Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]

Groups > sci.physics.relativity > #620554 > unrolled thread

[SR] Baby steps, my lord, baby steps!!!

Back to article view | Back to sci.physics.relativity

Contents

  [SR] Baby steps, my lord, baby steps!!! Richard Hachel <r.hachel@frite.fr> - 2023-09-20 12:50 +0000
    Re: [SR] Baby steps, my lord, baby steps!!! Athel Cornish-Bowden <athel.cb@gmail.com> - 2023-09-20 16:31 +0200
    Re: [SR] Baby steps, my lord, baby steps!!! Richard Hachel <r.hachel@frite.fr> - 2023-09-21 09:10 +0000

#620554 — [SR] Baby steps, my lord, baby steps!!!

New or revolutionary notions must be introduced into the world with great 
caution.
 
The main thing you need to do well is explain things correctly.
 
We must not be like Python, who, after having consumed three bottles of 
whiskey when I recommended coffee, begins to say incomprehensible things 
in public.
 
Is there “really” a relativity of times?
 
The question is very good.
 
You say no, and you take the example of the muon evolving at 0.995c (we 
can also take the meson evolving at 0.9995c with γ =31.63), there is no 
variation of the proper time, and in in a sense, you are right, if we 
admit that the half-life of a µ meson is 25.5 nanoseconds, it is 25.5 
nanoseconds
in all repositories.
 
It is absurd to think that placed elsewhere, let's say a rocket, it will 
start to claim, like Macron, that its half-life will increase to 26.6 
instead of 25.5 depending on the increase in the price of a barrel of oil, 
if on the contrary we postulated "all the frames of reference are equal 
and the laws of physics are the same".
 
The same thing for two rockets moving away from me, one to the right, the 
other to the left, at 0.5c.
 
We know that between them, they move away at Vo=0.8/c (law of composition 
of speeds), and therefore that they
should perceive events three times longer (let's not forget the 
longitudinal Doppler effect).
 
We know, however, that the same event in one rocket will obviously last 
the same time in both rockets (since they are the same). Everything 
therefore only acts in relative mode, and "in relation to".
 
In this sense, proper time is constant. Let us experiment wherever we 
want, the own half-life
of the µ meson will always be the same, and the film on my screen will 
always last 1h30.
 
I think everyone agrees on that.
 
But beware.
 
Yes and no.
 
Baby steps concept.
 
The great genius in relativity is knowing how to do little baby steps, and 
explaining things at the same time to the big names, physics professors, 
and to ordinary people, crank on the internet.
 
What about Stella's own time and Terrence's own time?
 
We know that for the two rockets evolving at 0.5c, and each making a loop 
on its side relative to me, will return with the same age. It's a truism. 
Even if they move at 0.8c between them.
 
But for Stella and Terrence, an anomaly will arise, because the two 
protagonists do not do the same thing, and for no possible examiner.
 
She comes back, she is 18 years old; he has 30.
 
There is indeed a real relativity of time there.
 
We need to clearly define what we are talking about. This is essential in 
relativity.
 
Otherwise it's all confusion. I was talking yesterday about the confusion 
that physicists make between a segment of time in a frame of reference 
observing an accelerated movement for this frame of reference, and a 
segment of time
in a frame of reference observing two accelerated movements between them.
 
There are two different equations to use for two different situations.
 
Physicists use the first for the second.
 
The first is:
ΔTo=ΔTr2.sqrt(1+2c²/ax2)-ΔTr1.sqrt(1+2c²/ax1)
the second is:
ΔTo=(Δx/c).sqrt(1+2c²/a[sqrt(x1)+sqrt(x2)]
 
The error, among them, is then colossal
 
The first equation should be used to, for example, calculate the time 
difference between
the return of a rocket moving over 11 light years, with an acceleration a, 
and the return of another rocket, moving with the same acceleration but 
over 12 light years.
 
In this their equation is good.
 
And physicists like Paul B. Andersen at sci.physics.relativity are right.
 
But if they imagine that the same equation is valid to calculate small 
segments of time or instantaneous speeds on the same route between any two 
points A and B, they are wrong.
 
You have to use the equation that I gave, and the instantaneous speeds 
that I gave.
 
Thank you for your distracted attention.
 
I say “distracted” because I know men, and their natural arrogance.
 
R.H.

[toc] | [next] | [standalone]

#620561

On 2023-09-20 12:50:24 +0000, Richard Hachel said:

> New or revolutionary notions must be introduced into the world with 
> great caution.
>  The main thing you need to do well is explain things correctly.

True enough. Why don't you try it some time?

> 
> -- 
> athel -- biochemist, not a physicist, but detector of crackpots

[toc] | [prev] | [next] | [standalone]

#620608

Le 20/09/2023 à 14:50, Richard Hachel a écrit :

> There are two different equations to use for two different situations.
>  
> Physicists use the first for the second.
>  
> The first is:
> ΔTo=ΔTr2.sqrt(1+2c²/ax2)-ΔTr1.sqrt(1+2c²/ax1)
> the second is:
> ΔTo=(Δx/c).sqrt(1+2c²/a[sqrt(x1)+sqrt(x2)]
>  
> The error, among them, is then colossal

 Paul B. Andersen propose :
"therefore dtau = sqrt(dt^2 - dx^2) = dt/sqrt(1 + (at)^2), whose integral 
is tau = (1/a)invsinh(at) = 3.139 years at t=12.915 years. Agreed?"

 No, absolutely not.

 Faire cette intégration, c'est ajouter, sans s'en rendre compte, des 
carottes et des navets. 

 BUT, did he try to integrate the second equation which seems more correct 
to use?

 R.H. 

[toc] | [prev] | [standalone]

Back to top | Article view | sci.physics.relativity

csiph-web