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May 21, 2023, 6:16:07 AM5/21/23

to

My animation

https://www.geogebra.org/m/rfa2m4ys

correctly represents the relativistic contraction of lengths?

https://www.geogebra.org/m/rfa2m4ys

correctly represents the relativistic contraction of lengths?

May 22, 2023, 3:09:33 AM5/22/23

to

of view of B (I'm not going to try to check your math), but the object

would also be foreshortened by the same amount. That is, the

object would no longer appear as a circle but as an ellipse. The

tangent point, where the dotted lines are tangent to the circle,

would always be the same point on the object C, not shifted

towards the sub-observer point (where the dashed line enters

the circle C) as your animation shows.

Rich L.

May 22, 2023, 3:10:03 AM5/22/23

to

you use to build the graph?! And how long shall we repeat just that??

(Which is more of a complaint about the moderation.)

Julio

May 22, 2023, 1:16:07 PM5/22/23

to

Julio Di Egidio il 22/05/2023 09:07:46 ha scritto:

>> My animation

>> https://www.geogebra.org/m/rfa2m4ys

>> correctly represents the relativistic contraction of lengths?

>

>> My animation

>> https://www.geogebra.org/m/rfa2m4ys

>> correctly represents the relativistic contraction of lengths?

>

> Where are your *formulas*, the ones you use to build the graph?!

In my animation
https://www.geogebra.org/m/rfa2m4ys

I added the formulas I used (Beta e Gamma).

Distance is contracted in proportion to the gamma factor.

May 22, 2023, 1:16:37 PM5/22/23

to

Richard Livingston il 22/05/2023 09:07:46 ha scritto:

>> My animation

>> https://www.geogebra.org/m/rfa2m4ys

>> correctly represents the relativistic contraction of lengths?

>

> Sorry, no. It may correctly show the object C closer from the point

> of view of B (I'm not going to try to check your math), but the object

> would also be foreshortened by the same amount. That is, the

> object would no longer appear as a circle but as an ellipse.

Right: I fixed the animation.
>> My animation

>> https://www.geogebra.org/m/rfa2m4ys

>> correctly represents the relativistic contraction of lengths?

>

> Sorry, no. It may correctly show the object C closer from the point

> of view of B (I'm not going to try to check your math), but the object

> would also be foreshortened by the same amount. That is, the

> object would no longer appear as a circle but as an ellipse.

But for observers, body C always remains exactly circular because the

contraction is in the direction of motion from B to C (and not

transversely).

> The tangent point, where the dotted lines are tangent to the circle,

> would always be the same point on the object C, not shifted

> towards the sub-observer point (where the dashed line enters

> the circle C) as your animation shows.

>

> Rich L.

instantaneous situation when observer B (in motion) passes by observer

A.

At that moment, the Vel-A speed is always zero and the Vel-B speed is

the one chosen by the user.

If the Vel-B slider is at zero, observers A and B are both stationary

and body C is at distance 16 from both.

If the Vel-B slider is at velocity Vel-B=0.866c zero, body C is at

distance 16 from observer A (Vel-A=0) and at distance 8 (16/2) from

observer B, being gamma=2.

To each value of Vel-B corresponds a precise gamma and a precise

contraction of the BC distance.

Luigi.

May 23, 2023, 4:32:04 PM5/23/23

to

On Monday, 22 May 2023 at 19:16:07 UTC+2, Luigi Fortunati wrote:

the direction of motion. That said, computing the distance

to C is easy, but each and every distance transforms in the

same way, so you don't just get semi-circles, that should be

an ellipse... Indeed, here is what I meant by "formulas":

<https://www.desmos.com/calculator/e20mfh6ln0>

(I am not an expert: I hope I have not made any mistakes.)

But my objection to you is one of method and understanding

what is what: a simulation requires a (mathematical) model!

Your reasoning on pretty pictures, especially since you are just

starting and have not yet developed any "correct intuitions", is

simply misguided.

Julio

> Julio Di Egidio il 22/05/2023 09:07:46 ha scritto:

> > On Sunday, 21 May 2023 at 12:16:07 UTC+2, Luigi Fortunati wrote:

> > > My animation

> > > https://www.geogebra.org/m/rfa2m4ys

> > > correctly represents the relativistic contraction of lengths?

> >

> > On Sunday, 21 May 2023 at 12:16:07 UTC+2, Luigi Fortunati wrote:

> > > My animation

> > > https://www.geogebra.org/m/rfa2m4ys

> > > correctly represents the relativistic contraction of lengths?

> >

> > Where are your *formulas*, the ones you use to build the graph?!

>

>

> In my animation

> <https://www.geogebra.org/m/rfa2m4ys>

> I added the formulas I used (Beta e Gamma).

> Distance is contracted in proportion to the gamma factor.

For constant (relative) motion, lengths only contract along
> <https://www.geogebra.org/m/rfa2m4ys>

> I added the formulas I used (Beta e Gamma).

> Distance is contracted in proportion to the gamma factor.

the direction of motion. That said, computing the distance

to C is easy, but each and every distance transforms in the

same way, so you don't just get semi-circles, that should be

an ellipse... Indeed, here is what I meant by "formulas":

<https://www.desmos.com/calculator/e20mfh6ln0>

(I am not an expert: I hope I have not made any mistakes.)

But my objection to you is one of method and understanding

what is what: a simulation requires a (mathematical) model!

Your reasoning on pretty pictures, especially since you are just

starting and have not yet developed any "correct intuitions", is

simply misguided.

Julio

May 24, 2023, 1:33:38 PM5/24/23

to

Julio Di Egidio il 23/05/2023 08:30:16 ha scritto:

>> In my animation

>> <https://www.geogebra.org/m/rfa2m4ys>

>> I added the formulas I used (Beta e Gamma).

>> Distance is contracted in proportion to the gamma factor.

>

> For constant (relative) motion, lengths only contract along

> the direction of motion. That said, computing the distance

> to C is easy, but each and every distance transforms in the

> same way, so you don't just get semi-circles, that should be

> an ellipse... Indeed, here is what I meant by "formulas":

> <https://www.desmos.com/calculator/e20mfh6ln0>

> (I am not an expert: I hope I have not made any mistakes.)

You made no mistakes and your simulation is correct.
>> In my animation

>> <https://www.geogebra.org/m/rfa2m4ys>

>> I added the formulas I used (Beta e Gamma).

>> Distance is contracted in proportion to the gamma factor.

>

> For constant (relative) motion, lengths only contract along

> the direction of motion. That said, computing the distance

> to C is easy, but each and every distance transforms in the

> same way, so you don't just get semi-circles, that should be

> an ellipse... Indeed, here is what I meant by "formulas":

> <https://www.desmos.com/calculator/e20mfh6ln0>

> (I am not an expert: I hope I have not made any mistakes.)

Thanks for the suggestion.

I corrected my animation which is now this:

https://www.geogebra.org/m/dzuyjaz6

> But my objection to you is one of method and understanding

> what is what: a simulation requires a (mathematical) model!

something to conform to the "mathematical model" you are talking about?

> Julio

Luigi

May 25, 2023, 3:18:30 AM5/25/23

to

On Wednesday, 24 May 2023 at 19:33:38 UTC+2, Luigi Fortunati wrote:

> <https://www.geogebra.org/m/dzuyjaz6>

"*your* *modelling* of the problem at hand", and the need thereof.

That said, yes, by "eye inspection", you still get the tangents wrong.

Indeed, a "model" concretely is a collection of "specific formulas",

ideally together with proofs of properties of those formulas/of that

model, that ensure that the model is *correct* re the underlying

(physical, in this case) theory in question: e.g. as to how to compute

the tangents and points of tangency, as well as prove (with some

mathematical derivation) that "[t]he tangent point[s], where the dotted

lines are tangent to the circle, would always be the same point[s] on the

object C", as Richard Livingston has put it upthread (slightly adapted).

