Known physics defeated by simple puzzle?
in [Relativity]
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From: Inertial on 10 Apr 2010 00:51 "Ste" <ste_rose0(a)hotmail.com> wrote in message news:e1d473d9-bd92-4373-ac6c-4ec08f157f25(a)z6g2000yqz.googlegroups.com... > On 10 Apr, 02:26, "Inertial" <relativ...(a)rest.com> wrote: >> >> > Incidentally, if you have two clocks a certain distance apart, >> > synchronised (obviously, accounting for propagation delays), >> >> Yes .. that is always assumed. >> >> > and >> > stationary relative to each other, >> >> Yes >> >> > what happens when you accelerate >> > them towards each other. Does the distant clock appear to slow down, >> > or speed up? >> >> Acceleration complicates things uncesssarily .. so lets assume no >> acceleration for simplicity. > > Well, I asked the question precisely because I wanted an answer to > *that* scenario. If the clocks are already moving (and don't stop > before passing each other), then that really tells me nothing about > what I wanted to know. Then you don't understand the scenario > The purpose of having the clocks stopped at the > beginning and end is that it allows a 'simple' correction for any > propagation delay when testing for synchronisation. That is exactly WHY my scenario below has clocks stopped at the beginning >> An equivalent set up is this, with no acceleration invovled, where A, B, >> A' >> and B' are all clocks. >> >> A'->v B'<-v >> A o B >> >> Let A and B be our mutually-at-rest, synchronized clocks (as you >> mentioned) >> >> Let A' and B' be moving at the same speed (but opposite directions) >> relative >> to A and B. >> >> As A' passes A and B' passes B (at the same time according to A and B), >> we >> copy the reading from A clock to A' clock, and copy the reading from B >> clock >> to B' clock. A' and B' keep moving and arrive together at o, where there >> times are compared. > > As I say, I'd also like to discuss the specific scenario that I > raised. Same thing happens > Because at least in my scenario, we can agree that they are > both synchronised at the start, as in my simpler scenario > and at the end, as in my simpler scenario > and that we have > accounted for propagation delays when testing for synchronisation. as in my simpler scenario (that is implicit in the definition of syncrhonisation) > So > the question is how to interpret what happens in the middle, but > obviously I need you to describe what happens. Its the same as in my scenario. But in my case there is no need to talk about what happens during acceleration, nor specify how long they are accelerating for etc etc. The acceleration itself isn't the important concept here .. it is the change of rest inertial reference frames. >> After the pass A and B and have their clock readings adjusted, the an A' >> observer would measure B' as ticking slowly (as I described above), and a >> B' >> observer would measure A' as ticking slowly. They will both also see the >> other as showing the 'wrong' time (in particular, from what I think (not >> done the calculations), each frame will measure the other clock as being >> ahead of their own). > > Each frame will measure the other clock as being *ahead*? That is what I said > By any > chance, does the other clock always appear slow I just said it would measure it as ahead. We are not talking about appearances and optical illusions > when they are sailing into the distance away from each other, > and always appear fast when approaching each other? We are not talking about appearances and optical illusions >> > And do their times match when they meet up and are >> > brought back to stationary again? >> >> Yes. > > Ok.
From: Peter Webb on 10 Apr 2010 01:05 "Ste" <ste_rose0(a)hotmail.com> wrote in message news:e1d473d9-bd92-4373-ac6c-4ec08f157f25(a)z6g2000yqz.googlegroups.com... On 10 Apr, 02:26, "Inertial" <relativ...(a)rest.com> wrote: > > > Incidentally, if you have two clocks a certain distance apart, > > synchronised (obviously, accounting for propagation delays), > > Yes .. that is always assumed. > > > and > > stationary relative to each other, > > Yes > > > what happens when you accelerate > > them towards each other. Does the distant clock appear to slow down, > > or speed up? > > Acceleration complicates things uncesssarily .. so lets assume no > acceleration for simplicity. Well, I asked the question precisely because I wanted an answer to *that* scenario. If the clocks are already moving (and don't stop before passing each other), then that really tells me nothing about what I wanted to know. The purpose of having the clocks stopped at the beginning and end is that it allows a 'simple' correction for any propagation delay when testing for synchronisation. _______________________________________________ For SR, you can assume that the change in speed is instantaneous, eg the speed goes from V to 0 or -V in zero time. While physical objects can't do that, reference frames certainly can. Contrary to what I suspect you believe, the "twin's paradox" is *not* caused by acceleration of the travelling twin directly; it is caused by changes in the reference frame in which the elapsed time is measured, as any non-crank site on the twins paradox will explain. So you can assume that changes in velocity are instantaneous. > An equivalent set up is this, with no acceleration invovled, where A, B, > A' > and B' are all clocks. > > A'->v B'<-v > A o B > > Let A and B be our mutually-at-rest, synchronized clocks (as you > mentioned) > > Let A' and B' be moving at the same speed (but opposite directions) > relative > to A and B. > > As A' passes A and B' passes B (at the same time according to A and B), we > copy the reading from A clock to A' clock, and copy the reading from B > clock > to B' clock. A' and