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There is a square moving at a constant speed across an inertial frame of reference.

Relative to an observer at rest in the inertial frame of reference the square is moving diagonally.

The moving square is length contracted in its direction of motion.

(A diagonally moving square is diamond in its direction of motion. Contracted, it becomes a rhombus.)

And so, from the perspective of the observer, the height of the diagonally moving and length contracted square is the same when it is at rest or when it is in diagonal motion.

There is also a rectangle moving at a constant speed across this same inertial frame of reference.

Relative to the same observer the rectangle is moving vertically.

The moving rectangle is length contracted in its direction of motion.

And so, from the perspective of the observer, the height of the vertically moving rectangle is less when in vertical motion than when at rest.

From the perspective of the observer, in the inertial frame of reference of the observer, the height of the diagonally moving square in motion is the same as when it is at rest and the height of the vertically moving rectangle in motion is less than when it is when at rest.

The square is inside the rectangle. When at rest, the height of the square is slightly less than the height of the rectangle.

The square moves diagonally down and to the right. The rectangle moves straight down.

The vertical component of the square’s velocity is the same at the rectangle’s vertical velocity.

The moving square contracts in its direction of motion. The moving rectangle contracts in its direction of motion.

(The velocity of the square is more and so it contracts more in its direction of motion.)

The height of the square remains the same when at rest. And the height of the rectangle decreases.

And so, with the height of the square only slightly less than the height of the rectangle when at rest, then, assuming the velocities relative to the observer are considerable, the length contracted square will be beyond (break through, depending on the materials) the bounds of the length contracted rectangle.

The Special Theory of Relativity predicts, in the inertial frame of reference of the observer, the square will not remain within the walls of the rectangle.

1. From the perspective of the square at rest, where the rectangle and the observer are in motion, does the square remain within in the rectangle or does the square break through the rectangle?

2. From the perspective of the rectangle at rest, where the square and the observer are in motion, does the square remain within the rectangle or does the square break through the rectangle?

3. Does the square break through the rectangle in one inertial frame of reference while the square remains with the rectangle in other inertial frames of reference?

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This is a version of the pole in the barn and the barn doors or the train in the tunnel paradox.

The solution to all these types of apparent paradoxes is the same.

You need to take relativity of simultaneity into account when measuring distance between extremities.

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In the ladder/barn/pole "paradox" the ladder and the barn doors only encounter each other at a specific moment in time.

The ladder moves through the one door and then the ladder moves through the other door.

Here, time is not necessarily a factor.

This thought experiment could be modified to say "the square and rectangle are moving like this relative to one another over 1 billion year" or "the square and rectangle are moving like this relative to one another over an infinite amount of time"

If there is a way to get the square to break through the rectangle in the square's rest frame and if there is a way to get the squrae to break through the rectangle in the rectangle's rest frame with "time" and with "non-synchronous events" I haven't been able to do that yet.

But, perhaps you seem to have made it work.

I don't know.

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I have a different understanding of your conundrum. It appears to me that you are not transforming the motions (hence the geometry) correctly.

If your initial condition is a (1w x 1h) square moving diagonally and a (2w x 1h) rectangle moving vertically, the square will be a diamond sticking out of the compressed rectangle. OK

However, if you transform to a reference frame where the square is at rest, it will become a square again, OK, but the rectangle will have diagonal motion and be a parallelogram. You don’t show that. Determining the dimensions of that parallelogram will give you a figure into which you cannot fit the square.

I’m saying that on faith. I have not done the calculations. They are complicated. If you don’t feel up to doing them I could do them but I will need a few days.

The important part is that, given the initial conditions, there cannot be a reference frame where the square is a square and the rectangle is a rectangle both having vertical and horizontal sides.

Correction to this last statement......... There may actually be such a reference frame. But the overall dimensions of the two shapes will change such that there will be no paradox.

Edited by Mike-from-the-Bronx
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In the ladder/barn/pole "paradox" the ladder and the barn doors only encounter each other at a specific moment in time.

The ladder moves through the one door and then the ladder moves through the other door.

Here, time is not necessarily a factor.

This thought experiment could be modified to say "the square and rectangle are moving like this relative to one another over 1 billion year" or "the square and rectangle are moving like this relative to one another over an infinite amount of time"

If there is a way to get the square to break through the rectangle in the square's rest frame and if there is a way to get the squrae to break through the rectangle in the rectangle's rest frame with "time" and with "non-synchronous events" I haven't been able to do that yet.

