colp, why did AE use the word "relativity"?
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From: colp on 24 Jun 2010 17:56 On Jun 24, 10:42 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > colp says... > > >Coordinate systems are arbitrary conventions which are not required by > >the premises of SR. > > Right. But simultaneity (deciding when two different events > take place at the same time) is coordinate-dependent. > > >The paradox isn't about events that are simultaneous because it > >occurs when the twins return to the point that they started from. > > It's *not* a paradox. It is a paradox because: 1. The SR premise of observed time dilation applies to one twin as much as to the other. 2. Each twin observes the time dilation of the other on both the outgoing and return legs. 3. In no case does SR predict that a twin will observe the other's time to be compressed. 4. A twin must observe time compression of the other twin in order that his observations agree with the fact that the twins are the same age when they return to the starting point. > Every inertial coordinate system predicts > exactly the same results for the answer to the question: how > old is each question when they reunite? That is true, but observations do not have to be restricted to a single inertial coordinate system. > > Different coordinate systems only predict different answers to > questions of the form: Is event E1 simultaneous with event E2. > > >> Disagreement between coordinate systems is *not* a paradox. > > >In the symmetric twin paradox, SR predicts that each twin will see the > >other's clock run slow, > > No, it does not. Yes it does. <quote> 2. An ideal clock traveling at speed v for time period t will show an elapsed time of T = t square-root(1-(v/c)^2). </quote> > It predicts that each twin's clock runs slow > as measured in the coordinate system in which the other twin > is at rest. Yes, and that applies for both legs of the experiment. > What it predicts for what each twin sees is the > pattern of delayed and rushed signals that I've been over. That pattern included signal transit times which made it look like a twin saw time compression on the return leg because of the diminishing signal path length. It is important to differentiate between observation of non-local time and patterns of delayed and rushed signals. > > >but it must be seen to run fast in order to that the twin's > >clocks read the same time at the end of the experiment and > >avoid the paradox. > > I went through that. In the symmetric case, if each twin > is sending out signals to the other twin at the rate of > once per second (as measured by the sender's clock), then > those signals will be received delayed (less than one per > second) during part of the journey, they will be received > rushed (more than one per second) during the other part. Right. > If you add them up in the symmetric case, then the *average* > rate of pulses received from the other twin is exactly one > per second. > > There is no paradox. Repeating a denial adds nothing to your argument. > > There are two different ways of looking at it: (1) In terms > of coordinates, and (2) in terms of what is literally seen > by each twin. Pick one or the other, but there is no contradiction > in either case. By what is literally seen, you mean the received pulse rate, right? There is a contradiction between what is _observed_ by each twin, where the observation can be derived from the received pulse rate by taking the signal transit time into account. Signal transit times are an unnecessary complication which obscures the paradox. > > But in the case of what each twin literally sees, it is not > true that each twin sees the other twin's clocks slowed down. Ignoring signal transit times, the following points are true: 1. A twin sees the other's clock to be slowed on the outgoing leg. 2. A twin sees the other's clock to be slowed on the return leg. Which of these points do you disagree with? > > >The premises of SR specify observed time dilation, never time > >compression, so the paradox cannot be avoided. > > The premises talk about time dilation as measured in an *INERTIAL* > coordinate system. Yes. > The twins are *NOT* at rest in an inertial > coordinate system. In the inertial coordinate system in which one twin is stationary, the other is moving, and vice versa. > > If you stick to any specific inertial coordinate system, > the time dilation formula correctly predicts the ages of > the two twins when they get back together. Right. > The only way > to get a contradiction out of it is if you erroneously > pretend that each twin is at rest in an inertial coordinate > system throughout the journey. No, you get the contradiction from an inference which is based on valid observations which are made in two different inertial coordinate