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

Groups > sci.physics.relativity > #389598 > unrolled thread

My mistake or Einstein's

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

Contents

  My mistake or Einstein's sepp623@yahoo.com - 2016-08-09 11:04 -0700
    Re: My mistake or Einstein's The Starmaker <starmaker@ix.netcom.com> - 2016-08-09 12:26 -0700
    Re: My mistake or Einstein's pcardinale@volcanomail.com - 2016-08-09 12:57 -0700
    Re: My mistake or Einstein's pcardinale@volcanomail.com - 2016-08-09 13:04 -0700
    Re: My mistake or Einstein's "Paul B. Andersen" <relativity@paulba.no> - 2016-08-09 22:39 +0200
      Re: My mistake or Einstein's sepp623@yahoo.com - 2016-08-09 13:56 -0700
        Re: My mistake or Einstein's "Paul B. Andersen" <relativity@paulba.no> - 2016-08-10 10:51 +0200
    Re: My mistake or Einstein's JanPB <filmart@gmail.com> - 2016-08-09 15:15 -0700
      Re: My mistake or Einstein's sepp623@yahoo.com - 2016-08-09 15:38 -0700
        Re: My mistake or Einstein's JanPB <filmart@gmail.com> - 2016-08-09 15:53 -0700
          Re: My mistake or Einstein's sepp623@yahoo.com - 2016-08-09 16:21 -0700
            Re: My mistake or Einstein's JanPB <filmart@gmail.com> - 2016-08-09 23:05 -0700
              Re: My mistake or Einstein's Maciej Woźniak <mlwozniak@wp.pl> - 2016-08-10 11:56 +0200
            Re: My mistake or Einstein's bartekltg <bartekltg@gmail.com> - 2016-08-10 14:17 +0200
      Re: My mistake or Einstein's The Starmaker <starmaker@ix.netcom.com> - 2016-08-10 01:39 -0700
    Re: My mistake or Einstein's The Starmaker <starmaker@ix.netcom.com> - 2016-08-09 17:26 -0700
      Re: My mistake or Einstein's sepp623@yahoo.com - 2016-08-09 17:38 -0700
        Re: My mistake or Einstein's The Starmaker <starmaker@ix.netcom.com> - 2016-08-09 17:53 -0700
          Re: My mistake or Einstein's sepp623@yahoo.com - 2016-08-09 17:59 -0700
            Re: My mistake or Einstein's The Starmaker <starmaker@ix.netcom.com> - 2016-08-09 18:26 -0700
              Re: My mistake or Einstein's sepp623@yahoo.com - 2016-08-09 19:05 -0700
                Re: My mistake or Einstein's moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-08-10 02:59 +0000
                Re: My mistake or Einstein's The Starmaker <starmaker@ix.netcom.com> - 2016-08-09 20:06 -0700
    Re: My mistake or Einstein's Sylvia Else <sylvia@not.at.this.address> - 2016-08-10 13:16 +1000
      Re: My mistake or Einstein's sepp623@yahoo.com - 2016-08-09 21:11 -0700
        Re: My mistake or Einstein's Sylvia Else <sylvia@not.at.this.address> - 2016-08-10 14:15 +1000
          Re: My mistake or Einstein's sepp623@yahoo.com - 2016-08-09 21:31 -0700
            Re: My mistake or Einstein's Sylvia Else <sylvia@not.at.this.address> - 2016-08-10 14:48 +1000
              Re: My mistake or Einstein's sepp623@yahoo.com - 2016-08-09 22:56 -0700
                Re: My mistake or Einstein's The Starmaker <starmaker@ix.netcom.com> - 2016-08-10 01:36 -0700
    Re: My mistake or Einstein's Sylvia Else <sylvia@not.at.this.address> - 2016-08-10 20:35 +1000
    Re: My mistake or Einstein's bartekltg <bartekltg@gmail.com> - 2016-08-10 14:04 +0200
      Re: My mistake or Einstein's Maciej Woźniak <mlwozniak@wp.pl> - 2016-08-10 14:29 +0200
        Re: My mistake or Einstein's bartekltg <bartekltg@gmail.com> - 2016-08-10 15:29 +0200
          Re: My mistake or Einstein's sepp623@yahoo.com - 2016-08-10 06:36 -0700
            Re: My mistake or Einstein's Sylvia Else <sylvia@not.at.this.address> - 2016-08-11 11:16 +1000
          Re: My mistake or Einstein's Maciej Woźniak <mlwozniak@wp.pl> - 2016-08-10 15:36 +0200
      Re: My mistake or Einstein's bartekltg <bartekltg@gmail.com> - 2016-08-10 15:49 +0200
        Re: My mistake or Einstein's Maciej Woźniak <mlwozniak@wp.pl> - 2016-08-10 16:14 +0200

Page 1 of 2  [1] 2  Next page →

#389598 — My mistake or Einstein's

This scenario of this simple problem ends up with contradictory results. Please identify the mistake.

In this problem, I use c = 3 * 10**8 meters/second as the speed of light.

Consider an inertial reference frame, F0, that has an object A moving along the x-axis at -2.8 * 10**8 meters/second. If this object starts accelerating in the positive x direction at a constant rate of 28 meters / second**2 as measured in F0, how long does it take for this object to reach a speed of 2.8 * 10**8 meters/second as measured in F0? When I do the calculation, I find that it takes 2 * 10**7 seconds. This based on the simple formula v = a * t

Now, when this object accelerates from -2.8 * 10**8 meters/second to 2.8 * 10**8 meters/second at the constant rate of 28 meters / second**2 as measured in F0, how far does this object travel along the x-axis during this time interval as measured in frame F0? Since the acceleration rate is constant, I used the formula d = 0.5 * (a * t**2) to determine the distance, along with the initial velocity before the acceleration starts of -2.8 * 10**8 meters / second. This resulted in:

d = ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters


I run into a problem when I use Einstein's simultaneous events concept in conjunction with these numbers.

