Galilean transformation equations
in [Relativity]
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From: Inertial on 17 Jul 2010 08:24 "rbwinn" wrote in message news:f67fd82f-41a0-4534-be1b-e939bd623bbb(a)w35g2000prd.googlegroups.com... On Jul 4, 5:09 am, artful <artful...(a)hotmail.com> wrote: > On Jun 13, 11:31 pm, 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 theGalilean > > transformation equations, that slower clock does not show t'. Time on > > the slower clock has to be represented by some other variable if the > >Galileantransformation 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 theGalilean > > 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? > >> OK .. so RBWINN is now (finally) claiming there is an absolute frame, >> S, in which the center of mass of the universe is at rest. >> >> He is also claiming that clocks in motion relative to that absolute >> frame the will run slow. >> >> Q1: Does EVERYTHING in motion relative to that frame run slow, or only >> some clocks? >> >> Q2: Are clock on earth all running slow then? >> >> Q3: If time is the same everywhere (as RBWINN agreed is the case due >> to t'=t) then why not just set all clocks to show the time t? Then >> there is no slow clocks and Gallilean transforms apply. > >The fastest clock would be at the center of gravity of the universe. >all other clocks are slower than that clock. t'=t applies to only two >frames of reference at a time. And not to clocks, it appears .. or else they would not run slow > For instance, if you are talking about >the earth and the moon, time on a clock on the moon would be n' and >time on the earth would be t. And vice versa, of course. . which means clocks run both faster than each other and slow than each other. A contradiction > If you are talking about the earth and >the sun, time on a clock on earth would be n', and time on the sun >would be t. And vice versa, of course. . which means clocks run both faster than each other and slow than each other. A contradiction So .. if you have two clocks, A and B, where A moves at v relative to B, and so B at -v relative to A, which is faster?
From: PD on 17 Jul 2010 12:52 On Jul 16, 6:07 pm, rbwinn <rbwi...(a)gmail.com> wrote: > On Jul 15, 11:16 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On Jul 14, 9:10 pm, rbwinn <rbwi...(a)gmail.com> wrote: > > > > On Jul 6, 7:58 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Jul 5, 9:13 am, rbwinn <rbwi...(a)gmail.com> wrote: > > > > > > On Jul 4, 12:03 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On Jul 3, 6:52 pm, rbwinn <rbwi...(a)gmail.com> wrote: > > > > > > > > On Jul 3, 3:17 pm, "Inertial" <relativ...(a)rest.com> wrote: > > > > > > > > > "rbwinn" wrote in message > > > > > > > > >news:c34cba53-2a43-453f-936b-7088df7d2bef(a)j7g2000prj.googlegroups.com... > > > > > > > > > On Jul 3, 1:01 am, "Inertial" <relativ...(a)rest.com> wrote: > > > > > > > > > > "rbwinn" wrote in message > > > > > > > > > >news:7a91960b-b849-4b8f-b358-0aceb2d1b712(a)i9g2000prn.googlegroups.com... > > > > > > > > > > >On Jun 28, 10:25 pm, "Inertial" <relativ...(a)rest.com> wrote: > > > > > > > > > >> It sounds like perhaps you are proposing something similar to LET > > > > > > > > > > >> In LET, reality isGalilean. Space doesn't contract and time doesn't > > > > > > > > > >> slow > > > > > > > > > >> down. TheGalileantransforms apply. > > > > > > > > > > >> However, in that simple 3D galillean universe, what happens is clocks > > > > > > > > > >> (and > > > > > > > > > >> all processes) run slower and rulers (and all matter and fields) > > > > > > > > > >> contract > > > > > > > > > >> due to absolute motion. > > > > > > > > > > >> They do so in such a way that the MEASUREMENTS made with such clocks > > > > > > > > > >> and > > > > > > > > > >> rulers are no longer related byGalileantransforms, but by Lorentz > > > > > > > > > >> transforms. > > > > > > > > > > >> It seems you are proposing the instead, we just have clock running slow > > > > > > > > > >> so > > > > > > > > > >> that the relation ship between what we MEASURE clocks (and processes) > > > > > > > > > >> to > > > > > > > > > >> do > > > > > > > > > >> is related by > > > > > > > > > > >> x'=x-vt > > > > > > > > > >> y'=y > > > > > > > > > >> z'=z > > > > > > > > > >> t'=t(1-v/c) > > > > > > > > > > >> Only you are using n for the measured time, there is no need for that. > > > > > > > > > >> If > > > > > > > > > >> you are talking about what is measured, you can just use x,y,z,t. > > > > > > > > > > >Those equations do not work. > > > > > > > > > > I know your equations are wrong. Glad to hear you admit it > > > > > > > > > > > They require a different reference for > > > > > > > > > > time in S' than in S. TheGalileantransformation equations require > > > > > > > > > > t' to equal t. > > > > > > > > > > And so your equation using t(1-v/c) for time in S' is wrong. > > > > > > > > > >> So .. given that the definition of a correct clock is