what is relativity of size that Einstein overlooked?
He needed to address parallax in dimension.
But how is distance a relative? It is not...

Mitchell Raemsch

Le 24/01/2022 à 19:59, "mitchr...@gmail.com" a écrit :
> what is relativity of size that Einstein overlooked?
> He needed to address parallax in dimension.
> But how is distance a relative? It is not...
>
> Mitchell Raemsch

Distances are relative in the sense that times are.

Take the size of a rocket, it measures thirty meters.

If I observe it at rest, it measures thirty meters.

But let's say it's moving towards me with a significant speed.

When the head of the rocket sends me a photon, and simultaneously, the
tail, the photon of the head reaches me before the other.

In short, it reaches me at the same time as another photon which had left
before, when the rocket was further away.

The rocket seems longer.

This is called the longitudinal Doppler effect.

And conversely the rocket seems shorter when it moves away.

Except that for me, the rocket is actually shorter or longer. This is NOT
a Doppler effect due to the limiting speed of light (it moves
instantaneously at home) but an effect due to spatial anisochrony.

Another effect, even more complex will occur: the transverse Doppler
effect which makes the rocket seem to shrink EVEN in the case of a simple
transverse passage.

The elasticity equation is given as:
L'=L.sqrt(1-v²/c²)/(1+cosµ.v/c).

The rocket can therefore be larger or smaller depending on how you look at
it. That is to say according to its position and its relative speed.

And you know everything.

You UNDERSTOOD everything!!!

R.H.

Richard Hachel wrote:

> When the head of the rocket sends me a photon, and simultaneously, the
> tail, the photon of the head reaches me before the other.
>
> In short, it reaches me at the same time as another photon which had
> left before, when the rocket was further away.

no shit Sherlock.

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"International Criminals"
https://www.bitchute.com/video/kcqVTleRn2sR/

but I can tell you the entire story. In capitalist europe, it sharp
forbidden selling a product in connection with another product.

yet, they sold each and every fucking computer with Windoze preinstalled.
Now guess, the uneducated mass murderer bill gaytes, involved the
governments to buy and inject each and every person in those countries
with his injections, called "vaccines". Fuck you all. If you can't stand
up against obvious government crimes against humanity, fuck you all.

don't go out protesting *singing*, you have to arrest those bankers sons
of a bitches. When I see them go out in thousands singing, makes me sick.

Richard Hachel wrote:

> Take the size of a rocket, it measures thirty meters.
>
> If I observe it at rest, it measures thirty meters.
>
> But let's say it's moving towards me with a significant speed.
>
> When the head of the rocket sends me a photon, and simultaneously, the
> tail, the photon of the head reaches me before the other.
>
> In short, it reaches me at the same time as another photon which had left
> before, when the rocket was further away.

It doesn’t.

> The rocket seems longer.

No, _shorter_ (if you could see it from the side somehow; otherwise the
concept “length” is meaningless). This is called “length contraction”.

It is an important consequence of the special principle of relativity and
the Lorentz transformation that follows from it (here simplified for motion
of the primed frame parallel to the [so chosen] x-axis of the unprimed-
frame):

x' = γ (x − v t)

From the linearity of the equation/transformation follows for uniform motion
for position x₁ at t₁, and x₂ at t₂:

x₂' − x₁' =: Δx' = γ (Δx + v Δt) ≔ γ [(x₂ − x₁) + v (t₂ − t₁)].

We (already intuitively) define the length of an object in a frame as the
spatial distance of its end points at the same time in that frame. Here we
wish to know the situation as observed in the unprimed frame, so we assume
Δt = 0

⇒ Δx' = γ Δx.

L₀ ≔ Δx' is the proper length of the rocket, i.e. the length measured in a
frame in which the rocket is at rest (here: the primed, moving frame).
L ≔ Δx is the observed length of the rocket in another frame (here: in the
unprimed, stationary frame). Therefore,

Δx = Δx'/γ

⇔ L = L₀/γ.

Further,

γ(v) = 1/√(1 − v²/c²) ≥ 1,

therefore

L < L₀.

An equivalent way to write it is

L = L₀ √(1 − v²/c²),

where, since 0 < √(1 − v²/c²) < 1, 0 < L < L₀.

IOW: Moving objects are shorter in the direction of motion than they are at
(relative) rest.

