Galilean transformation equations
in [Relativity]
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From: artful on 19 Jul 2010 00:18 On Jul 19, 9:41 am, rbwinn <rbwi...(a)gmail.com> wrote: > On Jul 18, 4:02 pm, "whoever" <whoe...(a)whereever.com> wrote: > > > > > > > "rbwinn" wrote in message > > >news:7d43cfe8-cbfa-4621-95ef-1995720826b8(a)a14g2000pro.googlegroups.com.... > > > >On Jul 17, 5:24 am, "Inertial" <relativ...(a)rest.com> wrote: > > >> 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. > > > No .. Its where YOUR 'theory' falls apart. > > > Your theory says that for any two frames (S and S') that the clock in S' > > ticks slower. > > > How do you explain that and resolve YOUR theory with reality? And please > > answer the question: > > > 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? > > > --- news://freenews.netfront.net/ - complaints: n...(a)netfront.net --- > > Suppose that clock A is in frame of reference S, and clock B is in > frame of reference S' and is slower than clock A. And if you suppose that clock A is in frame of reference S', and clock B is in frame of reference S, then clock A is slower than clock B So if you name the frame that clock A is in as S you get B as slower, and if you label it as S', then A is slower So it all depends on whether you label the rest frame of a clock as S or S' ... ie it depends on an arbitrary assigning of a label. That's not physics .. that's nonsense. Try again
From: artful on 19 Jul 2010 00:19 On Jul 19, 9:43 am, rbwinn <rbwi...(a)gmail.com> wrote: > On Jul 11, 10:10 pm, "Inertial" <relativ...(a)rest.com> wrote: > > > > > > > "rbwinn" wrote in message > > >news:3466e4f4-d655-4959-b499-315061eb04bf(a)m17g2000prl.googlegroups.com.... > > > >On Jul 7, 5:06 pm, "whoever" <whoe...(a)whereever.com> wrote: > > >> "rbwinn" wrote in message > > > >>news:ac4b310d-cef8-43aa-a588-73ead74ae6af(a)y12g2000prb.googlegroups.com... > > > >> > Can you answer that honestly? I doubt it. Prove me wrong. > > > >> > n'=t(1-v/c) > > > >> > The clock in S' is slower as observed from either frame of > > >> >reference. > > > >> So if we use n and n' for the time shown on clocks at rest in S and S > > >> respectively we have > > > >> in frames S > > >> n = t > > >> ie the clock shows the correct time in S > > >> and in frame S' > > >> n' = t(1-v/c) = n(1-v/c) > > >> ie the clock runs slow in S' > > > >> So clocks that move will not show the 'correct' time (similar to LET > > >> where > > >> clocks the move slow down and do not show the correct time) > > > >> Correct? > > > >> --- news://freenews.netfront.net/ - complaints: n...(a)netfront.net --- > > > > Why wouldn't the time be correct? > > > Because you said the clock is slow. If it is slow, it is, by definition, no > > correct. That is your claim .. not mine. You say the clocks runs slow .. > > so the time shown on the clock is not correct, yes? > > > > It is the molecules that compose > > > the clock that slow down to keep the speed of light at c in the moving > > > frame of reference. > > > So clocks that move will not show the correct time .. they will show a > > slowed time n', instead of the correct time t'. Yes? > > They will show a time that keeps light at a speed of 300,000 km/sec in > S'. You didn't answer the question So clocks that move will not show the correct time .. they will show a slowed time n', instead of the correct time t'. Yes?
From: artful on 19 Jul 2010 00:20 On Jul 19, 12:40 am, 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. > > Clocks on earth are slower than a clock at the center of the > universe. t'=t only applies to two frames of reference at a time. How do you know which pair of the infinite number of inertial reference frames that exist that the gallilean transforms apply to at a given time? What transform applies to the REST of the inertial reference frames in the universe?
From: PD on 19 Jul 2010 12:14 On Jul 18, 9:44 am, rbwinn <rbwi...(a)gmail.com> wrote: > On Jul 17, 9:52 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > 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? > > I am not going to try to measure it. I did not bother to calculate > it. I am just guessing what it would be. If you want to be more > accurate than that, go ahead and do the math. Ah, good. So you often state you believe things you just guess at? PD
From: PD on 19 Jul 2010 12:16
On Jul 18, 9:28 am, rbwinn <rbwi...(a)gmail.com> wrote: > 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, Not always, Robert. No. That's what MEASUREMENTS say -- not always. > which is what the Galilean > transformation equations show. Equations don't show anything. They claim something, but that claim needs to be put to experimental test. And in this case, Robert, the Galilean transformations don't match experimental measurement. In science, measurements always trump equations. You didn't know that? > Scientists are saying, we are > confused, so all people are required to be confused. |