HTH,

Julio

> <https://www.geogebra.org/m/dzuyjaz6>

> After this last correction of mine, is my animation still missing

> something to conform to the "mathematical model" you are talking

> about?

A meaningless statement, as already noted: I am talking about
> something to conform to the "mathematical model" you are talking

> about?

"*your* *modelling* of the problem at hand", and the need thereof.

That said, yes, by "eye inspection", you still get the tangents wrong.

Indeed, a "model" concretely is a collection of "specific formulas",

ideally together with proofs of properties of those formulas/of that

model, that ensure that the model is *correct* re the underlying

(physical, in this case) theory in question: e.g. as to how to compute

the tangents and points of tangency, as well as prove (with some

mathematical derivation) that "[t]he tangent point[s], where the dotted

lines are tangent to the circle, would always be the same point[s] on the

object C", as Richard Livingston has put it upthread (slightly adapted).

HTH,

Julio

May 29, 2023, 2:56:51 AM5/29/23

to

Julio Di Egidio il 25/05/2023 09:18:24 ha scritto:

>> <https://www.geogebra.org/m/dzuyjaz6>

>> After this last correction of mine, is my animation still missing

>> something to conform to the "mathematical model" you are talking

>> about?

>

> A meaningless statement, as already noted: I am talking about

> "*your* *modelling* of the problem at hand", and the need thereof.

My animation is modeled according to the dictates of Special
>> <https://www.geogebra.org/m/dzuyjaz6>

>> After this last correction of mine, is my animation still missing

>> something to conform to the "mathematical model" you are talking

>> about?

>

> A meaningless statement, as already noted: I am talking about

> "*your* *modelling* of the problem at hand", and the need thereof.

Relativity.

> That said, yes, by "eye inspection", you still get the tangents wrong.

> Julio

Luigi

May 30, 2023, 3:10:08 AM5/30/23

to

Op 24/05/2023 om 19:33 schreef Luigi Fortunati:

Your simulation may be more or less correct, but it doesn't help much in

one's understanding it correctly, it even may induce erroneous

understanding.

First, it should be made clear in which direction the moving observer is

supposed to move. My first impression was that he should move vertically

which didn't make sense.

Furthermore, it doesn't help to superpose both 'circles' in a same

graph, which suggests a same reference frame, where in fact different

reference frames are superposed. Now it gives the impression that "there

are two circles in space", one for each observer.

The fact is that there is only one object, but that both observers "fill

in" the space towards it, ánd the time values, differently! In order to

respect the singleness of the object, your method is not a very proper

one. There are better alternatives. One is, to make two graphs, each

with its reference frame. Another one, to display only one object, say

in the rest system, and superpose the different reference frame of the

moving observer upon that, showing the length contraction, and possibly,

time dilation effects.

Lastly, this is not what the observers are going to *see*! Einstein's

Lorentz equations don't describe objects *as seen*, they describe

*measured*, backcalculated, positions of simultaneity. What the

observers are going to see, is what the light photons are telling them,

as they arrive at each observer, come from the different parts of the

distant object. In other words, they include Doppler distortions!

There are nice examples of this distinction and of relativistic Doppler

watching around the net. When I find them back, I'll post some

links. It's also a main topic of my SRT pages

wugi's SRT world:

https://www.wugi.be/qbRelaty.html

https://www.wugi.be/paratwin.htm

https://www.youtube.com/playlist?list=PL5xDSSE1qfb6zyVKJbe8POgj-8ijmh5o0

--

guido wugi

one's understanding it correctly, it even may induce erroneous

understanding.

First, it should be made clear in which direction the moving observer is

supposed to move. My first impression was that he should move vertically

which didn't make sense.

Furthermore, it doesn't help to superpose both 'circles' in a same

graph, which suggests a same reference frame, where in fact different

reference frames are superposed. Now it gives the impression that "there

are two circles in space", one for each observer.

The fact is that there is only one object, but that both observers "fill

in" the space towards it, ánd the time values, differently! In order to

respect the singleness of the object, your method is not a very proper

one. There are better alternatives. One is, to make two graphs, each

with its reference frame. Another one, to display only one object, say

in the rest system, and superpose the different reference frame of the

moving observer upon that, showing the length contraction, and possibly,

time dilation effects.

Lastly, this is not what the observers are going to *see*! Einstein's

Lorentz equations don't describe objects *as seen*, they describe

*measured*, backcalculated, positions of simultaneity. What the

observers are going to see, is what the light photons are telling them,

as they arrive at each observer, come from the different parts of the

distant object. In other words, they include Doppler distortions!

There are nice examples of this distinction and of relativistic Doppler

watching around the net. When I find them back, I'll post some

links. It's also a main topic of my SRT pages

wugi's SRT world:

https://www.wugi.be/qbRelaty.html

https://www.wugi.be/paratwin.htm

https://www.youtube.com/playlist?list=PL5xDSSE1qfb6zyVKJbe8POgj-8ijmh5o0

--

guido wugi

May 30, 2023, 12:48:57 PM5/30/23

to

Op 25/05/2023 om 9:18 schreef Julio Di Egidio:

for b can be interpreted as a 'rotation' in a plane through the

horizontal axis and our eyes. So, the tangent lines should remain tangent.

--

guido wugi

> On Wednesday, 24 May 2023 at 19:33:38 UTC+2, Luigi Fortunati wrote:

>

>> <https://www.geogebra.org/m/dzuyjaz6>

>

>> <https://www.geogebra.org/m/dzuyjaz6>

>> After this last correction of mine, is my animation still missing

>> something to conform to the "mathematical model" you are talking

>> about?

>

>> something to conform to the "mathematical model" you are talking

>> about?

>

> A meaningless statement, as already noted: I am talking about

> "*your* *modelling* of the problem at hand", and the need thereof.

>

> "*your* *modelling* of the problem at hand", and the need thereof.

>

> That said, yes, by "eye inspection", you still get the tangents wrong.

> Indeed, a "model" concretely is a collection of "specific formulas",

> ideally together with proofs of properties of those formulas/of that

> model, that ensure that the model is *correct* re the underlying

> (physical, in this case) theory in question: e.g. as to how to compute

> the tangents and points of tangency, as well as prove (with some

> mathematical derivation) that "[t]he tangent point[s], where the dotted

> lines are tangent to the circle, would always be the same point[s] on the

> object C", as Richard Livingston has put it upthread (slightly adapted).

Looking at your own simulation, the 'proof' seems easy: the animation
> ideally together with proofs of properties of those formulas/of that

> model, that ensure that the model is *correct* re the underlying

> (physical, in this case) theory in question: e.g. as to how to compute

> the tangents and points of tangency, as well as prove (with some

> mathematical derivation) that "[t]he tangent point[s], where the dotted

> lines are tangent to the circle, would always be the same point[s] on the

> object C", as Richard Livingston has put it upthread (slightly adapted).

for b can be interpreted as a 'rotation' in a plane through the

horizontal axis and our eyes. So, the tangent lines should remain tangent.

--

guido wugi

Jun 2, 2023, 4:17:07 PM6/2/23

to

wugi il 29/05/2023 22:36:39 ha scritto:

>> I corrected my animation which is now this:

>> https://www.geogebra.org/m/dzuyjaz6

>

>> I corrected my animation which is now this:

>> https://www.geogebra.org/m/dzuyjaz6

>

> Your simulation may be more or less correct, but it doesn't help much in

> one's understanding it correctly, it even may induce erroneous

> understanding.

>

> First, it should be made clear in which direction the moving observer is

> supposed to move. My first impression was that he should move vertically

> which didn't make sense.

In my simulation:
> one's understanding it correctly, it even may induce erroneous

> understanding.

>

> First, it should be made clear in which direction the moving observer is

> supposed to move. My first impression was that he should move vertically

> which didn't make sense.

