Galilean transformation equations
in [Relativity]
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From: YBM on 4 Jul 2010 02:33 rbwinn a �crit : > [...] With regard to > motion, there is always some force which causes the motion. What you wrote there is aristotelician (ancient greece, you know) physics, known to be wrong since Galileo and Newton.
From: YBM on 4 Jul 2010 02:39 rbwinn a �crit : > 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) Such a clock won't account for a light with equation of motion x=-ct in S to have a speed c in S' too. x=-ct => x' = x - vt = -ct-vt = -(c+v)t = -[(c+v)/(1-v/c)]*n' = -[ c*(c+v)/(c-v) ]*n' NOT EQUAL TO -cn' So your equation failed at conserving light speed between frames.
From: artful on 4 Jul 2010 08:09 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 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? 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.
From: mpc755 on 4 Jul 2010 09:01 On Jul 4, 8:09 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 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? > > 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? > Everything is affected by the aether pressure in which it exists. Absent gravity, similar clocks will run slower the greater their momentum relative to the aether. The greater their momentum relative to the aether the more aether the clocks displace the greater the pressure exerted by the displaced aether towards the clock the slower the clock ticks. > Q2: Are clock on earth all running slow then? > Slower then they would absent the pressure exerted by the aether displaced by the earth. > 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. Time is a concept. Just asking the question 'is time the same everywhere' shows you do not understand time. If the Observers on the train and the Observers on the embankment know their state with respect to the aether and set their clocks accordingly then there is no relativity of simultaneity. There is simultaneity of relativity. If the Observers on the embankment and the Observers on the train determine the state of the aether as almost at rest with respect to the surface of the Earth and set their clocks accordingly, events which are determined to be simultaneous by the Observers on the embankment will also be determined to be simultaneous by the Observers on the train. The analogy is the embankment and train to exist in water at rest with respect to the embankment and for the train to be full of flat bed cars that ride under and do not disturb the water. If the Observers on the train and the Observers on the embankment know they exist in water at rest with respect to the embankment and lightning strikes occur at A/A' and B/B' and the light from the lightning strikes arrive at M on the embankment simultaneously then both the Observers on the embankment and the Observers on the train conclude the lightning strikes were simultaneous even though the light from the lightning strikes do not reach the Observer at M' simultaneously. In this analogy, the light propagates with respect to the water. Light propagates with respect to the state of the aether. The state of which is determined by its connections with the matter and the state of the aether in neighboring places. Which is the aether's state of displacement.
From: PD on 4 Jul 2010 15:03
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 reason for this is > that the train cannot move the earth. When the wheels of the train > are turned by the engine, the train moves, not the earth. This is > just a scientific fact. That means the clock on the train is slower > than the clock by the side of the railroad track. A clock in a > satellite is slower than a clock on earth. A clock on Mercury is > slower than time on the sun. If you cannot understand this, just keep > asking about it. I can explain reality to you as many times as is > necessary.- Hide quoted text - > > - Show quoted text - |