Known physics defeated by simple puzzle?
in [Physics]
|
From: JT on 18 Feb 2010 19:17 On 14 Feb, 20:42, Ste <ste_ro...(a)hotmail.com> wrote: > No takers for this simple question then? > > Consider this setup: > > S1 D2 > > D1 S2 > > We've got sources S1 and S2, paired with detectors D1 and D2. They're > all mechanically connected, so that a movement in one of them > produces > a movement in all the others - in other words, their relative > distances are always maintained. Each source is transmitting a > regular > pulse of light to its counterpart detector (so S1 is transmitting to > D1, etc.), and both sources are transmitting simultaneously with each > other. > > Now, we calculate that a pulse has just been emitted from both > sources, and we suddenly accelerate the whole setup "upwards" (i.e. > relative to how it's oriented on the page now) to near the speed of > light, and we complete this acceleration before the signals reach > either detector. > > Now, do both detectors *still* receive their signals simultaneously, > or does one receive its signal before the other? And are the signals > identical, or do they suffer from Doppler shifting, etc? Imagine this... Imagine object A and B travelling parallell vectors in space, A travels 0.1 c and B travel 0.9 c. For some reason the both pass lined up between sensor C and D at same moment x, when the sensor beams reach their front both ships emit one puls forward and one puls backward. The four lightpulses can not possible travel invariant thru the space of C and D, for them to travel invariant in C and D space the two backward pulses must travel aligned and parallell forever and so must the two front pulses. And if they do there something weird going on within A and B, especially the light do not spread uniform around B the light puls infront is contracted and expands at c-v=0.1 c relative restframe B in the space and the lightfront at back expands at 1.9 c relative B. The expansion of the two lightpulses is not uniform and invariant in frame B unless there is shorter meters at the front then at the back. At even higher velocities like 0.999... c the deformation is even clearer. For example consider that the two light pulses have been travelling for a year after B passed between C and D and emitted the two pulses now B suddenly come to halt/stop. Now anyone must surely realise that the pulses never traveled invariant at B and at speed c to begin with. One pulse is a lightyear away the other one is just in front off B. The assertions of SR is ridculous i would go so far to say they are a deliberate hoax. JT
From: Inertial on 18 Feb 2010 20:41 "Ste" <ste_rose0(a)hotmail.com> wrote in message news:42d85e4d-ca5c-464a-90ce-9a49386cd6dc(a)y33g2000yqb.googlegroups.com... > On 17 Feb, 21:15, "Inertial" <relativ...(a)rest.com> wrote: >> "Ste" <ste_ro...(a)hotmail.com> wrote in message >> >> >> Yes, and yes. >> >> > Ok. >> >> > Consider yet another setup: >> >> > D1 D3 >> >> > S1 D2 S2 D4 >> >> > In case it isn't clear, D1 and D2 are equidistant from S1, and D3 and >> > D4 are equidistant from S2. >> >> Again .. poorly 'drawn'. you show the distance S1..D2 as shorter than >> S2..D4. > > Indeed you are correct, but in any event that is why I included the > clarifying clause. Yeup .. just asking you to be a little more careful in your diagrams. >> You need to take a little more care. This is the second diagram that you >> have mis-drawn. > > I'm afraid I still can't see the problem with the first drawing. It > appears exactly as intended on my screen, although your requote of it > appears garbled. I looked at your original source .. you have an extra blank line in there .. that is all. Just need to be careful with ASCII diagrams .. They are difficult to draw, but it does help if you try to keep things consistent. Otherwise the diagram confuses rather than clarifies. >> > The S1 group (i.e. comprising S1, D1, and D2) are always stationary in >> > the frame, and S1 is emitting a pulse towards both D1 and D2. There is >> > a similar setup for the S2 group, except that after the emission of >> > the pulse, the S2 group moves in the y+ direction, so that S2 is now >> > in the former place of D3 (again, the whole group accelerates rapidly, >> > and is stationary again by time of detection). >> >> > Now, I assume D1 and D2 receive their signals simultaneously. But what >> > of D3 and D4? By the previous answers, I presume D4 receives it's >> > signal before D3? >> >> D4 will not receive the signal at all, because it is no longer in line >> with >> the direction of the signal (which went from left to right from the >> original >> location of S2. > > Indeed. > > So let's assume that the S2 group was already moving at the time of > emission, and that it continued to move. Then that is fine > Would the pulse *then* also > continue in the x+ direction (i.e horizontally across the screen) > towards the location of D4 *at the time of emission*, or would it take > a diagonal (or even curved) path to catch D4? (Again, the frame is > that of the stationary S1 group) Diagonal .. light gets it direction from the source .. just not its speed. Some people think that SR claims light does not get any velocity from the source .. That is not correct .. it is only the speed it does not get. Even that is not the best way to think of it, as it makes light sound like it is somehow 'special' .. velocities of ANYTHING in SR compose via the velocity composition rule (not by simple vector addition). And if you compose any velocity (v<c) with something with speed c, you get a velocity with speed c. So even though the source velocity does compose with that of the light emitted from it, the result is still a speed of c. So SR would say that ANY ballistically emitted object, with speed c relative to its source (if that was possible), would have a speed of c measured in every inertial frame. SR also says (or at least it is a consequence of it) that it is not possible to have any object with mass travel at c relative to another object.