- observer B moves (instantaneously) to the right, i.e. towards body C

- observer A is stationary with respect to body C

- Observers A and B, at time t=0 of both, share the same space and the same time

- The time of C in the reference of A is different from the time of C in the reference of B.

> Furthermore, it doesn't help to superpose both 'circles' in a same

> graph, which suggests a same reference frame, where in fact different

> reference frames are superposed. Now it gives the impression that "there

> are two circles in space", one for each observer.

>

> The fact is that there is only one object, but that both observers "fill

> respect the singleness of the object, your method is not a very proper

> one. There are better alternatives. One is, to make two graphs, each

> with its reference frame. Another one, to display only one object, say

> in the rest system, and superpose the different reference frame of the

> moving observer upon that, showing the length contraction, and possibly,

> time dilation effects.

>

> Lastly, this is not what the observers are going to *see*! Einstein's

> Lorentz equations don't describe objects *as seen*, they describe

> *measured*, backcalculated, positions of simultaneity. What the

> observers are going to see, is what the light photons are telling them,

> as they arrive at each observer, come from the different parts of the

> distant object. In other words, they include Doppler distortions!

Ok, let's talk about what happens when photons of light arrive on the
> one. There are better alternatives. One is, to make two graphs, each

> with its reference frame. Another one, to display only one object, say

> in the rest system, and superpose the different reference frame of the

> moving observer upon that, showing the length contraction, and possibly,

> time dilation effects.

>

> Lastly, this is not what the observers are going to *see*! Einstein's

> Lorentz equations don't describe objects *as seen*, they describe

> *measured*, backcalculated, positions of simultaneity. What the

> observers are going to see, is what the light photons are telling them,

> as they arrive at each observer, come from the different parts of the

> distant object. In other words, they include Doppler distortions!

photographic film (the observer) and form the image.

Are Einstein's Lorentz equations able to predict whether the body C will

appear perfectly spherical or whether it will be squashed in the

direction of motion?

[[Mod. note -- Yes.

As wugi noted, (1) and (2) below are very different questions, with

very different answers:

(1) What are the coordinate positions of the objects measured (i.e.,

"backcalculated", as wugi quite correctly terms it) in different

(inertial) reference frames, as determined by the Lorenz transformation?

(2) What image(s) would taken by a (ideal) camera located at a certain

point given (1) together with differential light-travel-time effects

(i.e., the Lampa-Penrose-Terrell effect, often just called the Terrell

effect or Terrell rotation)?

(1) is what's usually meant when we ask what an observer "measures" in

special relativity. (2) is what you (Luigi) have now asked about.

See

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

for a nice introduction to (2). Reference 4 in that Wikipedia article

(written by Victor Weisskopf!) is a very clear exposition of the effect,

explicitly working out the Terrell rotation for a moving cube and then

generalizing it to an arbitrary-shaped body. The original 1960 paper is

(still) behind a paywall :(, but as of a few minutes ago google scholar

finds a free copy.

-- jt]]

Jun 4, 2023, 11:54:01 PM6/4/23

to

Op 24/05/2023 om 19:33 schreef Luigi Fortunati:

> Julio Di Egidio il 23/05/2023 08:30:16 ha scritto:

>>> In my animation

>>> <https://www.geogebra.org/m/rfa2m4ys>

>>> I added the formulas I used (Beta e Gamma).

>>> Distance is contracted in proportion to the gamma factor.

>>

>> For constant (relative) motion, lengths only contract along

>> the direction of motion. That said, computing the distance

>> to C is easy, but each and every distance transforms in the

>> same way, so you don't just get semi-circles, that should be

>> an ellipse... Indeed, here is what I meant by "formulas":

>> <https://www.desmos.com/calculator/e20mfh6ln0>

>> (I am not an expert: I hope I have not made any mistakes.)

>

> You made no mistakes and your simulation is correct.

>

> Thanks for the suggestion.

>

> Julio Di Egidio il 23/05/2023 08:30:16 ha scritto:

>>> In my animation

>>> <https://www.geogebra.org/m/rfa2m4ys>

>>> I added the formulas I used (Beta e Gamma).

>>> Distance is contracted in proportion to the gamma factor.

>>

>> For constant (relative) motion, lengths only contract along

>> the direction of motion. That said, computing the distance

>> to C is easy, but each and every distance transforms in the

>> same way, so you don't just get semi-circles, that should be

>> an ellipse... Indeed, here is what I meant by "formulas":

>> <https://www.desmos.com/calculator/e20mfh6ln0>

>> (I am not an expert: I hope I have not made any mistakes.)

>

> You made no mistakes and your simulation is correct.

>

> Thanks for the suggestion.

>

>> But my objection to you is one of method and understanding

>> what is what: a simulation requires a (mathematical) model!

>

>> what is what: a simulation requires a (mathematical) model!

>

> After this last correction of mine, is my animation still missing

> something to conform to the "mathematical model" you are talking about?

It is not showing the true 'embedding' of both 'circles' in spacetime,
> something to conform to the "mathematical model" you are talking about?

so it doesn't put in evidence how different events come into play for

configuring them. You show us "a" space wherein you drop two

space-visions (from two systems), but you don't show us time and the

different time-visions involved.

I've tried a space-time model of what you want to show, see here:

https://www.desmos.com/calculator/x1wfo7kobn?lang=nl

We have an x,ct rest and an x',ct' moving system; the y-axis has to be

added 'mentally' perpendicular to the screen; the (x,y) circles involved

are shown as ellipse projections.

The "rest circle" object is measured in the rest system by events AB, in

the moving system by events A'B', all comprised in its red world line

band. Notice how we can never have the "same" instantaneous circle,

measured "differently" in the two systems: different events in space and

time are involved!

To stress the reciprocity of this feature, I've added the corresponding

unit circle for the moving system, as the green world line band. Notice

how the 'moving' unit circle is measured equally length-contracted by

the rest system. Red and green circles represent the same units in their

respective system, but their intervening events don't mingle!

--

guido wugi

Jun 5, 2023, 2:58:37 AM6/5/23

to

Op 2/06/2023 om 22:17 schreef Luigi Fortunati:

Thank you. About Terrell rotation, if I remember correctly it would seem

to alledge that length contraction is part of the "seeing" effects, in

that the object, together with its backside becoming visible, would seem

rotated, and total length "seen" would be preserved. This is messing up

what I call "Lorentz" data (measuring, backcalculating: length

contraction) with "Einstein" data (*seeing, watching* , the back side

becoming visible).

But it is partly true. From my webside which I mentioned in another

reply, let's see some examples here.

How would a frontline that's approaching an observer at relativistic

speed be perceived? Here:

https://www.wugi.be/MySRT/frontline1.PNG

(fat dot = observer, dotted line = actual 'true' position, rightmost

curve = as seen 'now', others = previous positions.)

Same case, after having past by the observer:

https://www.wugi.be/MySRT/frontline2.PNG

(doppler 'blue shift' at approach, 'red shift' at regression)

A series of frontlines combined into a "squadron" of squares:

https://www.wugi.be/MySRT/Squad.gif

(dots = actual position of the squad)

Notice the following "seeing/watching" effects:

At approach, a "length dilation" is seen, not a contraction!

At regression, an extra length contraction is seen.

This is due to the fact that light needs the longest time to reach the

observer from the farmost parts of the moving object.

The squares appear more or less distorted, according to position and

distance from observer. This is analogous to, and more general than,

what Terrell rotation tries to tell us.

But observe this! ->

The true *Lorentz length contraction can be seen*, but only

perpendicular to the motion, and far enough away: in our picture, along

the vertical through the observer point, at the far distances up and

down (compare with dot positions).