But, perhaps you seem to have made it work.

I don't know.

It is still the same paradox and the issue is the same.

You have to reduce the time coordinates to place the corners of either figure at the same time in the same frame. In other words establish the relativity of simultaneity as I said.

If the motion is as slow as you now say it is, then both the length and time change effects will be tiny.

Both effects are changed by the same factors so if one is small or large, then so is the other.

Edited by studiot
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Hello Mike from the Bronx, I’m from Mesa.

“but the rectangle will have diagonal motion and be a parallelogram. You don’t show that.”

Right.

In this thought experiment, three frames of reference are noted: the observer, the square, and the rectangle.

If I were to add more frames of reference there would be a frame of reference in which the rectangle is moving diagonally and so it would contract diagonally in that frame of reference.

In the three frames of reference noted here, however, the rectangle happens to not be moving diagonally in any of them.

In the observer’s frame of reference the rectangle is moving vertically, and so it contracts vertically.

In the square’s frame of reference the rectangle is moving horizontally, and so the rectangle contracts horizontally (… and the square remains within the rectangle).

In the rectangle’s frame of reference the square is moving horizontally, and so the square contracts horizontally (… and the square remains within the rectangle).

The Special Theory of Relativity, as I’ve come to understand, leads us to some weird conclusions, … but … in the end whatever happens in one frame of reference (e.g. the square colliding with the rectangle) must also happen in every other frame of reference (e.g. the square cannot not collide with the rectangle).

We can look at this from the perspective of the rectangle moving diagonally. But, it seems to me just looking at this from the perspective of the observer vs from the perspective of the square (and rectangle) there is already a problem.

No?

I am always willing to learn!

Hello studiot!

In the ladder and barn “paradox” the two clocks in question are in motion (… from the perspective of someone at rest relative to the moving barn). And when clocks are in motion “leading clocks lag.” And so, the ladder is within the closed barn doors even though the ladder is too long to fit but it’s able to do so due to synchronized clocks in one inertial frame of reference not being synchronized in another inertial frame of reference.

In this thought experiment (the “diagonal motion paradox”) the height of the square is not being measured by moving clocks in the square’s inertial frame of reference.

The two clocks in the drawing above, … measuring the length contracted square … are necessarily in the observer’s frame of reference. The observer measures the square based on clocks in his inertial frame of reference and not based on clocks in another (the square’s) inertial frame of reference. And since the two clocks are in the observer’s inertial frame of reference, the Special Theory of Relativity allows for them to be synchronized (… for the observer).

No?

Edited by Thales et al
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Ok.

Looks like I did not understand the initial setup.

Let me try again.

Initial Condition:

Rectangle moving horizontally (width contracted, height unchanged).

Square at Rest (fits in the rectangle, no contraction).

Third observer moving diagonally (diagonally contracted).

Switch reference frames to the rectangle.

Rectangle at rest (no contraction).

Square moving horizontally (width contracted, still fits in rectangle)

Third observer now moving vertically (contraction changed).

That's not a very challenging problem.

Switch reference frames from initial setup to Third Observer.

Rectangle now moving diagonally(with additional contraction).

Square now moving diagonally (with contraction).

Third Observer at rest.

Let me stop here and make sure I am thinking what everybody else is thinking. If the third observer started out moving diagonally and we want to describe the world from the point of view of that observer, we are saying that observer is now at rest. Both shapes which previously had no diagonal velocity must now have some.

Is that what is being said?

P.S. With regard to the pole-in-the-barn paradox, there are some similarities and at first I thought, OK that's a good analogy. But there is no diagonally moving observer in pole-in-the-barn. A better analogy would be the meterstick-in-the-hole where diagonal motion is the focus.

P.S. And yes I see Relativity of Simultaneity being involved all over the place in this problem.

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Alright … first things first … the most important question first … are you really from the Bronx?

“Switch reference frames from initial setup to Third Observer.

Rectangle now moving diagonally(with additional contraction).

Square now moving diagonally (with contraction).

Third Observer at rest.”

In the third observer’s inertial frame of reference the rectangle is moving vertically (not diagonally) but the square is moving diagonally.

In the third observer’s inertial frame of reference the rectangle is contracted vertically and the square is contracted diagonally.

Well, at least that’s what I tried to say.

More questions?

And thank you for taking my idea here seriously.