From: Inertial on 18 Feb 2010 20:47 "Ste" <ste_rose0(a)hotmail.com> wrote in message news:739ba99d-f4b3-49bd-ace6-383980ee62a9(a)f8g2000yqn.googlegroups.com... > On 18 Feb, 18:40, PD <thedraperfam...(a)gmail.com> wrote: >> On Feb 18, 11:24 am, Ste <ste_ro...(a)hotmail.com> wrote: >> >> >> >> >> >> > On 18 Feb, 14:34, PD <thedraperfam...(a)gmail.com> wrote: >> >> > > On Feb 18, 8:25 am, Ste <ste_ro...(a)hotmail.com> wrote: >> >> > > > > Let's make this a little more concrete. Let's say that S1 and S2 >> > > > > flash, emitting pulses that radiate in all directions, and that >> > > > > the >> > > > > detectors intercept a portion of the light from those pulses. (If >> > > > > you >> > > > > want, think of S1 and S2 as firecrackers.) >> >> > > > No, because that needlessly complicates the matter. The important >> > > > thing is establishing where a single photon would land. >> >> > > Yes, it complicates the matter because you haven't specified which >> > > photon. Do you mean the one that is aimed at the detector when fired >> > > but does not hit it because of the acceleration of the detector, or >> > > do >> > > you mean the one that is not aimed at the detector when fired but >> > > DOES >> > > hit it because of the acceleration of the detector? This is >> > > important, >> > > you see, because if you restrict the attention to one of those, you >> > > might get the (false) impression that the detector sees no light at >> > > all. >> >> > Well I'm assuming that the source emits one photon at a time. And of >> > course I refer to the photon that is aimed at the detector when fired, >> > but does *not* hit due to the acceleration of the detector. >> >> Ah, ok, then I have no objection to your conclusion, although I have >> no idea what you hope to learn about relativity about such a scenario. > > I'm basically trying to figure out how light can possibly travel at a > constant speed, or at least *appear* to do so, when measured from any > frame. The geometry of our universe is such that c is the upper limit for speed that no matter can achieve. If you have something (massless .. ie not matter) going at c and try to 'add' velocity to it .. you still get c .. that is how the geometry works (an the formula for it is the velocity composition formula). If you insist that the universe has a 'simple' 3D geometry (eucliean/newtonian/gallilean), then you will never find a reason for light always travelling at c, because in such a world light does NOT do that. In such a world, where there is no limit to the speed at which a change can propagate, the speed of light (which is given by how fast changing eletromagnetic fields propagates) would have no limit either. There would not be a universal 'speed of light'. Unless you also propose an aether in which the fields are at 'rest'. But of course, we know from experiment that we do not live in such a universe as that. So it really isn't much point talking about it as though we did.
