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Help with acceleration algebra and relativity

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  Help with acceleration algebra and relativity "sepp623@yahoo.com" <sepp623@yahoo.com> - 2022-06-19 08:12 -0700
    Re: Help with acceleration algebra and relativity Mikko <mikko.levanto@iki.fi> - 2022-06-19 18:27 +0300
    Idiot David Swppala back at trolling "Dono." <eggy20011951@gmail.com> - 2022-06-19 08:33 -0700
    Re: Help with acceleration algebra and relativity Al Coe <coeal5136@gmail.com> - 2022-06-19 09:04 -0700
      Re: Help with acceleration algebra and relativity "sepp623@yahoo.com" <sepp623@yahoo.com> - 2022-06-19 11:35 -0700
        Re: Help with acceleration algebra and relativity Al Coe <coeal5136@gmail.com> - 2022-06-19 11:41 -0700
          Re: Help with acceleration algebra and relativity "sepp623@yahoo.com" <sepp623@yahoo.com> - 2022-06-19 14:16 -0700
            Re: Help with acceleration algebra and relativity Al Coe <coeal5136@gmail.com> - 2022-06-19 14:44 -0700
              Re: Help with acceleration algebra and relativity "sepp623@yahoo.com" <sepp623@yahoo.com> - 2022-06-19 15:03 -0700
                Re: Help with acceleration algebra and relativity Al Coe <coeal5136@gmail.com> - 2022-06-19 15:22 -0700
              Re: Help with acceleration algebra and relativity "sepp623@yahoo.com" <sepp623@yahoo.com> - 2022-06-19 15:28 -0700
                Re: Help with acceleration algebra and relativity Al Coe <coeal5136@gmail.com> - 2022-06-19 15:32 -0700
                  Re: Help with acceleration algebra and relativity "sepp623@yahoo.com" <sepp623@yahoo.com> - 2022-06-19 16:07 -0700
                    Re: Help with acceleration algebra and relativity Al Coe <coeal5136@gmail.com> - 2022-06-19 16:23 -0700
                      Re: Help with acceleration algebra and relativity "sepp623@yahoo.com" <sepp623@yahoo.com> - 2022-06-20 05:03 -0700
                    Re: Help with acceleration algebra and relativity Al Coe <coeal5136@gmail.com> - 2022-06-19 19:38 -0700
    Re: Help with acceleration algebra and relativity The Starmaker <starmaker@ix.netcom.com> - 2022-06-19 11:50 -0700
    Re: Help with acceleration algebra and relativity Sylvia Else <sylvia@email.invalid> - 2022-06-20 13:55 +1000
      Re: Help with acceleration algebra and relativity Maciej Wozniak <maluwozniak@gmail.com> - 2022-06-19 23:03 -0700
      Re: Help with acceleration algebra and relativity Richard Hachel <r.hachel@tiscali.fr> - 2022-06-20 09:26 +0000
      Re: Help with acceleration algebra and relativity Richard Hachel <r.hachel@tiscali.fr> - 2022-06-20 09:41 +0000
    Re: Help with acceleration algebra and relativity Richard Hachel <r.hachel@tiscali.fr> - 2022-06-20 09:28 +0000
    Re: Help with acceleration algebra and relativity "Paul B. Andersen" <paul.b.andersen@paulba.no> - 2022-06-21 20:48 +0200
      Re: Help with acceleration algebra and relativity The Starmaker <starmaker@ix.netcom.com> - 2022-06-21 14:42 -0700

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#587195 — Help with acceleration algebra and relativity

I get conflicting results when I use simple acceleration algebra and then apply concepts of relativity. I would appreciate it if someone pointed out my error.

