Ballistic Theory, Progress report...Suitable for 5yo Kids
in [Relativity]
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From: jgreen on 12 Aug 2005 06:53 Jeff Root wrote: > Jim Greenfield replied to Jeff Root: > > >> >> Jim Greenfield replied to Paul B. Andersen: > >> >> > >> >> >> > And how big a fool(or coward), does it take to rabbit on about > >> >> >> > EARTH rotation, when the issue has NOTHING to do with that? > >> >> >> > We are discussing the rotation of an axle perpendicular to the > >> >> >> > earth, and the earth rotation has nothing to do with the > >> >> >> > scenario; ONLY the difference in gravity. > >> >> >> > >> >> >> I don't know what YOU are discussing, but what I > >> >> >> was answering in the posting you responded to was > >> >> >> this particular challenge defined by Sue: > >> >> >> | You are of course welcome to advance an opinion > >> >> >> | about how an axel should behave if it were repeating > >> >> >> | a geosynchronous clock to the ground or if it were > >> >> >> | repeating a ground clock to a geosynchronous satellite. > >> >> >> | Neither you nor Bz seem able to interpret what Einstein's > >> >> >> | relativity say's the shaft should do. > >> >> >> > >> >> >> My answer is what GR "say's the shaft should do". > >> >> >> And whether you like it or not, the Earth IS rotating. > >> >> >> And the rotation of the Earth IS relevant to GR's > >> >> >> prediction of what "the shaft should do". > >> >> >> > >> >> >> However, this was not my main point with "one Earth > >> >> >> rotation" in my scenario, see below. > >> >> >> > >> >> >> The apparent paradox is this: > >> >> >> If the top of the axle (or shaft) rotates at a slower > >> >> >> rate than the bottom, the axle should be twisted, > >> >> >> and "wind up" more and more as time passes. > >> >> >> I show that this isn't so. > >> >> >> To do that we can compare the number of turns done > >> >> >> by the bottom and the top of the axle when it points > >> >> >> in two different directions relative to the distant stars. > >> >> >> If the number of turns are equal, it will not twist. > >> >> >> I have - somewhat arbitrarily - chosen to compare the number > >> >> >> of turns of ends of the axle each time the axle points > >> >> >> in the same direction, that is after "one Earth rotation". > >> >> >> > >> >> >> > I don't expect you to answer-----you cannot! > >> >> >> > >> >> >> But I did. Below is my answer again. > >> >> >> This is what GR say will happen. > >> >> >> I challenge you to find and point out an inconsistency. > >> >> >> Your opinion of GR is irrelevant. > >> >> >> The challenge is to point out an inconsitency > >> >> >> in GR, showing that there is a real paradox. > >> >> >> > >> >> >> Let there be a clock A on the ground at equator. > >> >> >> Let there be a clock B in geostationary orbit. > >> >> >> Let both clocks be on the same radius. > >> >> >> (on the same line through the center of the Erth) > >> >> >> > >> >> >> Let A measure the proper duration of one Earth rotation to be T. > >> >> >> Then, according to GR, B will measure the proper > >> >> >> duration of one Earth rotation to be longer, T + delta_T. > >> >> >> > >> >> >> Let there be an axle between the two clocks. > >> >> >> Let this axle rotate in such a way that there is no > >> >> >> mechanical stress in the axle. > >> >> >> Let the axle rotate N times during one Earth rotation. > >> >> >> > >> >> >> A will measure the rotational frequency to be f_g = N/T > >> >> >> while B will measure it to be f_s = N/(T + delta_T). > >> >> >> > >> >> >> So the ground clock will measure the axle to rotate > >> >> >> faster than the satellite clock will, but both will > >> >> >> agree that the axle rotates N times per Earth rotation. > >> >> >> > >> >> >> frequency * duration = number_of_rotations > >> >> >> f_g*T = N > >> >> >> f_s*(T + delta_T) = N > >> >> >> > >> >> >> Loosly said: > >> >> >> "The satellite clock will see the axle rotate slower, > >> >> >> but for a longer time." > >> >> >> > >> >> >> I do not expect you to point out an inconsistency, > >> >> >> because there are none. > >> >> >> I do however expect you to laugh at what you don't understands. > >> >> >> Fools do. > >> >> > > >> >> > After all your pathetic attempts to duck the question, it STILL > >> >> > remains!! > >> >> > You claim here that the axle does NOT twist under GR. Since two of the > >> >> > clocks are ATTACHED to the axle, they MUST read the SAME time. But GR > >> >> > says the high unconnected (different gravity) one runs differentlty. > >> >> > This is a CONTRADICTION!!!!!!!!!!!!!!!!!!! of the "theory" > >> >> > >> >> Jim, > >> >> > >> >> How can a prediction made by a theory contradict the theory? > > > > This is the point! > > The prediction is that the top clock, although CONNECTED to the axle, > > will read a different elapsed time than the bottom one. > > I now see that you have confused two different scenarios: > > 1) A rotating axle with a mark on each end (like clock hands) > to show its rotation. This is the case which you refer to the > clocks as being "connected to the axle". > > 2) A rotating axle with a clock at each end. This is the case > in which the clocks read different elapsed times. > > > Therefore, (which