correct representation of slower clocks
in [Relativity]
|
Prev: Attraction of opposite electric charges would form neutronium
Next: EINSTEIN TOPPLED BY MOVING OBSERVER
From: PD on 24 Jul 2010 10:29 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.
From: PD on 24 Jul 2010 10:32 On Jul 22, 11:47 pm, rbwinn <rbwi...(a)gmail.com> wrote: > There seems little doubt that the clock in the nosecone of the > Vanguard missile was slower than the clock on the ground by the amount > the equations showed, but we are still brought to a conflict of > interpretation by the fact that the equations used to make the > calculation of time for the slower clock incorporate a length > contraction which does not manifest itself in the parameters of the > experiment. And how do you know that length contraction did not manifest itself? Where in this experiment was the measurement of length that did not agree with a prediction of length contraction? If in an experiment you do not even bother to measure a length, does this tell you that length contraction did not occur? > For instance, suppose that the Vanguard missile had been > put in orbit around the earth instead of falling back to earth and > recovered. How do we then calculate the rate of the clock in the > nosecone? It would have involved a different calculation with the same principles.
From: PD on 24 Jul 2010 10:38 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? > From this he knows the exact > length of his orbit. He times one orbit with the clock in his > satellite and divides that time into the length of his orbit. Does he > get a length contraction or does he get a faster speed for his > satellite than an observer on the ground making the same calculation? > You cannot make this calculation with Einstein's theory of > relativity. Actually, you can. I'm shocked that you think it can't be done. > It requires a length contraction and the same speed > calculated from the satellite as observed from the ground. What on earth makes you say THAT, Robert? > So, > although Einstein's equations give an answer that agrees with > experimental data for time, the equations do not agree with reality > with regard to distance.
From: PD on 24 Jul 2010 10:41 On Jul 23, 6:08 pm, rbwinn <rbwi...(a)gmail.com> wrote: > On Jul 22, 11:14 pm, "Inertial" <relativ...(a)rest.com> wrote: > > > > > > > >"rbwinn" wrote in message > > >news:114f2da8-f6d4-4a98-8af4-79dd4e710f71(a)p22g2000pre.googlegroups.com.... > > > >On Jul 22, 10:31 pm, "Inertial" <relativ...(a)rest.com> wrote: > > >> "rbwinn" wrote in message > > > >>news:3f93b27a-8cb5-48fa-a85a-bd880bd73984(a)x20g2000pro.googlegroups.com... > > > >> > You did not prove anything > > > >> You proved Galilean transforms do not work .. in that they are > > >> contradicted > > >> by what we observe experimentally. I just pointed out that you did it. > > > >> > except that you are capable of saying, > > >> > Absolute time, absolute time, absolute time, just as I predicted. > > > >> Funny .. you predict that the thing the shows Galilean transforms are > > >> incorrect would be used to prove Galilean transforms are incorrect. Not > > >> a > > >> terribly clever prediction. > > > >> Nor does your prediction reduce in anyway the validity of the arguments > > >> that > > >> refute Galilean transforms, because those transforms mean that time is > > >> the > > >> same in all frames (and we know, and you admit, that time is NOT the same > > >> in > > >> all frames). > > > >> > I already proved what I set out to prove. > > > >> No .. You proved Galilean transforms do not work. That's not what you > > >> set > > >> out to prove. But you did it anyway. > > > >Whatever. > > > What a dishonest way of ignoring that you've just proved yourself wrong.. > > Galilean transforms mean time is the same everywhere, reality shows > > otherwise, so Galilean transforms do not work in realirty. Simple. And you > > just proved it yourself > > Well, I am just using the Galilean transformation equations to > describe the rotation of the earth. For you scientists, the earth > rotates every 24 hours. That's not quite so, Robert. For scientists, the Earth rotates in 24 hours only in a particular frame of reference. In other frames of reference, the rotation rate is different. You sound like you're shocked to hear that. Time has not been based on the rotation of the earth for a long, long time, and for good reason. > From S the earth rotates once every 24 > hours. From S', the earth rotates once every 24 hours. The Galilean > transformation equations show this with the equation t'=t. See, 24 > hours = 24 hours. Amazing how mathematics works.
From: PD on 24 Jul 2010 10:41
On Jul 23, 9:23 pm, rbwinn <rbwi...(a)gmail.com> wrote: > On Jul 23, 4:35 pm, "Inertial" <relativ...(a)rest.com> wrote: > > > > > > > "rbwinn" wrote in message > > >news:6b2e9130-354e-45aa-a6fb-64780df93f6e(a)y32g2000prc.googlegroups.com.... > > > >On Jul 22, 11:14 pm, "Inertial" <relativ...(a)rest.com> wrote: > > >> >"rbwinn" wrote in message > > >> What a dishonest way of ignoring that you've just proved yourself wrong. > > >> Galilean transforms mean time is the same everywhere, reality shows > > >> otherwise, so Galilean transforms do not work in realirty. Simple.. And > > >> you > > >> just proved it yourself > > > >Well, I am just using the Galilean transformation equations to > > >describe the rotation of the earth. > > > They say the time is the same everywhere, whether you talk about earth of a > > spaceship or the sun or some little green man on the other side of the > > universe. > > > > For you scientists, the earth > > > rotates every 24 hours. > > > Close enough. Galilean transforms say ALL correct clocks will agree on > > that. But in reality it depends on who measures it. As your own example > > shows. > > > > From S the earth rotates once every 24 > > > hours. From S', the earth rotates once every 24 hours. > > > Nope .. you just showed that from a missile it take less time to rotate.. > > Try to keep up. > > > > The Galilean > > >transformation equations show this with the equation t'=t. > > > Yes they do .. glad you agree. > > > > See, 24 > > >hours = 24 hours. Amazing how mathematics works. > > > And how it doesn't correspond to reality in this case. Because according to > > the time in a missile, it takes less than 24 hours. > > I am talking about reality, Inertial. The earth rotates every 24 > hours. In a particular reference frame. In others, not. > You claim it does not. Whether you are viewing the earth from > S or S', the earth rotates once every 24 hours. So now you have two > clocks running at different rates. One shows the earth rotating once > every twenty four hours, the other shows it rotating in less than 24 > hours. That would be the one in the nosecone of the missile. > How fast is the missile going? > If you measure the speed of the missile by t', the rotation of the > earth, it goes a certain number of miles during a certain number of > degrees of rotation. You get the same result from t using degrees of > rotation or the clock in S which shows t. The clock in S' shows a > faster speed because less time has elapsed on the clock in the > nosecone of the missile. These are just physical facts, Inertial. > Sorry. |