systems. In these two systems one twin is at rest and one is moving. > That is not the case, and > applying the rules for an inertial coordinate system is > just making a mathematical mistake. > > Blaming that mistake on SR is just wrong. > > ><quote> > >2. An ideal clock traveling at speed v for time period t will show an > >elapsed time of T = t square-root(1-(v/c)^2). > ></quote> > > You left out the premise: AS MEASURED in any inertial coordinate > system. This rule is not talking about what a twin sees, it's > talking about what is computed to be true, as expressed in an > inertial coordinate system. What is the difference between what the twin sees and what is computed to be true? > > >> >Since paradoxes do not exist in reality the only remaining conclusion > >> >is that there is a preferred frame reference. > > >> Arbitrarily calling one frame the preferred frame makes no difference, > >> whatsoever, to the issue of whether there is a paradox or not. > > >The preferred frame of reference is not determined arbitrarily. > > Then how are you proposing to determine it? Experimentally, or by logical induction. > What experiment > determines which twin is *REALLY* at rest at what times? An experiment like the symmetric twin experiment. In the experiment the older twin is closer to the the preferred frame of reference than the younger twin. > Because the experimental results are the same for *EVERY* > coordinate system. Actual physical experiments have not been conducted in *EVERY* coordinate system. > > > > >> In the twin paradox, you have the paradoxical situation where > >> (1) In the coordinate system of the stationary twin, the traveling > >> twin is younger. > >> (2) In the coordinate system of the traveling twin, the stationary > >> twin is older. > > >> You could introduce a preferred frame, and *arbitrarily* say that > >> the stationary twin's coordinate system is the preferred one, and > >> that the traveling twin's coordinate system is bogus. How does > >> that change anything? You want to call one twin's perspective > >> correct, and the other twin's perspective deluded? Fine. So > >> you change the words, to: > > >> (1) The traveling twin is *actually* younger than the stationary > >> twin. > > >> (2) The stationary twin *appears* to be younger then the traveling > >> twin, when viewed from a bogus coordinate system. > > >> That change is just words. It has made *no* difference to the > >> physics. > > >The issue can be resolved by eliminating the paradoxical cases > > What paradoxical case are you talking about? (copied from above) > >> In the twin paradox, you have the paradoxical situation where > >> (1) In the coordinate system of the stationary twin, the traveling > >> twin is younger. > >> (2) In the coordinate system of the traveling twin, the stationary > >> twin is older. > The paradox is that > according to two different coordinate systems: each coordinate > system measures clocks at rest in the other coordinate system > to be running slow. That's just a fact. Calling one coordinate > system "preferred" doesn't change that fact. Right. > > >and deducing that the preferred frame of reference in the case of the > >symmetric twins is the frame of reference in which the twins journey's > >are symmetric. > > You can choose *ANY* frame whatsoever, and run a symmetric > twin paradox for that frame. No, you don't get a paradox from the frame in which the twin's paths are symmetric. > So your notion of "preferred" > frame would lead to the conclusion that *every* frame is a > preferred frame. non sequitur
From: colp on 24 Jun 2010 18:12 On Jun 24, 10:42 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > Let's consider once again three frames: > > F1 = the frame of the Earth > F2 = the rest frame of a rocket traveling at velocity v > in the +x direction relative to F1. > F3 = the rest frame of a rocket traveling at velocity v > in the -x direction relative to F1. > > Now, let's introduce a 4th frame: > F4 = the rest frame of a rocket traveling at velocity > v in the +x direction relative to frame F2. > > Now, we can do a symmetric twin paradox from the point > of view of frame F1 *and* F2. Introduce 3 twins: > Twin A travels 100 seconds (according to his clock) > at rest in frame F2, turns around, and travels 100 > seconds (according to his clock) at rest in frame F3. > > Twin B travels 100 seconds in frame F3, and then > 100 seconds in frame F4 > > Twin C travels 100 seconds in frame F4, then 100 > seconds in frame F1. > > Twin D travels 100 seconds in frame F1, then 100 > seconds in frame F4. > > So A&B are symmetric twins from the point of > view of frame F1, while C&D are symmetric > twins from the point of view of frame F2. > > So which frame is preferred? F1, or very close to F1. The Earth is by far the most massive object in the