Let there be a second inertial reference frame, F1, that is moving with a relative velocity with magnitude of sqrt(3)/2 * c with respect to F0. And let there be a second object in space, object B, that initially has the same velocity as object A. When object B accelerates, its pattern of acceleration is identical to object A's pattern of acceleration. Let observers in frame F1 measure the distance between object A and object B to be sqrt(3) light-seconds. At time t0, observers in frame F1, simultaneously, start both object A and object B accelerating in the positive x direction. Observers in frame F1, measure that the distance between object A and object B remains constant at sqrt(3) light-seconds. The distance between object A and object B remains constant as measured in frame F1 because both objects had the same initial velocity (the relative velocity of object A to object B was zero before the acceleration started), they accelerate in the identical manner (just at different locations in space) and both objects started accelerating simultaneously (as measured in frame F1).

Now in frame F0, prior to the start of any acceleration, let object B have a greater x coordinate than object A at any point in time, as they both move with velocity -2.8 * 10**8 meters/second along the x-axis of F0. And let the direction of the acceleration of both objects be in the positive x direction when the acceleration of each object starts. Per Einstein, frame F0 measures that one of the objects starts accelerating 3 seconds before the other object starts accelerating. Let the direction of relative velocity between frame F0 and F1 be such that object A starts accelerating 3 seconds before object B starts accelerating, as measured in frame F0.

Since object A started accelerating 3 seconds before object B, object A gets closer and closer to object B as function of time. During the acceleration as the velocity of object A goes from -2.8 * 10**8 meters/second as measured in F0 to 2.8 * 10**8 meters/second, how close does object A get to object B as measured in frame F0? Previously I computed that during that acceleration object A moves a distance of 
((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters

During that same time interval, with object B starting its acceleration 3 seconds later, object B moves a distance of 
((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * ((2 * 10**7) - 3) * ((2 * 10**7) - 3) meters

The difference between A's change of position and B's change of position during that time interval is:
difference in position = (0.5 * 28) * (12 * 10**7 - 9) meters
or approximately 16.8 * 10**8 meters

So object A moves 16.8 * 10**8 meters closer to object B during this time interval. But using the transform equations, since F1 observers measured the separation between object A and object B to be sqrt(3) light-seconds, observers in frame F0 measure the separation between object A and object B before the acceleration starts to be:
2 * sqrt(3) * 3 * 10**8 = 10.39 * 10**8 meters

So object A crashes into object B during this acceleration. However frame F1 measures that object A and object B always have a distance between them of
sqrt(3) * 3 * 10**8 meters = 5.2 * 10**8 meters

So frame F1 observers say the two objects never crash.The initial velocity of object A equals the initial velocity of object B, the acceleration of both objects started simultaneously as measured by observers in F1, the acceleration pattern of object A is identical to the acceleration pattern of object B, and their initial separation was 5.2 * 10**8 meters and always remains constant.

So, where is the error?

Thanks
David Seppala
Bastrop TX

[toc] | [next] | [standalone]

#389603

sepp623@yahoo.com wrote:
> 
> This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> 
> In this problem, I use c = 3 * 10**8 meters/second as the speed of light.
> 
> Consider an inertial reference frame, F0, that has an object A moving along the x-axis at -2.8 * 10**8 meters/second. If this object starts accelerating in the positive x direction at a constant rate of 28 meters / second**2 as measured in F0, how long does it take for this object to reach a speed of 2.8 * 10**8 meters/second as measured in F0? When I do the calculation, I find that it takes 2 * 10**7 seconds. This based on the simple formula v = a * t
> 
> Now, when this object accelerates from -2.8 * 10**8 meters/second to 2.8 * 10**8 meters/second at the constant rate of 28 meters / second**2 as measured in F0, how far does this object travel along the x-axis during this time interval as measured in frame F0? Since the acceleration rate is constant, I used the formula d = 0.5 * (a * t**2) to determine the distance, along with the initial velocity before the acceleration starts of -2.8 * 10**8 meters / second. This resulted in:
> 
> d = ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> 
> I run into a problem when I use Einstein's simultaneous events concept in conjunction with these numbers.
> 
> 
> 
> Now in frame F0, prior to the start of any acceleration, let object B have a greater x coordinate than object A at any point in time, as they both move with velocity -2.8 * 10**8 meters/second along the x-axis of F0. And let the direction of the acceleration of both objects be in the positive x direction when the acceleration of each object starts. Per Einstein, frame F0 measures that one of the objects starts accelerating 3 seconds before the other object starts accelerating. Let the direction
> 
> Since object A started accelerating 3 seconds before object B, object A gets closer and closer to object B as function of time. During the acceleration as the velocity of object A goes from -2.8 * 10**8 meters/second as measured in F0 to 2.8 * 10**8 meters/second, how close does object A get to object B as measured in frame F0? Previously I computed that during that acceleration object A moves a distance of
> ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> 
> During that same time interval, with object B starting its acceleration 3 seconds later, object B moves a distance of
> ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * ((2 * 10**7) - 3) * ((2 * 10**7) - 3) meters
> 
> The difference between A's change of position and B's change of position during that time interval is:
> difference in position = (0.5 * 28) * (12 * 10**7 - 9) meters
> or approximately 16.8 * 10**8 meters
> 
> So object A moves 16.8 * 10**8 meters closer to object B during this time interval. But using the transform equations, since F1 observers measured the separation between object A and object B to be sqrt(3) light-seconds, observers in frame F0 measure the separation between object A and object B before the acceleration starts to be:
> 2 * sqrt(3) * 3 * 10**8 = 10.39 * 10**8 meters
> 
> So object A crashes into object B during this acceleration. However frame F1 measures that object A and object B always have a distance between them of
> sqrt(3) * 3 * 10**8 meters = 5.2 * 10**8 meters
> 
> So frame F1 observers say the two objects never crash.The initial velocity of object A equals the initial velocity of object B, the acceleration of both objects started simultaneously as measured by observers in F1, the acceleration pattern of object A is identical to the acceleration pattern of object B, and their initial separation was 5.2 * 10**8 meters and always remains constant.
> 
> So, where is the error?
> 
> Thanks
> David Seppala
> Bastrop TX



The error is the speed of light.
When light acts like a wave...it slows down.
When light acts like a particle...it speeds up.