the dumbest rat would understand), the > > axle twisted. I say it doesn't happen, > > You are right: The axle does not become twisted. Notice that > everyone has agreed all along that the axle does not become > twisted. > > >> Didn't you want to answer that question? > >> > >> You claimed that the prediction made by GR contradicts GR. > >> Obviously it is impossible for a prediction of a theory to > >> contradict the theory which made it. Clearly, what you said > >> was not what you meant. > > > > As above. Is English your second language? Comprehension > > difficulties? > > You talked about two clocks connected to the axle. That was > misleading. What you really meant was the axle itself being > used like a clock. I responded to what you said, rather than > what you meant. I will accept that, when YOU accept that two clocks cannot be driven/read by the same light ray......which is when Hell freezes for a DHR, right? > > Since the axle is a single, solid, rotating object, it can > serve as a clock. One clock. The fact that it has a "hand" > at each end doesn't make it two clocks. Is the clock in your computer the SAME ONE as that in mine? they are both physically connected! > > >> >> Why do you say that because the two clocks are attached to the > >> >> axle, they must read the same? > >> > > >> > Because of the axle in my car. > >> > >> Really? > >> > >> Your car has only one axle? Assuming that you are referring > >> to the axles on which the wheels turn, a car with four wheels > >> has four axles so each can turn at a different speed. > >> See http://auto.howstuffworks.com/differential.htm > >> > >> I'd think a simpler example would be better. Though I don't > >> know why you needed an example at all. An axle is an axle, and > >> an axle by itself is an extremely simple thing. All it is in > >> this case is a single, solid rotating body. > >> > >> > A clock driven at each end mechanically will read the same > >> > for eternity > >> > >> That is what you need to show. You haven't shown it, merely > >> asserted it. It seems obvious that the two clocks would read > >> the same, but what seems obvious isn't always true. GR predicts > >> that in certain circumstances, the two clocks would not read the > >> same. If you think that is wrong, you need to show why. > > > > "A clock driven at each end of a rigid axle, will read the same > > for eternity!!!!!!!!!" > > That is still poorly expressed, but essentially correct. > And very obvious, of course. > > >> > Gr is trying to tell me, that if I lay the old girl on the side, > >> > time is passing differently at each end due to gravity disparity. > >> > >> Yes. > >> > >> > So how are the rotations per time at each end the same, > >> > if a clock at a higher altitude reads differently, due to > >> > gravity difference? > >> > >> They aren't. GR predicts that the rotational speed of the > >> bottom of the shaft, measured with a clock at the bottom of > >> the shaft, is greater than the rotational speed of the top > >> of the shaft, as measured by a clock at the top of the shaft. > > > > Go and have a THINK, or is that disalowed in DHR Land? > > All the "magic" can't make it happen- the clocks are CONNECTED > > via the axle. > > What I think is that you described the setup poorly. > > You meant that there is one "clock" and it *is* the axle. Nope! Deffinitely two clocks, or as many as you like spread along the diminishing grav field. > > >> You think that means the shaft must become twisted. Paul > >> showed that it doesn't mean that. > > > > Rubbish! Paul rabbitted on about the earth's spin; now I've > > suggested to try the experiment in a non-rotating (moon) > > situation, he has disappeared down the burrow. > > Paul simply used the rotation of the Earth-- one day-- as a > convenient reference against which to count the rotations > of the axle. It makes no difference what reference is used > as long as it is accessible at both ends of the axle. > > >> > For Paul to maintain the claim of time passing differently due > >> > to gravity, he must disconnect! > >> > The free (call it a GPS based clock) and the one connected to > >> > the (very high) axle, CANNOT read differently, WITHOUT a torque > >> > on the axle. > >> > >> Paul showed that they do read differently without a torque. > > > > sigh- they are CONNECTED!!!!!!!!!!!!!!! > > Fine. The two ends of the axle rotate together. Over the > course of any specified number of rotations of the Earth, the > axle rotates some number of times, which observers at both > ends of the axle agree on. That is as Paul said. > > >> > THIS is the contradiction. : The "Theory" says they will, and > >> > the axle proves they won't. > >> > >> The entire body of the axle is rotating at constant speed. > >> What that speed is measured to be depends on where in the > >> gravity field the measurement is made. That idea is very > >> surprising to almost everyone, including me. To some people, > >> it is incomprehensible. They reject the idea as being absurd, > >> even though the time difference is directly observed every day > >> in careful measurements. Those people sometimes reject the > >> observations because they contradict their own beliefs. > > > > Like me looking