I'd promised some other links to the effects of 'seeing' relativistic

motion. Here are some:

https://www.physicsforums.com/threads/terrell-revisited-the-invisibility-of-the-lorentz-contraction.520875/page-4

(with nice animations by J. Doolin)

https://www.tempolimit-lichtgeschwindigkeit.de/ueberblick/1

(with links to more videos)

https://www.spacetimetravel.org/tompkins/tompkins.pdf

(with other cases)

https://timms.uni-tuebingen.de/tp/UT_20040806_001_sommeruni2004_0001

(a lecture with animation videos of previous cases)

--

guido wugi

to alledge that length contraction is part of the "seeing" effects, in

that the object, together with its backside becoming visible, would seem

rotated, and total length "seen" would be preserved. This is messing up

what I call "Lorentz" data (measuring, backcalculating: length

contraction) with "Einstein" data (*seeing, watching* , the back side

becoming visible).

But it is partly true. From my webside which I mentioned in another

reply, let's see some examples here.

How would a frontline that's approaching an observer at relativistic

speed be perceived? Here:

https://www.wugi.be/MySRT/frontline1.PNG

(fat dot = observer, dotted line = actual 'true' position, rightmost

curve = as seen 'now', others = previous positions.)

Same case, after having past by the observer:

https://www.wugi.be/MySRT/frontline2.PNG

(doppler 'blue shift' at approach, 'red shift' at regression)

A series of frontlines combined into a "squadron" of squares:

https://www.wugi.be/MySRT/Squad.gif

(dots = actual position of the squad)

Notice the following "seeing/watching" effects:

At approach, a "length dilation" is seen, not a contraction!

At regression, an extra length contraction is seen.

This is due to the fact that light needs the longest time to reach the

observer from the farmost parts of the moving object.

The squares appear more or less distorted, according to position and

distance from observer. This is analogous to, and more general than,

what Terrell rotation tries to tell us.

But observe this! ->

The true *Lorentz length contraction can be seen*, but only

perpendicular to the motion, and far enough away: in our picture, along

the vertical through the observer point, at the far distances up and

down (compare with dot positions).

I'd promised some other links to the effects of 'seeing' relativistic

motion. Here are some:

https://www.physicsforums.com/threads/terrell-revisited-the-invisibility-of-the-lorentz-contraction.520875/page-4

(with nice animations by J. Doolin)

https://www.tempolimit-lichtgeschwindigkeit.de/ueberblick/1

(with links to more videos)

https://www.spacetimetravel.org/tompkins/tompkins.pdf

(with other cases)

https://timms.uni-tuebingen.de/tp/UT_20040806_001_sommeruni2004_0001

(a lecture with animation videos of previous cases)

--

guido wugi

Jun 7, 2023, 12:52:06 PM6/7/23

to

Initially I was asking about the contraction of the distances between

the observers and body C, then the discussion shifted to the

contraction of body C.

I return to the initial question with my simulation

https://www.geogebra.org/m/ujwjmgt8

Is it correct to say that for the alien flying disk the Earth-Sun

distance is 4.15 light-minutes?

the observers and body C, then the discussion shifted to the

contraction of body C.

I return to the initial question with my simulation

https://www.geogebra.org/m/ujwjmgt8

Is it correct to say that for the alien flying disk the Earth-Sun

distance is 4.15 light-minutes?

Jun 8, 2023, 3:48:24 AM6/8/23

to

Op 7/06/2023 om 18:52 schreef Luigi Fortunati:

Perhaps, but again (and again) your picture is not a complete one, in

that the Sun's image is not representative for both reference systems.

Look at my own desmos file: both observers, Earth and alien, are "at the

same event" in O.

But to Earth, the simultaneous position of the Sun is determined by

events AB. To the alien, it is determined by other (later ones, in

Earth's system) events A'B'.

Your single picture of the Sun has to represent these two

non-simultaneous event cases.

Also, while Earth's x-axis can be laid out as such in your picture,

alien's x'-axis must be understood as covering events that are

non-simultaneous in Earth's system.

--

guido wugi

that the Sun's image is not representative for both reference systems.

Look at my own desmos file: both observers, Earth and alien, are "at the

same event" in O.

But to Earth, the simultaneous position of the Sun is determined by

events AB. To the alien, it is determined by other (later ones, in

Earth's system) events A'B'.

Your single picture of the Sun has to represent these two

non-simultaneous event cases.

Also, while Earth's x-axis can be laid out as such in your picture,

alien's x'-axis must be understood as covering events that are

non-simultaneous in Earth's system.

--

guido wugi

Jun 9, 2023, 4:55:00 AM6/9/23

to

And this is the other question: how do moving spherical bodies appear

in monitors and photographs?

In my animation

https://www.geogebra.org/m/kh38nfpd

where there is the body C in motion and the body D stationary in the

reference of the monitor, I tried to give an answer.

Is it true that on the monitor (and in the photos) body D appears

spherical and body C appears squashed in the direction of motion?

[[Mod. note -- I can't tell from your your wording whether you are

asking about

observers in different (inertial) reference frames?, or

(2) What image(s) would taken by (ideal) cameras located at some positions,

taking into account differential light-travel-time effects (i.e., the

but if you mean (2) then the answer to your question is "no".

-- jt]]

in monitors and photographs?

In my animation

https://www.geogebra.org/m/kh38nfpd

where there is the body C in motion and the body D stationary in the

reference of the monitor, I tried to give an answer.

Is it true that on the monitor (and in the photos) body D appears

spherical and body C appears squashed in the direction of motion?

[[Mod. note -- I can't tell from your your wording whether you are

asking about

(1) What are the coordinate positions of the objects measured (i.e.,

"backcalculated", as wugi quite correctly termed it) by special-relativity
observers in different (inertial) reference frames?, or

(2) What image(s) would taken by (ideal) cameras located at some positions,

taking into account differential light-travel-time effects (i.e., the

Lampa-Penrose-Terrell effect, often just called the Terrell effect or

Terrell rotation)?

Roughly speaking, if you mean (1) then the answer to your question is "yes",
Terrell rotation)?

but if you mean (2) then the answer to your question is "no".

-- jt]]

Jun 9, 2023, 5:28:26 AM6/9/23

to

The commander of the spaceship does not care how far the Sun is from

the Earth, he is only interested in how far the Sun is from *his*

spaceship (in *his* reference), in light minutes, at the moment in

which it passes close to the Earth (the only event in common).

Is there a single answer to his legitimate question?

And is the right answer 4.15 light-minutes or is it something else?

[[Mod. note -- The complications are inherent to the physical situation.

The phrase "at the moment in which it [the spaceship] passes close to

the Earth" specifies an event, and (if we treat the Earth as a point,

and idealise "close to the Earth" as passing through the Earth-point)

all observers can agree on this event.

But simultaneity is only local in special relativity, not global.

That is, different observers *disagree* about which event on the Sun's

worldline is at the same time as the spaceship-passing-through-the-Earth-point

event. To put it another way, if we imagine an ideal clock at (and

moving with) the Sun, then different observers will compute different

answers to the question

At the time when the spacecraft passes through the

Earth-point, what is the Sun-point's clock reading?

And since (in the spaceship inertial-reference-frame) the Sun is moving,

asking for its position requires specifying *when* to measure that position,

i.e., it requires implicitly or explicitly specifying the Sun-point clock

reading at which you want that position.

If I understand your question correctly, you've asked for the Earth-Sun

distance as measured in the spaceship inertial-reference-frame, i.e.,

you've asked for the Sun's distance from the Earth (or equivalently,

the spaceship) at the Sun-point clock reading which is

simultaneous-in-the-spaceship-inertial-reference-frame

to the spaceship-passing-through-the-Earth-point event.

Without working it out, I suspect the answer is 4.15 light-minutes.

-- jt]]

Jun 11, 2023, 3:18:03 AM6/11/23

to

Luigi Fortunati il 08/06/2023 20:54:56 ha scritto:

> And this is the other question: how do moving spherical bodies appear

> in monitors and photographs?

