Galilean transformation equations
in [Relativity]
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From: rbwinn on 17 Jun 2010 22:22 On Jun 17, 6:24 pm, YBM <ybm...(a)nooos.fr.invalid> wrote: > rbwinn a crit : > > > > > > > On Jun 17, 1:06 pm, PD <thedraperfam...(a)gmail.com> wrote: > >> On Jun 13, 8:31 am, rbwinn <rbwi...(a)gmail.com> wrote: > > >>> x'=x-vt > >>> y'=y > >>> z'=z > >>> t'=t > >>> Experiment shows that a clock in moving frame of reference S' is > >>> slower than a clock in S which shows t. According to the Galilean > >>> transformation equations, that slower clock does not show t'. Time on > >>> the slower clock has to be represented by some other variable if the > >>> Galilean transformation equations are to be used. We call time on the > >>> slow clock in S' by the variable n'. > >>> We can calculate time on the slow clock from the Galilean > >>> transformation equations because we know that it shows light to be > >>> traveling at 300,000 km per second in S'. Therefore, if > >>> |x'|=300,000 km/sec(n') and |x| =300,000km/sec(t), then > >>> cn'=ct-vt > >>> n'=t(1-v/c) > >>> We can now calculate orbits of satellites and planets without > >>> the problems imposed by the Lorentz equations and their length > >>> contraction. For instance, the speed of earth in its orbit around the > >>> sun is 29.8 km/sec. While a second of time takes place on earth, a > >>> longer time is taking place on the sun. > >>> n'(earth)=t(sun)(1-v/c) > >>> 1 sec.=t(sun)(1-29.8/300,000) > >>> t(sun)=1..0001 sec. > >>> Since the orbit of Mercury was the proof used to verify that > >>> Einstein's equations were better than Newton's for gravitation, we > >>> calculate how time on earth compares with time on Mercury. > >>> n'Mercury=t(sun)(1-v(Mercury)/c) > >>> n'(mercury)=1.0001sec(1-47.87 km/sec/ > >>> 300,000km/sec) > >>> n'(Mercury)=.99994 sec > >>> So a second on a clock on earth is .99994 sec on a clock on > >>> Mercury. The question now is where would this put the perihelion of > >>> Mercury using Newton's equations? > >> Amazing to see you back, Robert. Even more amazing to find that you've > >> done a reset and started with the very same nonsense you've put out > >> for years and years. I would have thought that you would have learned > >> something. > > >> So you are claiming that for clocks A and B, where B is moving > >> relative to A and runs slower than A, then A is measuring time (as > >> denoted by the quantity t), but B is not measuring time (as denoted by > >> the quantity t'). > > >> The problem of course is that A is moving relative to B and runs > >> slower than B. Your conclusion consistently would be that B is > >> measuring time but A is not. > > >> Therefore, according to you, A is measuring time and not measuring > >> time, and B is measuring time and not measuring time. > > >> PD > > > You are confusing measurement of time with transformation of > > coordinates. Time can be measured about any way imaginable. > > Coordinates can be transformed only with t' and t. > > The point is: when are you planning to go back to the hospital > poor stupid, disgusting, lost Robert? I have no plans that include medical science.
From: Inertial on 17 Jun 2010 22:23 "rbwinn" <rbwinn3(a)gmail.com> wrote in message news:a4cf3177-76db-41a9-9521-78a222a48ae2(a)v12g2000prb.googlegroups.com... > On Jun 17, 6:04 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: >> rbwinn wrote: >> >> [...] >> >> > You are confusing measurement of time with transformation of >> > coordinates. Time can be measured about any way imaginable. >> > Coordinates can be transformed only with t' and t. >> >> So is it your opinion that t and t' are just symbols devoid of physical >> meaning? > > t'=t has a physical meaning. It is what is called an identity in > algebra. It means that t' is time on a clock in S. No .. it means t is the time shown on a correct clock in S and t' is the time shown on a correct clock in S' .. and that these two times are always the same. Do you disagree ?
