correct representation of slower clocks
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From: Inertial on 31 Jul 2010 03:39 "rbwinn" wrote in message news:531f91bf-82e6-4697-bfaa-1649844f3915(a)x24g2000pro.googlegroups.com... >On Jul 30, 5:22 am, "Inertial" <relativ...(a)rest.com> wrote: >> "rbwinn" wrote in message >> >On Jul 28, 7:24 pm, "whoever" <whoe...(a)whereever.com> wrote: >> >> "rbwinn" wrote in message >> >> You're the one not answering the simple question (because you're an >> >> ignorant >> >> troll): What is the relationship between what a clocks at rest in one >> >> frame >> >> reads compared to that of a clock at rest in some other frame? We know >> >> it >> >> is >> >> not t' = t. >> >whoever, >> > I thought I told you once. There is no clock in the moving frame >> >of reference that shows t'. A clock in S shows t' because t'=t in the >> >Galilean transformation equations. The time on a clock going at any >> >other rate has to be converted to t' before it can be used in the >> >Galilean transformation equations. >> >> Still avoiding the simple questions eh. What is the relationship between >> what a clocks at rest in one frame reads compared to that of a clock at >> rest >> in some other frame? We know it is not t' = t. > >Well, here are the equations. t is the time on a clock in S, a frame >of reference at rest. n' is time on a clock in motion. > > > x'=x-vt > y'=y > z'=z > t'=t There is no n' there. What is the mathematical relationship between what a clock at rest in one (arbitrary) inertial frame compared to that of a clock at rest in some other (arbitrary) inertial frame? > Scientists show no consistency in the information they >disseminate. Lie > One scientist will claim that a moving clock is slower, > the next will claim that a moving clock is faster. Lie > What they are > saying is that n' is not t' because t'=t, the time on a clock in S. If you claim t' = t, but t' is NOT the time showing on a correct clock. Then what is t? And what is the mathematical relationship between what a clock at rest in one (arbitrary) inertial frame compared to that of a clock at rest in some other (arbitrary) inertial frame? >So in order to use the time on the moving clock, its time has to be >converted to the time shown by t'=t. Then it can be used in the >Galilean transformation equations. So from the information scientists >have given, we can say > > n'=F(t) > > Once it is determined what the relationship of n' is to t, then n' >can be converted to t', and the problem can be solved. So .. yet again you fail to answer.... what is the mathematical relationship between what a clock at rest in one (arbitrary) inertial frame compared to that of a clock at rest in some other (arbitrary) inertial frame?
From: PD on 31 Jul 2010 14:24 On Jul 29, 6:19 pm, rbwinn <rbwi...(a)gmail.com> wrote: > On Jul 24, 7:29 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Jul 22, 11:47 pm, rbwinn <rbwi...(a)gmail.com> wrote: > > > > The most famous experiment regarding relativity of time > > > conducted in my lifetime was in 1958 when scientists put a cesium > > > clock in the nosecone of a Vanguard missile and then retrieved it > > > after the flight of the missile to compare it with an identical clock > > > kept on the ground. They reported that the clock in the missile was > > > slower than the clock on the ground by exactly the amount predicted by > > > Einstein's theory of relativity. Since that time we have a multitude > > > of similar experiments using satellites, etc., all with the same > > > reported results. > > > The problem I see with this is that scientists used a set of > > > equations to represent relativity that require a length contraction. > > > Scientists who lived before 1887 such as Galileo and Newton would > > > probably have been able to solve the mathematics of this event > > > correctly if they had seen the experiment because they were using the > > > correct equations, the Galilean transformation equations, but with the > > > wrong interpretation of time. Had they seen an experiment proving > > > that velocity affected the times on clocks, they would doubtlessly > > > have tried to incorporate this information into the equations they > > > were using instead of abandoning the Galilean transformation equations > > > altogether the way more modern scientists did when absolute time did > > > not describe the results of the Michelson-Morley experiment. > > > Why do you think this is doubtlessly what they would have done? > > > Just because YOU would have done it that way doesn't mean anyone else > > would have. > > > I'm sure you've been told this about a great number of things in your > > life. > > Everything Newton and Galileo did indicates that they both knew how to > do mathematics. Of course. But this does not mean they would have used mathematics in the manner you suggested they would. PD