If there is a way to use “relativity of simultaneity” to resolve this “paradox,” I can’t see it.

1. Clearly in the square’s rest frame the square remains within the rectangle.

2. And clearly in the rectangle’s rest frame the square remains within the rectangle.

And so,

It must be that in the observer’s rest frame the diagonally moving square also remains in the vertically moving rectangle (… for there not to be a paradox).

And so I suppose you could say that the bottom of the diagonally moving square is inside the vertically moving rectangle but at an earlier time and the top of the diagonally moving square is also inside the vertically moving rectangle but at a later point in time?

But this leads to the contradiction of the diagonally moving square being both longer and shorter (in its y dimension) at the same time (in the inertial frame of reference of the observer). Using relativity of simultaneity in this way would mean contradicting basic relativistic length contraction.

As far as I can tell, in the inertial frame of reference of the observer the length contracted square is outside the bounds of the length contracted rectangle.

?

Edited by Thales et al
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Yes, I grew up in the Bronx, East 223rd Street, long time ago now. I know the flavor of lead paint very well. I used to suck on it.

With respect to the rectangular shape, yes, now I see that the OP intended that the angle of the velocity of the third observer contained a component that would result in the rectangle only having vertical velocity when the third observer was at rest. Good catch, twice.

So now I agree with the last diagram; Rectangle at rest, square moving horizontally, third observer moving vertically.

Last thing to correlate:

Revised..................................................

When we switch reference frames to the third observer, the rectangle goes from rest to having vertical velocity (height contracted). The square already has horizontal velocity (already contracted horizontally) and now it acquires some vertical velocity so that its total velocity is somewhat diagonal and its contraction is somewhat diagonal. So the concern is weather the vertically contracted rectangle can still hold the diagonally contracted square.

The key I would focus on is that a square contracted diagonally in 2 steps does not have the same shape as a square contracted diagonally in 1 step. That's how it still fits in the rectangle.

This would be a good exercise:

Start with a square and transform its shape to that observed by someone moving at .8c in the x-direction.

Answer: Square is contracted by 40% in width and moving at .8c in the x-direction.

Then:

Transform that shape to that observed by someone moving at .8c in the y-direction.

I'm too tired right now to do it. But the final velocity is not at 45 degrees and the shape does loose height.

Edited by Mike-from-the-Bronx
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In the observer’s frame of reference, the square’s motion is diagonal. However, we can talk about the vertical component of the square’s velocity and the horizontal component of the square’s velocity.

It is my understanding of the Special Theory of Relativity that length contraction only occurs in the moving body’s actual direction of motion. It is my understanding of the Special Theory of Relativity that length contraction does not occur in a moving body’s “component” directions.

No?

The square is moving diagonally and so it should contract diagonally.

The square has a component of its motion moving vertically, but it does not (… or so I believe …) contract vertically. Nor should it contract horizontally. “Moving bodies contract in their direction of motion.”

Am I right?

"The key I would focus on is that a square contracted diagonally in 2 steps does not have the same shape as a square contracted diagonally in 1 step. That's how it still fits in the rectangle."

Can we do length contraction like this? Am I wrong that length contraction occurs in the direction of overall motion of a body?

Edited by Thales et al
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In the observer’s frame of reference, the square’s motion is diagonal. However, we can talk about the vertical component of the square’s velocity and the horizontal component of the square’s velocity.

It is my understanding of the Special Theory of Relativity that length contraction only occurs in the moving body’s actual direction of motion. It is my understanding of the Special Theory of Relativity that length contraction does not occur in a moving body’s “component” directions.

No?

The square is moving diagonally and so it should contract diagonally.

The square has a component of its motion moving vertically, but it does not (… or so I believe …) contract vertically. Nor should it contract horizontally. “Moving bodies contract in their direction of motion.”

Am I right?

"The key I would focus on is that a square contracted diagonally in 2 steps does not have the same shape as a square contracted diagonally in 1 step. That's how it still fits in the rectangle."

Can we do length contraction like this? Am I wrong that length contraction occurs in the direction of overall motion of a body?

But your directions of motion are different in different frames and you are switching frames.

When you switch frames, the change of direction constitutes a rotation and a rotation is necessarily an acceleration.

As Mike observes, this is a double Lorenz transformation and follows different rules.

