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My mistake or Einstein's

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  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 2 of 2 — ← Prev page 1 [2]

#389648

On Tuesday, August 9, 2016 at 8:25:45 PM UTC-5, The Starmaker wrote:
> 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.

Your posting of a "fix" is wrong, and uses the * to mean two different things.  In my posting, if you go thru the numbers my equation is equal to zero.  
 ((-2.8 * 10**8) * (2 * 10**7)) = -56,000,000,000,000,000 meters
 (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) = 56,000,000,000,000,000 meters

When you use * to mean two different operations doesn't clarify things, and you substituted a / for a * in your fix.

David Seppala
Bastrop T
  

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#389653

sepp623@yahoo.com writes:

>When you use * to mean two different operations doesn't clarify things, and
> you substituted a / for a * in your fix.

Ignore Starfaker. She's just a troll and she has trolled you big time.

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#389654

sepp623@yahoo.com wrote:
> 
> On Tuesday, August 9, 2016 at 8:25:45 PM UTC-5, The Starmaker wrote:
> > 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 th
> > > > > > >
> > > > > > > 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.
> 
> Your posting of a "fix" is wrong, and uses the * to mean two different things.  In my posting, if you go thru the numbers my equation is equal to zero.
>  ((-2.8 * 10**8) * (2 * 10**7)) = -56,000,000,000,000,000 meters
>  (0.5 * 28 * (2 * 10**7) * (2 * 10**7)) = 56,000,000,000,000,000 meters
> 
> When you use * to mean two different operations doesn't clarify things, and you substituted a / for a * in your fix.
> 
> David Seppala
> Bastrop T
> 


My mistake...

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#389659

On 10/08/2016 4:04 AM, sepp623@yahoo.com wrote:
> This scenario of this simple problem ends up with contradictory results. Please identify the mistake.

Your mistake is failing to read up on Bell's Spaceship Paradox.

https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox

Or, alternatively, reading up on it, and then completely 
misunderstanding it.

Sylvia.

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#389662

On Tuesday, August 9, 2016 at 10:16:32 PM UTC-5, Sylvia Else wrote:
> On 10/08/2016 4:04 AM, sepp623@yahoo.com wrote:
> > This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> 
> Your mistake is failing to read up on Bell's Spaceship Paradox.
> 
> https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox
> 
> Or, alternatively, reading up on it, and then completely 
> misunderstanding it.
> 
> Sylvia.
Bell's paradox and that wikipedia link talk about a physical object connected between two accelerating objects. They talk about the length contraction of that connecting object and relativistic stress, etc.
  In the problem I posted, there are just two accelerating objects, with nothing connecting them.  Using the coordinates in one inertial reference frame F0, the two objects crash into each other during the time in the problem. In the other inertial reference frame F1, the two objects always remain a fixed distance apart. The never crash. Both views cannot be correct.

David Seppala
Bastrop TX

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#389663

On 10/08/2016 2:11 PM, sepp623@yahoo.com wrote:
> On Tuesday, August 9, 2016 at 10:16:32 PM UTC-5, Sylvia Else wrote:
>> On 10/08/2016 4:04 AM, sepp623@yahoo.com wrote:
>>> This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
>>
>> Your mistake is failing to read up on Bell's Spaceship Paradox.
>>
>> https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox
>>
>> Or, alternatively, reading up on it, and then completely
>> misunderstanding it.
>>
>> Sylvia.
> Bell's paradox and that wikipedia link talk about a physical object connected between two accelerating objects. They talk about the length contraction of that connecting object and relativistic stress, etc.
>    In the problem I posted, there are just two accelerating objects, with nothing connecting them.  Using the coordinates in one inertial reference frame F0, the two objects crash into each other during the time in the problem. In the other inertial reference frame F1, the two objects always remain a fixed distance apart. The never crash. Both views cannot be correct.
>
> David Seppala
> Bastrop TX
>

Like I said - misunderstanding it.

Sylvia.

