Galilean transformation equations
in [Relativity]
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From: rbwinn on 18 Jun 2010 09:08 On Jun 16, 9:27 pm, "Inertial" <relativ...(a)rest.com> wrote: > "rbwinn" <rbwi...(a)gmail.com> wrote in message > > news:0e598f70-20d2-4b74-a35f-49051d91c33d(a)t26g2000prt.googlegroups.com... > > > > > > > On Jun 16, 5:04 pm, "Inertial" <relativ...(a)rest.com> wrote: > >> "rbwinn" <rbwi...(a)gmail.com> wrote in message > > >>news:2b2d79e3-ad4e-4786-8a75-9ad65827df01(a)k17g2000pro.googlegroups.com.... > > >> > 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? > > >> Do you mean you are really serious about the nonsense you post? You need > >> some counselling and education. > > > I am dead serious. t'=t is the equation for time coordinates in the > > Galilean transformation equations. > > So a correct clock in S' will show the time in S' which it t'. And t' = t' > according to Galilean transforms. So there is no slowing of moving correct > clocks (or rather, of measuring a single correct clock from two different > frames of reference .. which is what a Galilean transform tells you) Sorry, inertial, the equation is t'=t.
From: rbwinn on 18 Jun 2010 09:26 On Jun 17, 8:02 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > 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. I understand. You are claiming that S cannot be a preferred frame of reference.
From: PD on 18 Jun 2010 11:44 On Jun 17, 5:47 pm, rbwinn <rbwi...(a)gmail.com> wrote: > 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. t and t' stand for *measured* time, Robert. It really helps to know what the variable stand for in an algebraic expression. You can always write down any old algebraic expression and say that it's true. It's when you try to associate the variables in the algebraic expression with physical quantities that it becomes physics, and then the truth of the expression isn't a matter of algebra any more. It's a matter whether when you actually take measured values of those physical quantities and stick them in, the equality holds or not. If you stick measured values in and the equality doesn't hold, then the algebraic expression may be algebraically fine but physically worthless. Under some circumstances, such as in ordinary welding applications, if you use the Galilean transformation and check whether the measured values yield an equality, you find that the precision of the measurement is low enough that the equality holds. In this case, the Galilean transformation is "good enough". But in a large number of other circumstances, which are probably of little interest to welders, the precision is high enough or the circumstances sufficiently different, then the equality no longer holds. And then the Galilean transformation is no good. PD
From: PD on 18 Jun 2010 11:46 On Jun 18, 7:31 am, rbwinn <rbwi...(a)gmail.com> wrote: > On Jun 17, 11:39 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > > > rbwinn wrote: > > > [...] > > > So Robert, why exactly did you start posting this stuff again? > > > What was the thought process? > > Well, I wrote a book about scientists and their concept of relativity > as opposed to reality, and paid a self-publishing company to publish > the book, but they just took the money and did not publish it, using > behavior they had no doubt learned from scientists. I'm sure you have no doubt about where they learned it, Robert. It probably doesn't occur to you that they learned this behavior from previous suckers like you. > So then I started > discussing relativity with scientists in a science forum in Amazon.com > because I thought maybe Amazon would be a better company to get to > publish the book, and then I decided to post again in > sci.physics.relativity to see what was being discussed here. As I > suspected, nothing is being discussed here.
From: rbwinn on 18 Jun 2010 12:07
On Jun 18, 8:46 am, PD <thedraperfam...(a)gmail.com> wrote: > On Jun 18, 7:31 am, rbwinn <rbwi...(a)gmail.com> wrote: > > > On Jun 17, 11:39 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > > > > rbwinn wrote: > > > > [...] > > > > So Robert, why exactly did you start posting this stuff again? > > > > What was the thought process? > > > Well, I wrote a book about scientists and their concept of relativity > > as opposed to reality, and paid a self-publishing company to publish > > the book, but they just took the money and did not publish it, using > > behavior they had no doubt learned from scientists. > > I'm sure you have no doubt about where they learned it, Robert. It > probably doesn't occur to you that they learned this behavior from > previous suckers like you. > > > > > So then I started > > discussing relativity with scientists in a science forum in Amazon.com > > because I thought maybe Amazon would be a better company to get to > > publish the book, and then I decided to post again in > > sci.physics.relativity to see what was being discussed here. As I > > suspected, nothing is being discussed here. Well, as I said, scientists would be 100% in favor of what they did. |