If you take 20 people and
you tell them to hold hands and run...it kind 
of slows them down...but if one person runs
without holding hands...he moves faster than the wave.

Now, what is the speed of light if it changes from wave to particle between here and Pluto?? 
Do you account for the change in behavior of the light?

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

#389608

On Tuesday, August 9, 2016 at 11:04:14 AM UTC-7, sep...@yahoo.com wrote:
> This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
>

Why?  You've demonstrated dozens, if not hundreds of times that you can't learn from your mistakes.

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

#389610

To answer your question, the mistake is that you can't do math.
Someone who could do math would apply the L.T. and see that there's no contradiction.

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

#389612

On 09.08.2016 20:04, sepp623@yahoo.com wrote:
> []

Your mistake!

As always you have succeeded to confuse yourself by
making an unnecessary complicated scenario.
But your aim is always to confuse the reader, isn't it?
I won't bother to read it in detail.

But in this case your problem seems to be that you don't
know the difference between coordinate acceleration and
proper acceleration and how acceleration transform.

You might learn something from these:
https://paulba.no/pdf/TwinsByMetric.pdf
https://paulba.no/pdf/ClocksInMotion.pdf

-- 
Paul

https://paulba.no/

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

#389615

On Tuesday, August 9, 2016 at 3:39:50 PM UTC-5, Paul B. Andersen wrote:
> On 09.08.2016 20:04, sepp623@yahoo.com wrote:
> > []
> 
> Your mistake!
> 
> As always you have succeeded to confuse yourself by
> making an unnecessary complicated scenario.
> But your aim is always to confuse the reader, isn't it?
> I won't bother to read it in detail.
> 
> But in this case your problem seems to be that you don't
> know the difference between coordinate acceleration and
> proper acceleration and how acceleration transform.
> 
> You might learn something from these:
> https://paulba.no/pdf/TwinsByMetric.pdf
> https://paulba.no/pdf/ClocksInMotion.pdf
> 
> -- 
> Paul
> 
> https://paulba.no/

Paul,
   If you read the problem, you will see that in frame F1, two identical objects, having the same initial velocity, separated by a distance of sqrt(3) light-seconds as measured in F1, start their identical accelerations simultaneously.  Therefore, the distance between these two objects remains constant.
   Using the Lorentz tranform where frames F0 and F1 have a relative velocity of magnitude sqrt(3)/2*c, and where F1 measures the separation between object A and object B to be sqrt(3) light-second, this gives a 3 second delay as to when F0 measures these accelerations. The acceleration of object A and object B are both identical, but delayed by 3 seconds as measured by observers in frame F0.
   Using the initial velocities setup in the problem, this time delay in the start of the accelerations as measured by observers in frame F0 results in object A crashing into object B during the time interval stated in the problem, as measured by observers in frame F0.  As measured by the observers in frame F1, object A and object B never crash into each other.

David Seppala
Bastrop TX

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

#389680

On 09.08.2016 22:56, sepp623@yahoo.com wrote:
> On Tuesday, August 9, 2016 at 3:39:50 PM UTC-5, Paul B. Andersen wrote:
>> On 09.08.2016 20:04, sepp623@yahoo.com wrote:
>>> []
>>
>> Your mistake!
>>
>> As always you have succeeded to confuse yourself by
>> making an unnecessary complicated scenario.
>> But your aim is always to confuse the reader, isn't it?
>> I won't bother to read it in detail.
>>
>> But in this case your problem seems to be that you don't
>> know the difference between coordinate acceleration and
>> proper acceleration and how acceleration transform.
>>
>> You might learn something from these:
>> https://paulba.no/pdf/TwinsByMetric.pdf
>> https://paulba.no/pdf/ClocksInMotion.pdf
>>
>> --
>> Paul
>>
>> https://paulba.no/
>
> Paul,
>    If you read the problem, you will see that in frame F1,
>    two identical objects, having the same initial velocity,
>    separated by a distance of sqrt(3) light-seconds as measured in F1,
>    start their identical accelerations simultaneously.
>    Therefore, the distance between these two objects remains constant.
>    Using the Lorentz tranform where frames F0 and F1 have a relative
>    velocity of magnitude sqrt(3)/2*c, and where F1 measures the separation
>    between object A and object B to be sqrt(3) light-second, this gives a 3
>    second delay as to when F0 measures these accelerations.

There are only two objects which have identical proper acceleration.

Then this is given in chapter 2.1. here:
https://paulba.no/pdf/ClocksInMotion.pdf
Bells paradox.

>    The acceleration of object A and object B are both identical,
>    but delayed by 3 seconds as measured by observers in frame F0.


I told you above:
   "But in this case your problem seems to be that you don't
    know the difference between coordinate acceleration and
    proper acceleration and how acceleration transform."

1. If the coordinate accelerations of the objects are constant,
    simultaneous and identical in frame F1, then the proper
    accelerations of the objects are identical, but varying with time.
2. In frame F0 the coordinate accelerations of the objects will
    be _very_ different from their coordinate accelerations in frame F1
    and they will not be equal and not simultaneous.

Your gross error is that you think the coordinate accelerations
are the same in both frames.

>    Using the initial velocities setup in the problem,
>    this time delay in the start of the accelerations
>   as measured by observers in frame F0 results in object A
>   crashing into object B during the time interval stated in
>   the problem, as measured by observers in frame F0.
>   As measured by the observers in frame F1, object A and
>   object B never crash into each other.
>
> David Seppala
> Bastrop TX
>


-- 
Paul

https://paulba.no/

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

#389623

On Tuesday, August 9, 2016 at 11:04:14 AM UTC-7, sep...@yahoo.com wrote:
> This scenario of this simple problem ends up with contradictory results. Please identify the mistake.