>

> In my animation

> https://www.geogebra.org/m/kh38nfpd

> where there is the body C in motion and the body D stationary in the

> reference of the monitor, I tried to give an answer.

>

> Is it true that on the monitor (and in the photos) body D appears

> spherical and body C appears squashed in the direction of motion?

>

> [[Mod. note -- I can't tell from your your wording whether you are

> asking about

> (1) What are the coordinate positions of the objects measured (i.e.,

> "backcalculated", as wugi quite correctly termed it) by

> special-relativity observers in different (inertial) reference fram=
> And this is the other question: how do moving spherical bodies appear

> in monitors and photographs?

>

> In my animation

> https://www.geogebra.org/m/kh38nfpd

> where there is the body C in motion and the body D stationary in the

> reference of the monitor, I tried to give an answer.

>

> Is it true that on the monitor (and in the photos) body D appears

> spherical and body C appears squashed in the direction of motion?

>

> [[Mod. note -- I can't tell from your your wording whether you are

> asking about

> (1) What are the coordinate positions of the objects measured (i.e.,

> "backcalculated", as wugi quite correctly termed it) by

es?,

> or (2) What image(s) would taken by (ideal) cameras located at some

> positions,

Not from "cameras in some positions" but from a single and specific
> or (2) What image(s) would taken by (ideal) cameras located at some

> positions,

camera: that of observer B.

And since the body C is spherical, it can only appear as an ordinary

sphere or as a sphere squeezed in the direction of motion (and in no

other way).

So I repeat the question rigorously and clearly: how does the (moving)

body C appear on B's monitor, does it appear squashed or not squashed?

[[Mod. note -- I'm sorry, I still don't know what you're asking. That

is, I don't know if your "monitor" is supposed to show (1) or (2).

Your animation has the monitor showing C squeezed in the direction of

motion, which would be correct for (1).

But in your message you refer to a "camera", which makes me suspect you

are asking about (2), in which case the monitor (showing the image formed

by an ideal camera placed at position B) will show C as a sphere, rotated

in a somewhat unobvious (that's why the effect is often called Terrell

*rotation*). See

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

for an explanation.

-- jt]]

Jun 11, 2023, 3:21:36 AM6/11/23

to

Luigi Fortunati il 08/06/2023 21:28:21 ha scritto:

>> I return to the initial question with my simulation

>> https://www.geogebra.org/m/ujwjmgt8

> ...
>> I return to the initial question with my simulation

>> https://www.geogebra.org/m/ujwjmgt8

> [[Mod. note -- The complications are inherent to the physical situation.

>

> The phrase "at the moment in which it [the spaceship] passes close to

> the Earth" specifies an event, and (if we treat the Earth as a point,

> and idealise "close to the Earth" as passing through the Earth-point)

> all observers can agree on this event.

>

> But simultaneity is only local in special relativity, not global.

> That is, different observers *disagree* about which event on the Sun's

> worldline is at the same time as the

> spaceship-passing-through-the-Earth-point event. To put it another way, if

> we imagine an ideal clock at (and moving with) the Sun, then different

> observers will compute different answers to the question

> At the time when the spacecraft passes through the

> Earth-point, what is the Sun-point's clock reading?

You want to talk about times and not about spaces.
>

> The phrase "at the moment in which it [the spaceship] passes close to

> the Earth" specifies an event, and (if we treat the Earth as a point,

> and idealise "close to the Earth" as passing through the Earth-point)

> all observers can agree on this event.

>

> But simultaneity is only local in special relativity, not global.

> That is, different observers *disagree* about which event on the Sun's

> worldline is at the same time as the

> spaceship-passing-through-the-Earth-point event. To put it another way, if

> we imagine an ideal clock at (and moving with) the Sun, then different

> observers will compute different answers to the question

> At the time when the spacecraft passes through the

> Earth-point, what is the Sun-point's clock reading?

Okay.

In the position of my drawing, will the spacecraft arrive at the Sun

after 9.58 minutes of *its* time (8.3/0.866) or after 4.79 minutes

(4.15/0.866)?

I ask about the time of the spaceship, other times are of no interest.

[[Mod. note -- If I'm understanding things correctly, 4.79 minutes. -- jt]]

Jun 11, 2023, 4:59:24 PM6/11/23

to

"only squeezed in the direction of motion", so here you ask about the

Lorentz backcalculated *motion measurement*, not about *seeing the

moving sphere*.

> So I repeat the question rigorously and clearly: how does the (moving)

> body C appear on B's monitor, does it appear squashed or not squashed?

Here again about *seeing*.

> [[Mod. note -- I'm sorry, I still don't know what you're asking. That

> is, I don't know if your "monitor" is supposed to show (1) or (2).

>

> Your animation has the monitor showing C squeezed in the direction of

> motion, which would be correct for (1).

His animation is about Lorentz contraction measurement, not about

cameras recording motion "live".

> But in your message you refer to a "camera", which makes me suspect you

> are asking about (2), in which case the monitor (showing the image formed

> by an ideal camera placed at position B) will show C as a sphere, rotated

> in a somewhat unobvious (that's why the effect is often called Terrell

> *rotation*). See

> https://en.wikipedia.org/wiki/Terrell_rotation

> for an explanation.

> -- jt]]

His confusion remains...

FWIW, Luigi, your animation is about Lorentz contraction, not about what

cameras will record.

If you want a related example (I'm not going to redo your simulation),

look again at my picture of watching the approach, passing by, and

regression, of a squadron of squares:

https://wugi.be/paratwin.htm =>

https://wugi.be/MySRT/Squad.gif

The squares are Lorentz contracted (see "actual position" dots at the

right).

They are *seen* as quadrangles, ie distorted rectangles.

At approach there is blueshift, at regression redshift (my colors are

inverted, for a white background;).

At approach the speed seen is greater than the motion velocity v (it

becomes infinite for v=c), at regression it is less than v (it becomes

c/2 for v=c).

At approach there is a seen length expansion(! it becomes infinite for

v=c), at regression an extra length contraction(! it becomes L/2gamma

for v=c IIRC).

Now if you want to visualise spheres, or circles, fill in any of the

quadrangles with a corresponding ellipse-like figure, that's how the

distorted circles are going to be seen. If you want a symmetrical

circle, take 2*2 adjoining squares, 2 along either side of the axis of

symmetry through the observer (the small black circle).

--

guido wugi

Lorentz backcalculated *motion measurement*, not about *seeing the

moving sphere*.

> So I repeat the question rigorously and clearly: how does the (moving)

> body C appear on B's monitor, does it appear squashed or not squashed?

> [[Mod. note -- I'm sorry, I still don't know what you're asking. That

> is, I don't know if your "monitor" is supposed to show (1) or (2).

>

> Your animation has the monitor showing C squeezed in the direction of

> motion, which would be correct for (1).

cameras recording motion "live".

> But in your message you refer to a "camera", which makes me suspect you

> are asking about (2), in which case the monitor (showing the image formed

> by an ideal camera placed at position B) will show C as a sphere, rotated

> in a somewhat unobvious (that's why the effect is often called Terrell

> *rotation*). See

> https://en.wikipedia.org/wiki/Terrell_rotation

> for an explanation.

> -- jt]]

FWIW, Luigi, your animation is about Lorentz contraction, not about what

cameras will record.

If you want a related example (I'm not going to redo your simulation),

look again at my picture of watching the approach, passing by, and

regression, of a squadron of squares:

https://wugi.be/paratwin.htm =>

https://wugi.be/MySRT/Squad.gif

The squares are Lorentz contracted (see "actual position" dots at the

right).

They are *seen* as quadrangles, ie distorted rectangles.

At approach there is blueshift, at regression redshift (my colors are

inverted, for a white background;).

At approach the speed seen is greater than the motion velocity v (it

becomes infinite for v=c), at regression it is less than v (it becomes

c/2 for v=c).

