GETTING RID OF EINSTEIN RELATIVITY
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From: bz on 10 Aug 2007 03:43 HW@....(Henri Wilson) wrote in news:s86nb3l50j13ou9o6q8er9tc2t498togju(a)4ax.com: > George, even Paul wont enter this argument becasue he is so embarrassed > by your confusion. > George, this is physics. > Objects can move in an FoR. Objects can move relatively in an FoR. But your statements about the light moving at c+v wrt the source [as measured in the lab frame] makes as much sense as my saying that the telephone pole I see between my car and the distant mountain is moving at 60 mph wrt that mountain [as measured in the frame of my moving car]. The pole is NOT moving wrt the mountain when studied from the mountain. The mountain is not moving wrt the pole, when studied from the pole's location. And it would be silly from me to wait one minute and say that the pole had moved a mile wrt the mountain. But that is exactly the kind of thing you are doing when you do your BaTh. You are assuming that things which are moving at a constant velocity wrt each other[zero in my example] can be said to be moving at a different velocity wrt each other just because they appear to be moving differentially at that velocity from our viewpoint. You then expect that differential velocity to have a real effect on how long it takes the light to make the trip. It can not effect the transit time because the objects never REALLY move at that differential velocity. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+spr(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: George Dishman on 10 Aug 2007 12:17 "Henri Wilson" <HW@....> wrote in message news:s86nb3l50j13ou9o6q8er9tc2t498togju(a)4ax.com... > On Thu, 9 Aug 2007 09:16:14 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <HW@....> wrote in message >>news:rkocb3l62ps3fufdcv2mlpk6a9opupv58r(a)4ax.com... > >>>>You are just repeating a self-contradictory question >>>>that shows you don't know what "frame" and "wrt" mean. >>> >>> Answer the question George. >> >>The question is self-contradictory so has to be split: >> >>a) In the lab frame, the rays are seen to be moving at c. > > ...and at c+v wrt the source. No, that answer is in the lab frame, not the source frame. > Your statement is a postulate anyway. No, it is a statement of fact regarding a drawing on a web page. >>b) In the source frame ("wrt the source"), the rays would >> be seen to be moving at c. > > ....irrelevant relativist raving.... The answer is correct - you are forgetting you were asking about the "standard SR explanation". >>c) In the lab frame, the difference between the speed of >> the rays and the speed of the source is c+/-v. > > You finally got one right. All three are correct, only your comments regarding them are wrong. >>Which of those three answers do you wish to dispute? >> > >>>>> It clearly shows >>>>> the rays moving at c+/-v wrt that 'comoving frame. >>>>> >>>>> You obviously don't even understand your own theory. >>>> >>>>I can do the maths, you hav to stab in the dark. You >>>>got it wrong. >>> >>> YOU are in the wrong frame. >> >>There is no such thing as a "wrong frame", it is >>merely the coordinate scheme you have chosen. You >>still don't know what a frame is. > > George, even Paul wont enter this argument becasue he is so embarrassed by > your > confusion. No point, he knows what I said was correct and that trying to educate you is pointless. > George, this is physics. No Henry, it is semantics. The words "wrt the source" mean literally that you are making measurements "with respect to the source", or to put it another way, you are measuring from the source, or you are measuring using a coordinate system whose origin is the source. In this context, that is described as a measurement "in the source frame". Until you learn this terminology, you will just keep making these mistakes. > Objects can move in an FoR. Objects can move relatively in an FoR. Of course they can, but YOU need to learn the jargon to be able to say _which_ frame. >>>>The answer to the question is that you are wrong, the >>>>speed in the momentarily co-moving inertial frame >>>>("wrt the source" as you put it) is c, not c+v, and >>>>that is true in both ballistic theory and SR. If you >>>>want to continue making a fool of yourself by showing >>>>that you do not understand basic technical terms, feel >>>>free to carry on, and I will continue to laugh at you. >>> >>> George, what hapened to my question about the two cars. >> >>I answered it in a reply to that post. >> >>> If you are a bystander, is it or is it NOT legitimate for you to measure >>> an >>> quote the speed of one car wrt the other? >> >>Not. You can only measure the speed of the cars relative >>to your measuring instruments. You can then _calculate_ >>the speed of one car wrt the other, which is effectively >>predicting wat would be measured if your instrument were >>in one of the cars measuring the speed of the other, but >>that is a calculation that requires a coordinate