one that shows the > > > > > > > > >> time in the frame in which it is at rest .... what is the formula for the > > > > > > > > >> time shown on a correct clock at rest in S' as observed by an observer at > > > > > > > > >> rest in frame S ?? > > > > > > > > > >> Can you answer that honestly? I doubt it. Prove me wrong. > > > > > > > > > >The clock in S' is ticking slower than the clock in S as observed from > > > > > > > > >either frame of reference. A clock at rest in S' is moving with a > > > > > > > > >velocity of v relative to an observer in S. The time on the clock > > > > > > > > >would be > > > > > > > > > > n'=t(1-v/c) > > > > > > > > > >where t is time on a clock at rest in S. > > > > > > > > > You've still not answered .. just calling it 'S' doesn't say what the frame > > > > > > > > is. Are you at rest in this frame S now? Am I? Is anything? > > > > > > > > > Lets ask again .. see if you can answer this time > > > > > > > > > So in what frame of reference are the clocks ticking at the 'correct' > > > > > > > > rate, and not slowed by motion? What is the relationship between the > > > > > > > > time shown on some clock moving in that frame, and the actual time in > > > > > > > > that frame? > > > > > > > > > And a further question > > > > > > > > > If you have two frames moving relative to each other, and each with a clock > > > > > > > > at rest in them .. which clock runs slow and which runs fast? And why will > > > > > > > > they do that .. why don't the people at rest in those frames simply set the > > > > > > > > clocks to the correct rate .. why do they let their clocks run slow or fast? > > > > > > > > There are reasons why things happen, including motion. Now, I know > > > > > > > you scientists are all impressed by having a train stand still and the > > > > > > > railroad track moving. The problem with it is that it is not > > > > > > > reality. The train is still what is moving. > > > > > > > The Earth is not moving, Robert? Then why do the locations of the > > > > > > other planets move in the night sky? > > > > > > The earth is moving relative to the sun, PD. > > > > > Yes, it is. And so you see, the railroad tracks, which are connected > > > > to the moving earth, are also moving. Don't you think so too? > > > > So what? S represents the railroad track, and S' represents the > > > train. A clock on the train is slower than a clock on the ground by > > > the railroad track. > > > As measured from somone on the ground by the railroad track. > > > By the way, a clock on the ground by the railroad track is slower than > > a clock on the train, as measured by someone on the train. Did this come as a surprise to you? > > > > So all we have to do is put a clock on the ground > > > by the railroad track. What is supposed to be so difficult about that? > > > It's not at all difficult. Why do you suppose it is difficult? > > What do you think the size of the slowing-down effect is for a typical > > train speed, Robert? > > You took a year of college classes. Surely you can do arithmetic. > > There would be a difference after about ten decimal places, I believe. What speed are you using for the train, Robert? Let's suppose for an instant that you've done the calculation correctly (or at all). So what do you suppose would be involved in measuring a time difference in the tenth decimal place?
From: PD on 17 Jul 2010 12:53 On Jul 16, 6:06 pm, rbwinn <rbwi...(a)gmail.com> wrote: > On Jul 4, 5:09 am, artful <artful...(a)hotmail.com> wrote: > > > > > > > On Jun 13, 11:31 pm, 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 theGalilean > > > transformation equations, that slower clock does not show t'. Time on > > > the slower clock has to be represented by some other variable if the > > >Galileantransformation 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 theGalilean > > > 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? > > > OK .. so RBWINN is now (finally) claiming there is an absolute frame, > > S, in which the center of mass of the universe is at rest. > > > He is also claiming that clocks in motion relative to that absolute > > frame the will run slow. > > > Q1: Does EVERYTHING in motion relative to that frame run slow, or only > > some clocks? > > > Q2: Are clock on earth all running slow then? > > > Q3: If time is the same everywhere (as RBWINN agreed is the case due > > to t'=t) then why not just set all clocks to show the time t? Then > > there is no slow clocks and Gallilean transforms apply. > > The fastest clock would be at the center of gravity of the universe. What center of gravity of the universe? > all other clocks are slower than that clock. t'=t applies to only two > frames of reference at a time. For instance, if you are talking about > the earth and the moon, time on a clock on the moon would be n' and > time on the earth would be t. If you are talking about the earth and > the sun, time on a clock on earth would be n', and time on the sun > would be t.
