Known physics defeated by simple puzzle?
in [Physics]
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From: jbriggs444 on 17 Feb 2010 11:21 On Feb 17, 10:13 am, Ste <ste_ro...(a)hotmail.com> wrote: > On 16 Feb, 14:31, PD <thedraperfam...(a)gmail.com> wrote: > > > On Feb 16, 7:46 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > I'm not trying to prove anything. I'm trying to form some sort of > > > consistent picture in my own head of how this setup behaves under > > > different transformations. > > > Another terminology issue here. The setup as described is undergoing > > an acceleration or two. It therefore changes from being at rest in one > > reference frame to being at rest in another reference frame. > > "Transformations", as in the context Galilean transformations or > > Lorentz transformations, is the relationship between the descriptions > > of this system in those reference frames, but it does not mean the > > physical process of the acceleration of the system itself. > > > Do you understand the terminology? > > I'm not sure. If you're asking whether I considered "transformation" > synonymous with "acceleration", the answer is no. I basically meant > transformation in the sense that I have always know it, that is, "a > movement of some sort", including translation, rotation, reflection, > etc. That is not the sense in which physicists use the word. A "coordinate system transformation" does NOT involve any movement at all. Let us take a simple example. Little Red Riding Hood has a house in the woods. So does Grandma. There is a trail between the two houses. Assume that the trail is 1 mile long. It need not be straight. Little Red Riding Hood can put coordinates on the trail, starting with her house at coordinate value 0 and Grandma's house at coordinate value 1. Grandma can put coordinates on the trail starting with her house at coordinate value 1 and Little Red Riding Hood's house at coordinate value 0. The Big Bad Wolf can put coordinates on the trail starting from his hiding point at coordinate value 0 back to Little Red's house at coordinate value -0.3 and forward to Grandma's house at coordinate value +0.7 [Let us agree for the moment to ignore any shortcuts that the Wolf knows about] A "coordinate system transformation" is the process of taking coordinates expressed in one coordinate scheme and re-expressing them in a different scheme. When you switch from one coordinate system to another, coordinates change but nothing "moves". Red 0 0.3 1 Grandma 1 -0.7 0 Wolf -0.3 0 0.7 If, for instance, the Big Bad Wolf decides to adopt Red's coordinates instead of his own, he immediately goes from coordinate value 0.0 (Wolf) to coordinate value 0.3 (Red). But he's still sitting in his hiding place. He hasn't moved. He's just decided to use a different coordinate system. (This is an example of translation) And if Grandma decides to adopt Red's frame, she goes from coordinate value 0.0 (Grandma) to coordinate value 1.0 (Red). But she's still puttering around in the kitchen baking cookies. (This is an example of a reflection plus a translation) I've avoided "rotation" and "re-scaling" for simplicity. This is what coordinate systems and "frames of reference" are all about. Different ways of assigning coordinate values to positions (and times). Different ways of putting numbers on the same underlying physical reality.
From: dlzc on 17 Feb 2010 12:13 Dear J. Clarke: On Feb 17, 7:51 am, "J. Clarke" <jclarke.use...(a)cox.net> wrote: .... > Make some reasonable assumptions about the > difference in speed in different directions and > then calculate the effect on the measurement > based on the motion of Jupiter's moons and get > back to us. Ignorance of the impossibility of making one way light speed measurements is not my problem. Pull your head out, then get back to us. Not not. Posture more, and impress us all with your witticisms. David A. Smith
From: Ste on 17 Feb 2010 15:57 On 17 Feb, 15:53, PD <thedraperfam...(a)gmail.com> wrote: > On Feb 17, 9:13 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > Translations in relativity--or in fact, even in pure mathematics--are > > > > > very different things than rotations, and picking the detector up and > > > > > putting it somewhere else without changing the speed is a pure > > > > > translation. You don't even need relativity in this scenario, where > > > > > everything is at rest with respect to everything else. > > > > > Yes, but we're going to get to the bit where relativity is required in > > > > a moment. > > > > > Now, let us suppose we have two source and two detectors again: > > > > > D1 D2 D3 > > > > > S1 S2 S3 > > > > > S1 and D1 are stationary in the frame, and do not move. D2 is also > > > > stationary in the frame. S2, S3, and D3 are all moving in the y+ > > > > direction (i.e. same as the previous scenario) at a constant speed > > > > (which is close to 'c'). Just to be sure we understand, the same setup > > > > a few moments back in time would have looked like this: > > > > > D1 D2 > > > > > D3 > > > > > S1 > > > > > S2 S3 > > > > > Now, when all sources come into line with each other (as per the first > > > > illustration above), a pulse is emitted towards the respective > > > > detectors. After emission, S2 would continue towards D2, but in > > > > reality we remove S2 from the picture before any collision (and we've > > > > already established that any transformation of the sources after > > > > emission has no effect on photons already emitted). > > > > > Now, based on the previous scenario, I presume that in SR, D3 receives > > > > its pulse long after D1. However, this time, does D2 receive its pulse > > > > at the same time as D1? > > 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. 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?
From: J. Clarke on 17 Feb 2010 15:49 dlzc wrote: > Dear J. Clarke: > > On Feb 17, 7:51 am, "J. Clarke" <jclarke.use...(a)cox.net> wrote: > ... >> Make some reasonable assumptions about the >> difference in speed in different directions and >> then calculate the effect on the measurement >> based on the motion of Jupiter's moons and get >> back to us. > > Ignorance of the impossibility of making one way light speed > measurements is not my problem. Pull your head out, then get back to > us. Not not. Posture more, and impress us all with your witticisms. If the "two way effects" do not introduce significant error then what relevance do they have? And if you can't calculate the effects then you are a damned fool for pontificating on a physics forum.
From: Inertial on 17 Feb 2010 16:10
"dlzc" <dlzc1(a)cox.net> wrote in message news:f2cd2caa-9ec8-4930-a824-9743d8a3f4eb(a)t17g2000prg.googlegroups.com... > Dear J. Clarke: > > On Feb 17, 7:51 am, "J. Clarke" <jclarke.use...(a)cox.net> wrote: > ... >> Make some reasonable assumptions about the >> difference in speed in different directions and >> then calculate the effect on the measurement >> based on the motion of Jupiter's moons and get >> back to us. > > Ignorance of the impossibility of making one way light speed > measurements is not my problem. Pull your head out, then get back to > us. Not not. Posture more, and impress us all with your witticisms. It is not impossible, as it has been done, but it does involve some assumptions about how to keep to clocks synchronized (eg move them apart with the same speed profile). The same is true, of course, for any speed measured with a pair of clocks. One wonders if there is a way to measure the speed of light without a pair of clocks (without any assumptions about time at different locations). We can, for example, measure the speed of a car with a speedometer that doesn't involve a pair of clocks (either directly in measurement, or in the calibration of the instrument). |