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

Groups > sci.physics.relativity > #594645 > unrolled thread

Where is the (real) error

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

Contents

  Where is the (real) error Richard Hachel <richard.hachel@invalid.fr> - 2022-11-03 22:17 +0000
    Re: Where is the (real) error Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2022-11-04 00:19 +0100
      Re: Where is the (real) error Richard Hachel <richard.hachel@invalid.fr> - 2022-11-03 23:54 +0000
        Re: Where is the (real) error Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2022-11-04 01:11 +0100
          Re: Where is the (real) error Richard Hachel <richard.hachel@invalid.fr> - 2022-11-04 00:19 +0000
            Re: Where is the (real) error Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2022-11-04 01:25 +0100
              Re: Where is the (real) error Richard Hachel <richard.hachel@invalid.fr> - 2022-11-04 00:43 +0000
                Re: Where is the (real) error Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2022-11-04 08:30 +0100
    Re: Where is the (real) error Stan Fultoni <fultonistan@gmail.com> - 2022-11-03 16:57 -0700
      Re: Where is the (real) error Richard Hachel <richard.hachel@invalid.fr> - 2022-11-04 00:14 +0000
        Re: Where is the (real) error Stan Fultoni <fultonistan@gmail.com> - 2022-11-03 17:45 -0700
          Re: Where is the (real) error Richard Hachel <richard.hachel@invalid.fr> - 2022-11-04 01:09 +0000
            Re: Where is the (real) error Stan Fultoni <fultonistan@gmail.com> - 2022-11-03 18:49 -0700
    Re: Where is the (real) error Mikko <mikko.levanto@iki.fi> - 2022-11-04 11:58 +0200
      Re: Where is the (real) error Maciej Wozniak <maluwozniak@gmail.com> - 2022-11-04 03:20 -0700
      Re: Where is the (real) error Richard Hachel <richard.hachel@invalid.fr> - 2022-11-04 12:32 +0000
        Re: Where is the (real) error Stan Fultoni <fultonistan@gmail.com> - 2022-11-04 06:16 -0700
          Re: Where is the (real) error Maciej Wozniak <maluwozniak@gmail.com> - 2022-11-04 06:18 -0700
        Re: Where is the (real) error Mikko <mikko.levanto@iki.fi> - 2022-11-04 18:14 +0200
          Re: Where is the (real) error Richard Hachel <richard.hachel@invalid.fr> - 2022-11-04 16:21 +0000
            Re: Where is the (real) error JanPB <filmart@gmail.com> - 2022-11-04 11:00 -0700
              Re: Where is the (real) error Stefan Russo <ftfr@urstsoto.ee> - 2022-11-04 18:10 +0000
                Re: Where is the (real) error JanPB <filmart@gmail.com> - 2022-11-04 14:23 -0700
                Re: Where is the (real) error Jim Pennino <jimp@gonzo.specsol.net> - 2022-11-04 14:49 -0700
              Origin of sci.physics.relativity Tom Roberts <tjoberts137@sbcglobal.net> - 2022-11-05 17:30 -0500
                Re: Origin of sci.physics.relativity JanPB <filmart@gmail.com> - 2022-11-05 17:04 -0700
                  Re: Origin of sci.physics.relativity whodat <whodaat@void.nowgre.com> - 2022-11-05 20:37 -0500
                    Re: Origin of sci.physics.relativity JanPB <filmart@gmail.com> - 2022-11-05 18:48 -0700
                  Re: Origin of sci.physics.relativity The Starmaker <starmaker@ix.netcom.com> - 2022-11-05 21:06 -0700
                    Re: Origin of sci.physics.relativity The Starmaker <starmaker@ix.netcom.com> - 2022-11-06 12:03 -0800

Page 1 of 2  [1] 2  Next page →

#594645 — Where is the (real) error

Where is the error?

Do you have a playful temperament?

Problem of Tau Ceti (accelerated repository).

To obtain the average relativistic speed in a very small segment (or a 
larger one), one can first calculate the observable time taken by the 
object according to the distance (or the real speed).
This is the first of three steps.
We then note:
To=(x/c).sqrt(1+2c²/ax)
Or To=Tr.sqrt(1+(1/4)Vr²/c²)
We then find the same result.
We therefore have the observable (measurable, improper) time up to point 
A.
And we do the same to have the observable time up to point B.

