rbwinn

In comparing the rotations of the two planets, Neptune and Jupiter, we find that if we divide a Jupiter day into the same divisions that an earth day has, 24 hours, with each hour having 60 minutes, and each minute having sixty seconds, and divide a Neptune day into twenty four hours, each hour having sixty minutes and each minute having sixty seconds, then a second on Jupiter is not the same amount of time as a second on Neptune. Jupiter rotates 1.62 times for every rotation of Neptune. We can apply this same principle to transitions of cesium isotope atoms. Scientists do not like for transitions of cesium isotope atoms to be considered this way. In fact, it seems very likely that scientists are going to impose a punishment for anyone who dares to consider transitions of cesium isotope atoms this way.

No, I would never attempt that. I assume the clock is slower because of gravitation and maybe other reasons, but I just use a different set of Galilean transformation equations for every different rate of time. As I said, we could divide up the time of rotation of any planet into days, hours, seconds, etc., the way we do a day on earth, but a second on Neptune is going to be a different amount of time than a second on Mars, and it is no different with transitions of cesium atoms that are oscillating at different rates. If you are going to say a certain number of oscillations of a cesium atom are the basis for all time in the universe, you are going to run into difficulties with relativity because a Neptune second is not the same as a Mars second, and a second in K' is not the same amount of time as a second in K. But scientists have all of these equations they have been using that base everything on a certain number of transitions of a cesium isotope atom. If that is what scientists want to do, that is what they are going to do, but they can't describe relativity accurately if they do it, and so they are not going to describe relativity accurately.

John Cuthber Is your real name Bill Clinton? uncool All right. They put a cesium clock on an airplane and flew it east. It was slower by some number of nanoseconds than an identical clock kept on the ground. They put a cesium clock on an airplane and flew it west. It was faster by some number of nanoseconds than an identical clock kept on the ground. You say the clocks are not faster or slower. It is my addition to relativity. No it is not my addition. It was what these two scientists said, Haile-bob and Keating or whatever their names were. They said the clocks on the airplanes had different times from each other and from clocks on the ground. So you say clocks are neither slower nor faster, they are (different). I think I have heard this before in the politics forum. I got kicked out of Amazon discussion forums for telling a homosexual that I did not want to be a homosexual. I already have people in this forum threatening to get me kicked out. Now as to what I can predict, If I do what Einstein did with the Lorentz equations and plug the 186,000*t and 186000*t2' into the Galilean transformation equations, I get a slower time in the moving frame of reference than you get with the Lorentz equations. So suppose I go to scientists and say, Look, I think a clock on an airplane would be slower than a clock on the ground, do you think they are going to run the experiment. No, I don't think so either. You already know all this when you say my equations were no good because they did not result in an experiment like the Lorentz equations did. Then the result of the experiment was not exactly what scientists predicted, so it is explained by combining the Lorentz equations and General Relativity together, and then the answer comes within a certain amount. As I pointed out, now that the experiment has been done, I can just put the experimental values into the Galilean transformation equations. How are you going to do that with the equations you have? The Galilean transformation equations do not assume absolute time. You assume absolute time. As I said before, the Galilean transformation equations are public domain. I will continue to use them Yes, thank you, I will explain. What Einstein did was to substitute two little equations x=ct and x'=ct' into the Lorentz equations, which he said proved the Lorentz equations kept the speed of light the same in two frames of reference. The very first thing I did twenty years ago was substitute the same two little equations into the Galilean transformation equations, ending up with t'=t(1-v/c). That did not bother scientists nearly as much as what I am doing now because it was easy for them to find a mistake in it. It took me about six or seven years before I could see that you cannot use the variable t' in the Galilean transformation equations to say t'=t and then say t'=t(1-v/c), so I started using n'. x'=cn'. That made scientists a little angry because they could not really find anything wrong with it. Then last year I saw that what I was really doing was using the Galilean transformation equations twice, once for time as seen in K and once for time as seen in K'. So the equations I use now are for K x'=x-vt y'=y z'=z t'=t t'=t means a cesium clock at rest in K is the time used in both frames of reference. Then for the equations using the slower clock in K' we have x=x'+v'(t2') y=y' z=z' t2=t2' t2' is the time of the slower clock at rest in K'. Now our substitution into the Galilean transformation equation would be ct2'=ct-vt t2'=(t-v/c) = (t-vct/c^2) which shows us that the Lorentz equations are really just a special case of this relationship in the Galilean transformation equations, made by imposing a length contraction. x=ct, x'=ct2', c=x/t=x'/t2'=(x-vt)/(t-vx/c^2)=(x-vt)gamma/(t-vx/c^2)gamma It is my opinion that neither this representation in the Galilean transformation equations nor the Lorentz equations give a correct representation of light, but they do indicate the difference in times that was shown by experiment.

