They're not true at once. They're different experiments, despite the deception of calling them both 'experiment 4'.Steve3007 wrote: ↑October 6th, 2018, 3:20 amSo, Halc, bearing in mind the above, when David asks you this:
what are your thoughts as to the answer?David Cooper wrote:Are you trying to have your cake and eat it by having D>L and E>S as well, or do you recognise the mathematical impossibility of both of those being true at once?
The March Philosophy Book of the Month is Final Notice by Van Fleisher. Discuss Final Notice now.
The April Philosophy Book of the Month is The Unbound Soul by Richard L. Haight
Does Special Relativity contain contradictions?
 Halc
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Re: Does Special Relativity contain contradictions?
 Halc
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Re: Does Special Relativity contain contradictions?
That said, I presume there was a prior single scenario proposed by you, not David's two entirely separate experiments.Steve3007 wrote: ↑October 6th, 2018, 3:20 amSo, Halc, bearing in mind the above, when David asks you this:
what are your thoughts as to the answer?David Cooper wrote:Are you trying to have your cake and eat it by having D>L and E>S as well, or do you recognise the mathematical impossibility of both of those being true at once?
So switching to standard definitions here, we can get rid of the Ether reference A which appears to be David's addition, and just have clocks CD CE CL CS. D E L and S have no meaning, we need to designate our tick rates as something like De which means the tick rate of CD in the frame in which CE is stationary.
Then there is no cake and eating it too. Dd > Ld does not mathematically contradict De < Le.

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Re: Does Special Relativity contain contradictions?
Even if the assumption is made explicit, I still see mistakes in the logic.Halc wrote:People making and denying the same claims make different unstated assumptions. I read through what David posted and it is quite consistent. I saw no mistakes, except for the lack of explicit statement of that unstated assumption.
Yes, strictly speaking, you're right. He says that both CL and CE are moving at speed v relative to CA, in the first leg. But he doesn't actually explicitly conclude from that that CL is moving at speed 0 relative to CE. Only via CA. But if we can't draw that conclusion from those two premises then we've departed from logic as I understand it. And I don't really see how calling CA "reality" would change that. If CL and CE are both moving at speed v relative to "reality", then I think we're still saying that CL is moving at speed 0 relative to CE.David does not say this. He never compares CL to CE. All references are to CA, which apparently represents reality. CL has the same velocity as E, yes, but never does he use wording that CA is stationary relative to CE.Steve3007 wrote:CL is declared to be stationary relative to CE during leg 1 by stating that they both move at speed v relative to CA during that leg.
I admire your attempts to try to work out what he's on about. But I'm still not getting it.
True again. He doesn't use that wording here. He doesn't acknowledge, in that particular sentence, that the phrase "...is stationary relative to..." is a synonym for "...has the same velocity as...". But he does elsewhere in his posts.David does not say this. He never compares CL to CE. All references are to CA, which apparently represents reality. CL has the same velocity as E, yes, but never does he use wording that CA is stationary relative to CE.Steve3007 wrote:CL is declared to be stationary relative to CE during leg 1 by stating that they both move at speed v relative to CA during that leg.
That's what I thought he appeared to be saying because it's the only explanation I can think of for the fact that he makes two opposite inequality claims about 4 identically moving clocks. I guess I just assumed that when he refers to (for example) CL in the version of experiment 4 containing CA and in the version containing CA2 that he's referring to the same clock each time. If he isn't then, again, he's departing from the logical rules of language as I understand them. If you said "1 + 1 = 2 and 2 + 2 = 4" I'd assume that the symbol '2' doesn't change its meaning between those two statements.Wow, I didn't get that at all. You're interpreting his words using your own definitions of the terms. CA and CA2 are the same clock, and its all the other ones that changed, despite not getting their names changed.Steve3007 wrote:What David appears to be claiming is that replacing CA with a clock that is moving differently from CA, relative to the other 4 clocks, (CA2) changes the things that he can say about those other 4 clocks' relative tick rates.
No, I don't accept that that is a valid analogy for this situation. If we had opposite definitions of the words "black" and "white" but we were both consistent in our own uses of those words, there would be no problem. That would be like us having opposite, but internally consistent, definitions of the letters 'E' and 'L' here. That's not the issue.So you say that this swan is white and not black, but David says that it is black, not white because he labels the colors differently, and thus your statement is a contradiction.Steve3007 wrote:If either speaker accuses the other of a logical contradiction, then this is not a function of their respective languages. Either they are or they aren't.
As I say, I disagree that the "consistently swapping the names of colours" analogy applies here.The mistake is David using his labels for the colors in interpreting your words, and your mistake is using your definitions of those colors when interpreting David's words. Contradictions only appear when not using the language of the author of the statements made. I found no contradictions in the experiment 4's as described in David's post.
I agree that there is no relationship between CA and CA2. He could also have added CA3, CA4 etc. They are separate experiments on CD, CE, CL and CS with those 4 clocks moving at the same relative speeds though. Since the mathematical relationships between tick rates that he states are all comparisons (equalities and inequalities), each between 2 of those 4 clocks, he acknowledges that it is these relative speeds that we are considering.David does not say this. The two experiments 4 and 42 are entirely separate experiments (I see now why they're both called 4) and there is no relation between CA and CA2 since they don't both appear in any one experiment.Steve3007 wrote:When we've distilled experiment 4, leg 1, down to a consideration of 2 clocks separating at speed v (with, or without, another clock, CA/CA2, flying around) David's claim is that there is some objective sense in which one of those clocks ticks faster than the other, or vice versa, and which depends on the movements of that other (A) clock relative to the clocks being considered.
