Galilean transformation equations
in [Relativity]
|
From: rbwinn on 17 Jun 2010 18:40 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 the Galilean transforms. > > 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. > > So Galilean transforms 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 that Galilean transforms do not apply, as clocks show the same > time regardless of movement of clock or observer. > > [snip rest as it is nonsense] Well, if you do not want to use the Galilean transformation equations, use whatever equations you want to use. I will use the Galilean transformation equations because they are the correct equations. The way to tell if a set of equations is correct is whether or not they have a length contraction. If they have a length contraction, they are not the correct equations.
From: rbwinn on 17 Jun 2010 18:47 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 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? > > 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.
From: eric gisse on 17 Jun 2010 21:04 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?
From: YBM on 17 Jun 2010 21:24 rbwinn a �crit : > 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 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? >> 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. The point is: when are you planning to go back to the hospital poor stupid, disgusting, lost Robert?
From: rbwinn on 17 Jun 2010 22:21
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. |