From: eric gisse on 17 Jun 2010 23:02 rbwinn wrote: > On Jun 17, 6:04 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: >> rbwinn wrote: >> >> [...] >> >> > You are confusing measurement of time with transformation of >> > coordinates. Time can be measured about any way imaginable. >> > Coordinates can be transformed only with t' and t. >> >> So is it your opinion that t and t' are just symbols devoid of physical >> meaning? > > t'=t has a physical meaning. It is what is called an identity in > algebra. It means that t' is time on a clock in S. Yes, but it also means that t is the time on a clock in S'. I know you've invested many years in claiming otherwise, though. So I'm not expecting you to understand.
From: rbwinn on 18 Jun 2010 00:51 On Jun 16, 9:17 pm, "Inertial" <relativ...(a)rest.com> wrote: > "rbwinn" <rbwi...(a)gmail.com> wrote in message > > news:cd80a2c1-36c2-4699-9cae-ba3d0ec8a676(a)s4g2000prh.googlegroups.com... > > > > > > > On Jun 16, 5:00 pm, "Inertial" <relativ...(a)rest.com> wrote: > >> "rbwinn" <rbwi...(a)gmail.com> wrote in message > > >>news:02c5f8f9-4bd6-40f3-b54b-f9a12f7e5036(a)j12g2000pri.googlegroups.com.... > > >> > On Jun 15, 7:52 pm, "Inertial" <relativ...(a)rest.com> wrote: > >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message > > >> >>news:063e7006-c295-4d91-a567-9a4a813beb0c(a)s4g2000prh.googlegroups.com... > > >> >> > On Jun 15, 6:56 pm, "Inertial" <relativ...(a)rest.com> wrote: > >> >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message > > >> >> >>news:6d5da435-3595-497f-b480-2586e3daaa16(a)z13g2000prh.googlegroups.com... > > >> >> >> > On Jun 15, 3:55 am, "Inertial" <relativ...(a)rest.com> wrote: > >> >> >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message > > >> >> >> >>news:ef70417f-5f09-4f25-9cf3-bbb9760e5548(a)q36g2000prg.googlegroups.com... > > >> >> >> >> > On Jun 13, 7:53 am, "Inertial" <relativ...(a)rest.com> wrote: > >> >> >> >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message > > >> >> >> >> >>news:702e22b2-1bc0-4a16-9f46-3e571612e517(a)z13g2000prh.googlegroups.com... > > >> >> >> >> >> > x'=x-vt > >> >> >> >> >> > y'=y > >> >> >> >> >> > z'=z > >> >> >> >> >> > t'=t > > >> >> >> >> >> Amazing .. you appear to know what a Galilean transform is.. > > >> >> >> >> >> > Experiment shows that a clock in moving frame of > >> >> >> >> >> > reference > >> >> >> >> >> > S' > >> >> >> >> >> > is > >> >> >> >> >> > slower than a clock in S which shows t > > >> >> >> >> >> As measured be S. Hence refuting Galilean transforms > > >> >> >> >> >> > According to the Galilean > >> >> >> >> >> > transformation equations, that slower clock does not show > >> >> >> >> >> > t'. > > >> >> >> >> >> No .. according to Galilean transforms it DOSE show t' = t. > >> >> >> >> >> And > >> >> >> >> >> so > >> >> >> >> >> Galilean > >> >> >> >> >> transforms are wrong > > >> >> >> >> >> > Time on > >> >> >> >> >> > the slower clock has to be represented by some other > >> >> >> >> >> > variable > >> >> >> >> >> > if > >> >> >> >> >> > the > >> >> >> >> >> > Galilean transformation equations are to be used. > > >> >> >> >> >> They can't. Because then you are no longer using Galilean > >> >> >> >> >> transforms > > >> >> >> >> >> [snip nonsense that follows] > > >> >> >> >> > Why are you no longer using the Galilean transformation > >> >> >> >> > equations? > > >> >> >> >> YOU aren't. No me. Its your nonsense, not mine. > > >> >> >> >> > The Galilean transformation equations treat all slower clocks > >> >> >> >> > the > >> >> >> >> > same. > > >> >> >> >> There are NO slower clocks in Galilean transforms .. time is the > >> >> >> >> same > >> >> >> >> everywhere. > > >> >> >> >> Learn some physics .. or how to understand maths. Or both. > > >> >> >> > I bought a clock that lost ten minutes per day > > >> >> >> Irrelevant. Transforms are not about faulty