They are wrong in any language because, when following the logical rules of language, they are logically contradictory. Similarly, the statements:How are they wrong in his languageSteve3007 wrote:They are wrong in any language.
One = Two
and
Une = Deux
are both wrong.
Similarly, if I decide that black means white and white means black, and I stick consistently to that, there is no problem. But that's not the situation here.
Yes "L>D and D>L" is a selfcontradiction. That would also be true if we swapped the meanings of those symbols (your black/white analogy) and said "D>L and L>D", or if we replaced them with other symbols. In the context of demonstrating David's logical contradictions I do not claim that L and D are undefined meaningless values. I provisionally accept them as meaningful and then show the logical contradiction.The terms have meaning there, and L>D is true in the 42 experiment, and it would indeed contradict with D>L. In our language, L and D are undefined meaningless values, so statements made about them are meaningless and cannot contradict.
 Halc
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Re: Does Special Relativity contain contradictions?
Or maybe he did. At the bottom he states (in a list of 6 such relationships) that "CE moves at speed 0 relative to CL during leg 1." and that these relationships 'apply to both frames', a very nonabsolute way of putting things.
That is still a little different than saying they're stationary relative to each other, but more than I caught on the first read.
The above applies to both frames. If you want to include the superfluous arena clocks for the two frames:
His logic seems to revolve around tick rates, not statements of relative speed. Same thing, but one layer removed, which is somehow more offensive...Only via CA. But if we can't draw that conclusion from those two premises then we've departed from logic as I understand it.
Still attempting it. He explicitly calls the two experiments 'different frames' for the same experiment, not two separate experiments. Perhaps I got that wrong.I admire your attempts to try to work out what he's on about. But I'm still not getting it.
You'll have to realize as well that I'm sort of reading this the first time. I've not been following the long posts very closely. It sounds like more was to be said about all these clocks moving everywhere, but things seem to have been mired in these early differences.
Well, I was clued off about this designation of E as the tick rate of clock CE. There is no such tick rate in the standard interpretation. It is frame dependent, so there is only Ee and Ed, which are different things.That's what I thought he appeared to be saying because it's the only explanation I can think of for the fact that he makes two opposite inequality claims about 4 identically moving clocks. I guess I just assumed that when he refers to (for example) CL in the version of experiment 4 containing CA and in the version containing CA2 that he's referring to the same clock each time. If he isn't then, again, he's departing from the logical rules of language as I understand them. If you said "1 + 1 = 2 and 2 + 2 = 4" I'd assume that the symbol '2' doesn't change its meaning between those two statements.
Maybe he does say this. That little summary describes CA and CA2 as different frames for the same experiment. I was just noting that everything was always described relative to CA or CA2, and never to anything else. That's a valid thing to do if they're just frames, and not representations of the current state of reality.Halc wrote:David does not say this. The two experiments 4 and 42 are entirely separate experiments (I see now why they're both called 4) and there is no relation between CA and CA2 since they don't both appear in any one experiment.
Sorry, but I seem to have lost your claims of David making logical contradictions. I thought David claims that we're the ones making logical contradictions.Yes "L>D and D>L" is a selfcontradiction. That would also be true if we swapped the meanings of those symbols (your black/white analogy) and said "D>L and L>D", or if we replaced them with other symbols. In the context of demonstrating David's logical contradictions I do not claim that L and D are undefined meaningless values. I provisionally accept them as meaningful and then show the logical contradiction.

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Re: Does Special Relativity contain contradictions?
That's all that happens during the first leg, but after a while CS chases after CE and CL returns to CD. We know from the final scores that D>L and E>S overall (and we don't know if D>E or E>D), but what we're interested in is the ticking rates during the first leg. The above shows a naming problem though which is enabling people to sow confusion, so let's rename some of the tick rates. D and E remain fine for both legs individually and combined as clocks CD and CE don't change their motion at any time during the experiment, but we need the names S1 and S2 for CS's tick rates on the 1st & 2nd legs, and SA for the average of both legs, and we need L1 and L2 for CL's tick rates on the 1st & 2nd legs, and LA for the average of both.
I can now clarify the final scores for the stretch of time between clocks separating and being reunited: D>LA and E>SA. These are facts  they do not vary with frame. The other facts we have are that D=S1 and that E=L1  they do not vary with frame either. Denying these facts would be denying the best measurements that can be made, and that would be a rejection of physics. In addition to the facts, we have measurements made by comparing the apparent ticking rates of clocks while they are separated by distance and not comoving, but these change depending on which frame is used to make those measurements, so they are not facts  they merely describe how things appear from specific frames.
Because E>SA, we know that either S1 < E or S2 < E (or that both S1 < E and S2 < E) .
Because D>LA, we know that either L1 < D or L2 < D (or that both L1 < D and L2 < D).
Those too are facts. It is mathematically impossible for them not to be correct. Denying them would be a rejection of mathematics. We are getting into conditional facts with "if"s and "else"s, but they are mathematical facts regardless of how unfamiliar you may be with the mathematical rules of logical reasoning. We can continue to derive further facts from the above:
If D > L1, then S1 > E.
