Galilean transformation equations
in [Relativity]
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From: rbwinn on 15 Jun 2010 22:52 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. 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. As a matter of fact, there is no clock in S' that shows t'=t. So the Galilean transformation equations regard all clocks the same in S'. Whatever they say has to be converted to t' before transforming coordinates with the Galilean transformation equations. 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. That is what I do. Sorry if it offends you. You shouldn't really be getting so offended by correct use of the Galilean transformation equations.
From: Inertial on 15 Jun 2010 22:52 "rbwinn" <rbwinn3(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.
From: Inertial on 15 Jun 2010 22:56 "rbwinn" <rbwinn3(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. > 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. > 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' > 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.
From: eric gisse on 15 Jun 2010 23:43 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.
From: rbwinn on 15 Jun 2010 23:55
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. |