clocks. They are > >> >> >> about > >> >> >> what > >> >> >> the time REALLY IS at the location. Ie what a CORRECTLY working > >> >> >> clock > >> >> >> would > >> >> >> show > > >> >> >> [snip irrelevance] > > >> >> > A correctly working clock in S' is slower than a correctly working > >> >> > clock in S. > > >> >> So t' <> t. > > >> >> A correctly working clock is one that shows the correct time. So a > >> >> correctly working clock at rest in S shows time t. a correctly > >> >> working > >> >> clock at rest in S' shows time t'. If it doesn't, it is not working > >> >> correctly . .by definition. > > >> >> > Consequently, you cannot use time from a correctly > >> >> > working clock in S' as time coordinates for the Galilean > >> >> > transformation equations. > > >> >> Of course you can .. by the definition of what a correctly working > >> >> clock > >> >> IS. > > >> >> What you CANNOT use is the Galilean Transforms for time. > > >> >> Please. . be honest about what you are doing. If you are using > >> >> Galilean > >> >> transforms for what you meaasure, then you are proven wrong by > >> >> experiment. > >> >> If you are not, then do not dishonestly claim that you are, and > >> >> instead > >> >> talk > >> >> about the transform you ARE using. > > >> > The transform I am using is the Galilean transform. > > >> > x'=x-vt > >> > y'=y > >> > z'=z > >> > t'=t > > >> Liar > > >> > All the transform requires is that t' be the time shown by a > >> > clock in S. If you can prove otherwise, prove it. > > >> WRONG .. It says a correct clockin S' show t'. You say it doesn't So > >> you > >> are not using Galilean Transforms. > > > The Galilean transformation equations say that a correct clock in S > > shows t'. > > A correct clock seen in S shows t. The same correct clock seen in S' shows > t' which is the same as t. A moving observer (or clock) does not change its > time Sorry, your idea does not match up with experimental data. Scientists report that a clock in S' is slower than a clock in S.
From: Inertial on 18 Jun 2010 01:42
"rbwinn" <rbwinn3(a)gmail.com> wrote in message news:03606340-adca-4532-939f-5eacc53a1daa(a)n37g2000prc.googlegroups.com... > On Jun 16, 9:17 pm, "Inertial" <relativ...(a)rest.com> wrote: >> "rbwinn" <rbwi...(a)gmail.com> wrote in message >> >> news:cd80a2c1-36c2-4699-9cae-ba3d0ec8a676(a)s4g2000prh.googlegroups.com... >> >> >> >> >> >> > On Jun 16, 5:00 pm, "Inertial" <relativ...(a)rest.com> wrote: >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message >> >> >>news:02c5f8f9-4bd6-40f3-b54b-f9a12f7e5036(a)j12g2000pri.googlegroups.com... >> >> >> > On Jun 15, 7:52 pm, "Inertial" <relativ...(a)rest.com> wrote: >> >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message >> >> >> >>news:063e7006-c295-4d91-a567-9a4a813beb0c(a)s4g2000prh.googlegroups.com... >> >> >> >> > On Jun 15, 6:56 pm, "Inertial" <relativ...(a)rest.com> wrote: >> >> >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message >> >> >> >> >>news:6d5da435-3595-497f-b480-2586e3daaa16(a)z13g2000prh.googlegroups.com... >> >> >> >> >> > On Jun 15, 3:55 am, "Inertial" <relativ...(a)rest.com> wrote: >> >> >> >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message >> >> >> >> >> >>news:ef70417f-5f09-4f25-9cf3-bbb9760e5548(a)q36g2000prg.googlegroups.com... >> >> >> >> >> >> > On Jun 13, 7:53 am, "Inertial" <relativ...