In an inertial reference frame, F0, there are two rockets that can each accelerate at a constant rate as measured by observers in F0 of 3 meters/second**2.  If these rockets are moving along the x-axis in the negative x direction with a velocity of 9/10 the speed of light as measured in F0, and they accelerate in the positive x direction at a rate of 3 meters/second**2 if, using the speed of light as 3*10**8meters/second, these rockets can accelerate for 1.8*10**8 seconds without exceeding the speed of light as measured by observers in F0.
Now since the acceleration rate is a constant 3 meters/second**2 as measured by observers in F0, the distance traveled during an acceleration is:
     d = 1/2*a*t**2
If one rocket starts accelerating at time t and accelerates from -0.9c to 0.9c and the other rocket starts accelerating one second later as measured by observers in F0, then when the first rocket reaches a speed of 0.9c, that rocket traveled a distance of 1/2 a*t**2 while the other rocket traveled a distance of 1/2 a*(t-1)**2.  The difference in the distance traveled is:
   1/2 a * ((t**2) - (t-1)**2) = 1/2 a * (-2t+1)
Plugging in the values gives the difference in distance traveled by the two rockets as 3 * 1.8*10**8 - 1.5 meters when the rocket that started accelerating first reaches V = 0.9c as measured in F0.
    If that algebra is correct then if I apply Einstein's concept of simultaneous events to the scenario, I get conflicting results.  Here's the scenario:
There is a second inertial reference frame F1, moving with magnitude of velocity |V|=c*sqrt(3)/2 relative to F0.  In F1, observers measure that the two rockets are a distance L apart. Observers in F1 start the accelerations of the two rockets simultaneously with a distance L so that the observers in F0 measure that the two accelerations started one second apart. Using the Lorentz transform:
     t2-t1 = 1 second = gamma*(c*sqrt(3)/2*L/c**2) where gamma = 2.
Therefore L = 3*10**8/sqrt(3) meters = 1.732*10*8 meters.  If this is the case before the first rocket reaches a speed of 0.9c as measured by observers in F0, the first rocket crashes into the second rocket that started its acceleration one second later.  However, since the two rockets started their identical accelerations simultaneously as measured in F1, they always maintain the same separation so they never crash.  Please point out which numbers I computed are wrong.
Thanks,
David Seppala
Bastrop TX

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#587196

On 2022-06-19 15:12:02 +0000, sepp623@yahoo.com said:

> I get conflicting results when I use simple acceleration algebra and 
> then apply concepts of relativity. I would appreciate it if someone 
> pointed out my error.
> 
> In an inertial reference frame, F0, there are two rockets that can each 
> accelerate at a constant rate as measured by observers in F0 of 3 
> meters/second**2.  If these rockets are moving along the x-axis in the 
> negative x direction with a velocity of 9/10 the speed of light as 
> measured in F0, and they accelerate in the positive x direction at a 
> rate of 3 meters/second**2 if, using the speed of light as 
> 3*10**8meters/second, these rockets can accelerate for 1.8*10**8 
> seconds without exceeding the speed of light as measured by observers 
> in F0.
> Now since the acceleration rate is a constant 3 meters/second**2 as 
> measured by observers in F0, the distance traveled during an 
> acceleration is:
>      d = 1/2*a*t**2
> If one rocket starts accelerating at time t and accelerates from -0.9c 
> to 0.9c and the other rocket starts accelerating one second later as 
> measured by observers in F0, then when the first rocket reaches a speed 
> of 0.9c, that rocket traveled a distance of 1/2 a*t**2 while the other 
> rocket traveled a distance of 1/2 a*(t-1)**2.  The difference in the 
> distance traveled is:
>    1/2 a * ((t**2) - (t-1)**2) = 1/2 a * (-2t+1)
> Plugging in the values gives the difference in distance traveled by the 
> two rockets as 3 * 1.8*10**8 - 1.5 meters when the rocket that started 
> accelerating first reaches V = 0.9c as measured in F0.
>     If that algebra is correct then if I apply Einstein's concept of 
> simultaneous events to the scenario, I get conflicting results.  Here's 
> the scenario:
> There is a second inertial reference frame F1, moving with magnitude of 
> velocity |V|=c*sqrt(3)/2 relative to F0.  In F1, observers measure that 
> the two rockets are a distance L apart. Observers in F1 start the 
> accelerations of the two rockets simultaneously with a distance L so 
> that the observers in F0 measure that the two accelerations started one 
> second apart. Using the Lorentz transform:
>      t2-t1 = 1 second = gamma*(c*sqrt(3)/2*L/c**2) where gamma = 2.
> Therefore L = 3*10**8/sqrt(3) meters = 1.732*10*8 meters.  If this is 
> the case before the first rocket reaches a speed of 0.9c as measured by 
> observers in F0, the first rocket crashes into the second rocket that 
> started its acceleration one second later.  However, since the two 
> rockets started their identical accelerations simultaneously as 
> measured in F1, they always maintain the same separation so they never 
> crash.  Please point out which numbers I computed are wrong.
> Thanks,
> David Seppala
> Bastrop TX

Problems like this are easier if you use year as the time unit and
light year as the distance unit. The specified acceleration is close
to 0.3 light years / year², but 1 light year / year might work, too.