As it happens, Mordred has just posted these rules in another thread. See post#40

http://www.scienceforums.net/topic/105118-lorentz-transformations-split-from-why-nothing-c/page-2

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Can we do length contraction like this? Am I wrong that length contraction occurs in the direction of overall motion of a body?

Actually, that only applies transforming from rest to some non-zero velocity. Going from some non-zero velocity to another results in non-intuitive geometry changes.

Here is a reference to a Tutorial I wrote;

http://www.relativitysimulation.com/Tutorials/TutorialMeterstickAndHole.html

The meter-stick-and-the-hole is a well analyzed "paradox" that was even written up in the American Journal of Physics. (Trying to find the reference now) When the problem was first given to me I went crazy. I was sure the paradox as real! What can I say. I'm a slow learner.

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Okay … maybe I am wrong.

If a square is moving diagonally across an inertial frame of reference is that square’s height knowable?

Is the square’s height one thing if we look at it one way and is the square’s height a different thing if we look at it a different way (at the same time in the same inertial frame of reference)?

If that’s what I’m hearing that doesn’t seem right to me.

It seems like the square’s height and the rectangle’s height should be knowable and solely based on their direction of motion relative to the given inertial frame of reference.

And so, it seems to me that in the inertial frame of reference of the observer the height of the diagonally moving square should stay the same and the height of the vertically moving rectangle should decrease.

But … that would mean a paradox.

I’ll keep studying!

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Okay … maybe I am wrong.

If a square is moving diagonally across an inertial frame of reference is that square’s height knowable?

Is the square’s height one thing if we look at it one way and is the square’s height a different thing if we look at it a different way (at the same time in the same inertial frame of reference)?

If that’s what I’m hearing that doesn’t seem right to me.

It seems like the square’s height and the rectangle’s height should be knowable and solely based on their direction of motion relative to the given inertial frame of reference.

And so, it seems to me that in the inertial frame of reference of the observer the height of the diagonally moving square should stay the same and the height of the vertically moving rectangle should decrease.

But … that would mean a paradox.

I’ll keep studying!

Reads like you are going down the same tortuous path I went down when I first learned this aspect of SR. I'm not sure I would want to describe how I finally figured it out. But here are some hints.

First recognize that velocities don't add like vectors in SR.

So adding .7c in the y-direction to an existing .7c in the x-direction does not give a resultant velocity of (vx, vy) = (.7, .7). That's .99 at a 45 degree angle. Symmetrical diamond.

It gives a velocity (vx, vy) of (.5, .7). So you don't get a symmetrical diamond.

You can get the original dimensions (a square) back if you do all the transformations in reverse order. But what if many transformations were cascaded one after another and you don't know what they were? You are observing this strange shape moving in some strange direction. Can you just transform to a reference frame in which the object is at rest and get the square?

Answer: Sort of. You will get a square but it will be rotated.

At this point I am over my head and I will leave it to others to explain that.

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What if the observer is inside the square ???.

In other words what if the square & rectangle are over & surround the observer (who is standing at 0,0) ???.

I always have difficulty trying to follow these sorts of thought experiments (referring now to the OP).

Does the observer see a moving square-diamond ??. How ??. Wouldn't the lights (from say the 4 corners) take 4 times (or at least 3) to reach the eye ??. ((Likewise for the moving rectangle-diamond)).

I suppose that the observer is looking somewhat down on the square-diamond, otherwise he will see 2 points joined by a line instead of 4 corners joined by 4 lines (or 4 curves). ((Likewise for the rectangle-diamond)).

Or does the observer deduce a moving square-diamond ??. Here he sees apparent information (perhaps a pix taken by a camera), & then calculates or reasons the corrected true information (eg the position of the 4 or 8 corners at a particular time).

Would his deducing need a fine grid of small rods marking out the xx's & yy's ??.

Or would it need lots of observers with lots of clocks. Or all three ??.

I am not sure whether Q1 Q2 Q3 & Q4 relate to what is seen by 3 or more observers (in which case 3 or more pixes will answer the questions), or whether the questions relate to what is deduced (ie what is apparent or what is true).

It might be that what is seen in a certain frame is not true in that frame. Especially perhaps because what is seen depends on light taking time to reach the eye.

The main question, Q5, isn't actually stated, it is inferred (Q5 -- STR says that 2 corners will break out of the rectangle, true or false).

Here again, does this mean seen or does it mean deduced.

Re Q4 -- I reckon that it is a paradox if some frames see breaking out & some don't.

