Galilean transformation equations
in [Relativity]
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From: rbwinn on 16 Jun 2010 19:30 On Jun 15, 7:48 pm, "Inertial" <relativ...(a)rest.com> wrote: > "rbwinn" <rbwi...(a)gmail.com> wrote in message > > news:b8fb8d9d-a202-4cb7-973a-9bd8640e9aa1(a)s6g2000prf.googlegroups.com... > > > > > > > On Jun 15, 6:55 pm, "Inertial" <relativ...(a)rest.com> wrote: > >> "rbwinn" <rbwi...(a)gmail.com> wrote in message > > >>news:d4c02153-8d91-46ae-b074-cb6cba68b01c(a)r5g2000prf.googlegroups.com.... > > >> > On Jun 15, 2:19 am, "Androcles" <Headmas...(a)Hogwarts.physics_z> wrote: > >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message > > >> >>news:27b77bb9-8131-4364-a5d1-e6873a7e2dac(a)r5g2000prf.googlegroups.com... > >> >> On Jun 13, 7:50 am, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > > >> >> > On Jun 13, 10:12 am, blackhead <larryhar...(a)softhome.net> wrote: > > >> >> > > On 13 June, 14:46, "Androcles" <Headmas...(a)Hogwarts.physics_z> > >> >> > > 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 > >> >> > > > | > >> >> > > > | Experiment shows that a clock in moving frame of reference S' > >> >> > > > is > >> >> > > > | slower than a clock in S which shows t. > > >> >> > > > Liar. > > >> >> > > Hafele Keating experiment. > > >> >> >http://www.search.com/reference/Problematic_physics_experiments > > >> >> > GPS including Sagnac and Pound Rebka have some credibility. > > >> >> > Attempts to show that real clock mechanisms can mimic > >> >> > the Einstein Synchronisation procedure are always > >> >> > entertaining so don't let me discourage you. ;-) > > >> >> >http://en.wikipedia.org/wiki/Coordinate_timehttp://en.wikipedia.org/w... > > >> >> > Sue... > > >> >> > > You're the liar. > > >> >> Well, since you are so entertained, you might want to show what you > >> >> find so entertaining. The Galilean transformation equations are the > >> >> correct equations because they are the only transformation equations > >> >> that do not have a length contraction. This is difficult to explain > >> >> to people who have been taught that a length contraction is necessary. > >> >> Robert B. Winn > >> >> ============================================= > >> >> Produce the evidence to show that "a clock in moving frame of > >> >> reference > >> >> S' is slower than a clock in S which shows t" as you claim, Winn, and > >> >> we'll be entertained. > > >> > Well, the experiment I remember best concerning this was where they > >> > put a clock in the nosecone of a Vanguard rocket in 1958 and said that > >> > it slowed down exactly as predicted by Einstein's equations. I just > >> > go by what scientists say they have done and what the results were. > > >> Gad you agree that Einstein equations (ie Lorentz transforms) are > >> experimentally shown to be valid. That same experiment REFUTES Galilean > >> transforms. They don't work (except approximately at v << c) > > > The Lorentz equations agree with my mathematics at 30 miles per > > second, the speed of the planet Mercury to six decimal places. > > Irrelevant > > > The > > Lorentz equations agreed with the Galilean transformation equations > > and absolute time to two decimal places at the same speed. > > Irrelevant > > > I use the > > Galilean transformation equations with a slower clock in S' shown as a > > slower clock, not as time for time coordinates. The time coordinates > > in the Galilean transformation equations are t' and t. > > Yes they are .. or any other set of symbols you chose to use .. which ones > you use is just a matter of convention .. as long as you are consistent and > don't claim a change in letter as anything more than what it is. > > Galilean transforms say that time does not vary with motion. Correctly > ticking clocks will always show the same time for all observers, regardless > of motion. So a correctly ticking clock put in an aircraft will show the > same time as a stay-at-home clock when the travelling clock returns. > > What do YOU say that is different? If it IS different, then you are no > longer using Galilean transforms .. so be honest enough to say that. If it > is NOT different, you are proven wrong experimentally. The Galilean transformation equations say that t'=t. t is time on a clock in S. If there is a clock running at some other rate which shows some other time than t, then that clock does not show t'.
From: rbwinn on 16 Jun 2010 19:41 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 All the transform requires is that t' be the time shown by a clock in S. If you can prove otherwise, prove it.