My equations would be more accurate than GR or SR because once these experiments were run, it was known what the experimental values for the times were. In the Hafele-Keating experiment the traveling cesium clock lost 59 nanoseconds on the eastward trip and gained 275 nanoseconds on the westward trip. So to be more accurate than GR or SR, all I would have to do is say t'=t=the time of the eastward trip according to a clock on the ground, t2=t2' = the time of the eastward trip according to a clock on the ground minus 59 nanoseconds. Then I would do the same thing for the westward trip. t'=t=the time of the westward trip according to a clock kept on the ground, and t2=t2' = the time of the westward trip according to a clock kept on the ground plus 273 nanoseconds. Why would I worry about GR or SR if the experimental values were known? The same principle applies to any other differences in time in any other experiments involving GR and SR. Once the experiments are run, I just put the time values in the Galilean transformation equations. Are you saying you think that they are going to change from one experiment to the next? swansont OK, I will stop using the Quote thing. I did not want to use it in the first place. You believe scientists are the most honest people in the world. That has not been my experience with them. How many times are you going to have to forbid me from using the Galilean transformation equations? As many times as you want to. I already told you I was going to keep using them. I know how proud you are of your cesium clocks. You forbid anyone to express relativity using the Galilean transformation equations and one of your cesium clocks unless they use absolute time. Then some other scientist wants me to use the results of an experiment involving cesium clocks. So I just go ahead and use results obtained from cesium clocks, knowing you will be upset about it, but I see that as your problem, not mine. I quit trying to make scientists happy a long time ago. As I pointed out before, the Galilean transformation equations are public domain.

Well, that is fine. We were told that the moving clocks were slower, which is the way I always thought of it. Well, scientists have to agree on something in the end, and they are going to agree on whatever seems most advantageous to scientists. If they are getting large sums of money from governments to do "research" using Einstein's theory, it seems to me they are going to be influenced to some extent by that. But it does make further predictions - namely, that the speed of light is not isotropic relative to every frame. Which has been experimentally tested. No, I was just pointing out that events with regard to a light ray directed in the -x direction do not agree with what scientists have been saying. I do not agree with scientists about relativity of simultaneity either. Well, you keep making this accusation. So go ahead and show the experiment that cannot be explained by the Galilean transformation equations as I use them.

I thought you said you did not believe in God. Science is for scientists just the same way law is for lawyers. I would not hire a lawyer for anything. Why should I treat scientists any different? I already told you that I have to work and do not have unlimited time like scientists do to post to this forum. So I have to make an assessment before I post, Do I want to go into some long involved discussion of what the Galilean transformation equations say when scientists from first to last have already said they do not understand them? I do not see it as a profitable method of discussion. This is a more difficult situation than Copernicus faced when he dared to oppose Ptolemaic astronomy, and it was difficult enough for Copernicus that he did not have his ideas published until after he died. The situation is different today. Scientists have used Einstein's theory as a cash cow for more than a hundred years. What scientist of today would dare to speak out against money being given by governments to scientists for "research"? Any scientist who dared to do that would be run out of science. Where would a scientist like that get a job? So scientists are the opposition to truth today, not state religions as was the case back in the time of Copernicus. Scientists of that time kept using Ptoleaic astronomy after it had been proven wrong because they were afraid of persecution from state religions. Galileo spent his last years under house arrest because he said that the earth rotated on its axis. Scientists of today are afraid of peer pressure, the opinions of other scientists, who have all said that use of the Galilean transformation equations without absolute time and their super accurate cesium clocks is prohibited. Since I am not a scientist, I just go ahead and use the Galilean transformation equations because they are the correct equations to represent relativity. There is no requirement for me to please scientists. All I need to do is get the mathematics right. Well, OK, tell me what you want me to figure out, and I will do it. I am not going to go on fool's errands all over the internet. If you want me to explain something, post it here, and I will do it.