At approach there is a seen length expansion(! it becomes infinite for

v=c), at regression an extra length contraction(! it becomes L/2gamma

for v=c IIRC).

Now if you want to visualise spheres, or circles, fill in any of the

quadrangles with a corresponding ellipse-like figure, that's how the

distorted circles are going to be seen. If you want a symmetrical

circle, take 2*2 adjoining squares, 2 along either side of the axis of

symmetry through the observer (the small black circle).

--

guido wugi

Jun 11, 2023, 4:59:42 PM6/11/23

to

Luigi Fortunati il 10/06/2023 19:17:57 ha scritto:

> [[Mod. note -- I'm sorry, I still don't know what you're asking. That

> is, I don't know if your "monitor" is supposed to show (1) or (2).

>

> Your animation has the monitor showing C squeezed in the direction of

> motion, which would be correct for (1).

>

> But in your message you refer to a "camera", which makes me suspect you

> are asking about (2), in which case the monitor (showing the image formed

> by an ideal camera placed at position B) will show C as a sphere, rotated

> in a somewhat unobvious (that's why the effect is often called Terrell

> *rotation*). See

> https://en.wikipedia.org/wiki/Terrell_rotation

> for an explanation.

> -- jt]]

Thanks, now I understand that, on the monitor and on the photos, the
> [[Mod. note -- I'm sorry, I still don't know what you're asking. That

> is, I don't know if your "monitor" is supposed to show (1) or (2).

>

> Your animation has the monitor showing C squeezed in the direction of

> motion, which would be correct for (1).

>

> But in your message you refer to a "camera", which makes me suspect you

> are asking about (2), in which case the monitor (showing the image formed

> by an ideal camera placed at position B) will show C as a sphere, rotated

> in a somewhat unobvious (that's why the effect is often called Terrell

> *rotation*). See

> https://en.wikipedia.org/wiki/Terrell_rotation

> for an explanation.

> -- jt]]

spherical body C of my animation appears perfectly spherical and

identical to the body D, without any relativistic contraction.

Luigi.

Jun 12, 2023, 3:09:42 AM6/12/23

to

I don't understand how they (monitors and cameras) can represent anything else.

I had believed that the contraction of a sphere could be "seen" and, instead, it is not so.

The moderator directed me to the right path which is the one represented by my last animation

https://www.geogebra.org/m/axxtdurx

where the moving body C is exactly equal to the stationary body D.

Here's the lesson: the contraction of a moving sphere is like the trick: it's there but you can't "see".

Luigi.

Jun 14, 2023, 6:45:00 AM6/14/23

to

Luigi Fortunati il 10/06/2023 19:21:31 ha scritto:

> You want to talk about times and not about spaces.

>

> Okay.

>

> In the position of my drawing, will the spacecraft arrive at the Sun=20
> You want to talk about times and not about spaces.

>

> Okay.

>

> after 9.58 minutes of *its* time (8.3/0.866) or after 4.79 minutes=20

> (4.15/0.866)?

>

> I ask about the time of the spaceship, other times are of no interest.

>

> [[Mod. note -- If I'm understanding things correctly, 4.79 minutes. -- =
>

> I ask about the time of the spaceship, other times are of no interest.

>

jt]]

Exact.

If the spacecraft (which is next to Earth) travels at a speed of 0.866c=20

and takes 4.79 minutes to reach the Sun, then the spacecraft's distance=20

from the Sun is d=3Dv*t=3D0.855*4.79=3D4.15 light-minutes, half the dista=

nce=20

to Earth-Sun (8.3 minutes-light).

Here's how you can *see* (with the eyes, with the camera or with the=20

monitor) the contraction of distances: observing the size of the image=20

of the Sun, as in my animation

https://www.geogebra.org/m/cux34wvb

where the ray of the Sun on the spaceship's monitor (4.15 light-minutes=20

away) is twice as large as it appears on the Earth's monitor (8.3=20

light-minutes away).

Jun 16, 2023, 2:54:11 PM6/16/23

to

Luigi Fortunati il 14/06/2023 12:44:55 ha scritto:

>> In the position of my drawing, will the spacecraft arrive at the Sun=20

>> after 9.58 minutes of *its* time (8.3/0.866) or after 4.79 minutes=20

>> (4.15/0.866)?

>>

>> I ask about the time of the spaceship, other times are of no interest.

>>

>> [[Mod. note -- If I'm understanding things correctly, 4.79 minutes. -- =

> jt]]

>

> Exact.

>

> If the spacecraft (which is next to Earth) travels at a speed of 0.866c

>> In the position of my drawing, will the spacecraft arrive at the Sun=20

>> after 9.58 minutes of *its* time (8.3/0.866) or after 4.79 minutes=20

>> (4.15/0.866)?

>>

>> I ask about the time of the spaceship, other times are of no interest.

>>

>> [[Mod. note -- If I'm understanding things correctly, 4.79 minutes. -- =

> jt]]

>

> Exact.

>

> If the spacecraft (which is next to Earth) travels at a speed of 0.866c

> and takes 4.79 minutes to reach the Sun, then the spacecraft's distance

> from the Sun is d=v*t=0.866*4.79=4.15 light-minutes, half the distance
> to Earth-Sun (8.3 minutes-light).

>

> Here's how you can *see* (with the eyes, with the camera or with the

>

> Here's how you can *see* (with the eyes, with the camera or with the

> monitor) the contraction of distances: observing the size of the image

> of the Sun, as in my animation

> https://www.geogebra.org/m/cux34wvb

> where the ray of the Sun on the spaceship's monitor (4.15 light-minutes

> https://www.geogebra.org/m/cux34wvb

> where the ray of the Sun on the spaceship's monitor (4.15 light-minutes

> away) is twice as large as it appears on the Earth's monitor (8.3

> light-minutes away).
Reflecting on my animation, the images that appear on the monitors concern the moment in which the light rays from the Sun arrive on the Earth and on the spaceship.

But if their arrival is contemporary in the two references, their departure cannot be.

And then, I try to calculate the departure and the path of the light rays in the two references.

In the terrestrial reference the calculation is easy, the rays arrive on the Earth after 8.3 minutes and, therefore, have traveled a distance of 8.3 light-minutes.

In the spaceship reference, however, the calculation is a bit more complicated.

In this frame, the spacecraft is stationary and the Sun is approaching at speed v=0.866c, while the light rays is approaching speed v=c.

> From my calculations, these light rays that arrive at the spaceship

have traveled for an enormously longer time than the 8.3 minutes of the rays that arrive on Earth: they arrive after almost 31 minutes!

Consequently, they will have traveled a distance of 31 light-minutes!

It seems to me a disproportionate time and space: can you tell me if my calculation is correct and, if not, how far the light rays that reach the spaceship have traveled when it passes close to the Earth?

Jun 16, 2023, 3:35:54 PM6/16/23

to

[[Mod. note -- I apologise for the delay in processing this article,

which arrived in the moderation system on 2023-06-12. -- jt]]

Op 12/06/2023 om 9:09 schreef Luigi Fortunati:

Which of course it isn't, how wrong can one be.

> Here's the lesson: the contraction of a moving sphere is like the trick: it's there but you can't "see".

Did you even have a single look at my picture?

https://wugi.be/MySRT/Squad.gif

The squares are *seen* as (very) distorted quadrangles. So would the

inscribed circles/spheres.

Now it may be that the latter would look *less* distorted, because their

*angle of vision* varies less than that of the squares' appearances

themselves, as the perpendicular (to motion) directions are not

contracted. Try imagining angles of vision to inscribed circles. Well

then, spheres might look less distorted, but their meridians (and other

circles) will certainly do more so!

As for the Terell rotation,

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

as I've said before,

that's just a special case of the distortions seen, valid only

perpendicular to the direction of motion, and far enough away, in my

picture: at the far end of the vertical through the observer.