Second step:
We subtract ToA from ToB, and we set ΔTo=ToB-ToA

Third step, we assume that the observable instantaneous speed (or the 
average observable speed in the segment AB) is Vo=Δx/ΔTo

The physics is really very simple.

All the physicists in the world do that, and I used to do that too.

And yet everything collapses, and everything is false on arrival. This 
results in a speed that is enormously greater than the reality of things.

In which of the three conditions is the blunder?

I don't think there's anyone in the world capable of answering that, and 
that, to shove the problem under the rug, everyone is going to say that 
there's no problem, and that it's is Richard Hachel who is very stupid.

I've already been hit when I explained the relativistic blunder that 
created the Langevin paradox in Galilean velocities.

No one believed me, so full of certainty and arrogance is everyone.

We will necessarily do it again.

R.H. 

[toc] | [next] | [standalone]

#594652

Richard Hachel wrote:

> Where is the error?

See below.
 
> Problem of Tau Ceti (accelerated repository).

“repository” is not the correct word.  A repository is a room (or, by 
extension, a data area) that one stores things in.

You might have been looking for the term “frame of reference”, shorter 
“reference frame”.  While we do say that something or someone is “in” a 
frame of reference, what we mean by that, though, is that we are using 
coordinates in which that object or person is at rest.
 
> To obtain the average relativistic speed

The relativistic speed is just the speed.  The adjective “relativistic” is 
merely used to indicate that it is a fast speed, a considerable fraction of 
c (v > 0.15 c is common as an indicator).

> in a very small segment (or a larger one),

A segment of what?

> one can first calculate the observable time taken by the
> object according to the distance (or the real speed).

The speed of an object does not depend on any time that would be observed in 
that object’s frame.  It solely depends on the difference in coordinates of 
that object in the frame in which we say that it is moving (in motion 
relative to that frame).

> This is the first of three steps.

No.

> We then note:

*You* do; we *don’t*.

> To=(x/c).sqrt(1+2c²/ax)
> Or To=Tr.sqrt(1+(1/4)Vr²/c²)
> We then find the same result.

The speed of an object is simply defined and calculated as

  v ≔ ‖dx⃗/dt‖₂,

where x⃗ and t are the position and time measured in the frame relative to 
which said object is considered moving.

> The physics is really very simple.

Yes.
 
> All the physicists in the world do that, and I used to do that too.

_No_ physicist in the world is calculating like you do because because it is 
simply nonsense.
 

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)

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

#594654

Le 04/11/2022 à 00:19, Thomas 'PointedEars' Lahn a écrit :

> _No_ physicist in the world is calculating like you do 

 It's true.


>because because it is simply nonsense.

 I spent 36 years of my life trying to understand relativistic kinematics.

Today I achieved it.

So I managed to understand "something else" than the accepted standards.

On your side, try not to confuse the words "strangeness and nonsense".

On the other hand, I find it unfortunate that you criticize the 
theoretical positions of another man without having previously understood 
them correctly.


> PointedEars

R.H. 

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

#594660

Richard Hachel wrote:

> Le 04/11/2022 à 00:19, Thomas 'PointedEars' Lahn a écrit :
>> _No_ physicist in the world is calculating like you do
> 
>  It's true.

Yes, it is.  There is literally no physicist who subscribes to your 
fantasies *because* they are merely *fantasies*.

>>because because it is simply nonsense.
> 
>  I spent 36 years of my life trying to understand relativistic kinematics.

Other people manage to do that in 3 years, so you should not consider yours 
a great accomplishment.  Instead, it shows how slow a thinker you are.
 

PointedEars
-- 
Q: What did the female magnet say to the male magnet?  
A: From the back, I found you repulsive, but from the front
   I find myself very attracted to you.
(from: WolframAlpha)

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

#594663

Le 04/11/2022 à 01:11, Thomas 'PointedEars' Lahn a écrit :
> Richard Hachel wrote:
> 
>> Le 04/11/2022 à 00:19, Thomas 'PointedEars' Lahn a écrit :
>>> _No_ physicist in the world is calculating like you do
>> 
>>  It's true.
> 
> Yes, it is.  There is literally no physicist who subscribes to your 
> fantasies *because* they are merely *fantasies*.
> 
>>>because because it is simply nonsense.
>> 
>>  I spent 36 years of my life trying to understand relativistic kinematics.
> 
> Other people manage to do that in 3 years, so you should not consider yours 
> a great accomplishment.  Instead, it shows how slow a thinker you are.