No, I don't believe your assessment of the experiments. They were always reported as being that the moving clock was slower, which proved Einstein's theory. I think that was what happened. The clocks on the planes showed less time than clocks kept on the ground. In any event, I don't get all excited by what scientists say about it. They were trying to match up Einstein's theory with experiments and did whatever it took to say the experiments matched. The way I use the Galilean transformation equations does not make any predictions about times. All it does is uses the Galilean transformation equations with every rate of time separately. So here is the experiment you need to run with the Lorentz equations. K' is moving with velocity v relative to K in the +x direction. When the origins of K and K' coincide, light is emitted at the origins. In the +x direction, the light reaches x=1 light sec in one second in K and x'=(1-v)gamma light sec in t'=(1-v)gamma sec. In the -x direction light reaches x=-1 light sec in 1 sec. and x'=(-1-v)gamma light sec in t'=(1+v)gamma sec. There is no other way I see to interpret it than that there is a faster clock in K' because the light travels farther in more time than in K.

If you are the great scientist you claim to be, you can figure the numbers out for yourself. I know scientists of today like to have someone hold their little hands and give them money every time they do something. I provided the equations. If you do not like them, show the numbers that prove they are wrong.

Well, I did not stop using the Galilean transformation equations just because scientists did. At any rate, all I do is use the Galilean transformation equations to show different rates of time. For instance, if a clock in K' is slower than a clock in K, then I just use the Galilean transformation equations twice, once for each rate of time. x'=x-vt y'=y z'=z t'=t t'=t means that the time of a clock in K is being used as the time of the equations. The time of a slower clock in K' does not pertain to this set of transforms. Now we use the Galilean transformation equations with the time of the slower clock in K'. We call the time of the slower clock t2' because t' was already defined to be t'=t, the time of the clock in K. x=x'+v'(t2') y=y' z=z' t2=t2' t2= t2' means that the time of the slower clock in K' is being used for the time of the equations. The time of the clock at rest in K does not pertain to this transform.

Well, I predicted that two different rates of time could be accounted for without a length contraction, which turned out to be true. As I understand it you are saying that all time has to be referenced to your cesium clocks. You sound a little proud of your clocks. Out here in reality, we do not even have any of your cesium clocks. For instance, a prisoner in a jail cell might tell time by the position of the reflection of the sun on his wall. The Galilean transformation equations are public domain. We common people can use them however we choose. We can also notice that if one of your cesium clocks is moving, it has a different rate of time than a cesium clock that is not moving. I account for that by using the Galilean transformation equations twice, once for each rate of time. That is the mathematically correct way to do it. You can complain about it all you want to. I will still keep using the Galilean transformation equations. As I said before, there is nothing you can do about it. The Galilean transformation equations are public domain. Well, in experiments they are faster or slower, starting with cesium clocks transported by propeller airplanes on transcontinental flights. The Lorentz equations show that light is traveling at c in all frames of reference. If that is what you mean by isotropic, then the equations are isotropic. The problem I see with them is that it is physically impossible for a clock in K' to be both slower and faster than a clock in K.