Look carefully, comparing with the 'actual position' dots at the right:

that's the only area where the Lorentz contraction can 'really' be *seen*!

More on looking at relativistic spheres:

https://www.spacetimetravel.org/tompkins/tompkins.pdf

https://www.spacetimetravel.org/tompkins/1 (look at sphere animations!)

https://www.tempolimit-lichtgeschwindigkeit.de/ueberblick/1 of which:

https://www.tempolimit-lichtgeschwindigkeit.de/fussball

https://www.tempolimit-lichtgeschwindigkeit.de/sphere/sphere.pdf

https://www.tempolimit-lichtgeschwindigkeit.de/sphere/1

though I prefer to 'look at' cubes:

https://www.tempolimit-lichtgeschwindigkeit.de/tompkins

https://www.tempolimit-lichtgeschwindigkeit.de/filme/wuerfelketten/wuerfelketten-xd-640x480.mp4

--

guido wugi

which arrived in the moderation system on 2023-06-12. -- jt]]

Op 12/06/2023 om 9:09 schreef Luigi Fortunati:

> Here's the lesson: the contraction of a moving sphere is like the trick: it's there but you can't "see".

https://wugi.be/MySRT/Squad.gif

The squares are *seen* as (very) distorted quadrangles. So would the

inscribed circles/spheres.

Now it may be that the latter would look *less* distorted, because their

*angle of vision* varies less than that of the squares' appearances

themselves, as the perpendicular (to motion) directions are not

contracted. Try imagining angles of vision to inscribed circles. Well

then, spheres might look less distorted, but their meridians (and other

circles) will certainly do more so!

As for the Terell rotation,

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

as I've said before,

that's just a special case of the distortions seen, valid only

perpendicular to the direction of motion, and far enough away, in my

picture: at the far end of the vertical through the observer.

Look carefully, comparing with the 'actual position' dots at the right:

that's the only area where the Lorentz contraction can 'really' be *seen*!

More on looking at relativistic spheres:

https://www.spacetimetravel.org/tompkins/tompkins.pdf

https://www.spacetimetravel.org/tompkins/1 (look at sphere animations!)

https://www.tempolimit-lichtgeschwindigkeit.de/ueberblick/1 of which:

https://www.tempolimit-lichtgeschwindigkeit.de/fussball

https://www.tempolimit-lichtgeschwindigkeit.de/sphere/sphere.pdf

https://www.tempolimit-lichtgeschwindigkeit.de/sphere/1

though I prefer to 'look at' cubes:

https://www.tempolimit-lichtgeschwindigkeit.de/tompkins

https://www.tempolimit-lichtgeschwindigkeit.de/filme/wuerfelketten/wuerfelketten-xd-640x480.mp4

--

guido wugi

Jun 16, 2023, 3:36:37 PM6/16/23

to

[[Mod. note -- I apologise for the delay in processing this article,

which arrived in the s.p.r moderation system on 2023-06-12. -- jt]]
Op 11/06/2023 om 22:59 schreef wugi:

> FWIW, Luigi, your animation is about Lorentz contraction, not about what

> cameras will record.

> If you want a related example (I'm not going to redo your simulation),

> look again at my picture of watching the approach, passing by, and

> regression, of a squadron of squares:

> https://wugi.be/paratwin.htm =>

> https://wugi.be/MySRT/Squad.gif

>

> The squares are Lorentz contracted (see "actual position" dots at the

> right).

> They are *seen* as quadrangles, ie distorted rectangles.

> At approach there is blueshift, at regression redshift (my colors are

> inverted, for a white background;).

> At approach the speed seen is greater than the motion velocity v (it

> becomes infinite for v=c), at regression it is less than v (it becomes

> c/2 for v=c).

> At approach there is a seen length expansion(! it becomes infinite for

> v=c), at regression an extra length contraction(! it becomes L/2gamma

> for v=c IIRC).

>

> Now if you want to visualise spheres, or circles, fill in any of the

> quadrangles with a corresponding ellipse-like figure, that's how the

> distorted circles are going to be seen. If you want a symmetrical

> circle, take 2*2 adjoining squares, 2 along either side of the axis of

> symmetry through the observer (the small black circle).

visualisation at the tachyon page

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

of how a *tachyon sphere* would be seen, that is, *if it existed*:

https://upload.wikimedia.org/wikipedia/commons/6/64/Tachyon04s.gif

In our subluminal (v < c) case, you can get a qualitative idea of the

equivalent visualisation,

- by applying length contraction for v to the grey ("actual") sphere,

- by noticing that the approaching ellipsoid (blue shifted) will be

moving in the same direction as the actual (grey), reaching the observer

together (the blue at a larger speed than v), and

- will proceed further as the red shifted ellipsoid, at a lower speed

than v, similar to this picture.

--

guido wugi

Jun 17, 2023, 5:01:41 PM6/17/23

to

wugi il 16/06/2023 07:35:50 ha scritto:

> ...

this discussion is very useful to me.

What you wrote seems correct to me: a spherical body in motion

maintains its sphericity visually (and on the monitor) only in one

particular case and appears crushed in the other cases.

So, my two animations

https://www.geogebra.org/m/axxtdurx (The contraction is there but you

can't see it on the monitor)

https://www.geogebra.org/m/grq2shgx (The contraction is there and

appears on the monitor)

are both correct, one in one case and the other in another.

Is that it?

> ...

> As for the Terell rotation,

> https://en.wikipedia.org/wiki/Terrell_rotation

> as I've said before,

> that's just a special case of the distortions seen, valid only

> perpendicular to the direction of motion, and far enough away, in my

> picture: at the far end of the vertical through the observer.

The discussions serve to improve the knowledge of the interlocutors and
> https://en.wikipedia.org/wiki/Terrell_rotation

> as I've said before,

> that's just a special case of the distortions seen, valid only

> perpendicular to the direction of motion, and far enough away, in my

> picture: at the far end of the vertical through the observer.

this discussion is very useful to me.

What you wrote seems correct to me: a spherical body in motion

maintains its sphericity visually (and on the monitor) only in one

particular case and appears crushed in the other cases.

So, my two animations

https://www.geogebra.org/m/axxtdurx (The contraction is there but you

can't see it on the monitor)

https://www.geogebra.org/m/grq2shgx (The contraction is there and

appears on the monitor)

are both correct, one in one case and the other in another.

Is that it?

Jun 18, 2023, 7:53:05 PM6/18/23

to

Op 17/06/2023 om 23:01 schreef Luigi Fortunati:

So yes, you don't see the contraction (except perpendicular to the=20

motion, and far away, Terell-wise).

But no, you don't see the spheres "as is", but really with different=20

distortions at different positions.

This is the "measuring" process.

So no, you won't see it "appear on the monitor". It will result from=20

measuring and calculating.

> are both correct, one in one case and the other in another.

>

> Is that it?

A last try, hoping to be clear:

I've adapted a bit my aforementioned picture to the case of passing=20

circles (spheres):

https://pin.it/3SpivNV

The "true" actual positions with length contraction are in grey, at the=20

right. See contracted circles.

The "seen" positions are in red=C2=B0 (approach) and bluegreen=C2=B0 (reg=

ression),=20

with the distorted circles as (approximately) seen. No "perfect" circles=20

there!

=C2=B0 (negative Doppler colors. I've used "correct" blue and red instead=

for=20

the circles)

I've added the Terell rotation case, "to north and south of observer,=20

far away", compare with contracted grey circles in actual positions at=20

the right.

Hope you'll understand at last that there can be no question of a=20

*constant shape*, circle or not, to be *seen* passing by.

--=20

guido wugi

> wugi il 16/06/2023 07:35:50 ha scritto:

>> ...