See also:

<https://en.wikipedia.org/wiki/Length_contraction>

Originally (translated):

Albert Einstein: “On the Electrodynamics of Moving Bodies”. “§2. On the
relativity of lengths and times” to at least “§4. The physical meaning of
the equations obtained concerning moving rigid bodies and moving clocks”.
Annalen der Physik 17 (1905): 891–921. Translation: Anna Beck et al.,
Princeton University Press, September 1989.
<https://einsteinpapers.press.princeton.edu/vol2-trans/157>

The fact that photons/light pulses from the rear of the rocket reach the
observer _later_ than photons/light pulses from the front of the rocket
leads to another (lesser known) effect when we consider the actual
observation where we must consider the light-travel time as well:

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

> This is called the longitudinal Doppler effect.

No. According to the longitudinal Doppler effect, the approaching rocket
appears _blueshifted_ instead (in addition to the effects mentioned above):

<https://en.wikipedia.org/wiki/Relativistic_Doppler_effect>

This is the "standard" relativistic Doppler effect with which we can
determine whether a celestial object is "moving" through space towards us
(blueshift) or away from us (redshift), e.g. how we can identify and
characterize binary star systems as such (even if the components cannot be
separated visually, but the absorption lines in their common spectrum can be
identified).¹

See also:

<http://gamelab.mit.edu/games/a-slower-speed-of-light/>

PointedEars
___________
¹ This is not to be confused with cosmological redshift (of the light of
other galaxies, on average) due to the expansion of space, or
gravitational redshift due to a larger spacetime curvature (e.g. near
black holes).
--
Q: What happens when electrons lose their energy?
A: They get Bohr'ed.

(from: WolframAlpha)

On Monday, January 24, 2022 at 4:33:46 PM UTC-8, Thomas 'PointedEars' Lahn wrote:
> Richard Hachel wrote:
>
> > Take the size of a rocket, it measures thirty meters.
> >
> > If I observe it at rest, it measures thirty meters.
> >
> > But let's say it's moving towards me with a significant speed.
> >
> > When the head of the rocket sends me a photon, and simultaneously, the
> > tail, the photon of the head reaches me before the other.
> >
> > In short, it reaches me at the same time as another photon which had left
> > before, when the rocket was further away.
> It doesn’t.
>
> > The rocket seems longer.
>
> No, _shorter_ (if you could see it from the side somehow; otherwise the
> concept “length” is meaningless). This is called “length contraction”.
>
> It is an important consequence of the special principle of relativity and
> the Lorentz transformation that follows from it (here simplified for motion
> of the primed frame parallel to the [so chosen] x-axis of the unprimed-
> frame):
>
> x' = γ (x − v t)
>
> From the linearity of the equation/transformation follows for uniform motion
> for position x₁ at t₁, and x₂ at t₂:
>
> x₂' − x₁' =: Δx' = γ (Δx + v Δt) ≔ γ [(x₂ − x₁) + v (t₂ − t₁)].
>
> We (already intuitively) define the length of an object in a frame as the
> spatial distance of its end points at the same time in that frame. Here we
> wish to know the situation as observed in the unprimed frame, so we assume
> Δt = 0
>
> ⇒ Δx' = γ Δx.
>
> L₀ ≔ Δx' is the proper length of the rocket, i.e. the length measured in a
> frame in which the rocket is at rest (here: the primed, moving frame).
> L ≔ Δx is the observed length of the rocket in another frame (here: in the
> unprimed, stationary frame). Therefore,
>
> Δx = Δx'/γ
>
> ⇔ L = L₀/γ.
>
> Further,
>
> γ(v) = 1/√(1 − v²/c²) ≥ 1,
>
> therefore
>
> L < L₀.
>
> An equivalent way to write it is
>
> L = L₀ √(1 − v²/c²),
>
> where, since 0 < √(1 − v²/c²) < 1, 0 < L < L₀.
>
> IOW: Moving objects are shorter in the direction of motion than they are at
> (relative) rest.
>
> See also:
>
> <https://en.wikipedia.org/wiki/Length_contraction>
>
> Originally (translated):
>
> Albert Einstein: “On the Electrodynamics of Moving Bodies”. “§2. On the
> relativity of lengths and times” to at least “§4. The physical meaning of
> the equations obtained concerning moving rigid bodies and moving clocks”.
> Annalen der Physik 17 (1905): 891–921. Translation: Anna Beck et al.,
> Princeton University Press, September 1989.
> <https://einsteinpapers.press.princeton.edu/vol2-trans/157>
>
>
> The fact that photons/light pulses from the rear of the rocket reach the
> observer _later_ than photons/light pulses from the front of the rocket
> leads to another (lesser known) effect when we consider the actual
> observation where we must consider the light-travel time as well:
>
> <https://en.wikipedia.org/wiki/Terrell_rotation>
> > This is called the longitudinal Doppler effect.
> No. According to the longitudinal Doppler effect, the approaching rocket
> appears _blueshifted_ instead (in addition to the effects mentioned above):
>
> <https://en.wikipedia.org/wiki/Relativistic_Doppler_effect>
>
> This is the "standard" relativistic Doppler effect with which we can
> determine whether a celestial object is "moving" through space towards us
> (blueshift) or away from us (redshift), e.g. how we can identify and
> characterize binary star systems as such (even if the components cannot be
> separated visually, but the absorption lines in their common spectrum can be
> identified).¹
>
> See also:
>
> <http://gamelab.mit.edu/games/a-slower-speed-of-light/>
>
>
> PointedEars
> ___________
> ¹ This is not to be confused with cosmological redshift (of the light of
> other galaxies, on average) due to the expansion of space, or
> gravitational redshift due to a larger spacetime curvature (e.g. near
> black holes).
> --
> Q: What happens when electrons lose their energy?
> A: They get Bohr'ed.
>
> (from: WolframAlpha)

