Galilean transformation equations
in [Relativity]
|
From: Sue... on 13 Jun 2010 11:44 On Jun 13, 9:31 am, rbwinn <rbwi...(a)gmail.com> wrote: [...] http://allthingsignant.files.wordpress.com/2010/02/danica-patrick.jpg > 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? Let's make a deal. I will contribute one wheel of a Formula 1 race car if you will contribute the other little bits to make it complete. Then we can take turns driving it on weekends. Anomalous Precessions http://www.mathpages.com/rr/s6-02/6-02.htm Sue...
From: rbwinn on 13 Jun 2010 15:39 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 Which one of the equations is not a Galilean transformation equation?
From: rbwinn on 13 Jun 2010 15:54 On Jun 13, 8:44 am, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > On Jun 13, 9:31 am, rbwinn <rbwi...(a)gmail.com> wrote: > [...]http://allthingsignant.files.wordpress.com/2010/02/danica-patrick.jpg > > > 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? > > Let's make a deal. I will contribute one wheel > of a Formula 1 race car if you will contribute > the other little bits to make it complete. > Then we can take turns driving it on weekends. > > Anomalous Precessionshttp://www.mathpages.com/rr/s6-02/6-02.htm > > Sue... I don't drive a car, sorry. Robert B. Winn
From: eric gisse on 13 Jun 2010 19:48 rbwinn wrote: [...] Oh for fucks sake.
From: Inertial on 13 Jun 2010 20:43
"rbwinn" <rbwinn3(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? |