I have a suggestion. You've been posting those very similar-looking questions
(which show no progress on your part) for about 20 years now (IIRC). Since you
are obviously running in circles, I strongly recommend hiring a teacher. You'll
need to straighten out the basics:

* linear algebra (and I don't mean just "matrix theory" by that),
* decent vector calculus in N dimensions (N = 3 is not enough),
* the relevant physics that went on before special relativity, esp. classical
electrodynamics.

What you are doing now was fine in 1996 but it's become completely stale
by now, you are wasting your time.

--
Jan

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

#389626

On Tuesday, August 9, 2016 at 5:15:20 PM UTC-5, JanPB wrote:
> On Tuesday, August 9, 2016 at 11:04:14 AM UTC-7, sep...@yahoo.com wrote:
> > This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> 
> I have a suggestion. You've been posting those very similar-looking questions
> (which show no progress on your part) for about 20 years now (IIRC). Since you
> are obviously running in circles, I strongly recommend hiring a teacher. You'll
> need to straighten out the basics:
> 
> * linear algebra (and I don't mean just "matrix theory" by that),
> * decent vector calculus in N dimensions (N = 3 is not enough),
> * the relevant physics that went on before special relativity, esp. classical
> electrodynamics.
> 
> What you are doing now was fine in 1996 but it's become completely stale
> by now, you are wasting your time.
> 
> --
> Jan
Hi Jan,
  I think that the physics professors at Princeton, UT, Dartmouth and other places that I exchange email with are knowledgeable enough.
  Many posters here think non-physics replies are not a waste of time, but unlike your great suggestion I don't find them helpful.
David Seppala
Bastrop TX

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

#389630

On Tuesday, August 9, 2016 at 3:39:01 PM UTC-7, sep...@yahoo.com wrote:
> On Tuesday, August 9, 2016 at 5:15:20 PM UTC-5, JanPB wrote:
> > On Tuesday, August 9, 2016 at 11:04:14 AM UTC-7, sep...@yahoo.com wrote:
> > > This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> > 
> > I have a suggestion. You've been posting those very similar-looking questions
> > (which show no progress on your part) for about 20 years now (IIRC). Since you
> > are obviously running in circles, I strongly recommend hiring a teacher. You'll
> > need to straighten out the basics:
> > 
> > * linear algebra (and I don't mean just "matrix theory" by that),
> > * decent vector calculus in N dimensions (N = 3 is not enough),
> > * the relevant physics that went on before special relativity, esp. classical
> > electrodynamics.
> > 
> > What you are doing now was fine in 1996 but it's become completely stale
> > by now, you are wasting your time.
> > 
> > --
> > Jan
> Hi Jan,
>   I think that the physics professors at Princeton, UT, Dartmouth and other places that I exchange email with are knowledgeable enough.

No, something is wrong. The problem is you keep asking the same questions
over and over for a couple of decades, after they have been explained
to you (people used to answer you in great detail). After all this time
you are still stuck not understanding the basics and forever coming up
with yet-another purported construction to square the circle.

It really won't work, it's a total waste of time at this point. You need
someone to help you reboot.

--
Jan

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

#389633

On Tuesday, August 9, 2016 at 5:53:09 PM UTC-5, JanPB wrote:
> On Tuesday, August 9, 2016 at 3:39:01 PM UTC-7, sep...@yahoo.com wrote:
> > On Tuesday, August 9, 2016 at 5:15:20 PM UTC-5, JanPB wrote:
> > > On Tuesday, August 9, 2016 at 11:04:14 AM UTC-7, sep...@yahoo.com wrote:
> > > > This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> > > 
> > > I have a suggestion. You've been posting those very similar-looking questions
> > > (which show no progress on your part) for about 20 years now (IIRC). Since you
> > > are obviously running in circles, I strongly recommend hiring a teacher. You'll
> > > need to straighten out the basics:
> > > 
> > > * linear algebra (and I don't mean just "matrix theory" by that),
> > > * decent vector calculus in N dimensions (N = 3 is not enough),
> > > * the relevant physics that went on before special relativity, esp. classical
> > > electrodynamics.
> > > 
> > > What you are doing now was fine in 1996 but it's become completely stale
> > > by now, you are wasting your time.
> > > 
> > > --
> > > Jan
> > Hi Jan,
> >   I think that the physics professors at Princeton, UT, Dartmouth and other places that I exchange email with are knowledgeable enough.
> 
> No, something is wrong. The problem is you keep asking the same questions
> over and over for a couple of decades, after they have been explained
> to you (people used to answer you in great detail). After all this time
> you are still stuck not understanding the basics and forever coming up
> with yet-another purported construction to square the circle.
> 
> It really won't work, it's a total waste of time at this point. You need
> someone to help you reboot.
> 
> --
> Jan
Of course, one of you scholars could simply post: "look at this line XXX in your posting.  That is your mistake."
 If any of you are that proficient or could see the error, that is what I would learn from.
  Reboot means to start over. Show me your erudition! Point out the error in this posting if you can, so I can reboot.
David Seppala
Bastrop TX

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

#389670

On Tuesday, August 9, 2016 at 4:21:37 PM UTC-7, sep...@yahoo.com wrote:
> On Tuesday, August 9, 2016 at 5:53:09 PM UTC-5, JanPB wrote:
> > On Tuesday, August 9, 2016 at 3:39:01 PM UTC-7, sep...@yahoo.com wrote:
> > > On Tuesday, August 9, 2016 at 5:15:20 PM UTC-5, JanPB wrote:
> > > > On Tuesday, August 9, 2016 at 11:04:14 AM UTC-7, sep...@yahoo.com wrote:
> > > > > This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> > > > 
> > > > I have a suggestion. You've been posting those very similar-looking questions
> > > > (which show no progress on your part) for about 20 years now (IIRC). Since you
> > > > are obviously running in circles, I strongly recommend hiring a teacher. You'll
> > > > need to straighten out the basics:
> > > > 
> > > > * linear algebra (and I don't mean just "matrix theory" by that),
> > > > * decent vector calculus in N dimensions (N = 3 is not enough),
> > > > * the relevant physics that went on before special relativity, esp. classical
> > > > electrodynamics.
> > > > 
> > > > What you are doing now was fine in 1996 but it's become completely stale
> > > > by now, you are wasting your time.
> > > > 
> > > > --
> > > > Jan
> > > Hi Jan,
> > >   I think that the physics professors at Princeton, UT, Dartmouth and other places that I exchange email with are knowledgeable enough.
> > 
> > No, something is wrong. The problem is you keep asking the same questions
> > over and over for a couple of decades, after they have been explained
> > to you (people used to answer you in great detail). After all this time
> > you are still stuck not understanding the basics and forever coming up
> > with yet-another purported construction to square the circle.
> > 
> > It really won't work, it's a total waste of time at this point. You need
> > someone to help you reboot.
> > 
> > --
> > Jan
> Of course, one of you scholars could simply post: "look at this line XXX in your posting.  That is your mistake."