From: Peter Webb on 18 Feb 2010 21:38 "Inertial" <relatively(a)rest.com> wrote in message news:4b7dec3c$0$27851$c3e8da3(a)news.astraweb.com... > > "Ste" <ste_rose0(a)hotmail.com> wrote in message > news:42d85e4d-ca5c-464a-90ce-9a49386cd6dc(a)y33g2000yqb.googlegroups.com... >> On 17 Feb, 21:15, "Inertial" <relativ...(a)rest.com> wrote: >>> "Ste" <ste_ro...(a)hotmail.com> wrote in message >>> >>> >> Yes, and yes. >>> >>> > Ok. >>> >>> > Consider yet another setup: >>> >>> > D1 D3 >>> >>> > S1 D2 S2 D4 >>> >>> > In case it isn't clear, D1 and D2 are equidistant from S1, and D3 and >>> > D4 are equidistant from S2. >>> >>> Again .. poorly 'drawn'. you show the distance S1..D2 as shorter than >>> S2..D4. >> >> Indeed you are correct, but in any event that is why I included the >> clarifying clause. > > Yeup .. just asking you to be a little more careful in your diagrams. > >>> You need to take a little more care. This is the second diagram that >>> you >>> have mis-drawn. >> >> I'm afraid I still can't see the problem with the first drawing. It >> appears exactly as intended on my screen, although your requote of it >> appears garbled. > > I looked at your original source .. you have an extra blank line in there > .. that is all. Just need to be careful with ASCII diagrams .. They are > difficult to draw, but it does help if you try to keep things consistent. > Otherwise the diagram confuses rather than clarifies. > >>> > The S1 group (i.e. comprising S1, D1, and D2) are always stationary in >>> > the frame, and S1 is emitting a pulse towards both D1 and D2. There is >>> > a similar setup for the S2 group, except that after the emission of >>> > the pulse, the S2 group moves in the y+ direction, so that S2 is now >>> > in the former place of D3 (again, the whole group accelerates rapidly, >>> > and is stationary again by time of detection). >>> >>> > Now, I assume D1 and D2 receive their signals simultaneously. But what >>> > of D3 and D4? By the previous answers, I presume D4 receives it's >>> > signal before D3? >>> >>> D4 will not receive the signal at all, because it is no longer in line >>> with >>> the direction of the signal (which went from left to right from the >>> original >>> location of S2. >> >> Indeed. >> >> So let's assume that the S2 group was already moving at the time of >> emission, and that it continued to move. > > Then that is fine > >> Would the pulse *then* also >> continue in the x+ direction (i.e horizontally across the screen) >> towards the location of D4 *at the time of emission*, or would it take >> a diagonal (or even curved) path to catch D4? (Again, the frame is >> that of the stationary S1 group) > > Diagonal .. light gets it direction from the source .. just not its speed. > > Some people think that SR claims light does not get any velocity from the > source .. That is not correct .. it is only the speed it does not get. > > Even that is not the best way to think of it, as it makes light sound like > it is somehow 'special' .. velocities of ANYTHING in SR compose via the > velocity composition rule (not by simple vector addition). And if you > compose any velocity (v<c) with something with speed c, you get a velocity > with speed c. So even though the source velocity does compose with that > of the light emitted from it, the result is still a speed of c. > > So SR would say that ANY ballistically emitted object, with speed c > relative to its source (if that was possible), would have a speed of c > measured in every inertial frame. SR also says (or at least it is a > consequence of it) that it is not possible to have any object with mass > travel at c relative to another object. > > You are "dumbing-down" and simplifying the argument too much here (even for Ste). The postulates of SR would work fine with some other constant maximum speed for information transfer - say d - replacing c. The reason that Einstein identified c as being this maximum speed is that it is needed for Galileo's principle of relativity to be maintained - that all inertial reference frames are equivalent. If d was not c, then the transformations for light and mass would be different, violating Galilean relativity. Indeed, as I have previously pointed out, SR can be derived from nothing other than Maxwell's equations and Galilean relativity. Of course, Ste would have no idea about how Maxwell's equations transform between different reference frames, so this whole argument would go straight over his head, this is more "for the record".