B' keep moving and arrive together at o, where there > times are compared. As I say, I'd also like to discuss the specific scenario that I raised. Because at least in my scenario, we can agree that they are both synchronised at the start, and at the end, and that we have accounted for propagation delays when testing for synchronisation. So the question is how to interpret what happens in the middle, but obviously I need you to describe what happens. ____________________________________________ As I understand your experiment, when the clocks are separating they each see the other clock as running more slowly, and when they are approach they see the other clock as ticking faster, and the percentage change is the same as that given by the formula for the Relativistic Doppler Shift. You can confirm that when you add the outbound and inbound changes together, they are both re-synchronised when they meet again, as a simple symmetry argument requires. > After the pass A and B and have their clock readings adjusted, the an A' > observer would measure B' as ticking slowly (as I described above), and a > B' > observer would measure A' as ticking slowly. They will both also see the > other as showing the 'wrong' time (in particular, from what I think (not > done the calculations), each frame will measure the other clock as being > ahead of their own). Each frame will measure the other clock as being *ahead*? By any chance, does the other clock always appear slow when they are sailing into the distance away from each other, and always appear fast when approaching each other? ______________________________ Yes. > > And do their times match when they meet up and are > > brought back to stationary again? > > Yes. Ok. _________________________________ Indeed.
From: Inertial on 10 Apr 2010 01:15 "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message news:4bc0072f$0$5523$afc38c87(a)news.optusnet.com.au... > > "Ste" <ste_rose0(a)hotmail.com> wrote in message > news:e1d473d9-bd92-4373-ac6c-4ec08f157f25(a)z6g2000yqz.googlegroups.com... > On 10 Apr, 02:26, "Inertial" <relativ...(a)rest.com> wrote: >> >> > Incidentally, if you have two clocks a certain distance apart, >> > synchronised (obviously, accounting for propagation delays), >> >> Yes .. that is always assumed. >> >> > and >> > stationary relative to each other, >> >> Yes >> >> > what happens when you accelerate >> > them towards each other. Does the distant clock appear to slow down, >> > or speed up? >> >> Acceleration complicates things uncesssarily .. so lets assume no >> acceleration for simplicity. > > Well, I asked the question precisely because I wanted an answer to > *that* scenario. If the clocks are already moving (and don't stop > before passing each other), then that really tells me nothing about > what I wanted to know. The purpose of having the clocks stopped at the > beginning and end is that it allows a 'simple' correction for any > propagation delay when testing for synchronisation. > > _______________________________________________ > For SR, you can assume that the change in speed is instantaneous, eg the > speed goes from V to 0 or -V in zero time. While physical objects can't do > that, reference frames certainly can. Contrary to what I suspect you > believe, the "twin's paradox" is *not* caused by acceleration of the > travelling twin directly; it is caused by changes in the reference frame > in which the elapsed time is measured, as any non-crank site on the twins > paradox will explain. So you can assume that changes in velocity are > instantaneous. > > > > >> An equivalent set up is this, with no acceleration invovled, where A, B, >> A' >> and B' are all clocks. >> >> A'->v B'<-v >> A o B >> >> Let A and B be our mutually-at-rest, synchronized clocks (as you >> mentioned) >> >> Let A' and B' be moving at the same speed (but opposite directions) >> relative >> to A and B. >> >> As A' passes A and B' passes B (at the same time according to A and B), >> we >> copy the reading from A clock to A' clock, and copy the reading from B >> clock >> to B' clock. A' and B' keep moving and arrive together at o, where there >> times are compared. > > As I say, I'd also like to discuss the specific scenario that I > raised. Because at least in my scenario, we can agree that they are > both synchronised at the start, and at the end, and that we have > accounted for propagation delays when testing for synchronisation. So > the question is how to interpret what happens in the middle, but > obviously I need you to describe what happens. > > ____________________________________________ > As I understand your experiment, when the clocks are separating they each > see the other clock as running more slowly, Ticking at the same rate, but see a later time. That is an illusion,of course > and when they are approach they see the other clock as ticking faster, There is some optical illusion of that, the actual measurement is slower. > and the percentage change is the same as that given by the formula for the > Relativistic Doppler Shift. You can confirm that when you add the outbound > and inbound changes together, they are both re-synchronised when they meet > again, as a simple symmetry argument requires. Yeup >> After the pass A and B and have their clock readings adjusted, the an A' >> observer would measure B' as ticking slowly (as I described above), and a >> B' >> observer would measure A' as ticking slowly. They will both also see the >> other as showing the 'wrong' time (in particular, from what I think (not >> done the calculations), each frame will measure the other clock as being >> ahead of their own). > > Each frame will measure the other clock as being *ahead*? By any > chance, does the other clock always appear slow when they are sailing > into the distance away from each other, and always appear fast when > approaching each other? > > ______________________________ > Yes. > > >> > And do their times match when they meet up and are >> > brought back to stationary again? >> >> Yes. > > Ok. > > _________________________________ > Indeed. >