If E > S1, then L1 > D.
From these, we can see that any attempt to have both D > L1 and E > S1 results in contradiction. This is a really simple mathematical proof. Your trick of changing frame to analyse the two parts of the same experiment is not legal mathematically because it generates contradictions, and it isn't legal in physics either for the same reason. No one in physics should be backing it.
In frame CD, D>L1 and S1>E.Steve3007 wrote: ↑October 5th, 2018, 8:07 pmSince, in leg 1, CD = CS (= CA) and CL = CE, the above is really just one inequality concerning 2 clocks separating at speed v. CD to the left. CL to the right. So you're simply claiming that there is some sense in which one of those clocks is ticking faster than the other. You've presumably tossed a coin and decided that it is the one that is moving to the left: CD....D>L, S>E.
In frame CE, E>S1 and L1>D.
The point is that if frame CD's account is true, then frame CE's account is false, and if frame CE's account is true, then frame CD's account is false. Tossing a coin doesn't determine which account is true, but pretending that there isn't an underlying reality can't make both accounts true. D>L1 and E>S1 are fundamentally incompatible  it's one or other; mathematically impossible for it to be both. You want it to be both, but you can't have that without rejecting mathematics, and if you do that, you've ceased to do real physics.
Why are you rejecting mathematics? You've been categorically proved wrong  every competent mathematician on and off the planet will back my position on this and tell you that you've got it wrong. The top physicists will also back my position on this if they are forced to choose, because to do otherwise would destroy their reputation. When the maths tells you you're wrong, you need to listen.Wrong.That was because you changed frame between experiments, and changing frame is an illegal move if you're going to combine measurements from two frames.
I am using frameinvariant measurements for my argument. In all frames, D=S1, E=L1, E>SA and D>LA. I don't depend on any other measurements at all for this. The fact that the relative movements of the clocks and frames are the same for the single experiment that we're doing here is no surprise at all  how could they be anything otherwise! Look at the maths and learn from it. The measurements from different frames which disagree with each other (most notably in relative tick rates) demonstrate that most of those measurements are misleading, so we cannot rely on any of them. The measurements that I rely on are the ones that are facts for all frames. I then take you out of your comfort zone into conditional facts, and that moves us into the mathematics of logical reasoning, although it is pretty standard maths when you say things like if a is a whole number less than 10 and greater than 8, it must be 9, so I don't see why you should be so troubled by it.The reference frames that matter are those against which measurements are made, not the one against which we decide to describe relative speeds. What I did in the post to which you objected was to show that the relative movements of those reference frames are the same for all 3 experiments.
You're trying to divert things into a place where the contradictions are hidden. That's why I keep pointing you back to the two places where you can't hide from them: the relative ticking rates of clocks and the unhappening/rehappening of events. Both of these things destroy the set 2 models. The reason you lot put up such a determined defence of these invalidated models though is that you depend on model mixing and muddying the waters in order to hide the fact that you don't have any functional SR model. The set 2 ones are gone, and the set zero ones don't contain any time at all  just a pretence of time in the form of a misnamed space dimension in a static block. None of those models are viable. The set 1 models have eventmeshing failures, but you can't even bring yourself to look at them, even though the key part of them in which some clocks travel more quickly into the future than others (no clocks run slow in these models) is often attributed to SR, but it can only happen that way in set 1 models, so again you rely on model mixing (mixing incompatible models to pretend that an attribute of set 1 models that's incompatible with all the other sets operates in set zero) to try to maintain the pretence that you have a working SR model. All you're left with is set 3 models with absolute frames, and one of those is LET. The other has a whole stack of superfluous junk in it and every point in Spacetime is zero distance away from every other point in Spacetime for light which never travels any distance at all and which therefore has no speed. Every single one of your models is a mathematical contrivance which adds complexity rather than reducing it.It's like this: Consider car A moving along road A at 60mph relative to that road. Consider car B moving along road B at 60mph relative to that road. If road A is road B, then the experiments are identical. If road A is moving at constant velocity relative to road B, then the experiments are identical. That is what you appear to be denying.
Of course they're the same as they were before  that's my point. You seemed to imagine that my argument depended on there being a difference there, but it doesn't. The contradictions don't show up there  they show up when you apply reasoning, investigating the conditional facts, and this is something you have failed to do, because if you had, you'd have found the asymmetries and contradictions for yourself.The relative velocities of CD, CE, CL and CS are the same in both legs as they were before. Various other clocks, moving relative to these clocks in various ways, may or may not be present. They are irrelevant to our measurements.With experiment 4, it becomes easy to see why changing frame is cheating. Let's change frame and see what it does to experiment 4. The change of frame forces us to use CA2 instead of CA in order to maintain full symmetry (with CA2 being at rest in this new frame):
// Experiment 4
CD moves at v relative to CA2 throughout.
CE stands still relative to CA2 throughout.
CL stands still relative to CA2 during leg 1 and moves faster (negatively) than v relative to CA2 during leg 2.
CS moves at v relative to CA2 during leg 1 and at v relative to CA2 during leg 2.