From: rbwinn on 18 Jul 2010 10:21 On Jul 17, 9:53 am, PD <thedraperfam...(a)gmail.com> wrote: > On Jul 16, 6:06 pm, rbwinn <rbwi...(a)gmail.com> wrote: > > > > > > > On Jul 4, 5:09 am, artful <artful...(a)hotmail.com> wrote: > > > > On Jun 13, 11:31 pm, 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 theGalilean > > > > transformation equations, that slower clock does not show t'. Time on > > > > the slower clock has to be represented by some other variable if the > > > >Galileantransformation 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 theGalilean > > > > 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? > > > > OK .. so RBWINN is now (finally) claiming there is an absolute frame, > > > S, in which the center of mass of the universe is at rest. > > > > He is also claiming that clocks in motion relative to that absolute > > > frame the will run slow. > > > > Q1: Does EVERYTHING in motion relative to that frame run slow, or only > > > some clocks? > > > > Q2: Are clock on earth all running slow then? > > > > Q3: If time is the same everywhere (as RBWINN agreed is the case due > > > to t'=t) then why not just set all clocks to show the time t? Then > > > there is no slow clocks and Gallilean transforms apply. > > > The fastest clock would be at the center of gravity of the universe. > > What center of gravity of the universe? > > > > > all other clocks are slower than that clock. t'=t applies to only two > > frames of reference at a time. For instance, if you are talking about > > the earth and the moon, time on a clock on the moon would be n' and > > time on the earth would be t. If you are talking about the earth and > > the sun, time on a clock on earth would be n', and time on the sun > > would be t. So you are saying that the universe does not have a center of gravity. Why would everything else have a center of gravity, but not the universe?
From: rbwinn on 18 Jul 2010 10:28
On Jul 17, 5:24 am, "Inertial" <relativ...(a)rest.com> wrote: > "rbwinn" wrote in message > > news:f67fd82f-41a0-4534-be1b-e939bd623bbb(a)w35g2000prd.googlegroups.com... > > On Jul 4, 5:09 am, artful <artful...(a)hotmail.com> wrote: > > > > > > > On Jun 13, 11:31 pm, 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 theGalilean > > > transformation equations, that slower clock does not show t'. Time on > > > the slower clock has to be represented by some other variable if the > > >Galileantransformation 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 theGalilean > > > 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? > > >> OK .. so RBWINN is now (finally) claiming there is an absolute frame, > >> S, in which the center of mass of the universe is at rest. > > >> He is also claiming that clocks in motion relative to that absolute > >> frame the will run slow. > > >> Q1: Does EVERYTHING in motion relative to that frame run slow, or only > >> some clocks? > > >> Q2: Are clock on earth all running slow then? > > >> Q3: If time is the same everywhere (as RBWINN agreed is the case due > >> to t'=t) then why not just set all clocks to show the time t? Then > >> there is no slow clocks and Gallilean transforms apply. > > >The fastest clock would be at the center of gravity of the universe. > >all other clocks are slower than that clock. t'=t applies to only two > >frames of reference at a time. > > And not to clocks, it appears .. or else they would not run slow > > > For instance, if you are talking about > >the earth and the moon, time on a clock on the moon would be n' and > >time on the earth would be t. > > And vice versa, of course. . which means clocks run both faster than each > other and slow than each other. A contradiction > > > If you are talking about the earth and > >the sun, time on a clock on earth would be n', and time on the sun > >would be t. > > And vice versa, of course. . which means clocks run both faster than each > other and slow than each other. A contradiction > > So .. if you have two clocks, A and B, where A moves at v relative to B, and > so B at -v relative to A, which is faster? Well, here is where reality disagrees with science. In reality, if you have one clock slower than the other, it is slower than the other from both frames of reference, which is what the Galilean transformation equations show. Scientists are saying, we are confused, so all people are required to be confused. |