This answer is insulting.

Besides, she's stupid, and doesn't honor you. Any sane man would have 
immediately thought that if after 36 years, I had managed to tie up a 
complete theory without paradox, I had necessarily understood something 
other than what is commonly accepted.

> PointedEars

R.H. 

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

#594665

Richard Hachel wrote:

> Le 04/11/2022 à 01:11, Thomas 'PointedEars' Lahn a écrit :
>> Richard Hachel wrote:
>>>  I spent 36 years of my life trying to understand relativistic
>>>  kinematics.
>> 
>> Other people manage to do that in 3 years, so you should not consider
>> yours a great accomplishment.  Instead, it shows how slow a thinker you
>> are.
> 
> This answer is insulting.
> 
> Besides, she's stupid, and doesn't honor you. Any sane man would have
> immediately thought that if after 36 years, I had managed to tie up a
> complete theory without paradox, I had necessarily understood something
> other than what is commonly accepted.

If you had; but you haven’t.  You are merely having delusions that you have 
accomplished anything, as the record shows.  If you consider that insulting, 
then apparently I have hit a nerve, and you have difficulties dealing with 
the truth (of your incompetence, which of course is the original reason why 
you are having delusions).

BTW:

“10 points [on the Crackpot Index] for beginning the description of your 
theory by saying how long you have been working on it. (10 more for 
emphasizing that you worked on your own.)”

<https://math.ucr.edu/home/baez/crackpot.html>


PointedEars
-- 
Q: Why is electricity so dangerous?  
A: It doesn't conduct itself.

(from: WolframAlpha)

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

#594666

Le 04/11/2022 à 01:25, Thomas 'PointedEars' Lahn a écrit :
> If you had; but you haven’t.  You are merely having delusions that you have 
> accomplished anything, as the record shows.  If you consider that insulting, 
> then apparently I have hit a nerve, and you have difficulties dealing with 
> the truth (of your incompetence, which of course is the original reason why 
> you are having delusions).

I don't think I have any delirium at home, and the problem is not delirium 
but religious belief at home.

You have the perfect religious belief that it is absolutely impossible for 
other people to tell truths.

It's a dogma your thing, and this dogma, it is very quickly stripped.

Just ask someone: "Do you understand what Richard Hachel is saying, and 
are you able to explain it on a board?"

The person will tell you NO.

It will then be very easy to answer: "No, but wait, I don't understand. If 
you haven't understood anything, how can you say that it's absurd or that 
it's wrong"?

Because it is dogma.

We build the dogma that Richard Hachel was wrong, without even having 
listened to or understood what he said and why he said it.

It's not scientific.

It's a religious dogma. 

R.H.

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

#594681

Richard Hachel wrote:

> Le 04/11/2022 à 01:25, Thomas 'PointedEars' Lahn a écrit :
>> If you had [made an accomplishment ]; but you haven’t.  You are merely 
>> having delusions that you have accomplished anything, as the record
>> shows.  If you consider that insulting, then apparently I have hit a
>> nerve, and you have difficulties dealing with the truth (of your
>> incompetence, which of course is the original reason why you are having
>> delusions).
> 
> I don't think I have any delirium at home, […[

One of your problems that lead (past tense) to and reinforce your self-
delusion is that you have never learned comprehensive reading.  I wrote 
“delusion”, not “delirium”.


PointedEars
-- 
I heard that entropy isn't what it used to be.

(from: WolframAlpha)

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

#594656

On Thursday, November 3, 2022 at 3:17:38 PM UTC-7, Richard Hachel wrote:
> To obtain the average speed for a trajectiry from A to B in terms 
> of a given coordinate system x,t one can first calculate the coordinate 
> time taken by the object.  We have for a trajectory of constant proper 
> acceleration "a" beginning from rest at the origin:   t = sqrt(x^2 + 2x/a).
> We therefore have the coordinate time tA at xA, and coordinate time tB 
> at xB.  We have Δt = tB - tA  and we set Δx = xB - xA, and the average
> velocity in terms of these cordinates is v = Δx/Δt.