transform >>so differs between theories. > > Coordinate transforms are unnecessary. To get from one frame to another you _must_ use a transform. > Relative speeds are the same in all > frames. Some transforms preserve speeds, some do not, and in reality they are not the same. >>> Whether or not that is the same answer one of the drivers would come up >>> with is >>> a completely irrelevant. >> >>Not at all, it is a _prediction_ of what the driver >>would _measure_ if he had your instrument in the >>car with him. That's what the words "with respect to" >>mean, they say you are using that object as the origin >>of the measurement coordinate system. If I ask "where >>is Chicago with respect to Washington?" you could >>give me distances west and north using Washington as >>the origin of those numbers. Do you get it yet? > > George, you stated above: > > ">c) In the lab frame, the difference between the speed of > "> the rays and the speed of the source is c+/-v." > > That means, "in the lab frame, the rays appear to be moving at c+/-v wrt > the > source. No it doesn't, you are still making the same basic layman mistake. Do try to learn a little physics Henry, your errors are getting tedious. George
From: George Dishman on 10 Aug 2007 12:21 "Henri Wilson" <HW@....> wrote in message news:k77nb3tqiivi90hjf5oh70nnb3irn560hb(a)4ax.com... > On Thu, 9 Aug 2007 08:59:47 +0100, "George Dishman" > <george(a)briar.demon.co.uk> wrote: >>"Henri Wilson" <HW@....> wrote in message >>news:iupcb3p0d72scb6sqb94peuima6m7f8tiq(a)4ax.com... >>> On Sat, 4 Aug 2007 18:02:53 +0100, "George Dishman" > >>>>Whatever you call it, the spin of a photon is similar >>>>to that of the javelin but with the proviso that it >>>>is always xactly aligned with the direction of motion. >>>>The quantisation of photon spin only allows two values >>>>which can be thought of as the javelin spinning either >>>>clockwise or anti-clockwise as seen by the thrower. >>> >>> The shaft isn't spinning George. >> >>Photons carry angular momentum Henry. >> >>> I am taking about the angle between the shaft and the direction of >>> motion. >> >>The angular momentum is quantised and the vector is >>parallel to the direction of propagation. >> >>However, none of that explains how a change of spin, >>or even a twist, changes the time taken for two >>javelins to travel the same distance to be different. >>That is what happens in the Sagnac experiment, photons >>sent in opposite directions round the ring do not >>arrive at the detector simultaneously. > > It does. Think about the reflections from each moving mirror. The incident speed is c so the reflected speed is c, it does not explain the difference in arrival times. We covered that in February 2004 from the file date of this! http://www.briar.demon.co.uk/Henri/speed.gif George
From: George Dishman on 10 Aug 2007 13:29 "Henri Wilson" <HW@....> wrote in message news:n47nb39nnvlektocgv4edt043q1dr1a5ji(a)4ax.com... > On Thu, 9 Aug 2007 08:30:15 +0100, "George Dishman" > <george(a)briar.demon.co.uk> wrote: >>"Henri Wilson" <HW@....> wrote in message >>news:1tdkb35gc07k9bp0s6gg30237su7a2pub2(a)4ax.com... >>> On Tue, 07 Aug 2007 11:48:58 -0700, Jerry >>> <Cephalobus_alienus(a)comcast.net> wrote: >>... >>> >>> I will try to explain Crank. >> >>Jerry is not "Crank". > > Crank likes to use female names "Minor" isn't female. > and secretly dress in womens' clothes. That would be Jerry, and I think she is quite open about dressing in womens' clothes.
From: George Dishman on 10 Aug 2007 13:33
"Henri Wilson" <HW@....> wrote in message news:gu6nb35pmkh8hje75gu65hil4e9hc4ed4t(a)4ax.com... > On Thu, 9 Aug 2007 08:30:15 +0100, "George Dishman" > <george(a)briar.demon.co.uk> wrote: >>"Henri Wilson" <HW@....> wrote in message >>news:1tdkb35gc07k9bp0s6gg30237su7a2pub2(a)4ax.com... >>> On Tue, 07 Aug 2007 11:48:58 -0700, Jerry >>> <Cephalobus_alienus(a)comcast.net> wrote: .... >>> Say I produce a certain brightness curve with a magnitude change of 1, >>> using maximum velocity 0.0001c, a distance 120LY and a period 0.4 years. >>> >>> Doubling the distance or velocity or halving the period will produce the >>> same result... >> >>Not true. If you double the distance and halve the >>velocity then the ADoppler term stays the same but >>the VDoppler term is halved (since it depends on >>velocity but is independent of distance) so both >>the shape and phase of the curve will alter. > > Yes for very close distances that is correct....but for nearly all distant > stars with periods greater than a few days, the VDoppler term is > negligible. No, VDoppler is dominant in every case we have examined. > I thought I made it clear that the VDoppler contribution will determine > the > phase shift between velocity and brightnes curves. No, you have been saying that velocity and luminosity (not brightness) should be almost the same shape and in phase. In fact the ratio of the two components determines the phase but ADoppler is always negligible in the real systems we have examined. George |