Path: csiph.com!weretis.net!feeder6.news.weretis.net!news.glorb.com!border1.nntp.dca1.giganews.com!nntp.giganews.com!buffer1.nntp.dca1.giganews.com!buffer2.nntp.dca1.giganews.com!nntp.earthlink.com!news.earthlink.com.POSTED!not-for-mail NNTP-Posting-Date: Tue, 09 Aug 2016 20:25:38 -0500 Message-ID: <57AA82C5.77F0@ix.netcom.com> Date: Tue, 09 Aug 2016 18:26:29 -0700 From: The Starmaker Reply-To: starmaker@ix.netcom.com X-Mailer: Mozilla 3.04Gold (WinNT; U) MIME-Version: 1.0 Newsgroups: sci.physics.relativity Subject: Re: My mistake or Einstein's References: <5497e551-1d89-4146-a1c6-0ddf12fa05ef@googlegroups.com> <57AA74AF.243A@ix.netcom.com> <95a809c1-2502-43e7-aa2f-8b2188ea95ff@googlegroups.com> <57AA7AF5.3727@ix.netcom.com> <7e517c9a-76d4-4cc0-8ff0-3f229503f383@googlegroups.com> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Lines: 97 X-Usenet-Provider: http://www.giganews.com NNTP-Posting-Host: 108.219.229.47 X-Trace: sv3-pZWFznYqL1m6qUzlUUsK0J0IZzsD8vDroE8A671+gMGjT8ootEDcV3fSk8llU5YjUS306BWdc4QCgIg!Hf44Y50XSFmBzlsepyBprFPORXMhffobLyIFZ+Vt9wwvGhq6k+KXgnEq5mfnaa2WyhsWjPywrQLW!dP+qMama0WM= X-Abuse-and-DMCA-Info: Please be sure to forward a copy of ALL headers X-Abuse-and-DMCA-Info: Otherwise we will be unable to process your complaint properly X-Postfilter: 1.3.40 X-Original-Bytes: 6921 Xref: csiph.com sci.physics.relativity: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.