From: rbwinn on 16 Jun 2010 19:46 On Jun 15, 7:56 pm, "Inertial" <relativ...(a)rest.com> wrote: > "rbwinn" <rbwi...(a)gmail.com> wrote in message > > news:adbb8478-79f8-4a0e-bfd1-8cb5fcceed94(a)6g2000prg.googlegroups.com... > > > > > > > On Jun 15, 7:10 pm, "Inertial" <relativ...(a)rest.com> wrote: > >> "rbwinn" <rbwi...(a)gmail.com> wrote in message > > >>news:4466492f-0a28-4aec-9a1b-05cce138c867(a)t34g2000prd.googlegroups.com.... > > >> > On Jun 15, 3:57 am, "Inertial" <relativ...(a)rest.com> wrote: > >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message > > >> >>news:ff4c8b77-ca8b-45a1-9d04-4e614e476447(a)s4g2000prh.googlegroups.com... > > >> >> > On Jun 13, 5:43 pm, "Inertial" <relativ...(a)rest.com> wrote: > >> >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message > > >> >> >>news:8b250e8c-7689-460d-83b3-e25bfb5c83e1(a)11g2000prw.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] > > >> >> >> > What do you mean I am no longer using the Galilean transformation > >> >> >> > equations? > > >> >> >> > x'=x-vt > >> >> >> > y'=y > >> >> >> > z'=z > >> >> >> > t'=t > > >> >> >> Because you said you are not using t' = t .. you are using > >> >> >> something > >> >> >> else. > >> >> >> So it is no longer a Galilean transform. You can't throw away your > >> >> >> cake > >> >> >> and > >> >> >> eat it too. > > >> >> >> > Which one of the equations is not a Galilean transformation > >> >> >> > equation? > > >> >> > I am using t'=t. t is time on a clock in S. t'=t is what is known > >> >> > in > >> >> > algebra as an identity. t' is time on a clock in S. Time on a > >> >> > clock > >> >> > in S' is not t'. > > >> >> Yes it is .. if you are using Galilean Transforms. It is is something > >> >> OTHER > >> >> than t', then you are NO LONGER using Galilean Transforms. Simple. > > >> >> > It has to be shown by some other variable. > > >> >> So you are no longer using t' for the time. And so you are no longer > >> >> using > >> >> Galilean Transforms. As I said. You have made up some DIFFERENT > >> >> transform > >> >> instead that treats time differently. > > >> >> That's fine if you want to do that ... but do not LIE by claiming you > >> >> are > >> >> using Galilean transforms. A bit of honesty goes a long way. A bit > >> >> of > >> >> physics goes even further. Start with the honesty. > > >> > Here are the Galilean transformation equations. Honest. > > >> > x'=x-vt > >> > y'=y > >> > z'=z > >> > t'=t > > >> Yes.. I know what they are.. And they are the ones you go on to NOT use. > > >> > Notice the equation that says t'=t. That kind of equation is > >> > called an identity in algebra. What it means is that the time in S' > >> > for transforming coordinates is t', and that t' is the time that is on > >> > a clock in S because t'=t. > > >> Yes .. so according to the transforms, all correctly working clocks tick > >> at > >> the same rate regardless of motion. But we know that they DO tick at > >> different rates due to motion. so Galilean Transforms do not apply. > > >> > You might want to try practicing with > >> > coordinates in S and S' using t'=t. I am sure you will find that the > >> > coordinates do transform. So the Galilean transformation equations do > >> > not transform coordinates in any way to a clock in S' that scientists > >> > say they have found to be slower than a clock in S. > > >> That's right .. Galilean transforms do not work in reality. > > >> > You cannot use > >> > the time on that clock as t' because t'=t, the time on the faster > >> > clock in S. > > >> Then you are no longer using Galilean transforsm .. you are using some > >> OTHER > >> transform that does not have the time in one frame the same as the time > >> in > >> another. > > >> > But the slower clock in S' shows light to be traveling at > >> > c=300,000 km/sec. Surely it must be t'. Not according to the > >> > Galilean transformation equations. > > >> Wrong .. According to the Galillean transforms it WILL be t'. Experiment > >> shows that it not the case. > > >> > t'=t, the time on a clock in S. > >> > But the fact that time on a clock in S' shows light to be traveling at > >> > c gives us a way to solve for time on that clock from the Galilean > >> > transformation equations > > >> No .. it doesn't > > >> > cn'=ct-vt > > >> > where n' is time on the slower clock in S'. > > >> > n'=t(1-v/c) > > >> So you have used a DIFFERENT equation for the time in S' to what > >> Galillean > >> transforms use. > > >> So .. as i said .. you are NO LONGER using Galilean transforms > > >> > As you seem to recognize, n' cannot be used with the Galilean > >> > transformation equations. > > >> If you are claiming n' is the time in S'. then that is just a change of > >> letter to use. You are REALYL showing (using conventional notation) > > >> t' = t ( 1 - v/c) > > >> Which is NOT the same as > > >> t' = t > > >> So you are NOT using Galilean transforms > > >> > In order to transform coordinates, you have > >> > to convert n' to t' and use the Galilean transformation equation, t'=t > >> > for time coordinates in S and S'. > >> > Since you claim so vehemently that you have found an error in > >> > this reasoning, go ahead and show the error you think you have found.. > > >> I have > > > Well, no, I am sorry, but you have not. > > Yes .. I have > > > Here is what you are > > claiming. You are saying that t' cannot equal t in S' because a clock > > shows some other time in that frame of reference. > > A correct cloak .. Yes. That is a fact by definition of what a correct > clock is. > > > As a matter of > > fact, there is no clock in S' that shows t'=t. > > So Galilean transforms are wrong. Or you are talking about clocks that are > wrong. > The clocks are fine. There just do not happen to be any that show t'. > > So the Galilean > > transformation equations regard all clocks the same in S'. > > Irrelevant > > > Whatever > > they say has to be converted to t' before transforming coordinates > > with the Galilean transformation equations. > > No .. they SHOW t'. That is what a correct clock does. Just as a clock at > rest in S shows the time t. Nothing needs converting. > See this equation? t'=t That kind of equation is what is known in algebra as an identity. It means that time on a clock in S is t'. > > Whenever the time of a > > clock running at any speed has been converted to t', then it can be > > used in the Galilean transformation equations. > > A correct clock IS showing t' Maybe according to Androcles. Scientists say otherwise. The say a clock in S' is slower than a clock in S. > > > That is what I do. Sorry if it offends you. > > Only your arrogance and lies offend me > > > You shouldn't really be > > getting so offended by correct use of the Galilean transformation > > equations. > > You aren't doing that. Prove it.
From: rbwinn on 16 Jun 2010 19:53 On Jun 16, 2:45 am, harald <h...(a)swissonline.ch> wrote: > On Jun 15, 11:20 am, rbwinn <rbwi...(a)gmail.com> wrote: > > > > > > > On Jun 13, 5:43 pm, "Inertial" <relativ...(a)rest.com> wrote: > > > > "rbwinn" <rbwi...(a)gmail.com> wrote in message > > > >news:8b250e8c-7689-460d-83b3-e25bfb5c83e1(a)11g2000prw.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] > > > > > What do you mean I am no longer using the Galilean transformation > > > > equations? > > > > > x'=x-vt > > > > y'=y > > > > z'=z > > > > t'=t > > > > Because you said you are not using t' = t .. you are using something else. > > > So it is no longer a Galilean transform. You can't throw away your cake and > > > eat it too. > > > > > Which one of the equations is not a Galilean transformation equation? > > > I am using t'=t. t is time on a clock in S. t'=t is what is known in > > algebra as an identity. t' is time on a clock in S. Time on a clock > > in S' is not t'. It has to be shown by some other variable. > > Here you show that you still do *not* know what a Galilean > transformation is. The symbol t' in the Galilean transformation refers > to clock time in S'. You can show clock time by some other variable, > but then you do not have a Galilean transformation anymore. > > Harald The Galilean transformation equations I use just say t'=t. t is time on a clock in S. That means according to the Galilean transformation equation t' is time on a clock in S. If you have another clock somewhere that shows some other time, you cannot use that time in the Galilean transformation equations until you convert it to t'.
From: rbwinn on 16 Jun 2010 19:55
On Jun 16, 1:18 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > rbwinn wrote: > > On Jun 16, 1:37 am, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > >> rbwinn wrote: > >> > On Jun 15, 8:43 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > >> >> rbwinn wrote: > > >> >> [...] > > >> >> > Well, every morning I see the sun rise and say, It is a new day. > >> >> > The fact that I do this does not diminish my mental capacity. When > >> >> > the sun comes up, it actually is a new day where I am. Posting the > >> >> > Galilean transformation equations is a similar process. There is > >> >> > really no harm in repeating anything that is true. > > >> >> So you are autistic. > > >> > I have been called a lot of things, but you are the first to call me > >> > autistic. > > >> If you were not autistic, or a sociopath, you would take a moment to > >> consider why people keep calling you names. > > >> The answer is not 'because I'm right'. > > > If people keep calling me names, it would appear that they are the > > sociopaths, not me. > > Thanks for playing. You think this is a game, Eric? |