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#389665

On Tuesday, August 9, 2016 at 11:15:30 PM UTC-5, Sylvia Else wrote:
> On 10/08/2016 2:11 PM, sepp623@yahoo.com wrote:
> > On Tuesday, August 9, 2016 at 10:16:32 PM UTC-5, Sylvia Else wrote:
> >> On 10/08/2016 4:04 AM, sepp623@yahoo.com wrote:
> >>> This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> >>
> >> Your mistake is failing to read up on Bell's Spaceship Paradox.
> >>
> >> https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox
> >>
> >> Or, alternatively, reading up on it, and then completely
> >> misunderstanding it.
> >>
> >> Sylvia.
> > Bell's paradox and that wikipedia link talk about a physical object connected between two accelerating objects. They talk about the length contraction of that connecting object and relativistic stress, etc.
> >    In the problem I posted, there are just two accelerating objects, with nothing connecting them.  Using the coordinates in one inertial reference frame F0, the two objects crash into each other during the time in the problem. In the other inertial reference frame F1, the two objects always remain a fixed distance apart. The never crash. Both views cannot be correct.
> >
> > David Seppala
> > Bastrop TX
> >
> 
> Like I said - misunderstanding it.
> 
> Sylvia.
Why not just tell me which numbers in my posting are incorrect if you think I'm misunderstanding Bell's Spaceship Paradox.  Can you do that?  No one else has.

David Seppala
Bastrop TX

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#389666

On 10/08/2016 2:31 PM, sepp623@yahoo.com wrote:
> On Tuesday, August 9, 2016 at 11:15:30 PM UTC-5, Sylvia Else wrote:
>> On 10/08/2016 2:11 PM, sepp623@yahoo.com wrote:
>>> On Tuesday, August 9, 2016 at 10:16:32 PM UTC-5, Sylvia Else wrote:
>>>> On 10/08/2016 4:04 AM, sepp623@yahoo.com wrote:
>>>>> This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
>>>>
>>>> Your mistake is failing to read up on Bell's Spaceship Paradox.
>>>>
>>>> https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox
>>>>
>>>> Or, alternatively, reading up on it, and then completely
>>>> misunderstanding it.
>>>>
>>>> Sylvia.
>>> Bell's paradox and that wikipedia link talk about a physical object connected between two accelerating objects. They talk about the length contraction of that connecting object and relativistic stress, etc.
>>>     In the problem I posted, there are just two accelerating objects, with nothing connecting them.  Using the coordinates in one inertial reference frame F0, the two objects crash into each other during the time in the problem. In the other inertial reference frame F1, the two objects always remain a fixed distance apart. The never crash. Both views cannot be correct.
>>>
>>> David Seppala
>>> Bastrop TX
>>>
>>
>> Like I said - misunderstanding it.
>>
>> Sylvia.
> Why not just tell me which numbers in my posting are incorrect if you think I'm misunderstanding Bell's Spaceship Paradox.  Can you do that?  No one else has.
>

We've been down that rabbit warren before. It takes a lot of effort, and 
endless patience, and all you ever do is come back with another mistaken 
analysis.

Anyone who doubts this need only look at the Google Groups archive.

Sylvia.

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#389669

On Tuesday, August 9, 2016 at 11:48:45 PM UTC-5, Sylvia Else wrote:
> On 10/08/2016 2:31 PM, sepp623@yahoo.com wrote:
> > On Tuesday, August 9, 2016 at 11:15:30 PM UTC-5, Sylvia Else wrote:
> >> On 10/08/2016 2:11 PM, sepp623@yahoo.com wrote:
> >>> On Tuesday, August 9, 2016 at 10:16:32 PM UTC-5, Sylvia Else wrote:
> >>>> On 10/08/2016 4:04 AM, sepp623@yahoo.com wrote:
> >>>>> This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> >>>>
> >>>> Your mistake is failing to read up on Bell's Spaceship Paradox.
> >>>>
> >>>> https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox
> >>>>
> >>>> Or, alternatively, reading up on it, and then completely
> >>>> misunderstanding it.
> >>>>
> >>>> Sylvia.
> >>> Bell's paradox and that wikipedia link talk about a physical object connected between two accelerating objects. They talk about the length contraction of that connecting object and relativistic stress, etc.
> >>>     In the problem I posted, there are just two accelerating objects, with nothing connecting them.  Using the coordinates in one inertial reference frame F0, the two objects crash into each other during the time in the problem. In the other inertial reference frame F1, the two objects always remain a fixed distance apart. The never crash. Both views cannot be correct.
> >>>
> >>> David Seppala
> >>> Bastrop TX
> >>>
> >>
> >> Like I said - misunderstanding it.
> >>
> >> Sylvia.
> > Why not just tell me which numbers in my posting are incorrect if you think I'm misunderstanding Bell's Spaceship Paradox.  Can you do that?  No one else has.
> >
> 
> We've been down that rabbit warren before. It takes a lot of effort, and 
> endless patience, and all you ever do is come back with another mistaken 
> analysis.
> 
> Anyone who doubts this need only look at the Google Groups archive.
> 
> Sylvia.
Didn't think you would post the error I made in the problem. No one else has.
David Seppala
Bastrop TX