Using the new names, that's D=S1, L1=E, E>S1, L1>D. (For the previous frame, we had D=S1, L1=E, D>L1, S1>E.)OK, so as above CD = CS and CL = CE. So you have 2 clocks (CD and CL) separating at speed v, as before. You've tossed your coin again and this time you've decided that it's the one moving to the right that is, in some sense, ticking faster. What is it? Heads = left, tails = right?The tick rates during the first leg are as follows: D=S, L=E, E>S, L>D.
You're the one who wants to toss the coin, but you also want it to land both heads and tails at the same time, and that's the place where you make your monumental error. In the underlying reality, only one of the two options can be right. It is mathematically impossible to have both E>S1 and D>L1, but so long as you deny this, you will continue to try to have your cake and eat it. You are tolerating contradictions.
We have frameindependent facts from measurements that show that D>LA and E>SA, so we know for fact that some clocks tick at faster rates than others. Denying that is denying the facts. I have not made any arbitrary decision about which clocks are ticking faster than which at any time  I cover all the possible options (including the possibility not mentioned directly here in which D=L1 and D>L2, and an equivalent case where E=S1 and E>S2). There is a set of possible realities and I have addressed ALL of them. That is what we do when we go through all the "if x then y else z" options. These are standard nuts and bolts of mathematics, and you really ought to be able to apply them.No you are the one asserting relative tick rates here. As I've said, when we gather together all clocks that are comoving and treat them as single clocks then it's simply 2 clocks separating at speed v in leg 1. You have arbitrarily decided that the left moving clock ticks fastest or the right moving clock ticks fastest with no physical difference between the two, accept for an irrelevant CA or CA2 (or any number of other irrelevant clocks) floating around.The reason your frame change is an illegal move is that you're taking D>L from the first frame and S<E from the second.

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Re: Does Special Relativity contain contradictions?
There's no deception  there's only one experiment in experiment 4, but we're looking at it through the "lens" of two different frames, one of which has CD at rest, while the other has CE at rest. We have one set of events with a single reality (which is necessarily the underlying reality).
I haven't pinned an aether to any frame here. I've left that entirely open (though its existence is discovered by this along the way without me needing to introduce it). I remind you: we have two key facts that D > LA and E > SA, which proves that some clocks tick more quickly than others. This gives us the right (and duty) to consider the possibilities as to how quickly the clocks were ticking relative to each other during the first or second legs. We know that it's impossible for CL to tick more slowly than CD overall without also ticking more slowly than CD on at least one of the two legs of its trip. We thus have a right (and duty) to consider the possibility that D > L1, and when we do that, we find that if D>L1, then E<S1. We also have the right (and duty) to consider the other two available options that D=L1 and that D<L1, so let's do those as well. If D<L1, then S1<E. If D=L1, then S1=E (and S2<E). None of these options allow both D>L1 and E>S1 to be true at once  if one is true, the other is necessarily false.Halc wrote: ↑October 6th, 2018, 9:21 amSo switching to standard definitions here, we can get rid of the Ether reference A which appears to be David's addition, and just have clocks CD CE CL CS. D E L and S have no meaning, we need to designate our tick rates as something like De which means the tick rate of CD in the frame in which CE is stationary.
 Halc
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Re: Does Special Relativity contain contradictions?
OK, I didn't read it that way first time. I'm not used to you actually referring to things relative to frames.David Cooper wrote: ↑October 6th, 2018, 5:47 pmThere's no deception  there's only one experiment in experiment 4, but we're looking at it through the "lens" of two different frames, one of which has CD at rest, while the other has CE at rest.
And as I said, I deny the meaningfulness of those so called facts. Yes, at the end when CD and CL are reunited, L has logged few ticks. That part is a fact, and observed in any frame. The actual tick rates L and D are meaningless without a frame, and nonconstant since L accelerates.I remind you: we have two key facts that D > LA and E > SAHalc wrote:D E L and S have no meaning, we need to designate our tick rates as something like De which means the tick rate of CD in the frame in which CE is stationary.
Which can be interpreted as some clocks ticking faster than others. Another interpretation is that they all tick at the same rate and there's just less duration between measurement events. Either interpretation seems valid. I prefer the latter because it prevents distant clocks from ticking backwards.which proves that some clocks tick more quickly than others.
I'm not sure why all the fancy frames and accelerations and such if that's all you wanted to demonstrate. Clearly two clocks simply on different floors of a building 'tick at different rates' as you put it, and in a frameindependent way no less. Even in your view this is true. I'm not disputing that. Clocks clearly don't measure the flow of time or this would not be the case. There might be flow, but nothing seems to measure/detect it.

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Re: Does Special Relativity contain contradictions?
Yes, and it is specifically to that leg that you have referred when you give us your proposed equalities (=) and inequalities (>). In that leg, you have claimed that in two identical experiments involving 2 pairs of clocks (CD and CS, CE and CL) the question of which pair ticks faster than the other pair depends on the movements of another clock (with an A in its name). That is your claim, in a nutshell.David Cooper wrote:That's all that happens during the first leg,Steve3007 wrote:You're just describing two clocks moving apart from each other at speed v. See earlier explanations.