experiment. > > >> One can state the principle of relativity in the following > >> way: > > >> There is no experiment that can allow us to determine which > >> coordinate systems is preferred, and which coordinate system > >> is bogus. > > >You can't prove a claim by negation. In other words, the fact that you > >haven't detected something doesn't mean that it doesn't exist. > > If you make such a detection, that means that SR is *WRONG*. > But until then, you haven't proved that SR is wrong. If I haven't proved that SR is wrong by showing that it can produce paradoxes, then why are you trying to show that the idea of the existence of a preferred frame of reference is wrong? > > You don't prove theories of physics correct. You test whether > they are correct by performing experiments. If the experiments > contradict the predictions, then the theory if wrong. Theories can be proved false. In that case the most likely theory is the simplest rational alternative (Occam's razor)
From: PD on 24 Jun 2010 18:26 On Jun 24, 5:12 pm, colp <c...(a)solder.ath.cx> wrote: > On Jun 24, 10:42 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) > wrote: > > > > > > > Let's consider once again three frames: > > > F1 = the frame of the Earth > > F2 = the rest frame of a rocket traveling at velocity v > > in the +x direction relative to F1. > > F3 = the rest frame of a rocket traveling at velocity v > > in the -x direction relative to F1. > > > Now, let's introduce a 4th frame: > > F4 = the rest frame of a rocket traveling at velocity > > v in the +x direction relative to frame F2. > > > Now, we can do a symmetric twin paradox from the point > > of view of frame F1 *and* F2. Introduce 3 twins: > > Twin A travels 100 seconds (according to his clock) > > at rest in frame F2, turns around, and travels 100 > > seconds (according to his clock) at rest in frame F3. > > > Twin B travels 100 seconds in frame F3, and then > > 100 seconds in frame F4 > > > Twin C travels 100 seconds in frame F4, then 100 > > seconds in frame F1. > > > Twin D travels 100 seconds in frame F1, then 100 > > seconds in frame F4. > > > So A&B are symmetric twins from the point of > > view of frame F1, while C&D are symmetric > > twins from the point of view of frame F2. > > > So which frame is preferred? > > F1, or very close to F1. The Earth is by far the most massive object > in the experiment. > > > > > >> One can state the principle of relativity in the following > > >> way: > > > >> There is no experiment that can allow us to determine which > > >> coordinate systems is preferred, and which coordinate system > > >> is bogus. > > > >You can't prove a claim by negation. In other words, the fact that you > > >haven't detected something doesn't mean that it doesn't exist. > > > If you make such a detection, that means that SR is *WRONG*. > > But until then, you haven't proved that SR is wrong. > > If I haven't proved that SR is wrong by showing that it can produce > paradoxes, then why are you trying to show that the idea of the > existence of a preferred frame of reference is wrong? You haven't shown that SR produces paradoxes. You've demonstrated that COLP's Oversimplified Relativity produces paradoxes, and I agree that COR should be tossed. It would do you well to learn the difference between COR and SR. > > > You don't prove theories of physics correct. You test whether > > they are correct by performing experiments. If the experiments > > contradict the predictions, then the theory if wrong. > > Theories can be proved false. In that case the most likely theory is > the simplest rational alternative (Occam's razor). That too is not a good strategy, especially if the simplest rational alternative has *also* been proven false by experimental test, as an aether theory with strong entrainment by the Earth has been. All that you learn by that exercise is that you do not have a viable theory that explains the evidence. However, this is not the case with SR. Summarizing: - COR is ruled out because it is internally inconsistent and produces paradoxes. - Aether theories with entrainment by the Earth have been ruled out by experimental evidence. - SR has not yet been ruled out, either by internal contradictions or by inconsistency with experimental evidence. Hmmm.... seems pretty straightforward to me. PD