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#389677

sepp623@yahoo.com wrote:
> 
> On Tuesday, August 9, 2016 at 11:48:45 PM UTC-5, Sylvia Else wrote:
> > On 10/08/2016 2:31 PM, sepp623@yahoo.com wrote:
> > > On Tuesday, August 9, 2016 at 11:15:30 PM UTC-5, Sylvia Else wrote:
> > >> On 10/08/2016 2:11 PM, sepp623@yahoo.com wrote:
> > >>> On Tuesday, August 9, 2016 at 10:16:32 PM UTC-5, Sylvia Else wrote:
> > >>>> On 10/08/2016 4:04 AM, sepp623@yahoo.com wrote:
> > >>>>> This scenario of this simple problem ends up with contradictory results. Please identify the mistake.
> > >>>>
> > >>>> Your mistake is failing to read up on Bell's Spaceship Paradox.
> > >>>>
> > >>>> https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox
> > >>>>
> > >>>> Or, alternatively, reading up on it, and then completely
> > >>>> misunderstanding it.
> > >>>>
> > >>>> Sylvia.
> > >>> Bell's paradox and that wikipedia link talk about a physical object connected between two accelerating objects. They talk about the length contraction of that connecting object and relativistic stress, etc.
> > >>>     In the problem I posted, there are just two accelerating objects, with nothing connecting them.  Using the coordinates in one inertial reference frame F0, the two objects crash into each other during the time in the problem. In the other inertial reference frame F1, the two objects always remain a fixed distance apart. The never crash. Both views cannot be correct.
> > >>>
> > >>> David Seppala
> > >>> Bastrop TX
> > >>>
> > >>
> > >> Like I said - misunderstanding it.
> > >>
> > >> Sylvia.
> > > Why not just tell me which numbers in my posting are incorrect if you think I'm misunderstanding Bell's Spaceship Paradox.  Can you do that?  No one else has.
> > >
> >
> > We've been down that rabbit warren before. It takes a lot of effort, and
> > endless patience, and all you ever do is come back with another mistaken
> > analysis.
> >
> > Anyone who doubts this need only look at the Google Groups archive.
> >
> > Sylvia.
> Didn't think you would post the error I made in the problem. No one else has.
> David Seppala
> Bastrop TX

Someone mention you have had this...problem for ten years...

is it possible you're dreaming
and everytime you wakeup
you're having the same dream?

and we are in your dream..
a dream where
"no one else has post the error"
dream?

It could be a nightmare you're having...

Certaintly would be a nightmare to me to have
a mathematically problem that 
"no one else has post the error"
for ten years.

I think it's time to
change your dream.

Take your formula
burn it up
and start a new dream..

it will end the nightmare
but there is no guarantee
you'll wake up.

It will be just a different dream.

Even Einstein made mistakes...

his unified theory never got finished..

that was his nightmare.

You were right 
with the first two words
of this subject heading..

How about a game of chess?

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#389688

On 10/08/2016 4:04 AM, 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.
>
> 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).
>
In your scenario, the continually accelerating objects will reach the 
speed of light three seconds apart in frame F0. In frame F1, since they 
are always travelling at the same speed, they will reach the speed of 
light simultaneously.

So regardless of the values of the acceleration, or the velocity, the 
three second difference in F0 will transform to nothing in F1.

This is inherently improbable, and makes the whole scenario suspect. 
What you've effectively done is to assert, without proof, that starting 
identical constant accelerations some time apart in one frame transforms 
into simultaneous identical (though not constant) accelerations in some 
other frame.