Here is the diagram I drew earlier to remind us of the relative movements of the clocks. Look at the lefthand column. Note that the relative movements of CD, CE, CL and CS are identical in the top and bottom rows and that the only reason the arrows look different is that in the diagram we are drawing their movements relative to clocks CA and CA2 respectively. The fact that we decide to draw this diagram like this does not change any relationship between CD, CE, CL and CS. Drawing diagrams to show speeds relative to various CAx clocks does not constitute "changing frames" or changing the experiment. If I had drawn the diagram showing relative speeds compared to another clock called CA3 which is moving at some other speed relative to the other 4, that would not change things either. Can you accept that?
And here are your claims as a reminder:
// Experiment 4
CD stands still relative to CA throughout .
CE moves at speed v relative to CA throughout.
CL moves at speed v relative to CA during leg 1 and at v relative to CA during leg 2.
CS stands still relative to CA during leg 1 and moves at a speed much higher than v relative to CA during leg 2.
The tick rates during the first leg are as follows: D=S, L=E, D>L, S>E.
From here:// Experiment 4
CD moves at v relative to CA2 throughout.
CE stands still relative to CA2 throughout.
CL stands still relative to CA2 during leg 1 and moves faster (negatively) than v relative to CA2 during leg 2.
CS moves at v relative to CA2 during leg 1 and at v relative to CA2 during leg 2.
The tick rates during the first leg are as follows: D=S, L=E, E>S, L>D.
viewtopic.php?p=321139#p321139

My own view as to what happens to 2 clocks that separate from each other at relative speed v and then reunite at relative speed v is covered in detail here:
viewtopic.php?p=321015#p321015

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Re: Does Special Relativity contain contradictions?
Returning to the subject of the Equivalence Principle between observers in uniform gravitational fields and observers in accelerated reference frames, and posts such as these:
viewtopic.php?p=321092#p321092
viewtopic.php?p=321136#p321136
and this article:
https://thecuriousastronomer.wordpress. ... elativity/
that I cited earlier. I've come to the conclusion that the article is essentially correct. The maths in it seems correct. Although the wording may not be ideal.
In the part entitled "General Relativity" at the end of this post...
viewtopic.php?p=321015#p321015
...I said that the "twin paradox" can be understood in terms of GR as well as being understood in terms of SR.
As I said, understanding it in terms of SR makes use of the idealization that the traveler instantaneously switches inertial reference frames and spends zero time in a noninertial reference frame. That really means that he enters a noninertial reference frame for a vanishingly small period of time during which his acceleration tends to infinity. There's nothing wrong with that line of reasoning. It uses the concept of a "limit", in which we say "it is possible to allow the time to go arbitrarily close to zero". Perfectly good physics. But another line of reasoning, using GR, doesn't need to allow the time spent in a noninertial reference frame to tend towards zero. It can consider what happens if the time spent in the noninertial reference frame, and the acceleration of that reference frame, are both finite  not zero or its reciprocal, infinity. In so doing, it makes use of the Equivalence Principle.
The important thing to note here, as explained in detail in that previous post, is that all of this is about what the observers calculate based on the definition of simultaneity. It is not about the raw sensory information that they observe directly. The analysis using SR shows, using a Minkowski diagram, that observer B calculates there to be a sudden jump of 15 years in the Aticks that he deems to be simultaneous in time with his Bticks. That calculation is based on the definition of simultaneity. The suddenness of this jump reflects the idealization that I described above  the discontinuous change in velocity; the zero time and infinite acceleration. When this discontinuity is replaced with a curve, the sudden jump is replaced by a rapid, but not infinitely rapid, change. The traveler (B) doesn't calculate the Aticks to jump from 2.5 to 17.5 years. He calculates them to transition rapidly between those two values. The greater the acceleration, the more rapid the transition. This rapid transition is a result of the accelerational equivalent of gravitational time dilation, due to the fact that B is accelerating.
In the GR analysis, the red and blue lines in the Minkowski diagram that represent the B frames' notion of simultaneity no longer suddenly change angle when going from red to blue. They do it rapidly, but not suddenly. So the effect of the acceleration is to progressively alter the difference in the line of simultaneity between observers A and B, not to suddenly change it.
This is the same as what happens in the accelerating rocket in the article that I cited. The equivalence principle means that this rocket can be regarded as stationary in a uniform gravitational field. In any real gravitational field (such as that of the Earth) this means being confined to a small space relative to the distance to the centre of the Earth. i.e. it means being "local". So that the difference in gravity between the top and bottom of the rocket can be neglected.
In this scenario, another clock that is in freefall next to the rocket is inertial; it is stationary WRT to an inertial frame. That is a consequence of this Equivalence Principle  that freefalling observers are inertial. So it's a bit like our observer A. If we imagine that clock falling past the rocket, we can see that, from its point of view, the clock in the top of the rocket is moving slower than the clock in the bottom of the rocket. i.e. those two clocks are instantaneously in two different inertial reference frames. And therefore the top one (by the freely falling clock's calculations) is calculated to be ticking faster than the bottom one. So the lines of simultaneity are different. The clock in the top of the rocket calculates the freely falling clock to be ticking more slowly than him. The clock in the bottom of the rocket calculates the freely falling clock to be ticking even more slowly than him.
I think this shows the way that our temporarily (but not momentarily) accelerating traveler (B) changes his calculations about simultaneity as he accelerates. With the result that the Aticks that he regards as being simultaneous with the small number of Bticks that he experiences while accelerating rapidly (but not instantly) transition from 2.5 to 17.5 years, while B directly experiences his own Bticks move through a small (but not infinitesimally small) range of numbers centred around 5 years.