From: PD on 31 Jul 2010 14:28 On Jul 28, 6:56 pm, rbwinn <rbwi...(a)gmail.com> wrote: > On Jul 28, 5:53 am, PD <thedraperfam...(a)gmail.com> wrote: > > > On Jul 27, 9:49 pm, rbwinn <rbwi...(a)gmail.com> wrote: > > > > On Jul 26, 7:15 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Jul 24, 2:57 pm, rbwinn <rbwi...(a)gmail.com> wrote: > > > > > > On Jul 24, 7:38 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On Jul 22, 11:47 pm, rbwinn <rbwi...(a)gmail.com> wrote: > > > > > > > > According to Galileo's principle of equivalence, if the > > > > > > > missile were put in orbit around the earth at the altitude of the > > > > > > > moon, then it would have the same speed in its orbit that the moon has > > > > > > > in its orbit. If the orbits were opposite in direction, then > > > > > > > scientists can calculate for themselves what their theory of > > > > > > > relativity would predict for times on the clock in the nosecone and a > > > > > > > clock on the moon. The Galilean transformation equations and Newton's > > > > > > > equations show that a clock on the moon and a clock in the nosecone > > > > > > > would read the same. > > > > > > > And indeed, the same would be predicted by relativity in the case you > > > > > > mention! > > > > > > > > Both clocks would be slightly slower than a > > > > > > > clock on earth. > > > > > > > Which is different than what the Galilean transformations and > > > > > > Newtonian mechanics predicts. > > > > > > Newton was in fact quite emphatic that time was absolute and > > > > > > immutable, regardless of where it is measured. > > > > > > > What happens to clocks in orbit actually agrees with relativity very > > > > > > well. > > > > > > > > So now let us consider a third satellite at the same > > > > > > > altitude that has an astronaut. > > > > > > > "Calculate your speed," the astronaut is instructed. The > > > > > > > astronaut knows his exact altitude. > > > > > > > How does he know his exact altitude, Robert? > > > > > > There are a number of ways it could be done. To avoid confusion, maybe > > > > > we should have scientists on the ground tell him what it is. > > > > > So, what you are suggesting is that rather than seeing if two > > > > different observers make actual measurements to see which set of > > > > transformations are correct, it's better if one observer just tells > > > > the other observer not to bother measuring at all, and just to take > > > > his word for it that the Galilean transformations are correct. Ah. > > > > > > Are you saying that the satellite has a different altitude in the > > > > > frame of reference of the satellite than is observed from the ground? > > > > > Yes, of course. > > > > Oh, well this is different. So you are saying that the satellite has > > > a lower altitude from the frame of reference of the satellite. You > > > are the first scientist I have seen say this. > > > Well, Bobby, you've not really read anything, have you? > > > > OK, then, I will explain what I believe. The slower clock does not > > > mean the altitude is lower. It means that the clock is slower and is > > > showing a faster speed for the satellite because the length of the > > > orbit is still the same. > > > Well, you believe all sorts of crazy things, Bobby. I don't really > > care what you *believe*. I care about what is supported by > > experimental measurement. I understand that you don't use expermiental > > measurement to help determine what you believe, because you choose to > > disbelieve the experiments too. That's fine for you, Bobby. There are > > all sorts of goofy folks that don't operate scientifically in the > > head, and you're one of them. > > Well, every other scientist I have discussed this with says the > altitude stays the same. You are the only one so far with a lower > altitude. I don't know what scientists you claimed to have talked to that have told you otherwise. Judging by the fact that you make up other statements out of thin air, like train clocks varying from ground clocks in the tenth decimal place and Etvos experiments in Vanguard missiles in 1958, I'm sure you can understand why I think you've just made this up too. PD
