Some Contradictory Claims of SR.
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From: Sam Wormley on 28 Jul 2010 21:05 On 7/28/10 9:58 AM, kenseto wrote: > Sure IRT is a super set of SR It includes the correct prediction of SR > that says that an observed clock can run slow. But it reject the > faulty SR assertion that all clocks moving wrt the observer are > running slow. > > Ken Seto Then IRT has to also predict: A and B are observers with identical clocks. That is A and B's clocks ticked synchronously when they were together. ∆t represent a time interval between tick of the clocks. Special relativity predicts that observer A will measure that ∆t_B' = γ ∆t_B where ∆t represent a time interval, v is the relative velocity between A and B, and γ = 1/√(1-v^2/c^2) . Furthermore, special relativity predicts that observer B will measure that ∆t_A' = γ ∆t_A Physics FAQ: What is the experimental basis of special relativity? http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
From: kenseto on 29 Jul 2010 09:48 On Jul 28, 3:24 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) wrote: > kenseto <kens...(a)erinet.com> writes: > >On Jul 27, 9:25 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) > >wrote: > >> "Inertial" <relativ...(a)rest.com> writes: > >> >BUT .. it must also predict EXACTLY the same things the SR DOES predict, > >> >otherwise SR is not a subset, but is disjoint from IRT. > >> >So everything SR says, IRT must say, and IRT must then also say addition > >> >things > >> >Otherwise the claim that SR is a subset of IRT is just another lie. > > >> If Frame B is in inertial motion relative to Frame A, an observer in Frame > >> A will measure a clock in Frame B as running slow. Ken apparently agrees. > >No SR says that A predicts B is running slow by a factor of 1/ > >gamma....I agree to that. > > Another reason why SR conflicts with Ken's dreck - SR predicts A will > measure B's clock as running slow. No measurement....SR just predicts. > Ken thinks B's clock IS running > slow...B will disagree. No that's not what I think or what IRT says. IRT says that: From A's point of view: 1. B runs slow by a factor of 1/gamma OR 2. B runs fast by a factor of gamma From B's point of view: 1. A runs slow by a factor of 1/gamma OR 2. A runs fast by a factor of gamma. > > >> SR states an observer in Frame B will measure a clock in Frame A as > >> running slow. > >This SR prediction is derived from the faulty SR assumption that every > >SR observer is in a state of rest and thus all clcoks moving wrt him > >are running slow. Since we already predicted that A is running faster > >than B then B must run slower than A and thus B cannot predict A runs > >slow....instead he must predict that A run fast. > > So you admit that your beliefs are in conflict with SR. No I admit no such thing. I said that if A is truly running faster than B then B must truly running slower than A. What this mean is that the SR concept of mutual time dialtion is wrong. >Therefore SR > cannot be any subset of your beliefs, as SR is in conflict with your > beliefs. SR math is a subset of IRT math because the SR math agrees with IRT math in certain situation....such as that an observed clock is in a higher state of absolute motion. > > >> Ken claims that the Frame B observer will see the Frame A > >> clock as running fast, conflicting with SR. > >It is not conflicting with SR...it corrects an faulty assumption of > >SR. > > Your first aentence and your second sentence conflict with each other. > Pick one or the other, they cannot both be true. No the SR prediction of mutual time dilation is wrong....A cannot predict B is slow and at the same time A predicts A is slow. This contradicts all logic. > > >>Therefore, any claim that > >> SR is a subset of IRT is false. > >SR is a subset of IRT because it got A's prediction correctly. > > ...and it gets B's prediction wrong. > > Read what Inertial wrote: > > "So everything SR says, IRT must say, and IRT must then also say > addition[al] things. Otherwise the claim that SR is a subset of IRT is > just another lie." > > >> In addition, the experimental results > >> agree with SR and not IRT, so we also know that IRT is at least partially > >> wrong. > >Sigh....experimental result agree with A's prediction only. Not B's > >prediction that A is running slow. > > Did you follow PD's link to Ashby's paper that he always tries to get > you to read, or the links to experimental verification of SR? Not light > reading; maybe if you ask nicely they'll help you understand it. > > Denying experimental results won't make them go away.
From: kenseto on 29 Jul 2010 09:52 On Jul 28, 3:30 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) wrote: > kenseto <kens...(a)erinet.com> writes: > >Sure IRT is a super set of SR. It includes the correct prediction of SR > >that says that an observed clock can run slow. But it reject the > >faulty SR assertion that all clocks moving wrt the observer are > >running slow. > > The first and third sentences here are in direct conflict with each other.. > Pick one or the other: > > 1) "IRT is a super set of SR." > 2) "[IRT] rejects the faulty SR assertion that all clocks moving wrt > the observer are running slow." Both of these sentences are correct. SR is a subset of IRT because it does not include the possibility that an observed clock can run faster than the observer's clock.
From: kenseto on 29 Jul 2010 10:02 On Jul 28, 9:05 pm, Sam Wormley <sworml...(a)gmail.com> wrote: > On 7/28/10 9:58 AM, kenseto wrote: > > > Sure IRT is a super set of SR It includes the correct prediction of SR > > that says that an observed clock can run slow. But it reject the > > faulty SR assertion that all clocks moving wrt the observer are > > running slow. > > > Ken Seto > >   Then IRT has to also predict: > >   A and B are observers with identical clocks. That is A and B's >   clocks ticked synchronously when they were together. > >   ât represent a time interval between tick of the clocks. > >   Special relativity predicts that observer A will measure that >    ât_B' = γ ât_A From A's point of view IRT predicts the following: ât_B' = γ ât_A OR ât_B' = (1/γ)ât_A > >   where ât represent a time interval, v is the relative velocity >   between A and B, and γ = 1/â(1-v^2/c^2) . > >   Furthermore, special relativity predicts that observer B will >   measure that >    ât_A' = γ ât_B From B's point of view IRT predicts the following: ât_A' = γ ât_B OR ât_A' = 1/γ ât_B As you can see IRT includes the predictions of SR and more and that's why IRT is a super set of SR. Ken Seto > >   Physics FAQ: What is the experimental basis of special relativity? >    http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
From: Peter Webb on 29 Jul 2010 10:08
"kenseto" <kenseto(a)erinet.com> wrote in message news:cf033934-4c4d-41f1-b322-42f0539c17b5(a)f33g2000yqe.googlegroups.com... On Jul 28, 3:30 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) wrote: > kenseto <kens...(a)erinet.com> writes: > >Sure IRT is a super set of SR. It includes the correct prediction of SR > >that says that an observed clock can run slow. But it reject the > >faulty SR assertion that all clocks moving wrt the observer are > >running slow. > > The first and third sentences here are in direct conflict with each other. > Pick one or the other: > > 1) "IRT is a super set of SR." > 2) "[IRT] rejects the faulty SR assertion that all clocks moving wrt > the observer are running slow." Both of these sentences are correct. SR is a subset of IRT because it does not include the possibility that an observed clock can run faster than the observer's clock. __________________________________ If in your theory the twin paradox doesn't occur, then your theory is wrong, as this is observed. So, does it? |