As I said, this had been done many times. At this point you'd have to pay me to spend
any time on this.

>  If any of you are that proficient or could see the error, that is what I would learn from.

Yeah, we've heard it all before...

--
Jan

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

#389686

Użytkownik "JanPB"  napisał w wiadomości grup 
dyskusyjnych:5d98d22c-bb72-499c-b6b8-beef490e902b@googlegroups.com...
> Of course, one of you scholars could simply post: "look at this line XXX 
> in your posting.  That is your mistake."

|As I said, this had been done many times. At this point you'd have to pay 
me to spend
|any time on this.


You have to understand, David, You're talking to GURU. He is like Bach...
His mumble is wise beyond your imagination...
Don't You understand it? What an  idiot You are. 

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

#389692

On 10.08.2016 01:21, sepp623@yahoo.com wrote:

> Of course, one of you scholars could simply post: "look at this line XXX in your posting.  That is your mistake."

And scholars and different eggheads have time to go through
you _messy_ example?

I have found your error (it is a silent assumption that B
have constant coordinate acceleration in F0, equal to
A's acceleration; it contradict the assumption of the
identical trajectories in F1) but almost gave up two times;-)

bartekltg

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

#389679

JanPB wrote:
> 
> On Tuesday, August 9, 2016 at 11:04:14 AM UTC-7, sep...@yahoo.com wrote:
> > This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> 
> I have a suggestion. You've been posting those very similar-looking questions
> (which show no progress on your part) for about 20 years now (IIRC). Since you
> are obviously running in circles, I strongly recommend hiring a teacher. You'll
> need to straighten out the basics:
> 
> * linear algebra (and I don't mean just "matrix theory" by that),
> * decent vector calculus in N dimensions (N = 3 is not enough),
> * the relevant physics that went on before special relativity, esp. classical
> electrodynamics.
> 
> What you are doing now was fine in 1996 but it's become completely stale
> by now, you are wasting your time.
> 
> --
> Jan


He's not wasting his time...it's his dream, or nightmare.



"matrix theory"??? wat does that mean?

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

#389638

sepp623@yahoo.com wrote:
> 
> This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> 
> In this problem, I use c = 3 * 10**8 meters/second as the speed of light.
> 
> Consider an inertial reference frame, F0, that has an object A moving along the x-axis at -2.8 * 10**8 meters/second. If this object starts accelerating in the positive x direction at a constant rate of 28 meters / second**2 as measured in F0, how long does it take for this object to reach a speed of 2.8 * 10**8 meters/second as measured in F0? When I do the calculation, I find that it takes 2 * 10**7 seconds. This based on the simple formula v = a * t
> 
> Now, when this object accelerates from -2.8 * 10**8 meters/second to 2.8 * 10**8 meters/second at the constant rate of 28 meters / second**2 as measured in F0, how far does this object travel along the x-axis during this time interval as measured in frame F0? Since the acceleration rate is constant, I used the formula d = 0.5 * (a * t**2) to determine the distance, along with the initial velocity before the acceleration starts of -2.8 * 10**8 meters / second. This resulted in:
> 
> d = ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> 
> I run into a problem when I use Einstein's simultaneous events concept in conjunction with these numbers.
> 
> 
> 
> Now in frame F0, prior to the start of any acceleration, let object B have a greater x coordinate than object A at any point in time, as they both move with velocity -2.8 * 10**8 meters/second along the x-axis of F0. And let the direction of the acceleration of both objects be in the positive x direction when the acceleration of each object starts. Per Einstein, frame F0 measures that one of the objects starts accelerating 3 seconds before the other object starts accelerating. Let the direction
> 
> Since object A started accelerating 3 seconds before object B, object A gets closer and closer to object B as function of time. During the acceleration as the velocity of object A goes from -2.8 * 10**8 meters/second as measured in F0 to 2.8 * 10**8 meters/second, how close does object A get to object B as measured in frame F0? Previously I computed that during that acceleration object A moves a distance of
> ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> 
> During that same time interval, with object B starting its acceleration 3 seconds later, object B moves a distance of
> ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * ((2 * 10**7) - 3) * ((2 * 10**7) - 3) meters
> 
> The difference between A's change of position and B's change of position during that time interval is:
> difference in position = (0.5 * 28) * (12 * 10**7 - 9) meters
> or approximately 16.8 * 10**8 meters
> 
> So object A moves 16.8 * 10**8 meters closer to object B during this time interval. But using the transform equations, since F1 observers measured the separation between object A and object B to be sqrt(3) light-seconds, observers in frame F0 measure the separation between object A and object B before the acceleration starts to be:
> 2 * sqrt(3) * 3 * 10**8 = 10.39 * 10**8 meters
> 
> So object A crashes into object B during this acceleration. However frame F1 measures that object A and object B always have a distance between them of
> sqrt(3) * 3 * 10**8 meters = 5.2 * 10**8 meters
> 
> So frame F1 observers say the two objects never crash.The initial velocity of object A equals the initial velocity of object B, the acceleration of both objects started simultaneously as measured by observers in F1, the acceleration pattern of object A is identical to the acceleration pattern of object B, and their initial separation was 5.2 * 10**8 meters and always remains constant.
> 
> So, where is the error?
> 
> Thanks
> David Seppala
> Bastrop TX



Well, I can show you "where is the error" in your math...