 Halc
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Re: Does Special Relativity contain contradictions?
The article seems to merely demonstrate that the two clocks in the rocket get out of sync due to being compared in a different inertial frame. That is explained by SR, not GR. They're dilated only because they're moving in other frames, and they're dilated equally since they both move at the same speed, never mind that they also accelerate equally.Steve3007 wrote: ↑October 7th, 2018, 7:36 amGeneral Relativity
Returning to the subject of the Equivalence Principle between observers in uniform gravitational fields and observers in accelerated reference frames, and posts such as these:
viewtopic.php?p=321092#p321092
viewtopic.php?p=321136#p321136
and this article:
https://thecuriousastronomer.wordpress. ... elativity/
that I cited earlier. I've come to the conclusion that the article is essentially correct. The maths in it seems correct. Although the wording may not be ideal.
I thought of a way to do the centrifuge experiment. Put the one clock on the equator (moving at 450km/sec linear speed relative to Earth) but not accelerating much. Put the centrifuge at the North Pole where linear speed is only from the centrifuge, and also 450km/sec, as tight a radius as we can build. Same speed and gravitational potential, so those should cancel out. The only difference would be the acceleration. I say the clocks stay in sync. The articles that seem to address this point say this, but there is a lot of misunderstanding in the forums, and I admit to not being an expert.
The article you quoted misstates the principle of equivalence: "Einstein’s principle of equivalence tells us that whatever is true for acceleration is true for a gravitational field."
That is just wrong, so I question the rest of the article. The principle says there is no local test to determine which situation you're in.
SR handles gradual acceleration just fine. It just requires more calculus to compute the net effect of the gradually changing reference frame. SR just doesn't handle gravity.As I said, understanding it in terms of SR makes use of the idealization that the traveler instantaneously switches inertial reference frames and spends zero time in a noninertial reference frame. That really means that he enters a noninertial reference frame for a vanishingly small period of time during which his acceleration tends to infinity.
SR does not posit any dilation during the limit of 'infinite acceleration to some different finite velocity in zero time'. When that acceleration happens, no clock has changed. All events are the same. We're just looking at the identical situation from a new reference frame. Syncedbutseparated clocks are not in sync in other frames, but that isn't dilation, and that's all that the article above seems to be noting.
The important thing to note here, as explained in detail in that previous post, is that all of this is about what the observers calculate based on the definition of simultaneity. It is not about the raw sensory information that they observe directly. The analysis using SR shows, using a Minkowski diagram, that observer B calculates there to be a sudden jump of 15 years in the Aticks that he deems to be simultaneous in time with his Bticks. That calculation is based on the definition of simultaneity.[/quote]
Right. Neither clock was dilated due to that change of abstract reference frames. The new frame simply pairs two different events as simultaneous.
Acceleration had nothing to do with it. It was the frame change. In the new frame, the same two events (U5/B17.5) are simultaneous whether or not the one ship actually accelerated to be at rest in that frame or not. What the ship did changed nothing.The suddenness of this jump reflects the idealization that I described above  the discontinuous change in velocity; the zero time and infinite acceleration.
I know what you're saying, but this isn't a real change to any clock like what gravity does to it. You can tell in this instance because the remote clock is running faster (a lot faster) in the accelerating frame of the ship, and gravitational dilation slows a clock down.When this discontinuity is replaced with a curve, the sudden jump is replaced by a rapid, but not infinitely rapid, change. The traveler (B) doesn't calculate the Aticks to jump from 2.5 to 17.5 years. He calculates them to transition rapidly between those two values. The greater the acceleration, the more rapid the transition. This rapid transition is a result of the accelerational equivalent of gravitational time dilation, due to the fact that B is accelerating.
Agree, but it is still a SR analysis. All this can be worked out with only SR rules.In the GR analysis, the red and blue lines in the Minkowski diagram that represent the B frames' notion of simultaneity no longer suddenly change angle when going from red to blue. They do it rapidly, but not suddenly. So the effect of the acceleration is to progressively alter the difference in the line of simultaneity between observers A and B, not to suddenly change it.
In this scenario, another clock that is in freefall next to the rocket is inertial; it is stationary WRT to an inertial frame. That is a consequence of this Equivalence Principle  that freefalling observers are inertial.[/quote]Free falling objects accelerate in a gravitational field, and are thus not inertial. The principle just says that the two observers cannot tell which is which: Stationary out in nowhere or accelerating to a gravity source. Local experiments behave the same in both cases.This is the same as what happens in the accelerating rocket in the article that I cited. The equivalence principle means that this rocket can be regarded as stationary in a uniform gravitational field. In any real gravitational field (such as that of the Earth) this means being confined to a small space relative to the distance to the centre of the Earth. i.e. it means being "local". So that the difference in gravity between the top and bottom of the rocket can be neglected.
How do you figure this one?? A clock falling past an accelerating rocket is stationary in some inertial frame in which the rocket had been stationary at some point, and at the point where the two ship clocks had been synced. In that frame, the two clocks maintain identical velocity the whole time, and stay perfectly synced the whole time, although their nonzero velocity causes them to both fall equally behind the stationary clock being passed.So it's a bit like our observer A. If we imagine that clock falling past the rocket, we can see that, from its point of view, the clock in the top of the rocket is moving slower than the clock in the bottom of the rocket. i.e. those two clocks are instantaneously in two different inertial reference frames.