So that's why you get a contradiction - you've assumed something that's 
not true.

Sylvia.

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#389690

On 09.08.2016 20:04, 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


This is the source of the error;-)
Seriously, it is all right, but it will lead to the error.

You can think of object that have constant "coordinate" acceleration,
but it is probably not what you think.
A passenger on such object will feel bigger and bigger "g force".
In his local (temporary) inertial reference frame the acceleration
will grown.

But until it reach c, such object could exist.

> 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

The idea is sound, I did not check your calcultaions.


> 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.

Og, lets remember this assumption (*).

> 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.

Yes...

> 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

Do not do this! I do not know if that is a new assumption.
I do not known if it contradicts earlier assumptions
(which starts earlier)
This is messy and unreadable!

Just create the scene, tell us that F1 have velocity v
in F0. And the sign of v will tell us if F1 is going in the direction
of greater or smaller 'x'. It is your task to think through
what you want to achieve and create a clear situation.

Tell us that in one of the freme they have such and such position,
The velocity is directed in that direction...
The rest have to be conclusions (preferbly backed by calculations:)


> 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.

Not 3s.

In F0 the starts have coordinates (t,x):
(0,0)
and
(0,3)

velocity of F1 is v = sqrt(3)/2,
so gamma = 1/sqrt(1-3/4) = 1/(1/2)=2

The first start is still at (0,0),
th second is at time
t' = gamma t +- gamma x v = +-2*sqrt(3)/2 *3 = +-3*sqrt(3) s



> 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?

The closest approach will be at the end. When both spaceship
stops accelerating, transfer to its reference frame and
compute the distance.

> 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


This is bunch of numbers, not equation?
What does it mean?

I guess it is s(t) = v0*t + 0.5 at^2

And We have the error.

You have assumed that in F1 both rockets have the same
'acceleration pattern' (so the same trajectory).
This was the assumption I labelled (*).

They do not have the same pattern in F0!

The coordinate acceleration of the second rocket is not
even constant in F0.


The A rocket have acceleration pattern
v = v0+a*t
s = s0 + v0*t+0.5 a t^2
in F0.

this mean, as I mentioned, that proper acceleration
(acceleration measured by instruments on rocket A)
is changing.

In F1 the movement is not v' = v0'+a'*t',
but both rocket have the same trajectory (only
shifted in space by 3 light seconds).

Proper acceleration is changing, but at the same time
in F1 both rocket have the same proper acceleration.

Now, we are going back to F0. The surface of simultaneity
"rotate" and now at the same time both rocket have
different proper accelerations!

So them have (most of the time:) different coordinate accelerations.

The B rocket do not have constant acceleration equal to
A's acceleration.


There is no crash, you have used wrong trajectory for rocket B.



You can get the trajectory of B in F0 by transfering
trajectory of A to F1, shifting (x'->x' +3) and
transfering back to F0.

The result is not a nice equation;-)


You can also use the motion with constant proper acceleration:
https://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)
It looks the same in every reference frame.


bartekltg

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#389693

Użytkownik "bartekltg"  napisał w wiadomości grup 
dyskusyjnych:nof57j$8ua$1@node2.news.atman.pl...

|A passenger on such object will feel bigger and bigger "g force".

Remember however, that when bartek says "passenger",
"observer" or alike, it has nothing in common with real
human really feeling something, or anything real at
all. It's just a sick imagination of a brainwashed idiot.

If You push this idiot a little, he will even admit it himself.
Proudly. Yes, he is THAT stupid.

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#389697

On 10.08.2016 14:29, Maciej Woźniak wrote:
>
>
> Użytkownik "bartekltg"  napisał w wiadomości grup
> dyskusyjnych:nof57j$8ua$1@node2.news.atman.pl...
>
> |A passenger on such object will feel bigger and bigger "g force".
>
> Remember however, that when bartek says "passenger",
> "observer" or alike, it has nothing in common with real
> human really feeling something, or anything real at

This is exactly what I mean in this sentence:
a person or an instrument that measures the acceleration.

> Yes, he is THAT stupid.