From: Inertial on 18 Feb 2010 21:54
"Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message news:4b7df9b0$0$10265$afc38c87(a)news.optusnet.com.au... > > "Inertial" <relatively(a)rest.com> wrote in message > news:4b7dec3c$0$27851$c3e8da3(a)news.astraweb.com... >> >> "Ste" <ste_rose0(a)hotmail.com> wrote in message >> news:42d85e4d-ca5c-464a-90ce-9a49386cd6dc(a)y33g2000yqb.googlegroups.com... >>> On 17 Feb, 21:15, "Inertial" <relativ...(a)rest.com> wrote: >>>> "Ste" <ste_ro...(a)hotmail.com> wrote in message >>>> >>>> >> Yes, and yes. >>>> >>>> > Ok. >>>> >>>> > Consider yet another setup: >>>> >>>> > D1 D3 >>>> >>>> > S1 D2 S2 D4 >>>> >>>> > In case it isn't clear, D1 and D2 are equidistant from S1, and D3 and >>>> > D4 are equidistant from S2. >>>> >>>> Again .. poorly 'drawn'. you show the distance S1..D2 as shorter than >>>> S2..D4. >>> >>> Indeed you are correct, but in any event that is why I included the >>> clarifying clause. >> >> Yeup .. just asking you to be a little more careful in your diagrams. >> >>>> You need to take a little more care. This is the second diagram that >>>> you >>>> have mis-drawn. >>> >>> I'm afraid I still can't see the problem with the first drawing. It >>> appears exactly as intended on my screen, although your requote of it >>> appears garbled. >> >> I looked at your original source .. you have an extra blank line in there >> .. that is all. Just need to be careful with ASCII diagrams .. They are >> difficult to draw, but it does help if you try to keep things consistent. >> Otherwise the diagram confuses rather than clarifies. >> >>>> > The S1 group (i.e. comprising S1, D1, and D2) are always stationary >>>> > in >>>> > the frame, and S1 is emitting a pulse towards both D1 and D2. There >>>> > is >>>> > a similar setup for the S2 group, except that after the emission of >>>> > the pulse, the S2 group moves in the y+ direction, so that S2 is now >>>> > in the former place of D3 (again, the whole group accelerates >>>> > rapidly, >>>> > and is stationary again by time of detection). >>>> >>>> > Now, I assume D1 and D2 receive their signals simultaneously. But >>>> > what >>>> > of D3 and D4? By the previous answers, I presume D4 receives it's >>>> > signal before D3? >>>> >>>> D4 will not receive the signal at all, because it is no longer in line >>>> with >>>> the direction of the signal (which went from left to right from the >>>> original >>>> location of S2. >>> >>> Indeed. >>> >>> So let's assume that the S2 group was already moving at the time of >>> emission, and that it continued to move. >> >> Then that is fine >> >>> Would the pulse *then* also >>> continue in the x+ direction (i.e horizontally across the screen) >>> towards the location of D4 *at the time of emission*, or would it take >>> a diagonal (or even curved) path to catch D4? (Again, the frame is >>> that of the stationary S1 group) >> >> Diagonal .. light gets it direction from the source .. just not its >> speed. >> >> Some people think that SR claims light does not get any velocity from the >> source .. That is not correct .. it is only the speed it does not get. >> >> Even that is not the best way to think of it, as it makes light sound >> like it is somehow 'special' .. velocities of ANYTHING in SR compose via >> the velocity composition rule (not by simple vector addition). And if >> you compose any velocity (v<c) with something with speed c, you get a >> velocity with speed c. So even though the source velocity does compose >> with that of the light emitted from it, the result is still a speed of c. >> >> So SR would say that ANY ballistically emitted object, with speed c >> relative to its source (if that was possible), would have a speed of c >> measured in every inertial frame. SR also says (or at least it is a >> consequence of it) that it is not possible to have any object with mass >> travel at c relative to another object. >> >> > > You are "dumbing-down" and simplifying the argument too much here (even > for Ste). Can't get too dumb for some people :) > The postulates of SR would work fine with some other constant maximum > speed for information transfer - say d - replacing c. Indeed they would. The speed of light travels at the speed c BECAUSE c is the maximum rate at which change (information) can propagate. Not the other way around. Whether the geometry gives you that maximum speed, or vice versa, is a philosophical matter :) > The reason that Einstein identified c as being this maximum speed is that > it is needed for Galileo's principle of relativity to be maintained - that > all inertial reference frames are equivalent. Light travels at c because that is the speed that change (information)propagates in vacuo. (I thought I posted equivalent to that in this thread .. must have been another thread I posted to today) Einstein chose c so that equations such as maxwell's would count as 'laws' under the PoR, and to be compatible with some experimental results. > If d was not c, then the transformations for light and mass would be > different, violating Galilean relativity. If wouldn't really matter to SR if c was less than d (ie that light travelled slower than the maximum). But the thing is, most places we use 'c' in physics really should be called 'd'. 'd' is more fundamental a notion than 'c'. However, Maxwell's equations would then no longer be a frame-independent 'law' under the PoR as they are when c is d. You could probably change all the 'c's to 'd's in Eintein's 1905 paper and, instead of light, talk about information travelling at the maximum possible speed, and have the second postulate say that information travelling at the maximum speed speed in empty space in one frame travels at that speed as measured in all frames. Then you would need to show that light travels at that speed :):) But I don't think that would be an improvements :) > Indeed, as I have previously pointed out, SR can be derived from nothing > other than Maxwell's equations and Galilean relativity. Indeed it can .. if you assume maxwell's equations work the same in all inertial frames, you also get SR. There are many ways of getting there that result in the same theory. Which is a 'good thing'. > Of course, Ste would have no idea about how Maxwell's equations transform > between different reference frames, so this whole argument would go > straight over his head, this is more "for the record". Yeup |