((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters


It's missing  X in  ((-2.8 * 10**8)

should be:


((-2.8 * x10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters

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

#389639

On Tuesday, August 9, 2016 at 7:25:40 PM UTC-5, The Starmaker wrote:
> sepp623@yahoo.com wrote:
> > 
> > This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> > 
> > In this problem, I use c = 3 * 10**8 meters/second as the speed of light.
> > 
> > Consider an inertial reference frame, F0, that has an object A moving along the x-axis at -2.8 * 10**8 meters/second. If this object starts accelerating in the positive x direction at a constant rate of 28 meters / second**2 as measured in F0, how long does it take for this object to reach a speed of 2.8 * 10**8 meters/second as measured in F0? When I do the calculation, I find that it takes 2 * 10**7 seconds. This based on the simple formula v = a * t
> > 
> > Now, when this object accelerates from -2.8 * 10**8 meters/second to 2.8 * 10**8 meters/second at the constant rate of 28 meters / second**2 as measured in F0, how far does this object travel along the x-axis during this time interval as measured in frame F0? Since the acceleration rate is constant, I used the formula d = 0.5 * (a * t**2) to determine the distance, along with the initial velocity before the acceleration starts of -2.8 * 10**8 meters / second. This resulted in:
> > 
> > d = ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> > 
> > I run into a problem when I use Einstein's simultaneous events concept in conjunction with these numbers.
> > 
> > 
> > 
> > Now in frame F0, prior to the start of any acceleration, let object B have a greater x coordinate than object A at any point in time, as they both move with velocity -2.8 * 10**8 meters/second along the x-axis of F0. And let the direction of the acceleration of both objects be in the positive x direction when the acceleration of each object starts. Per Einstein, frame F0 measures that one of the objects starts accelerating 3 seconds before the other object starts accelerating. Let the direction
> > 
> > Since object A started accelerating 3 seconds before object B, object A gets closer and closer to object B as function of time. During the acceleration as the velocity of object A goes from -2.8 * 10**8 meters/second as measured in F0 to 2.8 * 10**8 meters/second, how close does object A get to object B as measured in frame F0? Previously I computed that during that acceleration object A moves a distance of
> > ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> > 
> > During that same time interval, with object B starting its acceleration 3 seconds later, object B moves a distance of
> > ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * ((2 * 10**7) - 3) * ((2 * 10**7) - 3) meters
> > 
> > The difference between A's change of position and B's change of position during that time interval is:
> > difference in position = (0.5 * 28) * (12 * 10**7 - 9) meters
> > or approximately 16.8 * 10**8 meters
> > 
> > So object A moves 16.8 * 10**8 meters closer to object B during this time interval. But using the transform equations, since F1 observers measured the separation between object A and object B to be sqrt(3) light-seconds, observers in frame F0 measure the separation between object A and object B before the acceleration starts to be:
> > 2 * sqrt(3) * 3 * 10**8 = 10.39 * 10**8 meters
> > 
> > So object A crashes into object B during this acceleration. However frame F1 measures that object A and object B always have a distance between them of
> > sqrt(3) * 3 * 10**8 meters = 5.2 * 10**8 meters
> > 
> > So frame F1 observers say the two objects never crash.The initial velocity of object A equals the initial velocity of object B, the acceleration of both objects started simultaneously as measured by observers in F1, the acceleration pattern of object A is identical to the acceleration pattern of object B, and their initial separation was 5.2 * 10**8 meters and always remains constant.
> > 
> > So, where is the error?
> > 
> > Thanks
> > David Seppala
> > Bastrop TX
> 
> 
> 
> Well, I can show you "where is the error" in your math...
> 
> 
> ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> 
> 
> It's missing  X in  ((-2.8 * 10**8)
> 
> should be:
> 
> 
> ((-2.8 * x10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters

In my posting I use the symbol * to mean "times" and
I use the symbol ** to mean "to the power of"

I'm not certain but I think you are using the symbol "X" to mean "times"
If so, then there is no error in the calculation you are pointing out.
If not, what does the X represent?