The metric is basically two measures of distance.

The measurement effect is a thing, though it's often related with indeterminacy
in the quantum, when really it's just about the real wave-function collapse..

Bell's "paradox" and the rocket cutting the string (or, not, or, everywhere all
at once), has various readings.

The measurement effect is a thing, involved in observer and sampling effects.

There's only intromissive not extromissive theory, classical optics, but, dynamics
can introduce optics written with an extromissive component.

Thanks for your writing P.E.

Le 25/01/2022 à 01:33, Thomas 'PointedEars' Lahn a écrit :

> γ(v) = 1/√(1 − v²/c²) ≥ 1,
>
> therefore
>
> L < L₀.
>
> An equivalent way to write it is
>
> L = L₀ √(1 − v²/c²)

Yes, I understood that very well.
Those who read me believe that if I refute, it's because I didn't
understand, or because I'm stupid.
I very frequently receive messages telling me that I am stupid.
But I repeat it tirelessly, when someone tries to convince you of
something, it means that the thing is not clear.
Imagine being told at home, and having it repeated to you countless times:
"Madam, you are going to have a new neighbour, but don't be afraid for
your children, he is not a pedophile".
This is where you should start freaking out for your kids.
So beware of all those people who spend their time pointing out that
"Richard Hachel is very stupid". There must be something unclear behind
it.

L = L₀ √(1 − v²/c²)/(1+cosµ.v/c) is better.

And L < L₀ is not always true.

R.H.

On 25-Jan-22 5:59 am, mitchr...@gmail.com wrote:
> what is relativity of size that Einstein overlooked?
> He needed to address parallax in dimension.
> But how is distance a relative? It is not...
>
> Mitchell Raemsch

More gibberish from Mitchell.

Sylvia.

Richard Hachel <r.hachel@tiscali.fr> wrote:
> Le 25/01/2022 à 01:33, Thomas 'PointedEars' Lahn a écrit :
>
>> γ(v) = 1/√(1 − v²/c²) ≥ 1,
>>
>> therefore
>>
>> L < L₀.
>>
>> An equivalent way to write it is
>>
>> L = L₀ √(1 − v²/c²)
>
> Yes, I understood that very well.
> Those who read me believe that if I refute, it's because I didn't
> understand, or because I'm stupid.
> I very frequently receive messages telling me that I am stupid.
> But I repeat it tirelessly, when someone tries to convince you of
> something, it means that the thing is not clear.

Oh I think I understand you quite well.

“It is impossible that others say that physicists explain relativity well,
because to me they are not clear at all, and the bulge in my pants attests
to that also.”

“Some say that I am not well, but I am in fact a doctor and must be well to
treat others that are not well, and besides, I have a large penis.”

“The problem cannot be with me, because it would mean that I am stupid, and
this cannot be, because of my huge masculine beauty.”

Do I have that about right?

> Imagine being told at home, and having it repeated to you countless times:
> "Madam, you are going to have a new neighbour, but don't be afraid for
> your children, he is not a pedophile".
> This is where you should start freaking out for your kids.
> So beware of all those people who spend their time pointing out that
> "Richard Hachel is very stupid". There must be something unclear behind
> it.
>
> L = L₀ √(1 − v²/c²)/(1+cosµ.v/c) is better.
>
> And L < L₀ is not always true.
>
> R.H.
>

--
Odd Bodkin -- maker of fine toys, tools, tables

On 1/25/2022 6:26 AM, Richard Hachel wrote:
> Le 25/01/2022 à 01:33, Thomas 'PointedEars' Lahn a écrit :
>
>>   γ(v) = 1/√(1 − v²/c²) ≥ 1,
>>
>> therefore
>>
>>   L < L₀.
>>
>> An equivalent way to write it is
>>
>>   L = L₀ √(1 − v²/c²)
>
> Yes, I understood that very well.
> Those who read me believe that if I refute, it's because I didn't
> understand, or because I'm stupid.
> I very frequently receive messages telling me that I am stupid.

