Galilean transformation equations
in [Relativity]
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From: Inertial on 17 Jun 2010 00:18 "rbwinn" <rbwinn3(a)gmail.com> wrote in message news:bb06cc17-a2fa-4854-9095-56d063d63dc2(a)e34g2000pra.googlegroups.com... > On Jun 16, 5:00 pm, "Inertial" <relativ...(a)rest.com> wrote: >> "rbwinn" <rbwi...(a)gmail.com> wrote in message >> >> news:55dce4d8-29e1-4de5-b979-c97cb74b3b2c(a)y6g2000pra.googlegroups.com... >> >> >> >> >> >> > On Jun 15, 7:56 pm, "Inertial" <relativ...(a)rest.com> wrote: >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message >> >> >>news:adbb8478-79f8-4a0e-bfd1-8cb5fcceed94(a)6g2000prg.googlegroups.com... >> >> >> > On Jun 15, 7:10 pm, "Inertial" <relativ...(a)rest.com> wrote: >> >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message >> >> >> >>news:4466492f-0a28-4aec-9a1b-05cce138c867(a)t34g2000prd.googlegroups.com... >> >> >> >> > On Jun 15, 3:57 am, "Inertial" <relativ...(a)rest.com> wrote: >> >> >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message >> >> >> >> >>news:ff4c8b77-ca8b-45a1-9d04-4e614e476447(a)s4g2000prh.googlegroups.com... >> >> >> >> >> > On Jun 13, 5:43 pm, "Inertial" <relativ...(a)rest.com> wrote: >> >> >> >> >> "rbwinn" <rbwi...(a)gmail.com> wrote in message >> >> >> >> >> >>news:8b250e8c-7689-460d-83b3-e25bfb5c83e1(a)11g2000prw.googlegroups.com... >> >> >> >> >> >> > On Jun 13, 7:53 am, "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 >> >> >> >> >> >> >> Amazing .. you appear to know what a Galilean transform >> >> >> >> >> >> is. >> >> >> >> >> >> >> > Experiment shows that a clock in moving frame of >> >> >> >> >> >> > reference >> >> >> >> >> >> > S' >> >> >> >> >> >> > is >> >> >> >> >> >> > slower than a clock in S which shows t >> >> >> >> >> >> >> As measured be S. Hence refuting Galilean transforms >> >> >> >> >> >> >> > According to the Galilean >> >> >> >> >> >> > transformation equations, that slower clock does not >> >> >> >> >> >> > show >> >> >> >> >> >> > t'. >> >> >> >> >> >> >> No .. according to Galilean transforms it DOSE show t' = >> >> >> >> >> >> t. >> >> >> >> >> >> And >> >> >> >> >> >> so >> >> >> >> >> >> Galilean >> >> >> >> >> >> transforms are wrong >> >> >> >> >> >> >> > Time on >> >> >> >> >> >> > the slower clock has to be represented by some other >> >> >> >> >> >> > variable >> >> >> >> >> >> > if >> >> >> >> >> >> > the >> >> >> >> >> >> > Galilean transformation equations are to be used. >> >> >> >> >> >> >> They can't. Because then you are no longer using Galilean >> >> >> >> >> >> transforms >> >> >> >> >> >> >> [snip nonsense that follows] >> >> >> >> >> >> > What do you mean I am no longer using the Galilean >> >> >> >> >> > transformation >> >> >> >> >> > equations? >> >> >> >> >> >> > x'=x-vt >> >> >> >> >> > y'=y >> >> >> >> >> > z'=z >> >> >> >> >> > t'=t >> >> >> >> >> >> Because you said you are not using t' = t .. you are using >> >> >> >> >> something >> >> >> >> >> else. >> >> >> >> >> So it is no longer a Galilean transform. You can't throw >> >> >> >> >> away >> >> >> >> >> your >> >> >> >> >> cake >> >> >> >> >> and >> >> >> >> >> eat it too. >> >> >> >> >> >> > Which one of the equations is not a Galilean >> >> >> >> >> > transformation >> >> >> >> >> > equation? >> >> >> >> >> > I am using t'=t. t is time on a clock in S. t'=t is what is >> >> >> >> > known >> >> >> >> > in >> >> >> >> > algebra as an identity. t' is time on a clock in S. Time on >> >> >> >> > a >> >> >> >> > clock >> >> >> >> > in S' is not t'. >> >> >> >> >> Yes it is .. if you are using Galilean Transforms. It is is >> >> >> >> something >> >> >> >> OTHER >> >> >> >> than t', then you are NO LONGER using Galilean Transforms. >> >> >> >> Simple. >> >> >> >> >> > It has to be shown by some other variable. >> >> >> >> >> So you are no longer using t' for the time. And so you are no >> >> >> >> longer >> >> >> >> using >> >> >> >> Galilean Transforms. As I said. You have made up some >> >> >> >> DIFFERENT >> >> >> >> transform >> >> >> >> instead that treats time differently. >> >> >> >> >> That's fine if you want to do that ... but do not LIE by >> >> >> >> claiming >> >> >> >> you >> >> >> >> are >> >> >> >> using Galilean transforms. A bit of honesty goes a long way. A >> >> >> >> bit >> >> >> >> of >> >> >> >> physics goes even further. Start with the honesty. >> >> >> >> > Here are the Galilean transformation equations. Honest. >> >> >> >> > x'=x-vt >> >> >> > y'=y >> >> >> > z'=z >> >> >> > t'=t >> >> >> >> Yes.. I know what they are.. And they are the ones you go on to >> >> >> NOT >> >> >> use. >> >> >> >> > Notice the equation that says t'=t. That kind of equation is >> >> >> > called an identity in algebra. What it means is that the time in >> >> >> > S' >> >> >> > for transforming coordinates is t', and that t' is the time that >> >> >> > is >> >> >> > on >> >> >> > a clock in S because t'=t. >> >> >> >> Yes .. so according to the transforms, all correctly working clocks >> >> >> tick >> >> >> at >> >> >> the same rate regardless of motion. But we know that they DO tick >> >> >> at >> >> >> different rates due to motion. so Galilean Transforms do not >> >> >> apply. >> >> >> >> > You might want to try practicing with >> >> >> > coordinates in S and S' using t'=t. I am sure you will find that >> >> >> > the >> >> >> > coordinates do transform. So the Galilean transformation >> >> >> > equations >> >> >> > do >> >> >> > not transform coordinates in any way to a clock in S' that >> >> >> > scientists >> >> >> > say they have found to be slower than a clock in S. >> >> >> >> That's right .. Galilean transforms do not work in reality. >> >> >> >> > You cannot use >> >> >> > the time on that clock as t' because t'=t, the time on the faster >> >> >> > clock in S. >> >> >> >> Then you are no longer using Galilean transforsm .. you are using >> >> >> some >> >> >> OTHER >> >> >> transform that does not have the time in one frame the same as the >> >> >> time >> >> >> in >> >> >> another. >> >> >> >> > But the slower clock in S' shows light to be traveling at >> >> >> > c=300,000 km/sec. Surely it must be t'. Not according to the >> >> >> > Galilean transformation equations. >> >> >> >> Wrong .. According to the Galillean transforms it WILL be t'. >> >> >> Experiment >> >> >> shows that it not the case. >> >> >> >> > t'=t, the time on a clock in S. >> >> >> > But the fact that time on a clock in S' shows light to be >> >> >> > traveling >> >> >> > at >> >> >> > c gives us a way to solve for time on that clock from the >> >> >> > Galilean >> >> >> > transformation equations >> >> >> >> No .. it doesn't >> >> >> >> > cn'=ct-vt >> >> >> >> > where n' is time on the slower clock in S'. >> >> >> >> > n'=t(1-v/c) >> >> >> >> So you have used a DIFFERENT equation for the time in S' to what >> >> >> Galillean >> >> >> transforms use. >> >> >> >> So .. as i said .. you are NO LONGER using Galilean transforms >> >> >> >> > As you seem to recognize, n' cannot be used with the >> >> >> > Galilean >> >> >> > transformation equations. >> >> >> >> If you are claiming n' is the time in S'. then that is just a >> >> >> change >> >> >> of >> >> >> letter to use. You are REALYL showing (using conventional >> >> >> notation) >> >> >> >> t' = t ( 1 - v/c) >> >> >> >> Which is NOT the same as >> >> >> >> t' = t >> >> >> >> So you are NOT using Galilean transforms >> >> >> >> > In order to transform coordinates, you have >> >> >> > to convert n' to t' and use the Galilean transformation equation, >> >> >> > t'=t >> >> >> > for time coordinates in S and S'. >> >> >> > Since you claim so vehemently that you have found an error >> >> >> > in >> >> >> > this reasoning, go ahead and show the error you think you have >> >> >> > found. >> >> >> >> I have >> >> >> > Well, no, I am sorry, but you have not. >> >> >> Yes .. I have >> >> >> > Here is what you are >> >> > claiming. You are saying that t' cannot equal t in S' because a >> >> > clock >> >> > shows some other time in that frame of reference. >> >> >> A correct cloak .. Yes. That is a fact by definition of what a >> >> correct >> >> clock is. >> >> >> > As a matter of >> >> > fact, there is no clock in S' that shows t'=t. >> >> >> So Galilean transforms are wrong. Or you are talking about clocks >> >> that >> >> are >> >> wrong. >> >> > The clocks are fine. There just do not