There is no error in the above.  For example, with xA=3.0 and xB=3.1, we have tA=3.834504 and tB=3.937453, so we have Δx = 0.1 and Δt = 0.102949, and therefore the average velocity is v = Δx/Δt = 0.971356. 

> Where is the error?

Your error is in the statement

> Or t = T * sqrt[1 + (1/4)(aT)^2]

where "T" represents the elapsed proper time along the unaccelerated path from the origin to the event on the trajectory at time t.  This has been explained to you many times before.  Again, if "T" was the proper time along the accelerating trajectory, it would imply 1=0, which is absurd.

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

#594661

Le 04/11/2022 à 00:57, Stan Fultoni a écrit :
> On Thursday, November 3, 2022 at 3:17:38 PM UTC-7, Richard Hachel wrote:
>> To obtain the average speed for a trajectiry from A to B in terms 
>> of a given coordinate system x,t one can first calculate the coordinate 
>> time taken by the object.  We have for a trajectory of constant proper 
>> acceleration "a" beginning from rest at the origin:   t = sqrt(x^2 + 2x/a).
>> We therefore have the coordinate time tA at xA, and coordinate time tB 
>> at xB.  We have Δt = tB - tA  and we set Δx = xB - xA, and the average
>> velocity in terms of these cordinates is v = Δx/Δt.
> 
> There is no error in the above.  For example, with xA=3.0 and xB=3.1, we have 
> tA=3.834504 and tB=3.937453, so we have Δx = 0.1 and Δt = 0.102949, and 
> therefore the average velocity is v = Δx/Δt = 0.971356. 

  Il y a une erreur ici.

  Les relativistes se trompent.

> 
>> Where is the error?
> 
> Your error is in the statement
> 
>> Or To = Tr * sqrt[1+ (1/4)Vr²/c²]

 I Repeat:

 To=Tr.sqrt(1+(1/4)Vr²/c²)

 or,

 To=(x/c)sqrt(1+2c²/ax)

  or, 

  To=Tr.sqrt(1+(1/4)a²Tr²/c²)

 or,

  To=sqrt(Tr²+Et²)

or,
   To=Tr.sqrt(1+xa/2c²)

 You can replace To by t, and Tr by tau if you want. 

> 
> where "T" represents the elapsed proper time along the unaccelerated path from 
> the origin to the event on the trajectory at time t.  This has been explained to 
> you many times before.  Again, if "T" was the proper time along the accelerating 
> trajectory, it would imply 1=0, which is absurd.

 The theory of relativity is not absurd.

 It is only very misunderstood by men.

R.H. 

 

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

#594667

On Thursday, November 3, 2022 at 5:14:15 PM UTC-7, Richard Hachel wrote:
>> Your error is in the statement
>>    t = T * sqrt[1+ (1/4)(aT)²]
>> where "T" represents the elapsed proper time along the unaccelerated path from
>> the origin to the event on the trajectory at time t. This has been explained to
>> you many times before. Again, if "T" was the proper time along the accelerating
>> trajectory, it would imply 1=0, which is absurd.
>
> The theory of relativity is not absurd.

You mis-read.  I did not say special relativity is absurd... it is not.  I said your beliefs (which imply 1=0) are absurd.

Again, to obtain the average speed for a trajectory t = sqrt(x^2 + 2x/a) from A to B in terms of a given coordinate system x,t have the coordinate time tA at xA, and coordinate time tB at xB, and we have Δt = tB - tA and Δx = xB - xA, and the average velocity in terms of these cordinates is v = Δx/Δt.

For example, with xA=3.0 and xB=3.1, we have tA=3.834504 and tB=3.937453, so we have Δx = 0.1 and Δt = 0.102949, and therefore we have the average velocity v = Δx/Δt = 0.971356.
>
>There is an error here.

What is the error?