From: Inertial on 10 Apr 2010 02:11 "Ste" <ste_rose0(a)hotmail.com> wrote in message news:da85279f-24ef-481b-93d0-287ce74db45d(a)y14g2000yqm.googlegroups.com... > On 10 Apr, 04:49, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> > wrote: >> I have some sympathy for your problem. I have no sympathy for the fact >> that >> instead of trying to understand how the Universe actually works, your >> position is that SR is wrong and/or Inertial, PD and I am wrong because >> we >> agree with SR and SR is wrong. > > I do wish you'd stop framing the issue as "you're either with us or > against us". There is obviously some validity to SR - and I've said it > a million times. It cannot be only partly valid > For us to argue about what explanation underlies SR, > or how to interpret it, is not to say SR is "wrong". But you do not KNOW what Sr says. You make statements about it that are contradictory to what it actually says. There is no point talking about explanations when you don't knwo what is being explained to you > Incidentally, I > seem to remember that Terrell (of the "Terrell effect") commented that > a lot of scientists even in his time (i.e. 50 years after relativity > had been formulated) still had huge misconceptions about SR. And even > Newtonian mechanics has not been thrown out - merely reinterpreted to > show that it gives the correct answers only to certain circumstances > (which is not how it would have been interpreted in its heyday). It is approximately correct .. that is all. It is never completely correct except when there is no motion involved (which is rather pointless) > As I say Peter, you must stop equating my demands for an explanation > of SR, or disagreements with certain SR proponents here, as being a > sign of a fundamental opposition to SR. No .. it is a fundamental misunderstanding of SR. [ snip more useless rhetoric when your fundamental problems remain ]
From: Ste on 10 Apr 2010 02:16
On 10 Apr, 05:51, "Inertial" <relativ...(a)rest.com> wrote: > "Ste" <ste_ro...(a)hotmail.com> wrote in message > > >> > what happens when you accelerate > >> > them towards each other. Does the distant clock appear to slow down, > >> > or speed up? > > >> Acceleration complicates things uncesssarily .. so lets assume no > >> acceleration for simplicity. > > > Well, I asked the question precisely because I wanted an answer to > > *that* scenario. If the clocks are already moving (and don't stop > > before passing each other), then that really tells me nothing about > > what I wanted to know. > > Then you don't understand the scenario No, it's because I want specific answers to specific questions. There's no point responding to a scenario that I never asked questions about. > > The purpose of having the clocks stopped at the > > beginning and end is that it allows a 'simple' correction for any > > propagation delay when testing for synchronisation. > > That is exactly WHY my scenario below has clocks stopped at the beginning But I want them stopped at the end, too. > > So > > the question is how to interpret what happens in the middle, but > > obviously I need you to describe what happens. > > Its the same as in my scenario. But in my case there is no need to talk > about what happens during acceleration, nor specify how long they are > accelerating for etc etc. The acceleration itself isn't the important > concept here .. it is the change of rest inertial reference frames. It is not important to talk about how much acceleration or for how long - it suffices that they do accelerate for some period of time, at the start and at the end. The point is simply to talk in qualitative terms about what happens to "simultaneity" between the points at the start and at the end at which we agree that the clocks are synchronised. > >> After the pass A and B and have their clock readings adjusted, the an A' > >> observer would measure B' as ticking slowly (as I described above), and a > >> B' > >> observer would measure A' as ticking slowly. They will both also see the > >> other as showing the 'wrong' time (in particular, from what I think (not > >> done the calculations), each frame will measure the other clock as being > >> ahead of their own). > > > Each frame will measure the other clock as being *ahead*? > > That is what I said > > > By any > > chance, does the other clock always appear slow > > I just said it would measure it as ahead. > > We are not talking about appearances and optical illusions > > > when they are sailing into the distance away from each other, > > and always appear fast when approaching each other? > > We are not talking about appearances and optical illusions Just answer the questions Inertial - the posts will multiply beyond all conscionable length if you see every comma or full stop of mine as an invitation to reopen previous arguments or repeat your mantra. After all, of the four sentences you made above, consisting of two core meanings repeated, did either of them really have to be said? I know "what you said", and I responded with a question that incorporated what you said. It was also completely besides the point whether the effect here is apparent or real, illusion or no. So, whether it "appears" slow or is "really" slow, the question is does it slow down or speed up (whether really or apparently) depending on whether it is approaching or receding? In other words, if we moved the clocks in the opposite direction (starting near, and then receding to a point in the distance)? |