happen to be any that show t'. >> >> >> > So the Galilean >> >> > transformation equations regard all clocks the same in S'. >> >> >> Irrelevant >> >> >> > Whatever >> >> > they say has to be converted to t' before transforming coordinates >> >> > with the Galilean transformation equations. >> >> >> No .. they SHOW t'. That is what a correct clock does. Just as a >> >> clock >> >> at >> >> rest in S shows the time t. Nothing needs converting. >> >> > See this equation? >> > t'=t >> >> > That kind of equation is what is known in algebra as an identity. >> > It means that time on a clock in S is t'. >> >> >> > Whenever the time of a >> >> > clock running at any speed has been converted to t', then it can be >> >> > used in the Galilean transformation equations. >> >> >> A correct clock IS showing t' >> > Maybe according to Androcles. Scientists say otherwise. The say a >> > clock in S' is slower than a clock in S. >> >> >> > That is what I do. Sorry if it offends you. >> >> >> Only your arrogance and lies offend me >> >> >> > You shouldn't really be >> >> > getting so offended by correct use of the Galilean transformation >> >> > equations. >> >> >> You aren't doing that. >> >> > Prove it. >> >> I have. > > Well, no, you have not proven anything except that you do not like the > Galilean transformation equations. Wrong on both counts. What YOU have proven is that you do not understand Galilean transforms, not that you are not using them. Your ignorance is obvious
From: Inertial on 17 Jun 2010 00:26 "rbwinn" <rbwinn3(a)gmail.com> wrote in message news:8589aca5-b2b9-4db3-83fc-88d61285dec5(a)u20g2000pru.googlegroups.com... > On Jun 16, 5:04 pm, "Inertial" <relativ...(a)rest.com> wrote: >> "rbwinn" <rbwi...(a)gmail.com> wrote in message >> >> news:42a4cf8f-6e54-4dd9-88d5-6351aeecb6f8(a)z15g2000prh.googlegroups.com... >> >> >> >> >> >> > On Jun 16, 2:45 am, harald <h...(a)swissonline.ch> wrote: >> >> On Jun 15, 11:20 am, rbwinn <rbwi...(a)gmail.com> wrote: >> >> >> > On Jun 13, 5:43 pm, "Inertial" <relativ...(a)rest.com> wrote: >> >> >> > > "rbwinn" <rbwi...(a)gmail.com> wrote in message >> >> >> > >news:8b250e8c-7689-460d-83b3-e25bfb5c83e1(a)11g2000prw.googlegroups.com... >> >> >> > > > On Jun 13, 7:53 am, "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 >> >> >> > > >> Amazing .. you appear to know what a Galilean transform is. >> >> >> > > >> > Experiment shows that a clock in moving frame of >> >> > > >> > reference >> >> > > >> > S' is >> >> > > >> > slower than a clock in S which shows t >> >> >> > > >> As measured be S. Hence refuting Galilean transforms >> >> >> > > >> > According to the Galilean >> >> > > >> > transformation equations, that slower clock does not show t'. >> >> >> > > >> No .. according to Galilean transforms it DOSE show t' = t. >> >> > > >> And >> >> > > >> so >> >> > > >> Galilean >> >> > > >> transforms are wrong >> >> >> > > >> > Time on >> >> > > >> > the slower clock has to be represented by some other variable >> >> > > >> > if >> >> > > >> > the >> >> > > >> > Galilean transformation equations are to be used. >> >> >> > > >> They can't. Because then you are no longer using Galilean >> >> > > >> transforms >> >> >> > > >> [snip nonsense that follows] >> >> >> > > > What do you mean I am no longer using the Galilean >> >> > > > transformation >> >> > > > equations? >> >> >> > > > x'=x-vt >> >> > > > y'=y >> >> > > > z'=z >> >> > > > t'=t >> >> >> > > Because you said you are not using t' = t .. you are using >> >> > > something >> >> > > else. >> >> > > So it is no longer a Galilean transform. You can't throw away >> >> > > your >> >> > > cake and >> >> > > eat it too. >> >> >> > > > Which one of the equations is not a Galilean transformation >> >> > > > equation? >> >> >> > I am using t'=t. t is time on a clock in S. t'=t is what is known >> >> > in >> >> > algebra as an identity. t' is time on a clock in S. Time on a >> >> > clock >> >> > in S' is not t'. It has to be shown by some other variable. >> >> >> Here you show that you still do *not* know what a Galilean >> >> transformation is. The symbol t' in the Galilean transformation refers >> >> to clock time in S'. You can show clock time