swansont It is not a question of predicting anything. If scientists say that a clock in K' has a different rate than a clock in K, then how do you use the Galilean transformation equations to show the time of the clock in K'? Clocks can have different rates of time. It happens all the time. I went to Walgreen's drug store and bought a clock that lost ten minutes every day. Are you saying that the Galilean transformation equations cannot describe what my Walgreen's alarm clock says? I would just use the Galilean transformation equations twice, once for a GPS clock that scientists would agree with and once for my Walgreen's alarm clock, which is fairly useless except to show what a person using it would perceive if they did not know it was slow. For example, if they had the slow clock in a car and used milage markers and the slow clock to determine the speed of the car, they would get too fast a speed. Scientists were the people who said they had a slow clock in frame of reference K'. Whether they were lying, misinterpreted something, or actually had a slower clock, I have to decide. Since they seem to be consistent about this, I have to assume there is a slower clock. So I just apply the Galilean transformation equations the same way I would with any other slow clock. This does not break any laws of physics or mathematics as you claim it does. It just compares the rate of a slow clock to distances. Your claim is that this cannot be done. My claim is that I did it and will continue to do it. You say I am making claims about the Galilean transformation equations that are not true. What are the claims I am making that are not true? Greg H Well, all experimental results will match my theory because it does nothing but apply the results of experiments scientists say they have already made. I can predict what a moving clock will read by saying x=ct and x'=c(t2') and substituting those two values into the Galilean transformation equations to get the difference between the two times, but I do not say that it is more than an estimate. For values at the velocities I was considering, planets in the solar system, that substitution produced answers that agreed with the Lorentz equations to several decimal places, so I was not too concerned about their accuracy at that point. What I was trying to do was to find a way to use Isaac Newton's equations for orbits of planets with the different times Einstein's theory predicted. All my equations do is show the differences in rates of time, regardless of what the rates are. So if we are comparing rates of rotation of two planets and saying that one rotation of a planet is a day, then we would use the Galilean transformation equations twice, once for each rate of time. Where scientists and I differ is that I do the same thing with scientific time. I say that if transitions of a cesium isotope atom are slower in K' than transitions of a cesium isotope atom in K, then there are two different rates of time, just like with the rotations of two different planets, and so I apply the Galilean transformation equations twice, just as I did with the two planets. It is really nothing to argue about. Scientists have convinced me that they are not going to use the Galilean transformation equations in any circumstances. I can understand that. Einstein's theory was the greatest cash cow for scientists that ever came along. I would not really imagine that scientists are going to abandon something like that or even consider something a high school graduate was doing. You say you are turning science into a religion and are going to enforce dogma. I can see that is true. However, consider what is going to happen. Enforcement of scientific dogma only causes people to want to experiment, just as people did when Copernicus decided that the planets were orbiting the sun. Galileo spent the last years of his life under house arrest because he said that the earth rotated on its axis. Fortunately for us today, we are allowed freedom of religion, so we don't have to follow the teachings of yours.