>> As for the Terell rotation,

>> https://en.wikipedia.org/wiki/Terrell_rotation

>> as I've said before,

>> that's just a special case of the distortions seen, valid only

>> perpendicular to the direction of motion, and far enough away, in my

>> picture: at the far end of the vertical through the observer.

>

> The discussions serve to improve the knowledge of the interlocutors an=
>> ...

>> As for the Terell rotation,

>> https://en.wikipedia.org/wiki/Terrell_rotation

>> as I've said before,

>> that's just a special case of the distortions seen, valid only

>> perpendicular to the direction of motion, and far enough away, in my

>> picture: at the far end of the vertical through the observer.

>

d

> this discussion is very useful to me.

>

> What you wrote seems correct to me: a spherical body in motion

> maintains its sphericity visually (and on the monitor) only in one

> particular case and appears crushed in the other cases.

>

> So, my two animations

> https://www.geogebra.org/m/axxtdurx (The contraction is there but you

> can't see it on the monitor)

This is the *seeing* process.
> this discussion is very useful to me.

>

> What you wrote seems correct to me: a spherical body in motion

> maintains its sphericity visually (and on the monitor) only in one

> particular case and appears crushed in the other cases.

>

> So, my two animations

> https://www.geogebra.org/m/axxtdurx (The contraction is there but you

> can't see it on the monitor)

So yes, you don't see the contraction (except perpendicular to the=20

motion, and far away, Terell-wise).

But no, you don't see the spheres "as is", but really with different=20

distortions at different positions.

This is the "measuring" process.

So no, you won't see it "appear on the monitor". It will result from=20

measuring and calculating.

> are both correct, one in one case and the other in another.

>

> Is that it?

I've adapted a bit my aforementioned picture to the case of passing=20

circles (spheres):

https://pin.it/3SpivNV

The "true" actual positions with length contraction are in grey, at the=20

right. See contracted circles.

The "seen" positions are in red=C2=B0 (approach) and bluegreen=C2=B0 (reg=

ression),=20

with the distorted circles as (approximately) seen. No "perfect" circles=20

there!

=C2=B0 (negative Doppler colors. I've used "correct" blue and red instead=

for=20

the circles)

I've added the Terell rotation case, "to north and south of observer,=20

far away", compare with contracted grey circles in actual positions at=20

the right.

Hope you'll understand at last that there can be no question of a=20

*constant shape*, circle or not, to be *seen* passing by.

--=20

guido wugi

Jun 21, 2023, 5:08:23 AM6/21/23

to

Il giorno lunedě 19 giugno 2023 alle 01:53:05 UTC+2 wugi ha scritto:

> ....

by throughout the movement.

And I agree with you.

But the photos don't show any movement, they are instantaneous and are

the ones you see in my 2 animations when we press the "Position:C=D"

button.

We can say that (based on the position of the camera) in one case the

photo will look like in the animation

https://www.geogebra.org/m/grq2shgx

in another as that of animation

https://www.geogebra.org/m/axxtdurx

and in another will it be even different?

> ....

> Hope you'll understand at last that there can be no question of

> *constant shape*, circle or not, to be *seen* passing by.

Yes, I understood that you can't see a body of constant shape passing
by throughout the movement.

And I agree with you.

But the photos don't show any movement, they are instantaneous and are

the ones you see in my 2 animations when we press the "Position:C=D"

button.

We can say that (based on the position of the camera) in one case the

photo will look like in the animation

https://www.geogebra.org/m/grq2shgx

in another as that of animation

https://www.geogebra.org/m/axxtdurx

and in another will it be even different?

Jun 23, 2023, 12:20:14 PM6/23/23

to

Op 21/06/2023 om 11:08 schreef Luigi Fortunati:

> We can say that (based on the position of the camera) in one case the

> photo will look like in the animation

> https://www.geogebra.org/m/grq2shgx

If you agree with me, then why do you keep insisting on the assumption that length contraction could be "photographed" instantly?

> in another as that of animation

> https://www.geogebra.org/m/axxtdurx

> and in another will it be even different?

If you agree with me, then why do you keep insisting on the assumption that a relativistic body could be "photographed" unaltered?

But you inspired me to update my "relativistic squadron watching", so you can try it out here:

https://www.desmos.com/calculator/yey8cgrgat?lang=nl

Moreover you inspired me to modify it into "relativistic circle watching", precisely your thema here, with choices of radius and position, so you can try here, and find out that "perfect circles" are nowhere to be "seen":

https://www.desmos.com/calculator/yxmju4xjrd?lang=nl

--

guido wugi

> Il giorno lunedė 19 giugno 2023 alle 01:53:05 UTC+2 wugi ha scritto:

>> ....

>> Hope you'll understand at last that there can be no question of

>> *constant shape*, circle or not, to be *seen* passing by.

>

> Yes, I understood that you can't see a body of constant shape passing

> by throughout the movement.

>

> And I agree with you.

>

> But the photos don't show any movement, they are instantaneous and are

> the ones you see in my 2 animations when we press the "Position:C=D"

> button.

So the animation should be disposed of?
>> ....

>> Hope you'll understand at last that there can be no question of

>> *constant shape*, circle or not, to be *seen* passing by.

>

> Yes, I understood that you can't see a body of constant shape passing

> by throughout the movement.

>

> And I agree with you.

>

> But the photos don't show any movement, they are instantaneous and are

> the ones you see in my 2 animations when we press the "Position:C=D"

> button.

> We can say that (based on the position of the camera) in one case the

> photo will look like in the animation

> https://www.geogebra.org/m/grq2shgx

> in another as that of animation

> https://www.geogebra.org/m/axxtdurx

> and in another will it be even different?

But you inspired me to update my "relativistic squadron watching", so you can try it out here:

https://www.desmos.com/calculator/yey8cgrgat?lang=nl

Moreover you inspired me to modify it into "relativistic circle watching", precisely your thema here, with choices of radius and position, so you can try here, and find out that "perfect circles" are nowhere to be "seen":

https://www.desmos.com/calculator/yxmju4xjrd?lang=nl

--

guido wugi

Jun 24, 2023, 2:44:09 AM6/24/23

to

Op 23/06/2023 om 18:20 schreef wugi:

Squad

https://www.desmos.com/calculator/kefqjx5yul?lang=nl

Circle

https://www.desmos.com/calculator/eyxedvdu6c?lang=nl

I thought I had saved the same file (same name), but Desmos manages to make it different files with same name, which I don't understand, sorry.

--

guido wugi

> But you inspired me to update my "relativistic squadron watching", so you can try it out here:

> https://www.desmos.com/calculator/yey8cgrgat?lang=nl

>

> Moreover you inspired me to modify it into "relativistic circle watching", precisely your thema here, with choices of radius and position, so you can try here, and find out that "perfect circles" are nowhere to be "seen":

> https://www.desmos.com/calculator/yxmju4xjrd?lang=nl

After necessary corrections and some additional features:
> https://www.desmos.com/calculator/yey8cgrgat?lang=nl

>

> Moreover you inspired me to modify it into "relativistic circle watching", precisely your thema here, with choices of radius and position, so you can try here, and find out that "perfect circles" are nowhere to be "seen":

> https://www.desmos.com/calculator/yxmju4xjrd?lang=nl

Squad

https://www.desmos.com/calculator/kefqjx5yul?lang=nl

Circle

https://www.desmos.com/calculator/eyxedvdu6c?lang=nl

I thought I had saved the same file (same name), but Desmos manages to make it different files with same name, which I don't understand, sorry.

--

guido wugi

Jun 27, 2023, 3:20:28 PM6/27/23

to

wugi il 23/06/2023 18:20:10 ha scritto:

> So the animation should be disposed of?

No, my animation is certainly fine at the top showing the actual
> So the animation should be disposed of?

contraction.

The lower part (the monitor, the visual part) must be corrected only in

the form of the deformation which is not symmetrical as I have proposed

but is more ramshackle, as you correctly propose it with your

animations.

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