by some other variable, >> >> but then you do not have a Galilean transformation anymore. >> >> >> Harald >> >> > The Galilean transformation equations I use just say t'=t. t is time >> > on a clock in S. That means according to the Galilean transformation >> > equation t' is time on a clock in S. >> >> Yes .. so correct clocks all run at same rate. We know experimentally >> they >> do not. So cannot use Galillean transforms >> >> > If you have another clock >> > somewhere that shows some other time, you cannot use that time in the >> > Galilean transformation equations until you convert it to t'. >> >> Then it is NOT a correct clock, and so irrelevant. WE do not base >> physics >> on whether we have a good battery in our little bed-side clock. YOU are >> claiming that all correct clocks in S' do NOT show time t', but instead >> show >> some other time that you have called n' (which is your attempt to obscure >> what you are doing). That means you are NOT using Gallilean transforms >> for >> time in S'. > > I absolutely am using the Galilean transformation equations for time > in S'. Nope .. you use a time on a correct clock in S' as n' .. but as a correct clock in S' must show the time in S' (which is t') that means t'=n'. And that means t' <> t Case proven ' > t'=t is shown by a clock in S. There is no other way to > interpret the Galilean transformation equations. I know .. yet you try to do so. It doesn't work > Since you have shown > no proof that t' has to be the time on a clock in S', It is by definition of what t' is and what a correct clock it. You do not need to prove a definition .. and you can see the logic above > the Galilean > transformation equations are valid in this application. Nope. You don't know what a transform is or how to use it. .that much is obvious.
From: Inertial on 17 Jun 2010 00:27 "rbwinn" <rbwinn3(a)gmail.com> wrote in message news:0e598f70-20d2-4b74-a35f-49051d91c33d(a)t26g2000prt.googlegroups.com... > On Jun 16, 5:04 pm, "Inertial" <relativ...(a)rest.com> wrote: >> "rbwinn" <rbwi...(a)gmail.com> wrote in message >> >> news:2b2d79e3-ad4e-4786-8a75-9ad65827df01(a)k17g2000pro.googlegroups.com... >> >> >> >> >> >> > On Jun 16, 1:18 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: >> >> rbwinn wrote: >> >> > On Jun 16, 1:37 am, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: >> >> >> rbwinn wrote: >> >> >> > On Jun 15, 8:43 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: >> >> >> >> rbwinn wrote: >> >> >> >> >> [...] >> >> >> >> >> > Well, every morning I see the sun rise and say, It is a new >> >> >> >> > day. >> >> >> >> > The fact that I do this does not diminish my mental capacity. >> >> >> >> > When >> >> >> >> > the sun comes up, it actually is a new day where I am. >> >> >> >> > Posting >> >> >> >> > the >> >> >> >> > Galilean transformation equations is a similar process. There >> >> >> >> > is >> >> >> >> > really no harm in repeating anything that is true. >> >> >> >> >> So you are autistic. >> >> >> >> > I have been called a lot of things, but you are the first to call >> >> >> > me >> >> >> > autistic. >> >> >> >> If you were not autistic, or a sociopath, you would take a moment >> >> >> to >> >> >> consider why people keep calling you names. >> >> >> >> The answer is not 'because I'm right'. >> >> >> > If people keep calling me names, it would appear that they are the >> >> > sociopaths, not me. >> >> >> Thanks for playing. >> >> > You think this is a game, Eric? >> >> Do you mean you are really serious about the nonsense you post? You need >> some counselling and education. > > > I am dead serious. t'=t is the equation for time coordinates in the > Galilean transformation equations. So a correct clock in S' will show the time in S' which it t'. And t' = t' according to Galilean transforms. So there is no slowing of moving correct clocks (or rather, of measuring a single correct clock from two different frames of reference .. which is what a Galilean transform tells you)
From: Inertial on 17 Jun 2010 01:36 "rbwinn" <rbwinn3(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]
From: PD on 17 Jun 2010 16:06
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 |