uncool Your idea is that by making the time of a slower clock in K a time coordinate, you solve the problem of a light ray directed in the negative direction requiring a faster clock in K'. It does not get the job done. In fact, a length contraction makes the problem worse. To see whether or not I am right, start the light beam at (0,0) in K and (0,0) in K' and send it to one light sec in K in one sec. It will go a longer distance in K' in a longer time. If sent in the +x direction, it will go a shorter distance in less time. So I have to regard the Lorentz equations an insufficient explanation of relativity along with GR, whatever their success may have been in experiments. Contracting lengths does nothing to solve the problem in the negative direction. The Lorentz equations as transformation equations are no better than the Galilean transformation equations with regard to isotropy and have the added problem of a length contraction. As far as your concern about the coordinate x, the planet Mercury is the fastest moving thing with regard to solar system astronomy, which was what I was using in my thought process. The Galilean transformation equations have the same times at all coordinates in K and K'. I never regarded t2' as anything but an estimate. Perhaps the Lorentz equation does give a more accurate time, but it agrees with t2' in my equations to about six decimal places at the speed of Mercury, so I figured it was close enough in computing orbits of planets. The difference between scientists and me is that I do not claim to have explained all things in the universe. I just explained a few things that I could see wrong with what scientists were saying. John Cuthber Well, I post this over an over again, but scientists never seem to grasp it. Supose that there is an astronaut in orbit around the earth. The astronaut has a clock in his satellite that is slower than a clock on earth. Most scientists seem to agree with this except for some who will say his clock will be faster. At any rate, his clock shows a different time for an orbit than a clock on earth. So a scientist on earth times an orbit of the satellite and computes its speed. Then the astronaut is told, calculate your speed using your clock. The astronaut has a means of determining the radius of his orbit. I used to work on radar in the Navy. This can be done by radar or by other means. It is possible. Once the radius of orbit is known, the circumference of the orbit can be determined. So what do you think the result of this experiment is going to be, the astronaut is going to get the same speed for his orbit as the scientist on the ground gets? That is what the Lorentz equations and General Relativity say. I say that if the astronaut's clock is slower, he will get a faster speed of orbit. So using Isaac Newton's equations for orbits, I started figuring orbits using differences in rates of time, using the Galilean transformation equations instead of the length contraction equations that scientists use. For the solar system, then, the sun would have the fastest time, Mercury would have the slowest, and time of each planet would increase until the outer planets would have clocks that agreed more closely with sun time. You may not like my experiments, but they are what is available to me and were what interested me at the time. swansont The Galilean transformation equations predict what I observe in nature. As I said before, I use the solar system to prove what I say. The sun has the fastest rate of time because it is at rest relative to the solar system. Mercury has the slowest rate of time because it orbits the fastest. Pluto is no longer considered a planet, but it has the time in our solar system that most agrees with sun time because it is the object in orbit around the sun that is moving the slowest, other than comets, etc. I use the Galilean transformation equations to compute all of these times because the Galilean transformation equations do not have a length contraction. As I told one other scientist, the Galilean transformation equations are public domain. If there were any scientists on earth who wanted to use them, they are free to do it. The problem I see is that I am the only person on earth using them, and I am not a scientist. So I have to believe we are in for a couple more centuries of the same kind of scientific extremism.

Anyone is free to use the Galilean transformation equations. They are public domain. uncool And I have shown several times that the Lorentz equations have exactly the same flipping of frames of reference mine have with regard to a light ray going in the -x direction. That is why I do not regard the equation for time as much more than an estimate and have the same kind of belief about the Lorentz equations. swansont I do not believe in absolute time. The Galilean transformation will work for any rate of time measured in any way. The only restriction the Galilean transformation equations have is that you have to use the same measurement of time in both frames of reference. So if you have different rates of time in K and K', you cannot set them equal to each other in the Galilean transformation equations. But there is no restriction on what the two observers can agree to use as time. They can use the rotation of one of Saturn's moons, they can use the orbit of the planet Mercury, they can use the rotation of the sun, they can use transitions of a cesium isotope atom. What they cannot do is use the transitions of two different cesium isotope atoms oscillating at different rates. John Cuthber Well, yes, my theory does give better answers than GR. GR assumes that the speed calculated from one frame of reference is the same as the speed calculated from the other frame of reference. Reality shows us that if a clock in one frame of reference is slower that the clock in the other frame of reference, then an observer using that clock will get a faster speed between frames of reference than an observer using a faster clock. As for these experiments you want me to look up, why don't you just give a brief description of them, and I will give my opinion. I don't have all day to do this the way scientists do. I have to work during the day.

uncool I have abandoned the idea that scientists can explain the speed of light. Whether you use my equation for t2' or the Lorentz equation for t', which is the same equation with a length contraction, a light beam directed in the -x direction takes a longer time in K' to get to a set of coordinates than it does in K. That means, taking two events associated with the light beam, time is slower in K than in K'. So the Lorentz equations may be showing light traveling at c in all frames of reference, but they are flipping frames of reference to do it, depending on the direction of the light ray. So I just go with experiment. A clock in K' shows less time than a clock in K. My own theory is that light is energy that reacts with the elements in a frame of reference at a rate of c. So whether an interferometer is moving or standing still, it shows light to be traveling at c.

