emission of photons used to view the image is different also. Plus there's the effect of moving towards or away from the photon stream.

But put it together, and you get the same answer as calculated entirely in the observer's frame.

Cheers -- sylas