From: Daryl McCullough on 24 Jun 2010 21:31 colp says... > >On Jun 24, 10:42=A0pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> colp says... >> >The paradox isn't about events that are simultaneous because it >> >occurs when the twins return to the point that they started from. >> >> It's *not* a paradox. > >It is a paradox because: > >1. The SR premise of observed time dilation applies to one twin as >much as to the other. As I have explained, time dilation in SR does not apply to a relationship between clocks, it applies to the relationship between (1) a coordinate system, and (2) a clock that is moving relative to that coordinate system. So correctly understood, there is no paradox. You have two clocks, and two coordinate systems. According to one coordinate system, one of the clocks is running slower. According to the other coordinate system, the other clock is running slower. The fact that coordinate systems disagree is *not* a paradox. >2. Each twin observes the time dilation of the other on both the >outgoing and return legs. We went over this. Each twin sees signals for the other twin that are delayed, during part of the journey, and rushed, in the other part of the journey. >3. In no case does SR predict that a twin will observe the other's >time to be compressed. That is not true. We just went over this, several times. >4. A twin must observe time compression of the other twin in order >that his observations agree with the fact that the twins are the same >age when they return to the starting point. Yes, and we went over exactly what each twin observes about the other twin, and there is no paradox. Both twins agree on the age of the other twin when they reunite. So what paradox are you talking about? >> Every inertial coordinate system predicts >> exactly the same results for the answer to the question: how >> old is each question when they reunite? > >That is true, but observations do not have to be restricted to a >single inertial coordinate system. If you are trying to make a claim about what SR predicts, you have to stick to SR. If you make up your own rules, and you end up with a paradox, then that's your fault, not SR's. >> >In the symmetric twin paradox, SR predicts that each twin will see the >> >other's clock run slow, >> >> No, it does not. > >Yes it does. > ><quote> >2. An ideal clock traveling at speed v for time period t will show an >elapsed time of T = t square-root(1-(v/c)^2). ></quote> You left out the premise: AS MEASURED in an inertial coordinate system. If you leave that part out, then you are not doing SR, you're doing some theory of your own devising, and if you end up with a paradox, then it's *your* mistake, not SR's. >> It predicts that each twin's clock runs slow >> as measured in the coordinate system in which the other twin >> is at rest. > >Yes, and that applies for both legs of the experiment. F1 = frame of the Earth. F2 = the rest frame of an observer moving at speed v relative to the Earth, traveling in the +x direction. F3 = the rest frame of an observer moving at speed v relative to the Earth, traveling in the -x direction. *ALL* three coordinate systems predict exactly the same result for the symmetric twin paradox. >> What it predicts for what each twin sees is the >> pattern of delayed and rushed signals that I've been over. > >That pattern included signal transit times which made it look like a >twin saw time compression on the return leg because of the diminishing >signal path length. > >It is important to differentiate between observation of non-local time >and patterns of delayed and rushed signals. You *can't* make such a distinction for a twin that is accelerating. Time dilation is *NOT* a relationship between clocks, it is a relationship between a coordinate system and a clock that is moving relative to that coordinate system. >> >but it must be seen to run fast in order to that the twin's >> >clocks read the same time at the end of the experiment and >> >avoid the paradox. >> >> I went through that. In the symmetric case, if each twin >> is sending out signals to the other twin at the rate of >> once per second (as measured by the sender's clock), then >> those signals will be received delayed (less than one per >> second) during part of the journey, they will be received >> rushed (more than one per second) during the other part. > >Right. > >> If you add them up in the symmetric case, then the *average* >> rate of pulses received from the other twin is exactly one >> per second. >> >> There is no paradox. > >Repeating a denial adds nothing to your argument. Well, you've done no calculations whatsoever, and you claim to be able to derive a contradiction. In contrast, I've done the calculation from every possible point of view: 1. From the signals received by each twin. 2. From the point of view of someone on Earth. 