Some Contradictory Claims of SR.
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From: Sam Wormley on 29 Jul 2010 10:30 On 7/29/10 9:03 AM, kenseto wrote: > No idiot....it merely means that the SR interpretation of the SR math > is wrong. Talk about an illogical statement "SR interpretation of the SR math is wrong"! Here are the facts, Seto: o Special relativity was proposed in 1905. o Special relativity is mathematically self consistent. o Special relativity makes predictions and the predictions are confirmed by observation and experiment. http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html There has never been an observation that contradicts a prediction of special relativity. The only thing wrong is that you, Seto, have no understanding of special relativity... not even the very basics. Physics FAQ: Are There Any Good Books on Relativity Theory? http://math.ucr.edu/home/baez/physics/Administrivia/rel_booklist.html
From: Sam Wormley on 29 Jul 2010 10:40 On 7/29/10 9:02 AM, kenseto wrote: > 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. >> >> ∆t represent a time interval between tick of the clocks. >> >> Special relativity predicts that observer A will measure that >> ∆t_B' = γ ∆t_A > > From A's point of view IRT predicts the following: > ∆t_B' = γ ∆t_A > OR > ∆t_B' = (1/γ)∆t_A Seto, you consistently confuse observer and observed. Furthermore you have no understanding of special relativity with says that that A will observe time dilation is B's clock, ∆t_B' = γ ∆t_B . You also ALTERED what I wrote in my posting. Utterly dishonest! You did that above and you did that below. Despicable! How can you stoop so low as a human being? I originally posted: 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 > >> >> 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_B > > From B's point of view IRT predicts the following: > ∆t_A' = γ ∆t_B > OR > ∆t_A' = 1/γ ∆t_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: Michael Moroney on 29 Jul 2010 11:34 kenseto <kenseto(a)erinet.com> writes: >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: >> >> 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. Yes, and experiments do show A measuring B's clock as running slow. >> 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. Both "2" phrases are in conflict with what SR states will be measured. >> >> 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. So you admit that your beliefs are in conflict with SR, since SR predicts A will measure B as running slow, and B will measure A as running slow. >>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. "In certain situation[s]" implies a conflict in other situations, and you admit a conflict with mutual time dilation predictions, thus SR cannot be a subset of your claims, because of these conflicts. Just because it does *not* conflict "in certain situation[s]" does not mean it *never* conflicts with SR. >> >> 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 sentence 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. So, you seem to have picked the second sentence, thus IRT conflicts with SR and (supposedly) corrects it.
From: Michael Moroney on 29 Jul 2010 11:41 kenseto <kenseto(a)erinet.com> writes: >On Jul 28, 3:30 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) >wrote: >> 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. Because SR makes an explicit prediction about observed clocks, and these prediction *never* include an observed clock running faster than the observer's clock, there is an explicit conflict between SR and IRT. Thus SR cannot be a subset of IRT. We'll just have to add "subset" and "superset" to the list of words Ken has redefined.
From: kenseto on 29 Jul 2010 17:36
On Jul 29, 10:08 am, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > "kenseto" <kens...(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? There is no twin paradox in IRT. Why? Because it doesn't include mutual time dilation as in SR. Besides, absolute time exists in IRT.....what this mean is that a traveling clock second contains a larger amount of absolute time than a stay at home clock second. That means that you cannot compare a traveling clock second directly with a stay at home clock second to reach the conclusion that the traveling clock is younger. The accumulated traveling clock time must be converted into the stay at home clock time before making the comparison as follows: Deta(T_ts)= gamma*delta(T_t)= delta(T_s) Where Delta (T_st)=accumulated traveling clock time of the traveling clock converted to the stay at home clock time. Delts(T_t)=accumulated clock time of the traveling clock during the journey. Delta (T_s)= accumulated clock time on the stay at home clock for the complete journey of the traveling clock. As you can see there is no paradox: Gamma*Delta(T_t)=delta(T_s) That menas that they both aged the same in terms of absolute time. Ken Seto |