A constant speed of light in all reference frames? Surely you can't be serious.
in [Relativity]
|
From: Peter Webb on 6 Mar 2010 21:51 "Ste" <ste_rose0(a)hotmail.com> wrote in message news:651a713d-7ae4-4048-bafb-f1b3219ee4fc(a)v20g2000yqv.googlegroups.com... > On 6 Mar, 12:47, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> > wrote: >> > This should make perfect sense to you. If a clock is running 2% >> > slower, then it is running 2% slower regardless of distance. But if, >> > as a result of running 2% slower, it falls behind 6 minutes after >> > running a certain amount of time, then it will fall behind 12 minutes >> > after running for twice as long. >> >> Agreed. >> >> The question now is, if we agree that both clocks suffer time dilation >> in this way, then when they return to the start point, how do they >> each reconcile the fact that (after accounting for the effects of >> acceleration) it ought to be the other clock which is slow, when in >> fact one clock (the one that went furthest from the start point) will >> be slower than the other? And a third clock, left at the start point, >> will be running ahead of both? >> >> _________________________ >> They know that the operations were not symmetric. Only one clock remained >> in >> the same inertial reference frame throughout. The other two clocks spent >> different amounts of time in at least 3 different inertial reference >> frames. >> Everybody can see this is true, and so nobody expects that the clocks >> will >> remain synchronised. > > Yes, but the important question here is whether they agree *after* the > effects of acceleration are taken into account. I mean, if we said > that each travelling clock slows by 2% when moving away from the start > point at a certain speed, then by rights both travelling clocks should > slow equally. Yes? > As I understand your thought experiment, no. In SR, time dilation is a function of relative speed and the time for which they are moving at the speed. It is not a function of accleration. A doesn't move. B moves at speed v for time t, and its clock will read x behind A. C moves at speed v for time 2t, and its clock will read 2x behind A. > And yet, the assertion seems to be that each clock will consider > itself correct, while holding that the other clock has slowed by, > what, 4%? No, not as I understand your question, anyway. BTW the clocks do not *consider* themselves correct; unless they are broken they *are* correct. There is no absolute time to use as a reference.
From: BURT on 6 Mar 2010 22:18 On Mar 6, 6:51 pm, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > "Ste" <ste_ro...(a)hotmail.com> wrote in message > > news:651a713d-7ae4-4048-bafb-f1b3219ee4fc(a)v20g2000yqv.googlegroups.com... > > > > > > > On 6 Mar, 12:47, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> > > wrote: > >> > This should make perfect sense to you. If a clock is running 2% > >> > slower, then it is running 2% slower regardless of distance. But if, > >> > as a result of running 2% slower, it falls behind 6 minutes after > >> > running a certain amount of time, then it will fall behind 12 minutes > >> > after running for twice as long. > > >> Agreed. > > >> The question now is, if we agree that both clocks suffer time dilation > >> in this way, then when they return to the start point, how do they > >> each reconcile the fact that (after accounting for the effects of > >> acceleration) it ought to be the other clock which is slow, when in > >> fact one clock (the one that went furthest from the start point) will > >> be slower than the other? And a third clock, left at the start point, > >> will be running ahead of both? > > >> _________________________ > >> They know that the operations were not symmetric. Only one clock remained > >> in > >> the same inertial reference frame throughout. The other two clocks spent > >> different amounts of time in at least 3 different inertial reference > >> frames. > >> Everybody can see this is true, and so nobody expects that the clocks > >> will > >> remain synchronised. > > > Yes, but the important question here is whether they agree *after* the > > effects of acceleration are taken into account. I mean, if we said > > that each travelling clock slows by 2% when moving away from the start > > point at a certain speed, then by rights both travelling clocks should > > slow equally. Yes? > > As I understand your thought experiment, no. > > In SR, time dilation is a function of relative speed and the time for which > they are moving at the speed. It is not a function of accleration. > > A doesn't move. B moves at speed v for time t, and its clock will read x > behind A. C moves at speed v for time 2t, and its clock will read 2x behind > A. > > > And yet, the assertion seems to be that each clock will consider > > itself correct, while holding that the other clock has slowed by, > > what, 4%? > > No, not as I understand your question, anyway. > > BTW the clocks do not *consider* themselves correct; unless they are broken > they *are* correct. There is no absolute time to use as a reference.- Hide quoted text - > > - Show quoted text - If clocks slow down they must start slowing from some fastest tick. This would corespond to zero gravity and zero motion for the clock. The two universal rates must go fastest at the beginning of time. Mitch Raemsch