That might be off a bit. If the rocket gets up to a good speed, its length contracts (in the original halted frame) and technically the clock on the bottom has moved a little further to partly catch up with the top one. That slight velocity difference would have the bottom clock running a tiny bit slower.
I can't agree with any of this. It seems to make no sense, but that is hardly evidence that it's wrong. From the POV of the accelerating clocks, it would be quite helpful to say when these relative tick rate assessments are done, because due to acceleration of both these clocks, 'simultaneously' has no fixed meaning. It has meaning for the stationary clock being passed, and hence my answer above, but this one is ambiguous. So simultaneously as determined how? The tick rates of the clock being passed changes continuously, so there is no single calculation of this rate.And therefore the top one (by the freely falling clock's calculations) is calculated to be ticking faster than the bottom one. So the lines of simultaneity are different. The clock in the top of the rocket calculates the freely falling clock to be ticking more slowly than him. The clock in the bottom of the rocket calculates the freely falling clock to be ticking even more slowly than him.

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Re: Does Special Relativity contain contradictions?
Halc wrote:The article seems to merely demonstrate that the two clocks in the rocket get out of sync due to being compared in a different inertial frame. That is explained by SR, not GR.
Yes that's right. But GR didn't spring from nowhere. It developed by careful consideration of the implications of SR combined with the equivalence of inertial and gravitational mass. The gravitational time dilation in GR is a consequence of these kinds of considerations in SR. Like this:
In GR, the Schwarzschild solution to Einstein's field equations which tells us the gravitational time dilation in a radial gravitational field is:
1.
t_{0} = t_{f} sqrt(1  2GM / rc^{2})
In this equation t_{0} is the coordinate time an infinite distance from the gravitating body (e.g. Earth) and t_{f} is the coordinate time at distance r from the body's centre. (G = Universal Gravitational Constant. M = mass of the gravitating body.)
But it can be seen how it derives from things like equation 8 in that article I cited. That equation is:
2.
DeltaT_{B} = DeltaT_{A}(1  gh/c^{2})
It is (as you've said) the result of simply considering the implications of SR in the accelerating frame of the rocket.
We can use Integral Calculus to integrate the equation over a vertical, radial path through the gravitational field over which the value of 'g' varies according to Newton's Law of Universal Gravitation:
3.
g = G M/r^{2}
Integration, of course, means adding up a series of infinitesimal steps. As we know, the equivalence between acceleration and gravity works locally because the type of gravitational field to which acceleration is equivalent is a uniform one. So we take a series of infinitesimal sections, within each of which the gravitational field is uniform (because it is local), and integrate over a finite distance. When we do that, the infinitesimal quantity 'h' becomes the 'dr' in the integral over the finite r. i.e. it becomes the step in the integral; the amount by which we climb up through the gravitational field in each step of the integration. So, substituting from equation 3 for the 'g' in equation 2, we have:
4.
dt = 1  ((GM dr) / (c^{2}r^{2}))
So, we have a differential equation in t and r. i.e. we have an equation that tells us the rate at which the time dilation varies with respect to radial distance from the centre of the gravitating mass. Integrating over a vertical path in a radial gravitational field we get something that starts to look like the Schwarzschild solution in GR:
(Remembering that the integral of r^{2} is r^{1})
5.
t = 1 + GM / (rc^{2}) + C
That's similar to the first term in the binomial expansion of the square root in the Schwarzschild solution, equation 1.

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Re: Does Special Relativity contain contradictions?
The facts are extremely meaningful. The only reason you're able to justify saying they aren't is by pointing to other models, but as I've said dozens of times, this argument about the contradictions applies to set 2 and 3 models where time runs and where some clocks are able to run slower than others. When we are dealing with those models, the set 2 ones are invalidated by contradictions. The relative average tick rates between D and L for the whole trip are frame invariant.Halc wrote: ↑October 6th, 2018, 8:42 pmAnd as I said, I deny the meaningfulness of those so called facts. Yes, at the end when CD and CL are reunited, L has logged few ticks. That part is a fact, and observed in any frame. The actual tick rates L and D are meaningless without a frame, and nonconstant since L accelerates.I remind you: we have two key facts that D > LA and E > SA
...as with set 2 and 3 models.Which can be interpreted as some clocks ticking faster than others.which proves that some clocks tick more quickly than others.
That takes you into set 1 models where D = S1 = S2 = SA = L1 = L2 = LA = E = 1. The result of this is that the model has to tolerate eventmeshing failures. Switching to that set of models does not rescue set 2 models  they remain invalidated by contradictions.Another interpretation is that they all tick at the same rate and there's just less duration between measurement events. Either interpretation seems valid. I prefer the latter because it prevents distant clocks from ticking backwards.
The third interpretation takes you into set zero models where D = S1 = S2 = SA = L1 = L2 = LA = E = 0. Again, switching to that set of models does not rescue set 2 models  they remain invalidated by contradictions.