3. From the point of view of the coordinate system in which the first twin is at rest during the initial part of his trip. 4. From the point of view of the coordinate system in which the second twin is at rest during the initial part of his trip. Every single way of calculating the results gives the *same* answer. You claim that you can get a different answer, in some way. Tell me HOW. Do the calculation. But don't make up your own rules, and then pretend that you are doing SR. SR does *NOT* say that the time dilation formula works for a noninertial coordinate system. As a matter of fact, SR says exactly the opposite: It says that it *ONLY* works for an inertial coordinate system. If you apply it to a noninertial coordinate system, then you are not doing SR. >By what is literally seen, you mean the received pulse rate, right? > >There is a contradiction between what is _observed_ by each twin, >where the observation can be derived from the received pulse rate by >taking the signal transit time into account. Post the calculation. You are mistaken. >Signal transit times are an unnecessary complication which obscures >the paradox. The paradox is an error of your own making. >> But in the case of what each twin literally sees, it is not >> true that each twin sees the other twin's clocks slowed down. > >Ignoring signal transit times, the following points are true: > >1. A twin sees the other's clock to be slowed on the outgoing leg. >2. A twin sees the other's clock to be slowed on the return leg. > >Which of these points do you disagree with? Both. Both are wrong. As I said, SR doesn't say anything about what each twin *sees*. It says something about the measured rate of a clock in an inertial reference frame. >> If you stick to any specific inertial coordinate system, >> the time dilation formula correctly predicts the ages of >> the two twins when they get back together. > >Right. > >> The only way >> to get a contradiction out of it is if you erroneously >> pretend that each twin is at rest in an inertial coordinate >> system throughout the journey. > >No, you get the contradiction from an inference which is based on >valid observations which are made in two different inertial coordinate >systems. Look, the rules of SR do *NOT* justify doing that kind of mixing of different coordinate systems. If you do that, you are NOT doing SR, you're doing some calculation of your own pet theory that is vaguely similar to SR. This is really ridiculous. I wrote down some rules for applying SR. Do you agree that if you *FOLLOW* those rules, you will never get into a contradiction? You keep *BREAKING* the rules, then reaching a contradiction, and then blaming it on the rules. That's nonsense. Once again, the rules (for the purposes of these twin thought experiments) are: 1. As measured in any inertial coordinate system, light always travels in straight lines at speed c in all directions. 2. As measured in any inertial coordinate system, an ideal clock moving at speed v will show an elapsed time T that satisfies dT/dt = square-root(1-(v/c)^2), where t is the time coordinate of that coordinate system. 3. As measured in any inertial coordinate system, an ideal measuring stick that is moving at speed v in the direction of its length will have a length L given by: L = 1 meter * square-root(1-(v/c)^2) 4. If C is an inertial coordinate system, and C' is obtained from C by a Lorentz transformation, then C' is an inertial coordinate system. Now, do you claim that a contradiction can be derived from 1-4? If so, then post such a contradiction. Show a derivation, for the symmetric twin paradox, of a contradiction. If you cannot derive a contradiction, then that means that the rules 1-4 are *CONSISTENT*. They are not paradoxical, they are consistent. Now, you don't want to stick to rules 1-4. You want to extend the rules by allowing rule 2 to apply to a *NONINERTIAL* coordinate system, or to the case in which an observer switches from one frame to another. The rules 1-4 do *NOT* say that you can do that. So if you are doing that, then you are going beyond rules 1-4. If you reach a contradiction, that shows that you *CAN'T* consistently go beyond rules 1-4. So, in that case, it is perhaps just a matter of terminology. You have a theory that *YOU* are calling Special Relativity, and that theory is inconsistent. There is a *DIFFERENT* theory that everyone else calls "Special Relativity", which is *CONSISTENT*. It doesn't matter what you call things. The fact is that there is a consistent theory for handling objects moving at near the speed of light, and which predicts time dilation for moving clocks. >> ><quote> >> >2. An ideal clock traveling at speed v for time period t will show an >> >elapsed time of T =3D t square-root(1-(v/c)^2). >> ></quote> >> >> You left out the premise: AS MEASURED in any inertial coordinate >> system. This rule is not talking about what a twin sees, it's >> talking about what is computed to be true, as expressed in an >> inertial coordinate system. > >What is the difference between what the twin sees and what is computed >to be true? Look at what I said: I didn't say *ANYTHING* about what a twin sees. I said something what is true IN AN INERTIAL COORDINATE SYSTEM. >> What experiment determines which twin is *REALLY* at rest at what times? > >An