Thanks,
David Seppala
Bastrop TX

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

#389640

sepp623@yahoo.com wrote:
> 
> On Tuesday, August 9, 2016 at 7:25:40 PM UTC-5, The Starmaker wrote:
> > sepp623@yahoo.com wrote:
> > >
> > > This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> > >
> > > In this problem, I use c = 3 * 10**8 meters/second as the speed of light.
> > >
> > > Consider an inertial reference frame, F0, that has an object A moving along the x-axis at -2.8 * 10**8 meters/second. If this object starts accelerating in the positive x direction at a constant rate of 28 meters / second**2 as measured in F0, how long does it take for this object to reach a speed of 2.8 * 10**8 meters/second as measured in F0? When I do the calculation, I find that it takes 2 * 10**7 seconds. This based on the simple formula v = a * t
> > >
> > > Now, when this object accelerates from -2.8 * 10**8 meters/second to 2.8 * 10**8 meters/second at the constant rate of 28 meters / second**2 as measured in F0, how far does this object travel along the x-axis during this time interval as measured in frame F0? Since the acceleration rate is constant, I used the formula d = 0.5 * (a * t**2) to determine the distance, along with the initial velocity before the acceleration starts of -2.8 * 10**8 meters / second. This resulted in:
> > >
> > > d = ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> > >
> > > I run into a problem when I use Einstein's simultaneous events concept in conjunction with these numbers.
> > >
> > >
> > >
> > > Now in frame F0, prior to the start of any acceleration, let object B have a greater x coordinate than object A at any point in time, as they both move with velocity -2.8 * 10**8 meters/second along the x-axis of F0. And let the direction of the acceleration of both objects be in the positive x direction when the acceleration of each object starts. Per Einstein, frame F0 measures that one of the objects starts accelerating 3 seconds before the other object starts accelerating. Let the direct
> > >
> > > Since object A started accelerating 3 seconds before object B, object A gets closer and closer to object B as function of time. During the acceleration as the velocity of object A goes from -2.8 * 10**8 meters/second as measured in F0 to 2.8 * 10**8 meters/second, how close does object A get to object B as measured in frame F0? Previously I computed that during that acceleration object A moves a distance of
> > > ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> > >
> > > During that same time interval, with object B starting its acceleration 3 seconds later, object B moves a distance of
> > > ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * ((2 * 10**7) - 3) * ((2 * 10**7) - 3) meters
> > >
> > > The difference between A's change of position and B's change of position during that time interval is:
> > > difference in position = (0.5 * 28) * (12 * 10**7 - 9) meters
> > > or approximately 16.8 * 10**8 meters
> > >
> > > So object A moves 16.8 * 10**8 meters closer to object B during this time interval. But using the transform equations, since F1 observers measured the separation between object A and object B to be sqrt(3) light-seconds, observers in frame F0 measure the separation between object A and object B before the acceleration starts to be:
> > > 2 * sqrt(3) * 3 * 10**8 = 10.39 * 10**8 meters
> > >
> > > So object A crashes into object B during this acceleration. However frame F1 measures that object A and object B always have a distance between them of
> > > sqrt(3) * 3 * 10**8 meters = 5.2 * 10**8 meters
> > >
> > > So frame F1 observers say the two objects never crash.The initial velocity of object A equals the initial velocity of object B, the acceleration of both objects started simultaneously as measured by observers in F1, the acceleration pattern of object A is identical to the acceleration pattern of object B, and their initial separation was 5.2 * 10**8 meters and always remains constant.
> > >
> > > So, where is the error?
> > >
> > > Thanks
> > > David Seppala
> > > Bastrop TX
> >
> >
> >
> > Well, I can show you "where is the error" in your math...
> >
> >
> > ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> >
> >
> > It's missing  X in  ((-2.8 * 10**8)
> >
> > should be:
> >
> >
> > ((-2.8 * x10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> 
> In my posting I use the symbol * to mean "times" and
> I use the symbol ** to mean "to the power of"
> 
> I'm not certain but I think you are using the symbol "X" to mean "times"
> If so, then there is no error in the calculation you are pointing out.
> If not, what does the X represent?
> 
> Thanks,
> David Seppala
> Bastrop TX




I only put in one x, the others where there is the number 10 does not require the x, just the first 10



((-2.8 * x10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters

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

#389642

On Tuesday, August 9, 2016 at 7:52:25 PM UTC-5, The Starmaker wrote:
> sepp623@yahoo.com wrote:
> > 
> > On Tuesday, August 9, 2016 at 7:25:40 PM UTC-5, The Starmaker wrote:
> > > sepp623@yahoo.com wrote:
> > > >
> > > > This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> > > >
> > > > In this problem, I use c = 3 * 10**8 meters/second as the speed of light.
> > > >
> > > > Consider an inertial reference frame, F0, that has an object A moving along the x-axis at -2.8 * 10**8 meters/second. If this object starts accelerating in the positive x direction at a constant rate of 28 meters / second**2 as measured in F0, how long does it take for this object to reach a speed of 2.8 * 10**8 meters/second as measured in F0? When I do the calculation, I find that it takes 2 * 10**7 seconds. This based on the simple formula v = a * t
> > > >
> > > > Now, when this object accelerates from -2.8 * 10**8 meters/second to 2.8 * 10**8 meters/second at the constant rate of 28 meters / second**2 as measured in F0, how far does this object travel along the x-axis during this time interval as measured in frame F0? Since the acceleration rate is constant, I used the formula d = 0.5 * (a * t**2) to determine the distance, along with the initial velocity before the acceleration starts of -2.8 * 10**8 meters / second. This resulted in:
> > > >
> > > > d = ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> > > >
> > > > I run into a problem when I use Einstein's simultaneous events concept in conjunction with these numbers.
> > > >
> > > >
> > > >
> > > > Now in frame F0, prior to the start of any acceleration, let object B have a greater x coordinate than object A at any point in time, as they both move with velocity -2.8 * 10**8 meters/second along the x-axis of F0. And let the direction of the acceleration of both objects be in the positive x direction when the acceleration of each object starts. Per Einstein, frame F0 measures that one of the objects starts accelerating 3 seconds before the other object starts accelerating. Let the direct
> > > >
> > > > Since object A started accelerating 3 seconds before object B, object A gets closer and closer to object B as function of time. During the acceleration as the velocity of object A goes from -2.8 * 10**8 meters/second as measured in F0 to 2.8 * 10**8 meters/second, how close does object A get to object B as measured in frame F0? Previously I computed that during that acceleration object A moves a distance of
> > > > ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> > > >
> > > > During that same time interval, with object B starting its acceleration 3 seconds later, object B moves a distance of
> > > > ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * ((2 * 10**7) - 3) * ((2 * 10**7) - 3) meters
> > > >
> > > > The difference between A's change of position and B's change of position during that time interval is:
> > > > difference in position = (0.5 * 28) * (12 * 10**7 - 9) meters
> > > > or approximately 16.8 * 10**8 meters
> > > >
> > > > So object A moves 16.8 * 10**8 meters closer to object B during this time interval. But using the transform equations, since F1 observers measured the separation between object A and object B to be sqrt(3) light-seconds, observers in frame F0 measure the separation between object A and object B before the acceleration starts to be:
> > > > 2 * sqrt(3) * 3 * 10**8 = 10.39 * 10**8 meters
> > > >
> > > > So object A crashes into object B during this acceleration. However frame F1 measures that object A and object B always have a distance between them of
> > > > sqrt(3) * 3 * 10**8 meters = 5.2 * 10**8 meters
> > > >
> > > > So frame F1 observers say the two objects never crash.The initial velocity of object A equals the initial velocity of object B, the acceleration of both objects started simultaneously as measured by observers in F1, the acceleration pattern of object A is identical to the acceleration pattern of object B, and their initial separation was 5.2 * 10**8 meters and always remains constant.
> > > >
> > > > So, where is the error?
> > > >
> > > > Thanks
> > > > David Seppala
> > > > Bastrop TX
> > >
> > >
> > >
> > > Well, I can show you "where is the error" in your math...
> > >
> > >
> > > ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> > >
> > >
> > > It's missing  X in  ((-2.8 * 10**8)
> > >
> > > should be:
> > >
> > >
> > > ((-2.8 * x10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> > 
> > In my posting I use the symbol * to mean "times" and
> > I use the symbol ** to mean "to the power of"
> > 
> > I'm not certain but I think you are using the symbol "X" to mean "times"
> > If so, then there is no error in the calculation you are pointing out.
> > If not, what does the X represent?
> > 
> > Thanks,
> > David Seppala
> > Bastrop TX
> 
> 
> 
> 
> I only put in one x, the others where there is the number 10 does not require the x, just the first 10
> 
> 
> 
> ((-2.8 * x10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters

Never heard of such a rule.
Which line has the physics mistake?