From: BURT on 6 Mar 2010 22:47 On Mar 4, 10:14 am, bert <herbertglazie...(a)msn.com> wrote: > On Mar 4, 1:10 pm, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > On Mar 4, 1:03 pm, JT <jonas.thornv...(a)hotmail.com> wrote: > > > > On 4 mar, 18:49, mpalenik <markpale...(a)gmail.com> wrote: > > > > > On Mar 4, 12:46 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > On Mar 4, 11:17 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > On 4 Mar, 16:49, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > On Mar 4, 11:45 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > On 4 Mar, 16:32, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > On Mar 4, 11:28 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > On 4 Mar, 16:20, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your > > > > > > > > > > > > > > > > distance from the clock. The effect is entirely apparent in SR. > > > > > > > > > > > > > > > > You must just go through the entire thread and not pay any attention > > > > > > > > > > > > > > > to what anybody says. Ever. > > > > > > > > > > > > > > > > 1) What you've stated above is not an effect of SR. It is an effect > > > > > > > > > > > > > > > of propagation delay, which was used to calculate c from the motion of > > > > > > > > > > > > > > > the moons of jupiter hundreds of years ago. > > > > > > > > > > > > > > > Ok. > > > > > > > > > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run > > > > > > > > > > > > > > > faster. But SR says nothing about whether you are moving toward or > > > > > > > > > > > > > > > away from an object. > > > > > > > > > > > > > > > <suspicious eyebrow raised> Ok. > > > > > > > > > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT > > > > > > > > > > > > > > > from the amount that SR predicts the clock *actually* slows down > > > > > > > > > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR > > > > > > > > > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown? > > > > > > > > > > > > > > And for example, if we racked up the value of 'c' to near infinity, > > > > > > > > > > > > > > would SR still predict an "actual" slowdown, even though the > > > > > > > > > > > > > > propagation delays would approach zero? > > > > > > > > > > > > > > With what you have described, I checked just to be sure, even though I > > > > > > > > > > > > > was already pretty sure what the answer would be, the time you read > > > > > > > > > > > > > moving away the clock would be: > > > > > > > > > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x > > > > > > > > > > > > > > and when you move toward the clock > > > > > > > > > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x > > > > > > > > > > > > > > so moving away from the clock: > > > > > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > > > > > and toward > > > > > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > > > > > > Special relativity predicts that the moving clock will always slow > > > > > > > > > > > > > down as > > > > > > > > > > > > > dt2/dt = sqrt(1-v^2/c^2) > > > > > > > > > > > > > > What you *measure* is a combination of the actual slow down predicted > > > > > > > > > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation > > > > > > > > > > > > > delays (which depend on the direction of motion). > > > > > > > > > > > > > Ok. So let us suppose that we take two clocks. Separate them by a > > > > > > > > > > > > certain distance, synchronise them when they are both stationary, and > > > > > > > > > > > > then accelerate them both towards each other (and just before they > > > > > > > > > > > > collide, we bring them stationary again). Are you seriously saying > > > > > > > > > > > > that both clocks report that the other clock has slowed down, even > > > > > > > > > > > > though they have both undergone symmetrical processes? Because there > > > > > > > > > > > > is obviously a contradiction there. > > > > > > > > > > > > Yes, that is correct. Both will report a slow down.. And in fact, > > > > > > > > > > > which ever one breaks the inertial frame to match speed with the other > > > > > > > > > > > is the one that will be "wrong". This is still within the realm of > > > > > > > > > > > SR, not GR. > > > > > > > > > > > What if they both "break the inertial frame"? > > > > > > > > > > Then whichever frame they both accelerate into will be the one that > > > > > > > > > has measured the "correct" time dilation. > > > > > > > > > So in other words, the clocks will register the same time, but will > > > > > > > > have slowed in some "absolute sense"? > > > > > > > > Yes--assuming they both accelerated by the same amount (that is to > > > > > > > say, assuming they both broke the inertial frame in a symmetric way). > > > > > > > Otherwise, they will register different times. > > > > > > > Agreed. > > > > > > > So let's explore an extension of this scenario. Let's