experiment like the symmetric twin experiment. How does that tell you which twin is really at rest? You can do the symmetric twin thought experiment in *ANY* frame whatsoever. You get the same results in EVERY single frame. So the symmetric twin experiment does *NOT* tell you what the preferred frame is. >In the experiment the older twin is closer to the the preferred frame >of reference than the younger twin. In the symmetric twin case, both twins are the same age at the end. >> >The issue can be resolved by eliminating the paradoxical cases >> >> What paradoxical case are you talking about? > >(copied from above) >> >> In the twin paradox, you have the paradoxical situation where >> >> (1) In the coordinate system of the stationary twin, the traveling >> >> twin is younger. >> >> (2) In the coordinate system of the traveling twin, the stationary >> >> twin is older. Introducing a preferred frame *DOESN'T* change that. >> >and deducing that the preferred frame of reference in the case of the >> >symmetric twins is the frame of reference in which the twins journey's >> >are symmetric. >> >> You can choose *ANY* frame whatsoever, and run a symmetric >> twin paradox for that frame. > >No, you don't get a paradox from the frame in which the twin's paths >are symmetric. You can choose *ANY* frame to be that frame. >> So your notion of "preferred" >> frame would lead to the conclusion that *every* frame is a >> preferred frame. > >non sequitur Your theory leads to a CONTRADICTION, unlike SR. You don't think that's relevant? You said: the frame in which the twins travel symmetrically is the preferred frame. That doesn't make a bit of sense. We have three frames: F1 = the frame of the Earth. F2 = the frame in which an object traveling at speed v in the +x direction, relative to the Earth, is at rest. F3 = the frame in which an object traveling at speed v in the -x direction, relative to the Earth, is at rest. I can perform a symmetric twin experiment in *ANY* of those frames, and I will get *EXACTLY* the same answer in each case. So which frame is "preferred"? -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 24 Jun 2010 21:55
colp says... > >On Jun 24, 10:42=A0pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: > >> Let's consider once again three frames: >> >> F1 = the frame of the Earth >> F2 = the rest frame of a rocket traveling at velocity v >> in the +x direction relative to F1. >> F3 = the rest frame of a rocket traveling at velocity v >> in the -x direction relative to F1. >> >> Now, let's introduce a 4th frame: >> F4 = the rest frame of a rocket traveling at velocity >> v in the +x direction relative to frame F2. >> >> Now, we can do a symmetric twin paradox from the point >> of view of frame F1 *and* F2. Introduce 3 twins: >> Twin A travels 100 seconds (according to his clock) >> at rest in frame F2, turns around, and travels 100 >> seconds (according to his clock) at rest in frame F3. >> >> Twin B travels 100 seconds in frame F3, and then >> 100 seconds in frame F4 >> >> Twin C travels 100 seconds in frame F4, then 100 >> seconds in frame F1. >> >> Twin D travels 100 seconds in frame F1, then 100 >> seconds in frame F4. >> >> So A&B are symmetric twins from the point of >> view of frame F1, while C&D are symmetric >> twins from the point of view of frame F2. >> >> So which frame is preferred? > >F1, or very close to F1. On what *EXPERIMENTAL* basis are you saying that? You said that the symmetric twin experiment allows you to figure out which is the preferred frame. But the experiment gives the same answer in *EVERY* frame. >> If you make such a detection, that means that SR is *WRONG*. >> But until then, you haven't proved that SR is wrong. > >If I haven't proved that SR is wrong by showing that it can produce >paradoxes, You *HAVEN'T*. You haven't produced a SINGLE derivation that uses actual numbers. You haven't produced a single argument that anything predicted by SR leads to a contradiction. This is not a matter of opinion. You have not performed a SINGLE calculation. You KNOW that. You have not derived anything at all. So you certainly haven't derived a contradiction from the rules of SR. Look, I've shown you that every possible LEGITIMATE way of calculating ages for twins in every possible thought experiment leads to exactly the same conclusion about the ages when they get back together. You basically argue: "Oh, but I can perform a calculation that *DOES* lead to a contradiction." Okay, so you know how to be inconsistent. You seem to think that this contradictory theory of yours should be considered the TRUE Special Relativity, while the consistent theory that everyone else but you uses is somehow a pale, inferior version of SR. But I really prefer the consistent theory, not your inconsistent version. -- Daryl McCullough Ithaca, NY |