Yes, your trolling is that stupid.

correcting KF rules...
bartekltg

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#389702

On Wednesday, August 10, 2016 at 8:30:00 AM UTC-5, bartekltg wrote:
> On 10.08.2016 14:29, Maciej Woźniak wrote:
> >
> >
> > Użytkownik "bartekltg"  napisał w wiadomości grup
> > dyskusyjnych:nof57j$8ua$1@node2.news.atman.pl...
> >
> > |A passenger on such object will feel bigger and bigger "g force".
> >
> > Remember however, that when bartek says "passenger",
> > "observer" or alike, it has nothing in common with real
> > human really feeling something, or anything real at
> 
> This is exactly what I mean in this sentence:
> a person or an instrument that measures the acceleration.
> 
> > Yes, he is THAT stupid.
> 
> Yes, your trolling is that stupid.
> 
> correcting KF rules...
> bartekltg

I FOUND MY MISTAKE!
In the original posting, I miscalculated the initial distance between objects A and B. They had zero relative velocity but had a velocity relative to both the F0 and F1 inertial frames I was looking at. I used the length transform in error, as that only applies when two points have zero velocity relative to one of the frames.

David Seppala
Bastrop TX

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#389756

On 10/08/2016 11:36 PM, sepp623@yahoo.com wrote:
> On Wednesday, August 10, 2016 at 8:30:00 AM UTC-5, bartekltg wrote:
>> On 10.08.2016 14:29, Maciej Woźniak wrote:
>>>
>>>
>>> Użytkownik "bartekltg"  napisał w wiadomości grup
>>> dyskusyjnych:nof57j$8ua$1@node2.news.atman.pl...
>>>
>>> |A passenger on such object will feel bigger and bigger "g force".
>>>
>>> Remember however, that when bartek says "passenger",
>>> "observer" or alike, it has nothing in common with real
>>> human really feeling something, or anything real at
>>
>> This is exactly what I mean in this sentence:
>> a person or an instrument that measures the acceleration.
>>
>>> Yes, he is THAT stupid.
>>
>> Yes, your trolling is that stupid.
>>
>> correcting KF rules...
>> bartekltg
>
> I FOUND MY MISTAKE!

This seems so incredibly improbable that I have to wonder whether the 
post is a forgery.

Sylvia.

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#389703

Użytkownik "bartekltg"  napisał w wiadomości grup 
dyskusyjnych:nofa8m$j2b$1@node1.news.atman.pl...

On 10.08.2016 14:29, Maciej Woźniak wrote:
>
>
> Użytkownik "bartekltg"  napisał w wiadomości grup
> dyskusyjnych:nof57j$8ua$1@node2.news.atman.pl...
>
> |A passenger on such object will feel bigger and bigger "g force".
>
> Remember however, that when bartek says "passenger",
> "observer" or alike, it has nothing in common with real
> human really feeling something, or anything real at

|This is exactly what I mean in this sentence:

A lie, as expecteed from fanatic trash. How many times did you
you explain what does the word "observer" mean in your moronic
newspeak?
Now explain it to others. Please?

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#389706

On 10.08.2016 14:04, bartekltg wrote:
>
> The A rocket have acceleration pattern
> v = v0+a*t
> s = s0 + v0*t+0.5 a t^2
> in F0.
>
> this mean, as I mentioned, that proper acceleration
> (acceleration measured by instruments on rocket A)
> is changing.
>
> In F1 the movement is not v' = v0'+a'*t',

I have been too pessimistic.

If both rocket have the same trajectory in F1,
only shifted by x0, and the first trajectory is

x=a/2 t^2

the second rocket flight along:
x +x0 gamma = a/2 (T - v x0 gamma ) ^2

The same shape, but the orgin transformet in spacetime.

But in other post:

 >I FOUND MY MISTAKE!

:-)

bartekltg

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#389711

Użytkownik "bartekltg"  napisał w wiadomości grup 
dyskusyjnych:nofbcl$f71$1@node2.news.atman.pl...

On 10.08.2016 14:04, bartekltg wrote:
>
> The A rocket have acceleration pattern
> v = v0+a*t
> s = s0 + v0*t+0.5 a t^2
> in F0.
>
> this mean, as I mentioned, that proper acceleration
> (acceleration measured by instruments on rocket A)
> is changing.
>
> In F1 the movement is not v' = v0'+a'*t',

|I have been too pessimistic.

No. You were too hardly brainwashed by moronic
mumble of your idiot guru.

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