A constant speed of light in all reference frames? Surely you can't be serious.
in [Relativity]
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From: Androcles on 5 Mar 2010 00:27 "Bruce Richmond" <bsr3997(a)my-deja.com> wrote in message news:7806715d-93d2-49ff-ad67-6dac8ea64d8c(a)e7g2000yqf.googlegroups.com... On Mar 4, 10:48 am, PD <thedraperfam...(a)gmail.com> wrote: > On Mar 3, 10:59 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > > > > On Mar 3, 11:21 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Mar 2, 8:12 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > > > It is not a function of finite propagation speeds, this we know, > > > > > because we took into account the finite propagation speeds in our > > > > > procedure for determining simultaneity/nonsimultaneity. Do you not > > > > > remember that? > > > > > I beg to differ. It is not a "mere" or "simple" function of finite > > > > propagation speed, but it *is* a function of it IMO. RoS only took > > > > it > > > > into account by allowing us to use different time coordinates in > > > > each > > > > frame. If the speed of light was infinite there would be no RoS. > > > > I disagree. All that is needed in relativity of simultaneity is a > > > signal speed that can be VERIFIED to be the same from both events by > > > either observer. > > > Well you are going to have problems with that. There is no way to > > *know* that the speed is the same both ways. > > Yes, there is. That's what isotropy experiments have determined. I'm > surprised you weren't aware of this. > Do you mean like this one? http://mysite.verizon.net/cephalobus_alienus/papers/Gagnon_et_al_1988.pdf The author is kind enough to point out problems in some similar experiments, while failing to notice any in his own. For example, has he made any assumption about contraction of his equipment in the direction of motion? Tom Roberts has written posts in this group showing where some of these experiments are in effect two way measurements. > > > > > That is why Einstein > > wrote, "But it is not possible without further assumption to compare, > > in respect of time, an event at A with an event at B. We have so far > > defined only an ``A time'' and a ``B time.'' We have not defined a > > common ``time'' for A and B, for the latter cannot be defined at all > > unless we establish by definition that the ``time'' required by light > > to travel from A to B equals the ``time'' it requires to travel from B > > to A." > > > > Since the distance from the events to the observer is > > > equal, as verifiable at any time by each observer, we learn from this > > > that each observer KNOWS the propagation delays from each event to the > > > observer are equal. This acknowledges the propagation delays > > > completely, but simply allows for verification that they are the same. > > > Assuming the speed of light is the same in both directions. > An apple is an orange assuming both are fruit. [(c+v)/c] * [(c-v)/c] = (c+v)(c-v)/c^2 = [c^2 -v^2]/c^2 = c^2/c^2 - v^2/c^2 = 1 - v^2/c^2 > Which is an experimentally confirmed fact. Done well after Einstein's > comments on the matter, by the way. > Provide a link to an experiment and I'll take a look. > > > > > > > > Then the determination of simultaneity or nonsimultaneity of the > > > original events is completely unambiguous: If the observer receives > > > both signals at the same time, then (because the propagation delays > > > are the same) the original events were simultaneous; if the observer > > > receives both signals at different times, then (because the > > > propagation delays are the same) the original events were > > > nonsimultaneous. > > > > Then the frame-dependence of simultaneity follows directly from the > > > experimental *observation* that for the same pair of events, one > > > observer correctly and unambiguously concludes the events were > > > simultaneous, and the other observer correctly and unambiguously > > > concludes the events were nonsimultaneous. > > > > You've mentioned in the past that you found your disbelief in > > > relativity stems from being unable to find a good, understandable > > > explanation of it. I invite you to read back on this thread where I > > > was trying to explain to Ste (who has a similar complaint) how this > > > comes about. > > > No need. Lorentz showed how all frames could measure the speed of > > light to be c. That in effect confirms the second postulate, which is > > the stumbling block for many. > > Well then, you are just *choosing* what you would like to believe. In > this case, lodging a complaint against relativity that it is not well > explained, when you are not interested in pursuing a better > explanation, having settled on LET instead, is a bit on the > disingenuous side. You are reading more into that than what I wrote. I am not choosing LET over SR. They use the same math ================================================== I am not choosing division over multiplication, they are identical. l = l0 * sqrt(1-v^2/c^2) http://en.wikipedia.org/wiki/Lorentz_ether_theory xi = (x-vt) / sqrt(1-v^2) http://www.fourmilab.ch/etexts/einstein/specrel/www/ I am not choosing an idiot over an imbecile, they are different.