say you have two > > > > > > clocks, and you accelerate both of them up to a common speed, and > > > > > > after they have travelled a certain distance, you turn them around and > > > > > > return them to the starting point. The only difference is that one > > > > > > clock goes a certain distance, and the other clock goes twice that > > > > > > distance, but they *both* have the same acceleration profile - the > > > > > > only difference is that one clock spends more time travelling on > > > > > > inertia. > > > > > > > Obviously, one clock will return to the starting point earlier than > > > > > > the other. But when both have returned, are their times still in > > > > > > agreement with each other, or have they changed? > > > > > > Agreement. Both of them will agree, but will be showing a time earlier > > > > > than a third clock that was left behind at the starting point. > > > > > Wait, maybe I'm confused by Ste's setup. Didn't he say that one > > > > travels twice as far as the other? But then he also says that you > > > > turn them both around and return them to the start after traveling a > > > > certain distance. Have they moved different distances in his scenario > > > > or not?- Dölj citerad text - > > > > > - Visa citerad text - > > > > lol you framejumping grasshoppers have just have no idea what is > > > ***REALLY*** going on have you. Don't forget u can always use the > > > fudgefactor. > > > > JT > > > You figured me out! Damn it. I guess the days of the lie are over.. > > Pretty soon the physicists absolute control over government, politics, > > and economics will come to an end. Damn you for uncovering our > > secret. Damn you all to hell!- Hide quoted text - > > > - Show quoted text - > > Photon has set speed period TreBert- Hide quoted text - > > - Show quoted text - If light has a set speed of C it has kinetic energy of C always. But this is wrong. Its energy comes from its wave frequency and not its constant motion. Mitch Raemsch
From: Y.Porat on 7 Mar 2010 01:00 On Mar 4, 8:27 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Mar 4, 10:24 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 4 Mar, 15:54, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Mar 4, 1:03 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > This is what fundamentally sets apart things like creationism from > > > > > > science. Whatever other hoops creationism manages to jump through, it > > > > > > will never jump through the hoop of naturalism, and that is what > > > > > > *fundamentally* sets it apart from science. > > > > > > And also FUNDAMENTALLY distinguishes science from religion. Thanks. > > > > > Agreed, but then religion in general never claimed to be science, > > > > Agreed! And so science is not a religion in the same fashion. > > > No, but neither did one religion ever claim to be the other. > > > > > and > > > > traditional religion is almost immediately identifiable by its > > > > supernaturalism. Creationism is different in that it actually claims > > > > to be scientific in some essential respects. > > > > Ah, yes, but as has been demonstrated even to layfolk (Dover v > > > Kitsmiller), this is an unsupportable claim. > > > I agree. I'm glad you brought up that case. I just reviewed the > > judgment quickly, and apparently the court agrees that the defining > > essence of science is naturalism. > > Gee, I didn't read that into the judgment at all. ----------------- what is all that nonstop spamming about 'A constant speed of light ' !!! who is the crook behind it ??!! Y.P -------------------
From: Inertial on 7 Mar 2010 02:03
"Y.Porat" <y.y.porat(a)gmail.com> wrote in message news:4f714a7d-ddca-4f64-9063-ce901648e6a1(a)z35g2000yqd.googlegroups.com... > On Mar 5, 3:31 am, "Inertial" <relativ...(a)rest.com> wrote: >> "Ste" <ste_ro...(a)hotmail.com> wrote in message >> >> news:8c0ae071-8d13-491b-92d0-cd2e2727af1a(a)u9g2000yqb.googlegroups.com... >> >> > On 4 Mar, 12:19, "Inertial" <relativ...(a)rest.com> wrote: >> >> "Ste" <ste_ro...(a)hotmail.com> wrote in message >> >> >> > Not really, because if the total acceleration is small, then so is >> >> > the >> >> > speed. >> >> >> That is a nonsense argument. Acceleration can be small and speeds >> >> very >> >> large. >> >> > When I went to school, you could not have a large change of speed with >> > only a small amount of total acceleration. >> >> Then you were badly taught. >> >> a) if you start at speed 0.8c and acceleration at 0.00001 m/s/s .. then >> your >> speed is still large. you claimed small acceleration means small speed >> >> b) if you start at speed 0.0 and acceleration at 0.00001 m/s/s .. then >> your >> speed after a million years will be quite fast. Yet the acceleration was >> small and constant. >> >> You do realize that you cannot 'total' acceleration. and acceleration of >> 1m/s/s followed by an acceleration of 1m/s/s is still an acceleration of >> 1m/s/s > > ------------------ > psychopath > ----------------------------- Senile old stalker. Get a life anda hobby you are good at .. or die. One of the two. But you are a failure at physics |