correct representation of slower clocks
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From: rbwinn on 6 Aug 2010 11:12 On Jul 31, 11:39 am, PD <thedraperfam...(a)gmail.com> wrote: > On Jul 28, 8:39 pm, rbwinn <rbwi...(a)gmail.com> wrote: > > > > > > > On Jul 28, 6:28 pm, artful <artful...(a)hotmail.com> wrote: > > > > On Jul 29, 10:51 am, rbwinn <rbwi...(a)gmail.com> wrote: > > > > > On Jul 26, 7:11 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > On Jul 25, 10:40 pm, rbwinn <rbwi...(a)gmail.com> wrote: > > > > > > > On Jul 25, 7:44 pm, "Inertial" <relativ...(a)rest.com> wrote: > > > > > > > > >"rbwinn" wrote in message > > > > > > > >news:d9d01d61-d162-4090-b2c8-a1528ce45568(a)t5g2000prd.googlegroups.com... > > > > > > > > [snip] > > > > > > > > Lets see if RB is honest enough to clarify his position here with simple > > > > > > > direct answers to a couple of questions. Here's three multiple-choice > > > > > > > questions for you RB. > > > > > > > > 1) Are the measurements of the length of an object (in general) > > > > > > > a) always the same regardless of the motion of the observer measuring it > > > > > > > b) smaller if the observer measuring it is in motion wrt the object > > > > > > > c) larger if the observer measuring it is in motion wrt the object > > > > > > > d) smaller or larger depending on the motion, if the observer measuring > > > > > > > it (using his own rulers and clocks) is in motion wrt the object > > > > > > > > 2) Are the measurements of the ticking rate of a clock > > > > > > > a) always the same regardless of the motion of the observer measuring it > > > > > > > b) slower if the observer measuring it is in motion wrt the clock > > > > > > > c) faster if the observer measuring it is in motion wrt the clock > > > > > > > d) slower or faster depending on the motion, if the observer measuring > > > > > > > it is in motion wrt the clock > > > > > > > > 3) Are the differences in times shown on a pair of mutually at rest > > > > > > > separated clocks (in general) > > > > > > > a) always the same regardless of the motion of the observer measuring > > > > > > > them > > > > > > > b) different if the observer measuring them is in motion wrt the clocks > > > > > > > > NOTE: That in the above we assume that observer use their own clocks and > > > > > > > rulers, at rest wrt them, for making measurements. > > > > > > > > OK .. what are you answers ... no need for any lengthy explanations, or > > > > > > > ad-homs about scientists. I just want to know what your position is: > > > > > > > > 1) > > > > > > > 2) > > > > > > > 3) > > > > > > > Your questions are completely off-topic and irrelevant, but I will > > > > > > answer them anyway. > > > > > > 1. Measurements of length are the same in different frames of > > > > > > reference. That is what the Galilean transformation equations show. > > > > > > Equations do not show what the results of measurements are. > > > > > Measurements do. Actual measurements. > > > > > > > 2. Measurements of the ticking rate of a clock are slower if the clock > > > > > > is in motion relative to the frame of reference with the clock that > > > > > > shows t in the Galilean transformation equations. > > > > > > 3. If two clocks are at rest, they both show the same time regardless > > > > > > of the motion of an observer.- Hide quoted text - > > > > > > > - Show quoted text - > > > > > Actual measurements. I will have to remember that. What about > > > > measurements of rotations of moons of Jupiter? > > > > You're nothing but an ignorant trol > > > Ignorant troll. I will have to remember that. Well, I have pretty > > much decided that I will use the Galilean transformation equations. > > You scientists decide for yourselves what you are going to do. > > I'm not surprised that you would decide this. You come on the group > every now and again to say that you think people should be doing A. > When you find out that scientists are doing B, something other than A, > this confirms in your mind your conviction to do A. > > By the way, scientists bathe regularly. This may account for your > decisions on this matter. So you claim I do not bathe often enough. Well, go ahead and specify how often you think would be enough. What you are saying is that if someone does not conform to your edicts about personal hygiene, they should not be allowed to discuss relativity. First of all, you do not know how often I bathe, and secondly, people in Europe during the time of Galileo did not bathe as often as people today, probably accounting for your bias against Galileo.
From: rbwinn on 6 Aug 2010 11:13 On Aug 3, 8:51 am, PD <thedraperfam...(a)gmail.com> wrote: > On Aug 2, 7:23 pm, rbwinn <rbwi...(a)gmail.com> wrote: > > > > > > > On Jul 31, 12:39 am, "Inertial" <relativ...(a)rest.com> wrote: > > > > "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? > > > I did not say anything about any frames of reference except S and S' > > as they are described by the Galilean transformation equations. > > > x'=x-vt > > y'=y > > z'=z > > t'=t > > > The equations say S is at rest and S' is moving. > > No, they don't. Whatever gave you that ridiculous idea? > > x'=x-vt
From: rbwinn on 6 Aug 2010 11:16 On Aug 2, 5:37 pm, "Inertial" <relativ...(a)rest.com> wrote: > "rbwinn" wrote in message > > news:adea05cf-c42a-4823-ac3e-db2d26980e4f(a)n19g2000prf.googlegroups.com... > > >> 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? > > >I did not say anything about any frames of reference except S and S' > > Which are *any* pair of arbitrarily chosen inertial frames of reference > > >as they are described by the Galilean transformation equations. > > Which apply to *any* pair of arbitrarily chosen inertial frames of reference > > > x'=x-vt > > y'=y > > z'=z > > t'=t > > >The equations say S is at rest and S' is moving. > > Nope. Wrong again > > [snip more avoiding of the question by dishonest rbwinn troll] > > 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? Well, I am not using the equations as arbitrary. I use them to represent reality. A train track is not moving, while a train is moving. The sun is not moving, while Mercury is moving.
From: rbwinn on 6 Aug 2010 11:17 On Jul 31, 11:36 am, PD <thedraperfam...(a)gmail.com> wrote: > On Jul 29, 6:30 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. > > > > > > > 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. > > > > > OK, make the calculation. How do you get a faster speed for the > > > > satellite using the Lorentz equations or General Relativity? They > > > > both say v is the same from either frame of reference. > > > > No, the Lorentz transforms and general relativity do NOT say v is the > > > same from either frame of reference. That would be true for an > > > inertial reference frame, but not for a satellite circling the earth. > > > > > > > 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? > > > > > v is the same from either frame of reference in Special or General > > > > Relativity. > > > > No, only for inertial reference frames, Bobby. > > > It would help if you would learn what special and general relativity > > > actually say. > > > They say that velocity is the same from either frame of reference. > > Who is "they", Robert? > > Scientists say that v is the same in two INERTIAL frames of reference. > The frames of reference for a satellite are not inertial. > > As I said, it would help if you learn what special and general > relativity actually say. At this point I do not care what they say. I have proven them wrong.
From: rbwinn on 6 Aug 2010 11:18
On Jul 31, 11:28 am, PD <thedraperfam...(a)gmail.com> wrote: > 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 Well, I spend all night lying awake worrying about what you think, PD. |