Mikko

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#587197 — Idiot David Swppala back at trolling

On Sunday, June 19, 2022 at 8:12:05 AM UTC-7, sep...@yahoo.com wrote:

> Now since the acceleration rate is a constant 3 meters/second**2 as measured by observers in F0, the distance traveled during an acceleration is: 
> d = 1/2*a*t**2 

No, it is not. You need to learn "hyperbolic motion". Go away and study. 

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#587198

On Sunday, June 19, 2022 at 8:12:05 AM UTC-7, sep...@yahoo.com wrote:
> Now since the acceleration rate is a constant 3 meters/second**2 as measured by 
> observers in F0, the distance traveled during an acceleration is:  d = 1/2*a*t**2 

No, you have specified that in terms of inertial coordinate system S0 the velocity at time t=0 is v(0)=-0.9c, so with constant coordinate acceleration "a" the distance at arbitrary time t of the object from its position at time t=0 is v(0)t + (1/2)at^2.  Also, don't confuse distance traveled with distance from starting point.

8th grade physics:  1 ....  Barnpole Dave:  0

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#587211

On Sunday, June 19, 2022 at 11:04:28 AM UTC-5, Al Coe wrote:
> On Sunday, June 19, 2022 at 8:12:05 AM UTC-7, sep...@yahoo.com wrote:
> > Now since the acceleration rate is a constant 3 meters/second**2 as measured by 
> > observers in F0, the distance traveled during an acceleration is: d = 1/2*a*t**2
> No, you have specified that in terms of inertial coordinate system S0 the velocity at time t=0 is v(0)=-0.9c, so with constant coordinate acceleration "a" the distance at arbitrary time t of the object from its position at time t=0 is v(0)t + (1/2)at^2. Also, don't confuse distance traveled with distance from starting point. 
> 
> 8th grade physics: 1 .... Barnpole Dave: 0
I didn't say that there is constant coordinate acceleration "a".  I said observers in inertial reference frame F0, measure that the acceleration of each ship is a constant 3 meters/second**2 as measured in the F0 inertial reference frame coordinate system.
David Seppala
Bastrop TX

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#587212

On Sunday, June 19, 2022 at 11:35:59 AM UTC-7, sep...@yahoo.com wrote:
> I didn't say that there is constant coordinate acceleration "a". I said observers in inertial reference frame F0, measure that the acceleration of each ship is a constant 3 meters/second**2 as measured in the F0 inertial reference frame coordinate system. 

That's what "constant coordinate acceleration" means, i.e., in terms of the coordinate system S0, you specified that each ship has the constant acceleration "a", meaning that d^2x/dt^2 = a = 3 msec^2.  Now do you understand what you said?

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#587239

On Sunday, June 19, 2022 at 1:41:55 PM UTC-5, Al Coe wrote:
> On Sunday, June 19, 2022 at 11:35:59 AM UTC-7, sep...@yahoo.com wrote: 
> > I didn't say that there is constant coordinate acceleration "a". I said observers in inertial reference frame F0, measure that the acceleration of each ship is a constant 3 meters/second**2 as measured in the F0 inertial reference frame coordinate system.
> That's what "constant coordinate acceleration" means, i.e., in terms of the coordinate system S0, you specified that each ship has the constant acceleration "a", meaning that d^2x/dt^2 = a = 3 msec^2. Now do you understand what you said?

Okay, so where are the errors in the calculation I did showing the time it takes for the spaceship to accelerate from -0.9c to 0.9c as measured by observers in F0 and the distance spanned during that acceleration?
David Seppala
Bastrop TX

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#587241

On Sunday, June 19, 2022 at 2:16:24 PM UTC-7, sep...@yahoo.com wrote:
> Where are the errors in the calculation I did...?