in a curved mirror, and rejecting the "observation" > > that I am thin. Observers can be mistaken, and misunderstand what is > > happening; in this case, the direction of the light has been changed. > > In the case of "apparent" time and length dilation, the VELOCITY of > > the light has changed. > > What light?? There was no light in your argument. Exactly- no light required! But in yours, the whole scenario DEPENDS on introducing the unwarranted arguements about c, because therein lies the MAGIC on which SR/GR is based. (A blind man can test my claims by FEELING the top and bottom hand positions) > Since the speed of light is always the same when carefully > and accurately measured (done thousands of times every minute, > all around the world), your assertion is proven wrong anyway. Save these false claims for school children and "true believers" > > Your analogy of looking at yourself in a curved mirror and > interpreting what you see as being misleading is apt. When > you look at the two ends of the axle, you see it rotating at > two different speeds. You know that can't be right, because > the axle is rotating as a solid body without becoming twisted. > What you see is misleading. You use general relativity to > calculate what is really happening-- what you would see if > spacetime wasn't curved, distorting your observations. Exactly!!!!!!! If I see something APPARENTLY doing something weird, I apply my brain, and calculate the TRUE situation by applying c'=c+v .....in the same way as I can reach the true situation as to my waist-line, by factoring in the angles and optics of the curved mirror. Anyway, your above is fallacious. At ground level, I see a delay of when the top moves, but the delay does NOT increase, unless the axle rotation was accellerating. That is not the case under discussion, and GR says the twist must appear to increase. So I see a twisted axle which remains the same. > > > Which do you think is more likely? > > 1) Millions of physicists, engineers, technicians, students, > and interested laymen have failed to notice glaringly obvious > contradictions in relativity over the last 80-some years. > You see the contradictions, but even when you explain them to > people smarter than you, they still don't see them. > > or > > 2) You don't like relativity because it seems wrong. So you > don't want to understand it and don't try to understand it. Millions of bomb chuckers are prepared to die, their beliefs in the forty virgins awaiting them being so strong. Are you one of these who think the emperor has a lovely cloak, or is he indeed starkers? (Hint: "Starkers" puts you in a small minority) Jim G c'=c+v
From: George Dishman on 12 Aug 2005 11:30 <jgreen(a)seol.net.au> wrote in message news:1123841949.837643.182940(a)g14g2000cwa.googlegroups.com... .... > If two photons are emitted from set positions at set times, and > approach each other, they interdict ie cause an interference, at point > A, due to distance covered at both velocities being "c". OK so far though "interdict" is an odd term to use for constructive interference. > But if one or the other is travelling +/- c due to its source velocity, > then the POINT of interdiction alters. Correct, though in the Sagnac experiemnt Ritz would say one was moving at c+v while the other moved at c-v. If you do the sums, you will find point A moves a distance v*T where T is the time of flight of each photon. > Voila'.......Sagnac machine works by identifying a change in the > interdiction position caused be c'=c+v Oops, you forgot that during the time of flight, the detector also moves a distance of v*T. The detector output should therefore be unaffected by the motion of the equipment if Ritz were right. In fact the output is proportional to the distance between the detector and point A, and the result tells us that point A doesn't move regardless of the speed v. How are you going to explain that Jim? George
From: George Dishman on 12 Aug 2005 12:36 "Jeff Root" <jeff5(a)freemars.org> wrote in message news:1123697847.177649.185930(a)g44g2000cwa.googlegroups.com... > George replied to Jeff: > >>> Interference occurs as if the light consisted of waves >>> something like surface waves on a liquid, even though only >>> a single photon is in the apparatus at any given time. >> >> Correct, and importantly even if the difference >> between the path lengths is many wavelengths. >> >>> QM doesn't provide any "explanation" of that, does it? >>> Only a way to calculate the effects? >> >> Scientifically speaking, what is the difference >> between an "explanation" and "a way to calculate >> the effects"? > > Understanding. > A computer can calculate the appearance of an interference > pattern without understanding it. A human can understand > what an interference pattern is, how it is caused, and what > it implies, without being able to calculate what it would > look like. The gas laws were initially empirical. Kinetic theory then "explained" them by providing a lower-level model. However, scientifically speaking, that explanation consists of the mathematical analysis of aggregates of particles which shows that the simpler laws can be derived statistically within certain constraints (an "ideal