And i reckon that it is a paradox if in some frame u see breaking out but deduce that in fact there is no breaking out.

But if all seeings & all deductions are all in agreement then there is no paradox.

But i don't know how this will end -- it might end up being a paradox, or not (dunno).

Plus, what if the observer is inside the square. If this makes much difference then here we might have another paradox.

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1)

The meter-stick-and-the-hole is a well analyzed "paradox" that was even written up in the American Journal of Physics. (Trying to find the reference now) When the problem was first given to me I went crazy. I was sure the paradox as real! What can I say. I'm a slow learner.

2)

Answer: Sort of. You will get a square but it will be rotated.

1) We had a long thread about a video of the companion 'train in the tunnel' version. I will try to find it unless someone else can.

2) I believe you are referring to 'Thomas precession'

https://en.wikipedia.org/wiki/Thomas_precession

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...

Answer: Sort of. You will get a square but it will be rotated.

...

1) We had a long thread about a video of the companion 'train in the tunnel' version. I will try to find it unless someone else can.

2) I believe you are referring to 'Thomas precession'

https://en.wikipedia.org/wiki/Thomas_precession

Or is it Terrell-Penrose rotation that Mike is referrng to (which is the combination of actual SR contraction and visual distortion due to finite/invariant speed of light transmission to observer). There was a superb animation explaining that posted a while back too. I will look for both

https://faraday.physics.utoronto.ca/PVB/Harrison/SpecRel/Flash/ContractInvisible.html Terrell rotation and the inability to actually observe length contraction but we actually see rotation

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Or is it Terrell-Penrose rotation that Mike is referrng to (which is the combination of actual SR contraction and visual distortion due to finite/invariant speed of light transmission to observer). There was a superb animation explaining that posted a while back too. I will look for both

https://faraday.physics.utoronto.ca/PVB/Harrison/SpecRel/Flash/ContractInvisible.html Terrell rotation and the inability to actually observe length contraction but we actually see rotation

I appreciate the suggestion, but Lord No. I stay away from visual effects in SR. I understand the existence of such effects but dealing with them is beyond me.

I was referring to the following phenomena in SR:

1. Start in an inertial reference frame for which an object is at rest. Note the proper dimensions and (proper) shape of the object as given by the coordinates.

2. Sequentially perform multiple transformations to other inertial reference frames with different relative velocities in 2D or 3D space.

3. Transform back to a reference frame for which the object is at rest. Note the proper dimensions and (proper) shape of the object as given by the final coordinates.

My experience (from doing shape transformations by the thousands) is that the final coordinates will define the original proper shape except that the shape will be rotated by some amount that is a function of the combined transformations. This SR behavior has been documented but I don't have a reference or name for it right now. studiot, in post above, suggested Thomas Precession.

P.S. For any laymen reading this, note that transforming a shape, like a square, in SR is not as straightforward as just applying the Lorentz Transformation to the 4 corners. The LT will give you 4 coordinates in space for 4 different times. Relativity of Simultaneity.

To determine the shape in the new reference frame, you have to adjust the spacial coordinates for one common time. If you are going to perform all 3 steps, you also have to calculate the relative velocity of the object with respect to each reference frame so that you know how to get back to zero relative velocity at the end.

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You are right that it takes time for light to reflect off an object and then reach the observer’s eyeballs.

I made the mistake in my OP of just talking about “the observer.”

In these kind of thought experiments the idea is not that there is only one observer or only one observer per frame of reference. I was using the term “the observer” to distinguish his frame of reference from the frame of references of the square and the rectangle.

What is missing from my OP is the idea of a “lattice of clocks.”

If we were trying to do something as simple as see how long a moving car is in the frame of reference of the road at rest and how long the car is in the frame of reference of the car at rest, we would need a lattice of clocks with an observer next to each clock. And we would need a lattice of clocks (with observers next to them) in the road’s rest frame and in the car’s rest frame.

(Obviously, no matter how close the observers are to each clock it will take an amount of time, very very small perhaps, for the light reflected clock readings to reach the observer’s eyes. But, it’s my understanding that the professional academics just kind of allow for this to be so small so as to be reasonably ignored.)

And, yes, just as in the drawing above where there are observers in both frames of reference, in the OP, I should have included observers in the square’s frame of reference and observers in the rectangle’s frame of reference (as well as more observers in, what I’ve been calling, “the observer’s” frame of reference).