Did you ever consider the possibility that they may be, in fact, correct?

> And  L < L₀ is not always true.

Did you look at the relationship for different values of v? If |v| > c,
then √(1 − v²/c²) is imaginary (square root of a negative number) and L
is purely imaginary. The only time L < L₀ isn't true is when v=0 and
therefore L = L₀. For L > L₀ v would have to be pure imaginary so v² is
negative, making √(1 − v²/c²) > 1.

√(1 − v²/c²)

Le 25/01/2022 à 14:11, Odd Bodkin a écrit :
> Oh I think I understand you quite well.
>
> “It is impossible that others say that physicists explain relativity well,
> because to me they are not clear at all, and the bulge in my pants attests
> to that also.”
>
> “Some say that I am not well, but I am in fact a doctor and must be well to
> treat others that are not well, and besides, I have a large penis.”
>
> “The problem cannot be with me, because it would mean that I am stupid, and
> this cannot be, because of my huge masculine beauty.”
>
> Do I have that about right?

No, I don't think you understood.

It's much finer than that the Hachel case.

Much, much more.

R.H.

Le 25/01/2022 à 16:28, Michael Moroney a écrit :
> On 1/25/2022 6:26 AM, Richard Hachel wrote:
>> Le 25/01/2022 à 01:33, Thomas 'PointedEars' Lahn a écrit :
>>
>>>   γ(v) = 1/√(1 − v²/c²) ≥ 1,
>>>
>>> therefore
>>>
>>>   L < L₀.
>>>
>>> An equivalent way to write it is
>>>
>>>   L = L₀ √(1 − v²/c²)
>>
>> Yes, I understood that very well.
>> Those who read me believe that if I refute, it's because I didn't
>> understand, or because I'm stupid.
>> I very frequently receive messages telling me that I am stupid.
>
> Did you ever consider the possibility that they may be, in fact, correct?

Yes.

Obviously.

But you?

LOL.

>> And  L < L₀ is not always true.
>
> Did you look at the relationship for different values of v? If |v| > c,
> then √(1 − v²/c²) is imaginary (square root of a negative number) and L
> is purely imaginary. The only time L < L₀ isn't true is when v=0 and
> therefore L = L₀. For L > L₀ v would have to be pure imaginary so v² is
> negative, making √(1 − v²/c²) > 1.
>
> √(1 − v²/c²)

? ? ?

R.H.

Ross A. Finlayson wrote:

> The metric is basically two measures of distance.

No, one. (Why two?)

> The measurement effect is a thing, though it's often related with
> indeterminacy in the quantum, when really it's just about the real
> wave-function collapse.

Quantum-mechanical effects have nothing to do with it. And “the quantum” is
just word salad.

PointedEars
--
“Science is empirical: knowing the answer means nothing;
testing your knowledge means everything.”
—Dr. Lawrence M. Krauss, theoretical physicist,
in “A Universe from Nothing” (2009)

Richard Hachel wrote:

> Le 25/01/2022 à 01:33, Thomas 'PointedEars' Lahn a écrit :
>> γ(v) = 1/√(1 − v²/c²) ≥ 1,
>>
>> therefore
>>
>> L < L₀.
>>
>> An equivalent way to write it is
>>
>> L = L₀ √(1 − v²/c²)
>
> Yes, I understood that very well.
> Those who read me believe that if I refute

You mean “refuse”. You have yet to refute those equations; your fantasies
are not refutations.

> it's because I didn't understand, or because I'm stupid.

I do not know whether you are not smart enough to understand it, but you
certainly have developed a mental blockage of some sort, probably caused
by a mental illness.

> L = L₀ √(1 − v²/c²)/(1+cosµ.v/c) is better.

No, it is not.

> And L < L₀ is not always true.

But it is; your fantasy formulae are irrelevant.

PointedEars
--
Two neutrinos go through a bar ...

(from: WolframAlpha)

On Tuesday, January 25, 2022 at 3:51:53 AM UTC-8, Sylvia Else wrote:
> On 25-Jan-22 5:59 am, mitchr...@gmail.com wrote:
> > what is relativity of size that Einstein overlooked?
> > He needed to address parallax in dimension.
> > But how is distance a relative? It is not...
> >
> > Mitchell Raemsch
> More gibberish from Mitchell.
>
> Sylvia.

Why are size appearances relative sylvia?
What does your hero have to say?

Mitchell Raemsch