John Cuthber No, if they used the equations I use, they would make them to be slower by about the same amount the Lorentz equations show for that velocity, the same to several decimal places. My difference in times would be t2'=(t-vt/c). The Lorentz equations difference would be t'=(t-vx/c^2)/sqrt(1-v^2/c^2). My difference in times is slightly slower than the Lorentz difference in times at that speed, but not enough to be considered consequential. The scientists who built the system knew from experiment the rate of the clocks compared to rates of clocks on earth. All I know is the incorrect information scientists give out. But I do know this: It does not matter what the rates of the clocks are, slow or fast. The Galilean transformation equations can describe those rates perfectly using them once for each rate of time.

Klaynos My predictions are not vague either. I gave an equation to show the difference in time. If scientists are pressed on this, they bring in gravitation, temperature effects, drift, etc., as was done in one of the preceeding posts, and some will state that a GPS satellite has a faster time than a clock on earth. I have no idea whether a GPS satellite has a faster or slower time. The equations I have agree with the ones Einstein had for special relativity, which Einstein described as relativity without gravitation, to several decimal places, so I am not all that concerned that I am saying anything Einstein did not say with Special Relativity, other than that he had a length contraction and I do not. If you say that I have not done the simple model experiment requested, then I have no idea what you are requesting. I showed that the Galilean transformation equations predict a slower time in the moving frame of reference. I showed that an observer using a clock in K' would get a faster speed between frames of reference than an observer in K using a clock in K, which agrees with experiment, while the scientific result of both observers getting the same speed does not agree with experiment. So, I do not know what else you want to know. Maybe if you were more specific, I could help you. uncool Yes, that is what the equations show. In order for light to be traveling at c in both frames of reference, there have to be both faster and slower clocks in K and K'. The Lorentz equations have this same problem. So I have abandoned all scientific ideas concerning transmission of light and just describe relativity, which I do by using the Galilean transformation equations twice, showing two different rates of time.

Strange I have posted this at least ten times. If light is going at a speed of c in K, then x=ct. If light is going at a speed of c in K', then x'=ct2'. x'=x-vt y'=y z'=z t'=t t' represents time coordinates in K' in the Galilean transformation equations and is defined to be the same as t in this transformation. t is time on a clock in K. Since t' is already defined to be time on a clock in K, we cannot use it to represent time on a clock in K' because a clock in K' has a slower rate of time than a clock in K. So we use a different variable t2' to represent time on a clock in K'. Now we use the Galilean transformation equations to describe relativity as seen from K' using the slower clock in K' as our reference for time. x=x'+v'(t2') y=y' z=z' t2=t2' t2 is a time coordinate in K and is shown by the time on the slower clock in K', not by a clock in K. The time of the slower clock is being used in both frames of reference. Now we substitute in the values for x and x' when light is traveling at c in both frames of reference. c(t2')=ct-vt t2'=(t-vt/c) The clock in K' is slower because t2' is less time than t. I can post this as long as there are scientists saying that I have not shown the clock in K' would be slower.

uncool I have posted this a couple of times also. A light beam directed in the -x direction has the coordinates on the x axis of x= -ct, x'= -ct2'. The first equation is its x coordinate according to a clock in K. The second equation is its x' coordinate according to a clock in K'. x'=x-vt -c(t2') = -ct-vt The light has a velocity of -186,000 mi/sec in both frames of reference. This is exactly what the Lorentz equations show even to the fact that light directed in this direction requires a slower clock in K and a faster clock in K' in order for this to happen. swansont Well, why wouldn't GPS work? I have already shown that The Galilean transformation equations as I use them give the same prediction for a slower clock in K' than in K. Secondly, GPS is adjusted continually. I already told you that I do not have a government borrowing trillions of dollars every year on national credit and giving me some of it for "research". What I do is send a small donation every year to the Bureau of the Public Debt and tell them to apply it toward payment of the debt. That way I can tell who is keeping things real, scientists or me. You have made some fairly definite statements. You say I am using mathematics incorrectly. You say I am getting incorrect answers. Now all you need to do is show where I am using mathematics incorrectly and which answers I am getting are incorrect. As soon as you do that , we will be making some headway in this conversation.

