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Re: My mistake or Einstein's

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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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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

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