And so, there are some times when things appear to perhaps violate the Special Theory of Relativity. For example, if someone had a laser pointer and moved it across the night’s sky really fast someone could try to make the claim that it is moving faster than the speed of light from one distant body in the night’s sky to the next. But, if you could make this argument, it’s only an argument about appearances. Nothing is actually moving faster than the speed of light.

If I understand Special Theory of Relativity “paradoxes” right, then they are only true paradoxes if the paradoxical thing is actually happening physically. If it only appears to be a parardox, it is not. And if it is actually a paradox (occurring in physical reality), but doesn’t appear that way, it still is a paradox.

I hope what I’ve said here addresses what you were getting at.

Bottom line: I should have included a whole bunch of clocks (with observers by each one) in the “observer’s” frame of reference as well as in the square’s and rectangle’s frame of references. And I should have stipulated that all of the clocks in the observer’s frame of reference are synchronized with one another and all of the clocks in the square’s frame of reference are synchronized with one another and all of the clocks in the rectangle’s frame of reference are synchronized with one another. (The Special Theory of Relativity allows for clocks in the same inertial frame of reference to be synchronized.)

Sorry for the confusion.

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What is missing from my OP is the idea of a “lattice of clocks.”

What is missing from that is the relativity of simultaneity.

ROS governs the synchronisation of your 'clock lattice'

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What is missing from that is the relativity of simultaneity.

ROS governs the synchronisation of your 'clock lattice'

Right.

I understand that now.

When I was first thinking about this idea, I was thinking that "relativity of simultaneity" had no relevance here (and I said as much in some previous posts).

I don't understand how this "paradox" is resolved yet, but I understand that it has to do with "switching frames" and the "relativity of simultaneity" and "the square/diamond losing its square/diamond" shape.

Again, I don't understand (yet) how this all works and how this "paradox" is then resolved ... but I will ... thank you (and everyone else) for the direction.

By "what is missing is a lattice of clocks" I believe is "what is missing" in my OP in terms of madmac's question.

-

"Although Thomas precession (net rotation after a trajectory that returns to its initial velocity) is a purely kinematic effect, it only occurs in curvilinear motion and therefore cannot be observed independently of some external force causing the curvilinear motion such as that caused by an electromagnetic field, a gravitational field or a mechanical force, so Thomas precession is usually accompanied by dynamical effects.[1"

In this thought experiment ("the diagonal motion paradox") there is no external force ... right?

When we go from looking at this from the "third observer's" frame of reference and then looking at this from the square's frame of reference there is a "shift" that takes place. It is a "shift" in perspective. There is no "external force" like that from an "electromagnetic field" or from an "gravitational field."

Does "Thomas precession" apply to this thought experiment?

I am confused.

-

-

I’m trying to understand “relativity of simultaneity” as it relates to this “paradox.”

One thing that can be noticed, is that if the square has four clocks, one on each corner, and if it is moving diagonally, then while the four clocks will be all synchronized in the square’s inertial frame of reference, with the diagonal motion, in the “third observer’s” inertial frame of reference the square’s two clocks perpendicular to the direction of the motion (A and B) will be synchronized while the leading clock © will be behind A and B and the trailing clock (D) will be ahead of A and B.

Am I on the right track for a resolution?

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I had a think. It is a paradox. And all one needs is 3 pix taken by 3 observers.

The pix taken by the original observer (pun intended) will always show 2 corners of the square outside the rectangle. Although the pix will not show the 8 corners at the same moment, due to the different distances travelled by the photons, but this isn't relevant -- the 2 corners will be outside at all times. And the pix will not show the sides as being straight -- but once again this is not relevant.

The pixes taken by the 2 observers in the other 2 frames will show the square inside the rectangle. Once again the 8 corners will not be shown at the same moment, & once again this is irrelevant -- the square will be inside at all times.

I suspect that Einsteinians are able to explain that this paradox is not a paradox because a proper analysis of what happens in the original observer's frame shows that the square does not break out of the rectangle.

Hmmmm, why did the OP specify that the frames are inertial & velocities constant -- it makes no difference to the paradox.

But having the original observer at the origin might be crucial -- if elsewhere he-she might see differently (paradox No2 ??).

The square & rectangle OP is in effect another version of our beloved train thought experiments.

A train & driver is rolling along rails (past a farmer), & another train & driver crosses on an overpass at such an angle & speed that the lower train appears to move diagonally.

