Galilean transformation equations
in [Relativity]
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From: Androcles on 15 Jun 2010 05:19 "rbwinn" <rbwinn3(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/wiki/Einstein_synchronisation > > 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.
From: rbwinn on 15 Jun 2010 05:20 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.
From: Inertial on 15 Jun 2010 06:55 "rbwinn" <rbwinn3(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.
From: Inertial on 15 Jun 2010 06:57 "rbwinn" <rbwinn3(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.
From: eric gisse on 15 Jun 2010 21:22
rbwinn wrote: [...] > 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. 15 years and you still don't understand the very equations you've been repeating that whole time. Nice. Are you autistic? |