uncool If you cannot remember what you said, go back and review it. I can remember what you said. So far you have said you were insulted because I said that uneducated people i work with every day can follow instructions better than scientists can. That is just a fact. I have not seen a scientist yet who can follow the most simple instructions. Here, put these values for x and x' into the Galilean transformation equation. It cannot be done. OK, I will do it for you. c(t2')=ct-vt It cannot be done. Explain why it cannot be done. It cannot be done. Why can't it be done? Because if it is done, it accounts for the experimental evidence that a clock in K' is slower than a clock in K, which scientists continue to say that the Galilean transformation equations cannot account for. So they repeat over and over, It cannot be done. I know how scientists converse. I have talked to scientists before. So I just avoid the endless conversations about irrelevancies and try to keep the conversation about what I want to discuss, the fact that the Galilean transformation equations can account for the slower clock in K' and light traveling at c in both frames of reference according to clocks in those frames of reference and the answer given by scientists, It cannot be done. It can be done because I did it. The fact that scientists are too helpless to put values for x and x' into the Galilean transformation equations does not mean it cannot be done. It means scientists refuse to do it. Bignose I have to go to work and do welding. I don't sit at a computer all day the way you do. So when I give a response, it cannot take me hours of contemplation the way your answers do. I just have to answer the irrelevancies scientists post the best I can. I posted the equations. No comment. I showed that t2' in the Galilean transformation equations as I used them is a slower rate of time than t. No comment. It cannot be done. Here, I will show you what the Lorentz equations are again. c=x/t=x'/t2'=(x-vt)/(t-vct/c)=(x-vt)/(t-vx/c^2)=(x-vt)gamma/(t-vx/c^2)gamma=x'(Lorentz equations)/t'(Lorentz equations) The Lorentz equation are what I have been showing you with a length contraction. It cannot be done. Well, if you do not want to talk about it, you do not want to talk about it. If you don't, then go find some of these nice scientists to talk with and you won't need to be all offended when I do not believe what you say.

uncool You ignore every answer I post and repeat the same unproven dogma over and over again. We will start with your statement that the Galilean transformation equations cannot show the speed of light to be isotropic. I have posted the proof that they can several times. The response of scientists to this is to repeat that the Galilean transformation equations cannot show the speed of light to be isotropic. So here is the proof once again. x=ct x'=ct2' t is time on a clock in K. t2' is time on a clock in K'. The clock in K' is slower than the clock in K. Scientists say they have shown this by experiment. x'=x-vt This is the equation for distances on the x axis in the Galilean transformation equations. Now we substitute the values for x and x' into the equation from the Galilean transformation equations. c(t2') = ct - vt t2'=(t-vt/c) t2' is less time than t, just as scientists have said they confirmed by experiment. The equations x=t and x'=c(t2') show that clocks in K and K' have the speed of light the same in both frames of reference. OK, now that I have shown this to you, what I expect will happen is that you will tell me once again that the Galilean transformation equations cannot show that speed of light is c in both frames of reference. I have tried to converse with scientists before. I know what they say. Now if I understand your objection, you are saying that if there is a third frame of reference going at some other speed relative to K, then the speed of light will not be isotropic to that frame of reference. Then the speed of light between the two moving frames of reference will not be isotropic, etc. You are wrong, but there is no reason to try to explain it to you when you cannot even understand the sixth grade level of mathematics required to account for only two frames of reference. I can see what your problem is. Scientists were using equations for relativity with Newton's absolute time and ether theory interpretation of the Galilean transformation equations which adapted right over into the Lorentz equations because the speed of one frame of reference relative to the other was kept the same as measured from either frame of reference. Since that does not agree with reality, as I have already shown numerous times, scientists are going to remain in la la land probably forever. That does not mean people who are not scientists are not allowed to discuss relativity, and there was no such rule given in this discussion group.