gas"). Our understanding of one level is then merely calculation at the next. Similarly why molecules of a gas behave as they do is explained by Coulomb or Van der Waals forces, so the explananation is "Qq/r^2". Then that is explained as a product of the interchange of virtual particles and again that is mathematical. At each level, a scientific explanation is a set of equations, the measurable quantities to which the parameters correspond and the rules for applying them. We might think that gives us an understanding, which I consider to be a consistent mental image, but of course that could be quite wrong even if the maths is correct. IMHO, understanding is associated philosophy rather than the scientific content. <snip> > I do not know what "understanding" is. It can be quantified > (at least crudely) by knowledge tests and intelligence tests. I'm not sure those test understanding rather than ones level of education and ability to learn. Many people know of black holes and many of those would be considered intelligent, yet I doubt more than a small fraction truly _understand_ their nature. I certainly don't. > I wonder whether it is amenable to explanation. > > QED doesn't provide any "explanation" of interference, > does it? The same as the classical understanding, interference occurs when two copies of a wave-like phenomenon with some relative phase displacement can reinforce or cancel (or anything in between) when recombined depending on whether a positive peak coincides with another positive peak or a negative one or some other part of the waveform. >>> Is the length of a photon involved in the calculation? >>> Does the result of the calculation depend on photon length? >>> Can the length be calculated from other givens? >> >> The most accurate theory is QED although a simpler >> classical analysis can give similar results in many >> cases. QED treats photons as point particles of exactly >> zero size so photons don't have a length at all. > > An explanation of why QED treats photons as point particles > might provide some of the understanding I desire. I'll pass on that. You really need to talk to someone with knowledge of the subject. > Although I would think that any action takes some amount > of time. It is natural for me to assume, for example, that > emission of a photon by an atom or charged particle takes > a time equal to one period of the resulting photon's wave. Consider again the fact that single photons can show interference effects when the difference in path lengths is many wavelengths. > (I had a hard time choosing the last word of that sentence. > I finally decided that if a photon can have a wavelength, > it can have a wave!) > >> IMHO, this question really shows how closely >> related the particle and wave views are, photons >> aren't really one or the other, they are somewhere >> between and it is our limited imaginations that try >> to insist on choosing. > > I think of light as having both wave and particle properties > simultaneousy, but I had long been under the impression that > they cannot be observed simultaneously. I'm not yet totally > convinced that they can. Maybe I need to see it done. A single photon has the maximum probability of hitting a point on a screen where the path difference is a multiple of the wavelength. Is that not exhibiting both properties? George
From: George Dishman on 12 Aug 2005 15:21 "sue jahn" <susysewnshow(a)yahoo.com.au> wrote in message news:42fa573d$0$18640$14726298(a)news.sunsite.dk... > ... If you can not get Feynman's book "QED" this might > scare up some diagrams. Jeff, if you aren't already aware of Sue's reference: http://www.amazon.com/exec/obidos/tg/detail/-/0691024170/102-9393346-0488961 George
From: jgreen on 13 Aug 2005 04:16
George Dishman wrote: > <jgreen(a)seol.net.au> wrote in message > news:1123841949.837643.182940(a)g14g2000cwa.googlegroups.com... > ... > > If two photons are emitted from set positions at set times, and > > approach each other, they interdict ie cause an interference, at point > > A, due to distance covered at both velocities being "c". > > OK so far though "interdict" is an odd term to use > for constructive interference. > > > But if one or the other is travelling +/- c due to its source velocity, > > then the POINT of interdiction alters. > > Correct, though in the Sagnac experiemnt Ritz would > say one was moving at c+v while the other moved at > c-v. If you do the sums, you will find point A moves > a distance v*T where T is the time of flight of each > photon. > > > Voila'.......Sagnac machine works by identifying a change in the > > interdiction position caused be c'=c+v > > Oops, you forgot that during the time of flight, the > detector also moves a distance of v*T. The detector > output should therefore be unaffected by the motion > of the equipment if Ritz were right. > > In fact the output is proportional to the distance > between the detector and point A, and the result > tells us that point A doesn't move regardless of > the speed v. How are you going to explain that Jim? Depends whether you measure from the airframe to which the sag is fixed, or the earth to who's view the entire machine rotates (as the plane circles) Cheers Jim G c'=c+v |