On Sunday, June 19, 2022 at 8:12:05 AM UTC-7, sep...@yahoo.com wrote:
> Now since the acceleration rate is a constant 3 meters/second**2 as measured by
> observers in F0, the distance traveled during an acceleration is: d = 1/2*a*t**2

No, you've specified that, in terms of inertial coordinate system S0, the velocity at time t=0 is v(0)=-0.9c, so with constant coordinate acceleration "a" in terms of S0 the distance at arbitrary time t of the object from its position at time t=0 is not (1/2)at^2, it is v(0)t + (1/2)at^2.   Do you understand this?

[I omit the second confusion identified previously, because your brain seems to have enough trouble grasping one bit of information, let alone two.]

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#587243

On Sunday, June 19, 2022 at 4:44:32 PM UTC-5, Al Coe wrote:
> On Sunday, June 19, 2022 at 2:16:24 PM UTC-7, sep...@yahoo.com wrote: 
> > Where are the errors in the calculation I did...?
> On Sunday, June 19, 2022 at 8:12:05 AM UTC-7, sep...@yahoo.com wrote: 
> > Now since the acceleration rate is a constant 3 meters/second**2 as measured by 
> > observers in F0, the distance traveled during an acceleration is: d = 1/2*a*t**2
> No, you've specified that, in terms of inertial coordinate system S0, the velocity at time t=0 is v(0)=-0.9c, so with constant coordinate acceleration "a" in terms of S0 the distance at arbitrary time t of the object from its position at time t=0 is not (1/2)at^2, it is v(0)t + (1/2)at^2. Do you understand this? 
> 
> [I omit the second confusion identified previously, because your brain seems to have enough trouble grasping one bit of information, let alone two.]
When you refer to "a" in terms of S0, what does S0 refer to?  I use F0 as the inertial coordinate system where the measurements are made.
David Seppala
Bastrop TX

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#587246

On Sunday, June 19, 2022 at 3:03:43 PM UTC-7, sep...@yahoo.com wrote:
> When you refer to "a" in terms of S0, what does S0 refer to? 

S0 denotes the standard inertial coordinate system at rest in you call the frame F0.  The word "frame" tends to carry connotation that are misleading to newbies, so it's best to be clear that we are specifying a system of coordinates, which is characterized not just by a trajectory but also by a temporal foliation and scale factors.  To uniquely specify velocities and accelerations, etc., it is not sufficient to specify a frame (elementary definition), you must specify a coordinate system.  I've noticed that you tend to forget this, so to help remind you of this, it's best to denote the coordinate system with S0, and if we let x,t denote those coordinates, the acceleration in terms of S0 is the second derivative a = d^2 x/dt^2.

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#587247

On Sunday, June 19, 2022 at 4:44:32 PM UTC-5, Al Coe wrote:
> On Sunday, June 19, 2022 at 2:16:24 PM UTC-7, sep...@yahoo.com wrote: 
> > Where are the errors in the calculation I did...?
> On Sunday, June 19, 2022 at 8:12:05 AM UTC-7, sep...@yahoo.com wrote: 
> > Now since the acceleration rate is a constant 3 meters/second**2 as measured by 
> > observers in F0, the distance traveled during an acceleration is: d = 1/2*a*t**2
> No, you've specified that, in terms of inertial coordinate system S0, the velocity at time t=0 is v(0)=-0.9c, so with constant coordinate acceleration "a" in terms of S0 the distance at arbitrary time t of the object from its position at time t=0 is not (1/2)at^2, it is v(0)t + (1/2)at^2. Do you understand this? 
> 
> [I omit the second confusion identified previously, because your brain seems to have enough trouble grasping one bit of information, let alone two.]
Yes, I understand that equation.  So show me what is wrong with my calculation.
Thanks,
David Seppala
Bastrop TX

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#587248

On Sunday, June 19, 2022 at 3:28:31 PM UTC-7, sep...@yahoo.com wrote:
> > > Now since the acceleration rate is a constant 3 meters/second**2 as measured by 
> > > observers in F0, the distance traveled during an acceleration is: d = 1/2*a*t**2 
> >
> > No, you've specified that, in terms of inertial coordinate system S0, the velocity at time t=0 is v(0)=-0.9c, so with constant coordinate acceleration "a" in terms of S0 the distance at arbitrary time t of the object from its position at time t=0 is not (1/2)at^2, it is v(0)t + (1/2)at^2. Do you understand this? 
> > 
> Yes, I understand that equation. So show me what is wrong with my calculation.