swansont No, we are assuming that the observer in K' is not a scientist and can therefore follow simple instructions. The instructions would be, Your clock is slow. You have to use the time of a clock in K in the Galilean transformation equations as we are using them. What I am doing in a science discussion forum is trying to explain to a scientist that scientific time involves different rates of time, and different rates of time cannot be used in the same set of Galilean transformation equations. You have to use a different set of Galilean transformation equations for each different rate of time. To reiterate once again, in K, the frame of reference at rest, x'=x-vt y'=y z'=z t'=t We are using the time on a clock in K for time coordinates in both frames of reference. Now we use the Galilean transformation equations again with time of a slower clock in K' for time coordinates in both frames of reference. x=x'+v'(t2') y=y' z=z' t2=t2" x in the second set of equations is the same as x in the first set of equations. x' in the second set of equations is the same as x' in the first set of equations. vt is the same distance as v'(t2'). That means that if t2' is less time than t, an observer in K' would believe his speed to be faster as computed from the slower clock in K' than an observer in K would believe the speed of K' to be computing the speed of K' with a clock in K. If we run this experiment in reality, we find what I say to be true. An observer in K' using a clock in K' will find the speed between frames of reference to be faster than and observer in K using a clock in K will find the speed between frames of reference to be. This is not really difficult. I would put it at about sixth grade level. So, once again, explain again why the Galilean transformation equations cannot account for a slower clock in K'.

AJB If you have two cesium clocks in the same frame of reference, they will both show the same rate of time. If you have one in K and one in K', a frame of reference moving relative to K, the one that is moving will show less time. That has nothing to do with the equations I gave. The Galilean transformation equations work with any rate of time. For instance, an observer in K could be using the Galilean transformation equations using units of time based on the rotation of Jupiter, and an observer in K' could be using the Galilean transformation equations using units of time based on the rotation of Neptune. Jupiter rotates faster than Neptune, just as a cesium isotope atom at rest oscillates faster than a cesium isotope atom in motion.

Look, I don't have time to follow all of the little rules scientists can make up. I have to go weld things, etc. I am not all excited about talking to scientists. I have talked to enough of them on the internet to know what to expect. Now I will post the equations again and say what they mean. x'=x-vt y'=y z'=z t'=t t'=t means that the time of a clock in K is being used for time coordinates in both frames of reference. A slower clock in K' is just a slower clock as far as these equations are concerned. A slower clock in K' does not show t' because t'=t, the time of a clock in K. x=x'+v'(t2') y=y' z=z' t2=t2' This time the time of the slower clock in K' is being used for time coordinates in both frames of reference. So now go ahead and explain how the Galilean transformation equations do not show that a clock in K' is slower than a clock in K.

Well, if I am using the correct equations, and scientists are not , then my observations are more important. It is like a scientist comparing observations of scientists to observations of of people like me who are not highly educated. So my suggestion is to wait a while and see whose observations are better.

Well, I am not the one who assumed there was no length contraction. Galileo and Isaac Newton had that idea. I just happen to agree with their thinking. James Clerk Maxwell did all of his calculations assuming there was no length contraction. So there really is no law that mandates belief in a length contraction yet, although scientists will probably eventually try to pass one as they attempt to turn present scientific beliefs into dogma. The Galilean transformation equations do not have a length contraction. That seems to be the primary complaint modern scientists have against the Galilean transformation equations. Frames of reference do not confuse me the way they do some people. I just am able to see that some frames of reference are more meaningful than others. The frame of reference of the earth is more meaningful than the frame of reference of a fly buzzing around in the air above the earth. Any theory of relativity says where physical objects are with respect to time, whether they are moving or standing still. That is what relativity means. I don't think it is very likely that I have misunderstood the equations. It is not the length contraction part of the Lorentz equations that keeps light traveling at c in both frames of reference. It is the part I showed you comes from the Galilean transformation equations and does the same thing in those equations without a length contraction. What you are saying is that since the Galilean transformation equations had the interpretation given to it by Isaac Newton, ether theory with absolute time, when scientists discarded those equations, no one is allowed to use them again or to have any other interpretation of how the Galilean transformation equations work.