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

#594670

Le 04/11/2022 à 01:45, Stan Fultoni a écrit :
> On Thursday, November 3, 2022 at 5:14:15 PM UTC-7, Richard Hachel wrote:
>>> Your error is in the statement
>>>    t = T * sqrt[1+ (1/4)(aT)²]
>>> where "T" represents the elapsed proper time along the unaccelerated path from
>>> the origin to the event on the trajectory at time t. This has been explained to
>>> you many times before. Again, if "T" was the proper time along the accelerating
>>> trajectory, it would imply 1=0, which is absurd.
>>
>> The theory of relativity is not absurd.
> 
> You mis-read.  I did not say special relativity is absurd... it is not.  I said 
> your beliefs (which imply 1=0) are absurd.
> 
> Again, to obtain the average speed for a trajectory t = sqrt(x^2 + 2x/a) from A 
> to B in terms of a given coordinate system x,t have the coordinate time tA at xA, 
> and coordinate time tB at xB, and we have Δt = tB - tA and Δx = xB - xA, and the 
> average velocity in terms of these cordinates is v = Δx/Δt.
> 
> For example, with xA=3.0 and xB=3.1, we have tA=3.834504 and tB=3.937453, so we 
> have Δx = 0.1 and Δt = 0.102949, and therefore we have the average velocity v = 
> Δx/Δt = 0.971356.
>>
>>There is an error here.
> 
> What is the error?

I told you before, but you make no effort.

However, if I tell you that you have to be very careful with the 
observable speeds, and that these are not the real speeds, you believe me.

If I tell you that proper times are not observable times, you believe me.

In short, when I tell you that you have to be very careful with the theory 
of relativity, and that it is very simple but stuffed with little 
pitfalls, in the end, you believe me.

When I tell you that two-speed clothing should not be added, you are 
telling me that you know it. And you believe me if I say that 
0.5c+0.5c=0.8c.

But if I tell you, beware, there are things you can't do in accelerated 
repositories either, you don't believe me anymore.

There is however a catastrophic error, which consists in believing that 
the observable times add or subtract without problem.

And that's what you all do.

You don't do this for observable velocities Vo, and that's fine.

But you do it for observable times.

Therein lies the catastrophic error, and the far too high speeds you 
predict.

Speeds moreover incompatible with the same joint Galilean speed, which is 
an absurdity.

If we take the right equation, all the problems disappear and we directly 
have v/c=[1+c²/2ax] at each point of the trajectory in x.

We must abandon the idea that To=To1+To2

It sounds amazing.

But this is only true for real times or proper times.
Tr=Tr1+Tr2
(tau=tau1+tau2)

But this is not true for observable times.

Therefore, Vo=Δx/ΔTo is true.

But To is wrong, it's not the right ΔTo.

R.H.

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

#594671

On Thursday, November 3, 2022 at 6:09:48 PM UTC-7, Richard Hachel wrote:
> > Again, to obtain the average speed for a trajectory t = sqrt(x^2 + 2x/a) from A 
> > to B in terms of a given coordinate system x,t have the coordinate time tA at xA, 
> > and coordinate time tB at xB, and we have Δt = tB - tA and Δx = xB - xA, and the 
> > average velocity in terms of these cordinates is v = Δx/Δt. 
> > 
> > For example, with xA=3.0 and xB=3.1, we have tA=3.834504 and tB=3.937453, so we 
> > have Δx = 0.1 and Δt = 0.102949, and therefore we have the average velocity v = 
> > Δx/Δt = 0.971356. 
> >> 
> >>There is an error here. 
> > 
> > What is the error?
>
> I told you before, but you make no effort. 

No, you did not tell me.  Please tell me what is the error?

> If I tell you the observable speeds are not the real speeds, you believe me. 

No, the speed in terms of the coordinates x,t is a real speed, in the sense that it is the distance traveled in the time traveled (in terms of x,t), which is the definition of velocity in terms of x,t.  This is what would be measured by a grid of standard rulers and clocks at rest and inertially synchronized in a given frame.

> If I tell you that proper times are not observable times, you believe me. 

No, proper times are perfectly observable... as are coordinate times.

> You believe me if I say that  0.5c+0.5c=0.8c. 