(a)rest.com> wrote: >> >> >> >> >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message >> >> >> >> >> >> >>news:702e22b2-1bc0-4a16-9f46-3e571612e517(a)z13g2000prh.googlegroups.com... >> >> >> >> >> >> >> > x'=x-vt >> >> >> >> >> >> > y'=y >> >> >> >> >> >> > z'=z >> >> >> >> >> >> > t'=t >> >> >> >> >> >> >> Amazing .. you appear to know what a Galilean transform >> >> >> >> >> >> is. >> >> >> >> >> >> >> > Experiment shows that a clock in moving frame of >> >> >> >> >> >> > reference >> >> >> >> >> >> > S' >> >> >> >> >> >> > is >> >> >> >> >> >> > slower than a clock in S which shows t >> >> >> >> >> >> >> As measured be S. Hence refuting Galilean transforms >> >> >> >> >> >> >> > According to the Galilean >> >> >> >> >> >> > transformation equations, that slower clock does not >> >> >> >> >> >> > show >> >> >> >> >> >> > t'. >> >> >> >> >> >> >> No .. according to Galilean transforms it DOSE show t' = >> >> >> >> >> >> t. >> >> >> >> >> >> And >> >> >> >> >> >> so >> >> >> >> >> >> Galilean >> >> >> >> >> >> transforms are wrong >> >> >> >> >> >> >> > Time on >> >> >> >> >> >> > the slower clock has to be represented by some other >> >> >> >> >> >> > variable >> >> >> >> >> >> > if >> >> >> >> >> >> > the >> >> >> >> >> >> > Galilean transformation equations are to be used. >> >> >> >> >> >> >> They can't. Because then you are no longer using Galilean >> >> >> >> >> >> transforms >> >> >> >> >> >> >> [snip nonsense that follows] >> >> >> >> >> >> > Why are you no longer using the Galilean transformation >> >> >> >> >> > equations? >> >> >> >> >> >> YOU aren't. No me. Its your nonsense, not mine. >> >> >> >> >> >> > The Galilean transformation equations treat all slower >> >> >> >> >> > clocks >> >> >> >> >> > the >> >> >> >> >> > same. >> >> >> >> >> >> There are NO slower clocks in Galilean transforms .. time is >> >> >> >> >> the >> >> >> >> >> same >> >> >> >> >> everywhere. >> >> >> >> >> >> Learn some physics .. or how to understand maths. Or both. >> >> >> >> >> > I bought a clock that lost ten minutes per day >> >> >> >> >> Irrelevant. Transforms are not about faulty clocks. They are >> >> >> >> about >> >> >> >> what >> >> >> >> the time REALLY IS at the location. Ie what a CORRECTLY working >> >> >> >> clock >> >> >> >> would >> >> >> >> show >> >> >> >> >> [snip irrelevance] >> >> >> >> > A correctly working clock in S' is slower than a correctly >> >> >> > working >> >> >> > clock in S. >> >> >> >> So t' <> t. >> >> >> >> A correctly working clock is one that shows the correct time. So a >> >> >> correctly working clock at rest in S shows time t. a correctly >> >> >> working >> >> >> clock at rest in S' shows time t'. If it doesn't, it is not >> >> >> working >> >> >> correctly . .by definition. >> >> >> >> > Consequently, you cannot use time from a correctly >> >> >> > working clock in S' as time coordinates for the Galilean >> >> >> > transformation equations. >> >> >> >> Of course you can .. by the definition of what a correctly working >> >> >> clock >> >> >> IS. >> >> >> >> What you CANNOT use is the Galilean Transforms for time. >> >> >> >> Please. . be honest about what you are doing. If you are using >> >> >> Galilean >> >> >> transforms for what you meaasure, then you are proven wrong by >> >> >> experiment. >> >> >> If you are not, then do not dishonestly claim that you are, and >> >> >> instead >> >> >> talk >> >> >> about the transform you ARE using. >> >> >> > The transform I am using is the Galilean transform. >> >> >> > x'=x-vt >> >> > y'=y >> >> > z'=z >> >> > t'=t >> >> >> Liar >> >> >> > All the transform requires is that t' be the time shown by a >> >> > clock in S. If you can prove otherwise, prove it. >> >> >> WRONG .. It says a correct clockin S' show t'. You say it doesn't So >> >> you >> >> are not using Galilean Transforms. >> >> > The Galilean transformation equations say that a correct clock in S >> > shows t'. >> >> A correct clock seen in S shows t. The same correct clock seen in S' >> shows >> t' which is the same as t. A moving observer (or clock) does not change >> its >> time > > Sorry, your idea Not mine.. what Glaillean transforms say > does not match up with experimental data. That's what I told you. > Scientists > report that a clock in S' is slower than a clock in S. And so prove that Galilean Transforms are wrong. Thanks for refuting yourself |