If you'd stop mixing models and trying to keep broken ones alive by pointing to unbroken equivalent aspects of rival, incompatible models, then you'd stand a better chance of getting the point. If you have a car with no engine, you can't magically make it function just by pointing at a different car (with no wheels) that does have an engine and saying, "Look  it works!" It isn't the same car, and the parts don't transfer because they're different makes of car.I'm not sure why all the fancy frames and accelerations and such if that's all you wanted to demonstrate.
When gravity slows the speed of light, a clock will run slow, underrecording the amount of time that has passed for it. If you want to claim that clocks at all heights in a gravity well tick at the same rate as each other and that some simply follow shorter paths into the future, that again takes you into set 1 models with eventmeshing failures  it does not apply to set 2 and 3 models where eventmeshing failures are banned. Any attempt to model things with no clocks running slow will display eventmeshing failures (unless you just look at a postconstruction block universe, but if you don't account for the construction phase, you're missing the most important part of the physics and your model is inadequate).Clearly two clocks simply on different floors of a building 'tick at different rates' as you put it, and in a frameindependent way no less. Even in your view this is true. I'm not disputing that.
Causality tells us that there is flow, and time is locked to causality, flowing with it.Clocks clearly don't measure the flow of time or this would not be the case. There might be flow, but nothing seems to measure/detect it.

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Re: Does Special Relativity contain contradictions?

 Posts: 5721
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 Location: UK
Re: Does Special Relativity contain contradictions?
As I explained, in order to switch between two inertial reference frames without spending finite (nonzero) time in a noninertial reference frame, the SR explanation uses the limit of a switch between those frames in a vanishingly small time period with an acceleration, during that time period, that tends to infinity. It's the kind of consideration of limits that is often used in physical theories that are special cases of a more general theory. In this case, the special case is SR and the general case is GR. That's why they have those names.Halc wrote:SR does not posit any dilation during the limit of 'infinite acceleration to some different finite velocity in zero time'.

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Re: Does Special Relativity contain contradictions?
Which clock with an A in its name? I'm not using CA and CA2  I only mentioned them because you wanted to introduce a start line clock, but I don't need them. CD and CE do the same job already, making CA and CA2 superfluous. It may be that you're referring to LA and SA, but these are tick rates for CL and CS. CL has tick rate L1 on the first leg of its trip, L2 on the second leg, and LA is its average tick rate for the whole trip. CS likewise has tick rate S1 on the first leg of its trip, S2 on the second leg, and SA is its average tick rate for the whole trip.Steve3007 wrote: ↑October 7th, 2018, 3:16 amYes, and it is specifically to that leg that you have referred when you give us your proposed equalities (=) and inequalities (>). In that leg, you have claimed that in two identical experiments involving 2 pairs of clocks (CD and CS, CE and CL) the question of which pair ticks faster than the other pair depends on the movements of another clock (with an A in its name). That is your claim, in a nutshell.David Cooper wrote:That's all that happens during the first leg,
When we compare clocks CD and CL at the end of the experiment, we find that D > LA (which means that the ticking rate of CA > the average tick rate of CL for the whole trip. In the same way, we have E > SA (which means that the ticking rate of CE > the average tick rate of CS for the whole trip. There are only four clocks in my thought experiment.
I've never claimed that doing anything with those superfluous clocks changes anything. We have a single experiment being done with only four clocks in it. When we change frame, none of the facts that my argument is built upon change  they are frame invariant.The fact that we decide to draw this diagram like this does not change any relationship between CD, CE, CL and CS. Drawing diagrams to show speeds relative to various CAx clocks does not constitute "changing frames" or changing the experiment. If I had drawn the diagram showing relative speeds compared to another clock called CA3 which is moving at some other speed relative to the other 4, that would not change things either. Can you accept that?
Don't take the D>L and S>E part of that as my claim  they are claims that are generated by using that frame and they cannot be trusted as they are not frameinvariant. Ditto for the E>S and L>D from the other frame. However, part of that can be trusted, because if D>L, then S>E. Also, if E>S, then L>D. Those conditional claims are mine, and they are mathematical requirements.And here are your claims as a reminder:
// Experiment 4
CD stands still relative to CA throughout .
CE moves at speed v relative to CA throughout.
CL moves at speed v relative to CA during leg 1 and at v relative to CA during leg 2.
CS stands still relative to CA during leg 1 and moves at a speed much higher than v relative to CA during leg 2.
The tick rates during the first leg are as follows: D=S, L=E, D>L, S>E.
The initial facts are:
D=S1
E=L1
D>LA
E>SA
The second pair of facts there depends on time running in a set 2 or 3 manner, so this proof is specific to those models. If you try to apply the facts to set 1 or set models, then using ">" and "<" is automatically wrong there, but that doesn't prevent this proof from destroying all set 2 models.
Mathematics allows us to derive the following further facts from the above ones:
If D>L, then S1>E (which necessarily means that: if D>L1, then not E>S1)
If E>S1, then L>D (which necessarily means that: if E>S1, then not D>L1)
You want both D>L1 and E>S1 to be true at the same time, but that is mathematically impossible (in set 2 or 3 models).
This argument about contradictions relates solely to the demolition of set 2 models in which time runs but clocks on different paths do not tick at the same rate as each other at all times (because that would take us into set 1). These models generate contradictions and are thus invalidated; removed from the enquiry. Extinct. We have a dead parrot.