No, that is insane.  0.5c + 0.5c = 1c.  Special relativity did not overthrow arithmetic.  You have been misled.  Your brain is filled with misinformation and misunderstanding.  Special relativity does NOT say 0.5c + 0.5c = 0.8c, it says (0.5c + 0.5c)/[1 + (0.5c)(0.5c)/c^2] = 0.8c.  Do you understand this?

> We must abandon [rationality and santiy].

No, we must not.  You need to discard everything you think you know about special relativity, and start over from the beginning.

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

#594685

On 2022-11-03 22:17:36 +0000, Richard Hachel said:

> Where is the error?
  ...
> Third step, we assume that the observable instantaneous speed (or the 
> average observable speed in the segment AB) is Vo=Δx/ΔTo
  ...

One error is here: that formula is not an assumption but a definition.
But it is not the definition of instantameous speed, it is the definiton
of the average speed of in that segment. From continuity we can infer that
it is the instantaneous speed at some point in that interval.

However, the real error that has confused you is probably elsewhere.
More likely, you have made so many errors that finding one does not
help.

Mikko

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

#594687

On Friday, 4 November 2022 at 10:58:42 UTC+1, Mikko wrote:
> On 2022-11-03 22:17:36 +0000, Richard Hachel said: 
> 
> > Where is the error? 
> ...
> > Third step, we assume that the observable instantaneous speed (or the 
> > average observable speed in the segment AB) is Vo=Δx/ΔTo
> ... 
> 
> One error is here: that formula is not an assumption but a definition. 

And what do you think a definition is, poor halfbrain?

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

#594694

Le 04/11/2022 à 10:58, Mikko a écrit :
> One error is here: that formula is not an assumption but a definition.
> But it is not the definition of instantameous speed, it is the definiton
> of the average speed of in that segment. From continuity we can infer that
> it is the instantaneous speed at some point in that interval.

In your post two things: an interesting comment and a bullshit comment.
> 
> However, the real error that has confused you is probably elsewhere.
> More likely, you have made so many errors that finding one does not
> help.

The first comment is interesting, because it differentiates an average 
speed and an instantaneous speed.

You say it's the same thing.

It is very true and very correct.

But do you really think I don't?

Now, nothing prevents me from choosing a very small segment in the 
evolution of the traveler of Tau Ceti (12ly). For example, a small segment 
of one hundred meters, if I speak in time, or a small segment of proper 
time, or a small segment of improper time, taking for example a small 
amount of time like thirty seconds.

I will therefore have an average speed between the segment AB which is 
similar to the instantaneous speed of the center of AB for example, 
whether we are talking about real speed (Vr) or observable speed (Vo=v).

Here I can simplify and practice equivalence.


On the other hand, I no longer agree with you at all when you say that my 
assertions are teeming with errors.

It's gratuitous and insulting.

I'm not saying that I can't do, sometimes mistakes, or huge blunders. But 
at least me, I try to correct them, and I manage to write superb things.

Relativists in general, unfortunately, are very arrogant, and don't even 
understand that sometimes they are outdone by me.

This arrogance is of the order of the religious doctrine "We can't be 
wrong, and if we were wrong, it's not that asshole Doctor Hachel who will 
teach us".

This behavior that I have known for years is not scientific, it is just 
human.

A real scientist would say to me: "Sir, sit down, we will listen to you 
with respect and attention".

This is NEVER the case.

Everyone criticizes me, but no one has ever understood anything.

I don't understand Chinese, and it wouldn't occur to me to criticize a 
poem by King-Foo-Yang-Tse.

With me, that's what all my opponents do.

It is not scientific or even humanly logical.

R.H. 

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

#594698

On Friday, November 4, 2022 at 5:32:07 AM UTC-7, Richard Hachel wrote:
> > To obtain the average speed for a trajectory t = sqrt(x^2 + 2x/a) from A
> > to B in terms of a given coordinate system x,t have the coordinate time tA at xA,
> > and coordinate time tB at xB, and we have Δt = tB - tA and Δx = xB - xA, and the
> > average velocity in terms of these cordinates is v = Δx/Δt.
> >
> > For example, with xA=3.0 and xB=3.1, we have tA=3.834504 and tB=3.937453, so we
> > have Δx = 0.1 and Δt = 0.102949, and therefore we have the average velocity v =
> > Δx/Δt = 0.971356.
> 
> There is an error here.