David Seppala
Bastrop TX

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

#389643

sepp623@yahoo.com wrote:
> 
> On Tuesday, August 9, 2016 at 7:52:25 PM UTC-5, The Starmaker wrote:
> > sepp623@yahoo.com wrote:
> > >
> > > On Tuesday, August 9, 2016 at 7:25:40 PM UTC-5, The Starmaker wrote:
> > > > sepp623@yahoo.com wrote:
> > > > >
> > > > > This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> > > > >
> > > > > In this problem, I use c = 3 * 10**8 meters/second as the speed of light.
> > > > >
> > > > > Consider an inertial reference frame, F0, that has an object A moving along the x-axis at -2.8 * 10**8 meters/second. If this object starts accelerating in the positive x direction at a constant rate of 28 meters / second**2 as measured in F0, how long does it take for this object to reach a speed of 2.8 * 10**8 meters/second as measured in F0? When I do the calculation, I find that it takes 2 * 10**7 seconds. This based on the simple formula v = a * t
> > > > >
> > > > > Now, when this object accelerates from -2.8 * 10**8 meters/second to 2.8 * 10**8 meters/second at the constant rate of 28 meters / second**2 as measured in F0, how far does this object travel along the x-axis during this time interval as measured in frame F0? Since the acceleration rate is constant, I used the formula d = 0.5 * (a * t**2) to determine the distance, along with the initial velocity before the acceleration starts of -2.8 * 10**8 meters / second. This resulted in:
> > > > >
> > > > > d = ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> > > > >
> > > > > I run into a problem when I use Einstein's simultaneous events concept in conjunction with these numbers.
> > > > >
> > > > >
> > > > >
> > > > > Now in frame F0, prior to the start of any acceleration, let object B have a greater x coordinate than object A at any point in time, as they both move with velocity -2.8 * 10**8 meters/second along the x-axis of F0. And let the direction of the acceleration of both objects be in the positive x direction when the acceleration of each object starts. Per Einstein, frame F0 measures that one of the objects starts accelerating 3 seconds before the other object starts accelerating. Let the di
> > > > >
> > > > > Since object A started accelerating 3 seconds before object B, object A gets closer and closer to object B as function of time. During the acceleration as the velocity of object A goes from -2.8 * 10**8 meters/second as measured in F0 to 2.8 * 10**8 meters/second, how close does object A get to object B as measured in frame F0? Previously I computed that during that acceleration object A moves a distance of
> > > > > ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> > > > >
> > > > > During that same time interval, with object B starting its acceleration 3 seconds later, object B moves a distance of
> > > > > ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * ((2 * 10**7) - 3) * ((2 * 10**7) - 3) meters
> > > > >
> > > > > The difference between A's change of position and B's change of position during that time interval is:
> > > > > difference in position = (0.5 * 28) * (12 * 10**7 - 9) meters
> > > > > or approximately 16.8 * 10**8 meters
> > > > >
> > > > > So object A moves 16.8 * 10**8 meters closer to object B during this time interval. But using the transform equations, since F1 observers measured the separation between object A and object B to be sqrt(3) light-seconds, observers in frame F0 measure the separation between object A and object B before the acceleration starts to be:
> > > > > 2 * sqrt(3) * 3 * 10**8 = 10.39 * 10**8 meters
> > > > >
> > > > > So object A crashes into object B during this acceleration. However frame F1 measures that object A and object B always have a distance between them of
> > > > > sqrt(3) * 3 * 10**8 meters = 5.2 * 10**8 meters
> > > > >
> > > > > So frame F1 observers say the two objects never crash.The initial velocity of object A equals the initial velocity of object B, the acceleration of both objects started simultaneously as measured by observers in F1, the acceleration pattern of object A is identical to the acceleration pattern of object B, and their initial separation was 5.2 * 10**8 meters and always remains constant.
> > > > >
> > > > > So, where is the error?
> > > > >
> > > > > Thanks
> > > > > David Seppala
> > > > > Bastrop TX
> > > >
> > > >
> > > >
> > > > Well, I can show you "where is the error" in your math...
> > > >
> > > >
> > > > ((-2.8 * 10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> > > >
> > > >
> > > > It's missing  X in  ((-2.8 * 10**8)
> > > >
> > > > should be:
> > > >
> > > >
> > > > ((-2.8 * x10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> > >
> > > In my posting I use the symbol * to mean "times" and
> > > I use the symbol ** to mean "to the power of"
> > >
> > > I'm not certain but I think you are using the symbol "X" to mean "times"
> > > If so, then there is no error in the calculation you are pointing out.
> > > If not, what does the X represent?
> > >
> > > Thanks,
> > > David Seppala
> > > Bastrop TX
> >
> >
> >
> >
> > I only put in one x, the others where there is the number 10 does not require the x, just the first 10
> >
> >
> >
> > ((-2.8 * x10**8) * (2 * 10**7)) + (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) meters
> 
> Never heard of such a rule.
> Which line has the physics mistake?
> 
> David Seppala
> Bastrop TX


The whole line has too many syntax errors....here I'll fix it for you:


((-(28/10) /10*8)*(2*10*7))+((5/10)*28*(2*10*7)*(2*10*7))


that should work.

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

Page 1 of 2  [1] 2  Next page →

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

csiph-web