See above.  Do you understand your mistake?

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#587251

On Sunday, June 19, 2022 at 5:32:33 PM UTC-5, Al Coe wrote:
> On Sunday, June 19, 2022 at 3:28:31 PM UTC-7, sep...@yahoo.com wrote: 
> > > > Now since the acceleration rate is a constant 3 meters/second**2 as measured by 
> > > > observers in F0, the distance traveled during an acceleration is: d = 1/2*a*t**2 
> > > 
> > > No, you've specified that, in terms of inertial coordinate system S0, the velocity at time t=0 is v(0)=-0.9c, so with constant coordinate acceleration "a" in terms of S0 the distance at arbitrary time t of the object from its position at time t=0 is not (1/2)at^2, it is v(0)t + (1/2)at^2. Do you understand this? 
> > >
> > Yes, I understand that equation. So show me what is wrong with my calculation.
> See . Do you understand your mistake?
No, show me.  I was looking at the difference in distances the two ships travel, so since they always have the same initial velocity (V= -0.9c) that canceled out in the calculation.  So compute, the differences in distances if one ship accelerates from -0.9c to 0.9c in t seconds with an acceleration rate of 3 meters/second**2 and the other ship accelerates from -0.9c to (-0.9c +  (t-1)*3 meters/second**2).
David Seppala
Bastrop TX

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#587253

On Sunday, June 19, 2022 at 4:07:28 PM UTC-7, sep...@yahoo.com wrote:
> > > > > Now since the acceleration rate is a constant 3 meters/second**2 as measured by 
> > > > > observers in F0, the distance traveled during an acceleration is: d = 1/2*a*t**2 
> > > > 
> > > > No, you've specified that, in terms of inertial coordinate system S0, the velocity at time t=0 is v(0)=-0.9c, so with constant coordinate acceleration "a" in terms of S0 the distance at arbitrary time t of the object from its position at time t=0 is not (1/2)at^2, it is v(0)t + (1/2)at^2. Do you understand this? 
>
> No...

You don't?  In the previous message you said you understood (were you lying?), but then without correcting anything in your original post you repeated your request to point out an error in your post.  Naturally I pointed out the same error, because you haven't corrected it.  Now you say you don't understand it after all.

Look, just go ahead and correct the error in your original post, and if you still get the wrong answer *after correcting that mistake*, go ahead and post your revised calculation and I'll be happy to point out the next error.

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#587278

On Sunday, June 19, 2022 at 6:23:56 PM UTC-5, Al Coe wrote:
> On Sunday, June 19, 2022 at 4:07:28 PM UTC-7, sep...@yahoo.com wrote: 
> > > > > > Now since the acceleration rate is a constant 3 meters/second**2 as measured by 
> > > > > > observers in F0, the distance traveled during an acceleration is: d = 1/2*a*t**2 
> > > > > 
> > > > > No, you've specified that, in terms of inertial coordinate system S0, the velocity at time t=0 is v(0)=-0.9c, so with constant coordinate acceleration "a" in terms of S0 the distance at arbitrary time t of the object from its position at time t=0 is not (1/2)at^2, it is v(0)t + (1/2)at^2. Do you understand this? 
> >
> > No... 
> 
> You don't? In the previous message you said you understood (were you lying?), but then without correcting anything in your original post you repeated your request to point out an error in your post. Naturally I pointed out the same error, because you haven't corrected it. Now you say you don't understand it after all. 
> 
> Look, just go ahead and correct the error in your original post, and if you still get the wrong answer *after correcting that mistake*, go ahead and post your revised calculation and I'll be happy to point out the next error.
Al,
    Thanks!!  I got it. I was using equations from a different problem where the initial velocities were zero. Didn't realize that the one ship in this problem moved V*1sec away from the coordinates showing the separation of the acceleration starting points were versus the separation of the ships at that time.  Thanks again.
David Seppala
Bastrop TX

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#587256

On Sunday, June 19, 2022 at 4:07:28 PM UTC-7, sep...@yahoo.com wrote:
> If this is the case before the first rocket reaches a speed of 0.9c as measured by observers in F0, the first rocket crashes into the second rocket that started its acceleration one second later. However, since the two rockets started their identical accelerations simultaneously as measured in F1, they always maintain the same separation so they never crash. Please point out which numbers I computed are wrong.

