A constant speed of light in all reference frames? Surely you can't be serious.
in [Relativity]
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From: Inertial on 5 Mar 2010 06:51 "Y.Porat" <y.y.porat(a)gmail.com> wrote in message news:868926cb-233d-417e-86c8-cd8987c43419(a)q16g2000yqq.googlegroups.com... > On Mar 4, 7:44 pm, PD <thedraperfam...(a)gmail.com> wrote: >> On Mar 4, 11:09 am, Ste <ste_ro...(a)hotmail.com> wrote: >> >> >> >> > On 4 Mar, 16:48, mpalenik <markpale...(a)gmail.com> wrote: >> >> > > On Mar 4, 10:19 am, Ste <ste_ro...(a)hotmail.com> wrote: >> >> > > > 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. >> >> > > The problem is your use of the term "total acceleration". If by >> > > total >> > > acceleration, you mean integral(a dt), then yes, you are correct. >> > > However, there is already a word for integral(a dt) -- it's called >> > > "the change in velocity". The term "total acceleration" isn't >> > > actually defined. Acceleration is defined, velocity is defined, >> > > deltav is defined. But "total acceleration is not". >> >> > Essentially, I'm defining "total acceleration" as something akin to >> > total force, so that even though the force may be small, if it >> > continues for a long time then the total force will be the same as if >> > a large force was applied for a short period of time. In this way, if >> > the application of force is what is causing either part or the whole >> > of the time dilation effect, then it is the final speed that counts, >> > not how quickly the object reached that speed. >> >> Indeed. This should tell you that it is not the details of the >> acceleration that matter. >> The overly simplistic statement would be, "Yes, you see that is why >> SR's effects are based on speed, not on acceleration." >> >> In fact, there is a speed time dilation effect on GPS satellites, >> which are going around in a circular path at constant speed, relative >> to earth clocks, and accounting for this is crucial to their proper >> operation. This is the same speed dilation effect, though different >> size, as seen in muons in a circulating ring. (Since, by the way, the >> GPS satellites are certainly not inside a magnetic ring but still >> experience time dilation properly calculated by SR, this is another >> good way to be sure that the magnetic ring is not what's responsible >> for the time dilation of the muons.) >> >> Regarding something I alluded to earlier, though, what really matters >> is how straight the path through spacetime is. We're used to thinking >> that the shortest path through space is the straight one (and that's >> right), but the straightest path through spacetime yields the LONGEST >> duration. Any change in motion (such as an acceleration) introduces a >> kink in this path (something that can be illustrated visually very >> easily) and so lowers the duration. Why this is, has to do with the >> structure of spacetime and we could discuss that. But this is perhaps >> the most intuitive way (once these concepts are explained) to >> understand why the traveling twin returns younger. >> >> >> >> > > Also, you could just be dealing with a system where the velocity >> > > started out high and you never measured any acceleration. >> >> > Indeed. > > ---------------------- > (:-) > to mix **biologic process** with > inorganic physics > is ridiculous!!! He didn't .. there was no biologic process mentioned in the above. > (i said it in a big understatement ...(:-) > Y.Porat > -----------------------
From: mpalenik on 5 Mar 2010 09:46 On Mar 5, 3:55 am, Ste <ste_ro...(a)hotmail.com> wrote: > 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)?- Hide quoted text - > > - Show quoted text - Right. The sources send out one pulse at one particular point in time. The only thing that matters is where they were located when they sent out that pulse. That location is the "source" of the pulse.
From: mpalenik on 5 Mar 2010 09:49 On Mar 5, 3:55 am, Ste <ste_ro...(a)hotmail.com> wrote: > 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)?- Hide quoted text - > Well, it's really only the relative velocity that matters. Distance doesn't enter into the equations at all. It's just that time slows down for the moving observer. So, lets say that an observer is moving so that his clock has slowed down by a factor of two. After 10 seconds, his clock is 5 seconds behind. After 20 seconds, his clock is 10 seconds behind. After 30 it is 15 seconds behind, etc. Distance is not part of the equations, although the longer you are moving, also, the farther you will have gone with respect to a stationary observer. And yes, it is only the relative velocities that matter.
From: jem on 5 Mar 2010 09:58 mpalenik 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. Actually, according to SR, the two (ideal) clocks will be synchronized when they meet, regardless which clock "breaks the inertial frame" via an instantaneous acceleration. That final acceleration, which occurs when the clocks are co-located, has no effect on the clock readings. Reported clock "slow downs" will be exactly offset by clock speed-ups during the initial accelerations*. * For clarity, both effects are purely observational - SR presumes (ideal) clock mechanisms are completely unaffected by a clock's motion.
From: PD on 5 Mar 2010 10:59
On Mar 5, 4:55 am, "Inertial" <relativ...(a)rest.com> wrote: > "Ste" <ste_ro...(a)hotmail.com> wrote in message > > news:057b5351-82a4-4487-8501-6308451c921a(a)x22g2000yqx.googlegroups.com... > > > > > 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. > > You certainly are taking the long and painful route (for yourself and us) to > learn the basics of physics. Ste: This is exactly what I was telling you earlier, that people will be less inclined to teach things on your terms, using your language and indulging your lack of skills, and will advise you that it is more efficient in the long run to teach after you've acquired some relevant skills and vocabulary. You didn't seem to think this was the case, and here you have others telling you the same thing. Reconsider? > It would have helped if either: > > a) you came here asking questions in order to help you learn > > b) you understood enough of physics before making bold statements, mostly > wrong, and criticizing science. > > Do you understand yet how time dilation and length contraction are both just > 'side effects' (or consequences) of simultaneity not being absolute. > > I can give you a simple example of how length contraction and time dilation > comes about simply from synchronizing clocks differently if you like. |