Galilean transformation equations
in [Relativity]
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From: eric gisse on 18 Jun 2010 16:18 rbwinn wrote: [....] > The Galilean transformation equations predict that a clock in S' will > be slower than a clock in S. Bobby, what do you think the "=" symbol in "t' = t" means? [...]
From: rbwinn on 18 Jun 2010 22:13 On Jun 18, 1:18 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > rbwinn wrote: > > [....] > > > TheGalileantransformation equations predict that a clock in S' will > > be slower than a clock in S. > > Bobby, what do you think the "=" symbol in "t' = t" means? > > [...] It means that t' is time on a clock in S. If there is a slower clock somewhere, it does not show t'.
From: rbwinn on 18 Jun 2010 22:21 On Jun 16, 10:36 pm, "Inertial" <relativ...(a)rest.com> wrote: > "rbwinn" <rbwi...(a)gmail.com> wrote in message > > news:702e22b2-1bc0-4a16-9f46-3e571612e517(a)z13g2000prh.googlegroups.com... > > > x'=x-vt > > y'=y > > z'=z > > t'=t > > They are theGalileantransforms. > > However, this and your subsequent posts you seen to have no idea what such a > transform means > > It means we have two observer frames of reference. S and S', where S' is > moving at v relative to S in the x-direction. > > For any event at a location (x,y,z,t) in the S frame, that same event has > coordinates (x' = x-vt, y' = y, z' = z, t' = t). > > If the events are the ticking of a correctly working clock .. then both > frames will observer the clock ticks to happen at the same times, so the > clock ticking rate is the same in both frames, and the clock will show the > correct time in both frames (which is the same). > > It also means that if you have two correctly working clocks .. one moving in > S and at rest in S' and the other moving in S' and at rest in S .. both > clocks will show the same time at all times. > > SoGalileantransforms mean that ALL correctly working clocks will show the > same time readings at the same time regardless of their motion or that of > their observers, and so must tick at the same rate as well. > > > Experiment shows that a clock in moving frame of reference S' is > > slower than a clock in S which shows t. > > This means thatGalileantransforms do not apply, as clocks show the same > time regardless of movement of clock or observer. > > [snip rest as it is nonsense] If t'=t, then a clock in S ticks at the same time in both frames of reference. A clock in S' ticks at the same time in both frames of reference. The problem you have is that a clock in S' does not tick at the same time a clock in S ticks. Your statement about the clocks is incorrect. A clock that does not move relative to S will be faster than a clock that moves relative to S.
From: rbwinn on 18 Jun 2010 22:29 On Jun 18, 1:00 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > rbwinn wrote: > > 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. > > I rather much doubt that. > > > You are claiming that S cannot be a preferred frame of > > reference. > > Since the subject is apparently being changed, let's change to one that > interests me. Why did you start posting here again? Do you feel that I have broken some scientific rule or principle by posting here again?
From: rbwinn on 18 Jun 2010 22:38
On Jun 18, 11:02 am, PD <thedraperfam...(a)gmail.com> wrote: > On Jun 18, 11:18 am, rbwinn <rbwi...(a)gmail.com> wrote: > > > > > > > On Jun 18, 8:44 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > 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 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? > > > > > > 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 theGalileantransformation 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, theGalileantransformation 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 theGalileantransformation is no good. > > > > PD > > > It does not seem to occur to you that you are measuring time two > > different ways in S'. > > But I'm not. I'm measuring it with a clock. > > I don't make measurements with a transformation, Robert. > Measurements are made with instruments, not transformations. > If you have the right transformation, it will tell you what the > relationship will be between the measurement in S and the measurement > in S'. If you have the wrong transformation, it will not tell you the > right relationship. It's so simple that a welder would understand it. > > > First, you are measuring time by the motion of > > S' relative to S. TheGalileantransformation equations account for > > this. The time used to compute the velocity is t, the time in S. > > Second, you are measuring time by the transitions of a cesium isotope > > molecule, which get slower the faster S' is moving. You claim that > > the way to resolve this difference is to say that the slower clock > > shows the same speed as the faster clock in S and compensate by having > > a length contraction. > > I say that the correct way to resolve the difference is to say > > that a slower clock will show a higher speed. That is what reality > > would dictate. No, we do not want reality, say scientists. We want > > to live in a fantasy world. > > OK, live in a fantasy world. > > I do not care if you live in a fantasy world. Why should you > > care if I use the correct equations? > > I don't care at all if you use the correct equations, Robert. You do > as you please. If you want to use theGalileantransformations, please > be my guest. We physicists will use them when it is appropriate to use > them, given the sensitivity of the measurements being made and the > observational circumstances, and will use others when it is > appropriate to use the others. It seems to work better that way. Well, that is wonderful, PD. It is nice to see you becoming so tolerant. Now, with regard to the motion of S' relative to S, were you aware that measurement of this motion constitutes what is known as a clock? If S' is moving at 10 feet per minute, what would be the time when S' has moved 20 feet? See, you could put 2 minutes at 20 feet, 3 minutes at 30 feet, and so on, and you would have a working clock. Then with regard to t', since t' equals t, you could put a mark on S' and marks every ten feet in S and keep time the same way. I know how amazing this concept must seem to a scientist. It is described by the Galilean transformation equations. |