Since Barnpole Dave has run away again, let's just close this out succinctly.  With c=1, in terms of S0 let V denote the initial speed of the two rockets, v the speed of S1, and let the rockets be at (0,0) and (X,T) when their accelerations begin. Then in terms of S0 the crash allegedly “occurs” at time (1 – vV + (a/2)Xv^2)/(av), and at this event the velocity of the two rockets would be 1/v + (a/2)vX and 1/v – (a/2)vX, independent of V. Thus at least one of the rockets' velocities exceeds 1 before the alleged crash would occur.

This has been explained to Barnpole Dave before, as have all of his fallacious canards, and yet he never absorbs the explanations, and keeps cycling back through the same confusions, over and over.  His strategy of always running away as soon as the explanation looms into view seems to help him preserve his state of confusion.  Very strange.

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#587214

sepp623@yahoo.com wrote:
> 
> I get conflicting results when I use simple acceleration algebra and then apply concepts of relativity. I would appreciate it if someone pointed out my error.


Your first mistake is...Acceleration is not often considered in special
relativity.


For example...if i'm driving my car eventually it will run out of gas. 

That is why you get conflicting results. You ran out of gas.









-- 
The Starmaker -- To question the unquestionable, ask the unaskable,
 to think the unthinkable, mention the unmentionable, say the unsayable,
and challenge
 the unchallengeable.

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#587258

On 20-June-22 1:12 am, sepp623@yahoo.com wrote:
> I get conflicting results when I use simple acceleration algebra and then apply concepts of relativity. I would appreciate it if someone pointed out my error.
> 
Liar.

This is just another of your futile attempts to find a contradiction in 
special relativity.

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#587266

On Monday, 20 June 2022 at 05:55:25 UTC+2, Sylvia Else wrote:
> On 20-June-22 1:12 am, sep...@yahoo.com wrote: 
> > I get conflicting results when I use simple acceleration algebra and then apply concepts of relativity. I would appreciate it if someone pointed out my error. 
> >
> Liar. 
> 
> This is just another of your futile attempts to find a contradiction in 
> special relativity.

A contradiction in this insane mumble is obvious,
considering the definition of second valid in 1905.

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#587273

Le 20/06/2022 à 05:55, Sylvia Else a écrit :
> On 20-June-22 1:12 am, sepp623@yahoo.com wrote:
>> I get conflicting results when I use simple acceleration algebra and then apply 
>> concepts of relativity. I would appreciate it if someone pointed out my error.
>> 
> Liar.
> 
> This is just another of your futile attempts to find a contradiction in 
> special relativity.

There are contradictions and paradoxes in the theory of relativity.

It's not that this theory is wrong, it's that being extremely poorly 
explained, it invariably leads to all kinds of blunders.

I've been saying it for decades now. The blunders, I denounced them and 
explained them.

The problem is then purely human and of a religious order: "We do not want 
this man to reign over us".

I beg you to believe me: it is as simple as that.

In good intelligence (ie not hateful), it would take a few seconds of 
understanding and three days of global reflection for us to say: Hachel is 
right. The theory is very misunderstood by ALL.

But the problem, I repeat, I repeat and repeat again, it is human and 
religious.

We prefer a badly put together dogma, rather than lowering our pants in 
front of a man.

This is one of my greatest surprises in life.

However, it would only take THREE minutes to show that Dr. Hachel is 
right.

It would be enough to show that in apparent speeds a "Langevin" quickly 
becomes absurd if one does not take the correct equations of Hachel.

Three minutes.

But we never do.

Why?

Because it's scary.

Why is it scary?

Because the human being is a narcissistic patient, and the narcissist is 
systematically the other who does not think like us, even if he thinks in 
a much more logical and coherent way.

It is the history of the world.

R.H. 

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