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# Length contraction

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### Luigi Fortunati

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?

### Richard Livingston

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

### Julio Di Egidio

May 22, 2023, 3:10:03 AM5/22/23
to
You expect an answer "by eye"?? Where are your *formulas*, the ones
you use to build the graph?! And how long shall we repeat just that??
(Which is more of a complaint about the moderation.)

Julio

### Luigi Fortunati

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?
>
> 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.

### Luigi Fortunati

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.

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.

My animation does not represent a movement in time but the
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.

### Julio Di Egidio

May 23, 2023, 4:32:04 PM5/23/23
to
On Monday, 22 May 2023 at 19:16:07 UTC+2, Luigi Fortunati wrote:
> 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?
> >
> > 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
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

### Luigi Fortunati

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.)

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!

After this last correction of mine, is my animation still missing
something to conform to the "mathematical model" you are talking about?

> Julio
Luigi

### Julio Di Egidio

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>
> After this last correction of mine, is my animation still missing
> something to conform to the "mathematical model" you are talking

"*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

HTH,

Julio

### Luigi Fortunati

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

My animation is modeled according to the dictates of Special
Relativity.

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

Can you show me how you would draw the tangents?

> Julio
Luigi

### wugi

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

--
guido wugi

### wugi

May 30, 2023, 12:48:57 PM5/30/23
to
Op 25/05/2023 om 9:18 schreef Julio Di Egidio:
> On Wednesday, 24 May 2023 at 19:33:38 UTC+2, Luigi Fortunati wrote:
>
>> <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
>
> "*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
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

### Luigi Fortunati

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
>
> 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:
- 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
> in" the space towards it, and 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!

Ok, let's talk about what happens when photons of light arrive on the
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
(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

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]]

### wugi

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.)
>
>
> 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!
>
> 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,
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

### 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:
(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:

(with nice animations by J. Doolin)

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

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

### Luigi Fortunati

Jun 7, 2023, 12:52:06 PM6/7/23
to
the observers and body C, then the discussion shifted to the
contraction of body C.

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?

### wugi

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

### Luigi Fortunati

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

### Luigi Fortunati

Jun 9, 2023, 5:28:26 AM6/9/23
to
Why all these complications?

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
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.

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]]

### Luigi Fortunati

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
> (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=
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
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]]

### Luigi Fortunati

Jun 11, 2023, 3:21:36 AM6/11/23
to
Luigi Fortunati il 08/06/2023 21:28:21 ha scritto:
>> 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?

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]]

### wugi

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?

> [[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]]
His confusion remains...

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 =>

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

### Luigi Fortunati

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
spherical body C of my animation appears perfectly spherical and
identical to the body D, without any relativistic contraction.

Luigi.

### Luigi Fortunati

Jun 12, 2023, 3:09:42 AM6/12/23
to
The monitors and photos capture the images present on the place and, therefore, show exactly what the eyes see.

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.

### Luigi Fortunati

Jun 14, 2023, 6:45:00 AM6/14/23
to
Luigi Fortunati il 10/06/2023 19:21:31 ha scritto:
>
> Okay.
>
> 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=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).

### Luigi Fortunati

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
> 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
> 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
> 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?

### wugi

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?
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

### 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:

> 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 =>
>
> 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).

FWIW, apart from the other visualisations I proposed, there is a nice
visualisation at the tachyon page
https://en.wikipedia.org/wiki/Tachyon
of how a *tachyon sphere* would be seen, that is, *if it existed*:

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

### Luigi Fortunati

Jun 17, 2023, 5:01:41 PM6/17/23
to
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 and
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?

### wugi

Jun 18, 2023, 7:53:05 PM6/18/23
to
Op 17/06/2023 om 23:01 schreef Luigi Fortunati:
> 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=
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.
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.

> https://www.geogebra.org/m/grq2shgx (The contraction is there and
> appears on the monitor)
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

### Luigi Fortunati

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:
> ....
> 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?

### wugi

Jun 23, 2023, 12:20:14 PM6/23/23
to
Op 21/06/2023 om 11:08 schreef Luigi Fortunati:
> 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?

> 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

### wugi

Jun 24, 2023, 2:44:09 AM6/24/23
to
Op 23/06/2023 om 18:20 schreef 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/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