The question being -- does the train derail??

Warning -- this is a trick question.

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Right.

I understand that now.

When I was first thinking about this idea, I was thinking that "relativity of simultaneity" had no relevance here (and I said as much in some previous posts).

I don't understand how this "paradox" is resolved yet, but I understand that it has to do with "switching frames" and the "relativity of simultaneity" and "the square/diamond losing its square/diamond" shape.

Again, I don't understand (yet) how this all works and how this "paradox" is then resolved ... but I will ... thank you (and everyone else) for the direction.

By "what is missing is a lattice of clocks" I believe is "what is missing" in my OP in terms of madmac's question.

-

"Although Thomas precession (net rotation after a trajectory that returns to its initial velocity) is a purely kinematic effect, it only occurs in curvilinear motion and therefore cannot be observed independently of some external force causing the curvilinear motion such as that caused by an electromagnetic field, a gravitational field or a mechanical force, so Thomas precession is usually accompanied by dynamical effects.[1"

In this thought experiment ("the diagonal motion paradox") there is no external force ... right?

When we go from looking at this from the "third observer's" frame of reference and then looking at this from the square's frame of reference there is a "shift" that takes place. It is a "shift" in perspective. There is no "external force" like that from an "electromagnetic field" or from an "gravitational field."

Does "Thomas precession" apply to this thought experiment?

I am confused.

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I’m trying to understand “relativity of simultaneity” as it relates to this “paradox.”

One thing that can be noticed, is that if the square has four clocks, one on each corner, and if it is moving diagonally, then while the four clocks will be all synchronized in the square’s inertial frame of reference, with the diagonal motion, in the “third observer’s” inertial frame of reference the square’s two clocks perpendicular to the direction of the motion (A and B) will be synchronized while the leading clock © will be behind A and B and the trailing clock (D) will be ahead of A and B.

Am I on the right track for a resolution?

Well, I would not take that approach. I spent some time doing the full calculations. Link below. Interestingly, there was no rotation as I had previously predicted. I know the rotation shows up sometimes but, apparently, not for this problem. If you browse the document you will see that it is not entry level stuff.

http://www.relativitysimulation.com/Documents/Square%20in%20Rectangle.htm

Added this comment: I hope I copied all the numbers correctly.

Edited by Mike-from-the-Bronx
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That cant be correct -- the square-diamond can only contract on its diagonal, ie in the direction of apparent movement (45dg here).

Its the same thing as a square-diamond momentarily "sitting" directly on the yy axis (with top & bottom corners on the yy axis) & moving parallel to the xx axis (observer at (0,0)). Surely no Einsteinian would say that there was any yy contraction -- all of the contraction would be in the xx direction (the direction of apparent motion).

In a pix the top corner & bottom corner could not both be directly on the yy axis because of the light delay -- but both corners would nonetheless always be the same vertical yy distance apart (in every pix, at all times).

None of this necessarily applies if the (stationary) observer is inside the (moving) square (a slightly different scenario i admit).

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I am satisfied that the analysis I provided in my link in post #23 is correct and consistent with SR. But nobody’s perfect and I’ll try to stay open to the possibility that I made a mistake. I’ll be glad to go over the analysis step by step with anyone who wants. But I would ask that the requester first demonstrate their level of proficiency in SR by at least trying to solve the following 2 step problem.

Given a 2D shape that is a square with proper dimensions 2 light-sec on each side when at rest with respect to reference frame S. One set of space time coordinates (common time for the 4 corners) with respect to S are;

Corner a: (t, x, y) = (0, +1, +1)

Corner b: (t, x, y) = (0, -1, -1)

Corner c: (t, x, y) = (0, +1, -1)

Corner d: (t, x, y) = (0, -1, +1)

Where t is in seconds and x and y are in light seconds.

Given also that there are 2 other reference frames S’ and S’’.

S’ has velocity (vx, vy) = (-.6c, 0) with respect to S.

S’’ has velocity (vx, vy) = (-.6c, .6c) with respect to S.

1. Transform the shape to S’. Provide the velocity of the shape with respect to S’ and one set of space time coordinates with common time for the 4 corners.
2. Transform the shape from S’ to S’’. Provide the velocity of the shape with respect to S’’ and one set of space time coordinates with common time for the 4 corners.

Note: Task 2 is not a request to transform from S to S’’. I is a request to transform from S’ to S’’.

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