From: Peter Webb on 5 Mar 2010 01:04 > You are reading more into that than what I wrote. I am not choosing > LET over SR. They use the same math > ================================================== Well, e = mc^2 is maths. It appears in SR, but not LET. So I guess you were wrong, and they don't use the same maths.
From: Ste on 5 Mar 2010 03:55 On 4 Mar, 18:17, mpalenik <markpale...(a)gmail.com> wrote: > On Mar 4, 1:12 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > On 4 Mar, 17:20, mpalenik <markpale...(a)gmail.com> wrote: > > > > On Mar 4, 12:02 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 4 Mar, 16:29, 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. > > > > > > I don't know what happened to this thread, or why google registers > > > > > several different copies of this thread, each with only a few messages > > > > > in it, but I'm going to describe the thought experiment I posted > > > > > before in greater detail. > > > > > > In order for the ^ to receive the pulse of light from both emitters at > > > > > the same time and half way between the two emitters, the light must be > > > > > emitted BEFORE the ^ is half way between the two emitters. > > > > > Indeed. > > > > > > From the stationary frame, the two emitters emit light at the same > > > > > time and the ^ moves vertically until it is half way between the two > > > > > of them, when it receives both pulses. > > > > > Ok. > > > > > > But from the moving frame, the ^ sees the emitters emit light when it > > > > > is not half way between them, and then it sees the emitters move until > > > > > he is exactly half way between them. However, the speed of light is > > > > > the same in all frames, and unaffected by the motion of the emitters. > > > > > > So the the emitter emit while one is closer to him than the other.. > > > > > The fact that they are moving does not affect the way light > > > > > propagates. If the light signal from the closer one reaches him at > > > > > the same time as the light signal from the farther one, that means the > > > > > farther one must have emitted first. The fact that he is half way > > > > > between the emitters at the end doesn't matter, because in his frame, > > > > > it is the emitters that are moving, which does not affect the speed of > > > > > light. > > > > > This seems implausible. You cannot possibly (that is, physically) have > > > > the situation you describe, in the way you describe it. The receiver, > > > > when it receives the pulse of light, cannot possibly be in more than > > > > one position relative to the two sources. > > > > It isn't. > > > > > It either receives when the > > > > sources are equidistant, or it receives when one source is closer. > > > > The sources are equidistant. > > > > > It > > > > cannot receive when it is equistant *and* when one source is closer, > > > > for that would be an obvious contradiction. > > > > What you don't understand is the fact that the sources were not > > > equidistant when they emitted the pulse. > > > Indeed, but it's neither here nor there where the sources where at the > > time of emission. > > Not in the rest frame. But it is in the moving frame. > > > > > The question is where the sources are at the time of reception, > > Not in the ^ frame. The fact that the sources have moved does not > affect the way the light propagates. They could run over the moons of > jupiter and back, and it wouldn't affect the way the light propagates. > > Lets say that you have two sets of emitters, A and B. One set is > moving and the other set is not. If they both emit > > They both emit like this > A B > > A B > > But the B set is moving like this > > A > B > > A > B > > The light from the A set and B set propagates exactly the same way. > It doesn't matter that the B set has moved. So what you're saying (and I had recognised this problem before you said it) is that it is the "original" position of emission that matters? And the "original" position changes depending on the frame (i.e. in the source frame, the source does not move, whereas in the receiver frame, the sources are constantly moving from their "original" positions)?
From: Ste on 5 Mar 2010 03:55 On 4 Mar, 18:12, PD <thedraperfam...(a)gmail.com> wrote: > On Mar 4, 12:04 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > On 4 Mar, 17:46, 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. > > > Oh dear. Mark contends otherwise. > > Right. I misunderstood. He's right. I was wrong. Ok. So what you're (both) saying is that time dilation (in SR) is a simple function of speed and distance, so that the quicker you travel the more time dilates, and the further you travel the more time dilates? And, to boot, you're saying that it's only *relative* distance and speed that counts (i.e. there is no absolute measure of movement in space)?
From: Ste on 5 Mar 2010 03:57
On 5 Mar, 01:31, "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 In any event, we've resolved the meaning of "total acceleration" - Mark suggests using the concept of "impulse" instead. |