What is the error?

> If I tell you the observable speeds are not the real speeds, you believe me.

No, the speed dx/dt in terms of the coordinates x,t is a real speed, the distance traveled in the time traveled (in terms of x,t), which is the definition of velocity in terms of x,t. This is what would be measured by a grid of standard rulers and clocks at rest and inertially synchronized in a given frame.

> If I tell you that proper times are not observable times, you believe me.

No, proper times are perfectly observable... as are coordinate times.  See above.

> You believe me if I say that 0.5c+0.5c=0.8c.

No, that's insane.  The fact is that 0.5c + 0.5c = 1c. Special relativity did not overthrow arithmetic. You have been misled. Your brain is filled with misinformation and misunderstanding. Special relativity does NOT say 0.5c + 0.5c = 0.8c, it says (0.5c + 0.5c)/[1 + (0.5c)(0.5c)/c^2] = 0.8c.  This is the composition (not the addition) of speeds in terms of two different systems of standard inertial coordinates.  Do you understand this?

> We must abandon...

No, we must not abandon sanity and rationality.  You are terribly misinformed, and laboring under severe misunderstandings. 
 You need to discard everything you think you know about special relativity, and start over from the beginning.  And you should stop running away from the truth.  Agreed?

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

#594705

On Friday, 4 November 2022 at 14:16:12 UTC+1, Stan Fultoni wrote:
> On Friday, November 4, 2022 at 5:32:07 AM UTC-7, Richard Hachel wrote: 
> > > To obtain the average speed for a trajectory t = sqrt(x^2 + 2x/a) from A
> > > to B in terms of a given coordinate system x,t have the coordinate time tA at xA, 
> > > and coordinate time tB at xB, and we have Δt = tB - tA and Δx = xB - xA, and the 
> > > average velocity in terms of these cordinates is v = Δx/Δt. 
> > > 
> > > For example, with xA=3.0 and xB=3.1, we have tA=3.834504 and tB=3.937453, so we 
> > > have Δx = 0.1 and Δt = 0.102949, and therefore we have the average velocity v = 
> > > Δx/Δt = 0.971356. 
> > 
> > There is an error here. 
> 
> What is the error?
> > If I tell you the observable speeds are not the real speeds, you believe me.
> No, the speed dx/dt in terms of the coordinates x,t is a real speed, the distance traveled in the time traveled (in terms of x,t), which is the definition of velocity in terms of x,t. This is what would be measured by a grid of standard rulers and clocks at rest and inertially synchronized in a given frame.

A grid of rulers; you guys are really funny with your 
real differently delusions.

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

#594712

On 2022-11-04 12:32:05 +0000, Richard Hachel said:

> Relativists in general, unfortunately, are very arrogant, and don't 
> even understand that sometimes they are outdone by me.
> 
> This arrogance is of the order of the religious doctrine "We can't be 
> wrong, and if we were wrong, it's not that asshole Doctor Hachel who 
> will teach us".

Don't generalize too much, but it is true and obvious that you are so
arrogant. It wouldn't matter if you were right, but sometinmes you aren't.

In this group everyone is presumed ignorant and stupid until proven
otherwise.

Mikko

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

#594713

Le 04/11/2022 à 17:14, Mikko a écrit :
> On 2022-11-04 12:32:05 +0000, Richard Hachel said:
> 
>> Relativists in general, unfortunately, are very arrogant, and don't 
>> even understand that sometimes they are outdone by me.
>> 
>> This arrogance is of the order of the religious doctrine "We can't be 
>> wrong, and if we were wrong, it's not that asshole Doctor Hachel who 
>> will teach us".
> 
> Don't generalize too much, but it is true and obvious that you are so
> arrogant. It wouldn't matter if you were right, but sometinmes you aren't.
> 
> In this group everyone is presumed ignorant and stupid until proven
> otherwise.
> 
> Mikko

Not just in this group. I think the phenomenon is general, and I've always 
had a hard time understanding why.

Many people questioned by this question that I asked them gave me the same 
answer: "Richard, you do not consider human nature enough".

R.H. 

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

Page 1 of 2  [1] 2  Next page →

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

csiph-web