From: PD on 31 Jul 2010 14:33 On Jul 30, 2:27 am, rbwinn <rbwi...(a)gmail.com> wrote: > On Jul 29, 2:18 pm, "Paul B. Andersen" <some...(a)somewhere.no> wrote: > > > > > On 29.07.2010 01:56, rbwinn wrote: > > > >>>>>> On Jul 22, 11:47 pm, rbwinn<rbwi...(a)gmail.com> wrote: > > >>>>>>> According to Galileo's principle of equivalence, if the > > >>>>>>> missile were put in orbit around the earth at the altitude of the > > >>>>>>> moon, then it would have the same speed in its orbit that the moon has > > >>>>>>> in its orbit. > > > Close, but not quite. > > Due to the mass of the Moon the speeds would be slightly different. > > Objects fall at the same speed only if their masses are negligible > > compared to the mass of the gravitating body (the Earth). > > The mass of the Moon isn't negligible. > > > >>>>>>> If the orbits were opposite in direction, then > > >>>>>>> scientists can calculate for themselves what their theory of > > >>>>>>> relativity would predict for times on the clock in > > >>>>>>> the nose cone and a clock on the moon. > > > Quite. > > And here is what they would calculate: > > > Look at this animation:http://home.c2i.net/pb_andersen/Satellites.html > > Choose the scenario: "Circ. Moon orbit + Moon orbit". > > The red satellite is in Moon orbit. > > The relative rate difference is 6.808E-10 at aphelion > > and 6.783E-10 at perihelion. > > The rate varies slightly because of the eccentricity, but it is > > always _fast_. > > > Now, look at this animation: > > (Not quite finished and probably never will be)http://home.c2i.net/pb_andersen/EarthMoon.html > > Choose the sceneario "High altitude satellite". > > Ignore the green satellite, we are only interested in > > the dark grey Moon. > > The "Moons clock rel. rate difference" is the rate of a clock on > > the Moon's surface, facing the Earth. > > It is 6.494E-10 at aphelion and 6.468E-10 at perihelion. > > > The rate is slightly less than for the satellite clock. That is because > > of the Moon's gravity; the gravitational potential difference is less > > for the Moon clock than for the satellite clock. > > > >>>>>>>The Galilean transformation equations and Newton's > > >>>>>>> equations show that a clock on the moon and a clock in the nosecone > > >>>>>>> would read the same. > > > According to Galilean relativity all clocks run at the same rate. > > But they don't. > > So what can we conclude from that fact? > > > >>>>>>> Both clocks would be slightly slower than a > > >>>>>>> clock on earth. > > > Nope. Faster. > > > [..] > > > -- > > Paul > > >http://home.c2i.net/pb_andersen/ > > Uh huh. Well, I am certain that scientists of today can convince > themselves that Einstein's theory explains all things just the way > Einstein said it did. The trouble I see with what they are doing is > that they worked the arithmetic wrong. > It would not matter what experiment shows, scientists of today > can find a way to make Einstein's theory match the experimental > results. If you need to make clocks go faster, you can make them go > faster. If they need to go slower, you can make them go slower. Well, this is the problem, Bobby. You say that it does not matter what experiments show. To a scientist, this is exactly backwards. Everything depends on what experiments show. You cannot "prove" anything with mathematics in science if experiments say otherwise. > Well, I have decided to use the Galilean transformation equations, > and scientists can do whatever they decide to do. Yes, of course you've decided that. No one is stopping you. It's a wholly unscientific decision to do that, and you will get wrong answers under some circumstances, but you don't care about that. Scientists have good reasons for not taking the path you've chosen. PD
From: PD on 31 Jul 2010 14:34
On Jul 30, 2:32 pm, rbwinn <rbwi...(a)gmail.com> wrote: > On Jul 30, 5:24 am, YBM <ybm...(a)nooos.fr.invalid> wrote: > > > rbwinn a écrit : > > > > Well, I have decided to use the Galilean transformation equations, > > > and scientists can do whatever they decide to do. > > > This is pointless given that you don't consider any real experiments but > > the fake ones you made up in your ill mind. > > Well, tell me about a real experiment, YBM. The only ones I know > about are done by scientists, which makes them suspect. And who does experiments that you trust? |