Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Paul B. Andersen on 2 Aug 2005 11:01 Henri Wilson skrev: > On Fri, 29 Jul 2005 12:08:58 +0200, "Paul B. Andersen" > <paul.b.andersen(a)deletethishia.no> wrote: > > >Henri Wilson wrote: > >> On Thu, 28 Jul 2005 12:10:03 +0200, "Paul B. Andersen" > >> <paul.b.andersen(a)deletethishia.no> wrote: > >> > >> > >>>Henri Wilson wrote: > >>> > >>>>On Wed, 27 Jul 2005 10:18:04 +0200, "Paul B. Andersen" > >>>><paul.b.andersen(a)deletethishia.no> wrote: > >>>> > >> > >> > >>>>>There is no way you can fail to see that the consequence of unifying > >>>>>"the speed of all light traveling in any particular direction" > >>>>>is that all light coming from any particular direction have > >>>>>the same speed. > >>>> > >>>> > >>>>1) 'tends to unify' does not mean 'co,plete unification'.2) the volumes of > >>>>space that are responsible for the tendency towards unity might themselves be > >>>>moving wrt little planet Earth. > >>>>So the final speed wrt Earth EVEN WITH 100% UNIFICATION can have any value. > >>> > >>>So we can conclude that all the light coming from any particular > >>>direction is red shifted by the same amount, but the amount can have any value. > >> > >> > >> Paul, I'm sure you weren't quite this confused before your recent holiday. > >> > >> > >>>>>So I repeat. > >>>>>We can conclude that all the light coming from any particular > >>>>>direction is red shifted by the same amount. > >>>>> > >>>>>Or maybe you can explain why this does not folow from your claim? > >>>> > >>>> > >>>>Paul, according to my very plausible and obviously correct theory, light loses > >>>>a minute amount of momentum every time it drags an atom along. > >>>> > >>>>Do you deny that a photon loses momentum when it drags an atom along? > >>>> > >>>>If the momentum lost is, on average proportional to momentum (all wrt the > >>>>source frame) then the decrease would be an exponential one. As you know small > >>>>sections of an exponential can appear fairly linear. Hence the resultant > >>>>redshift (wrt source frame) is virtually proportional to distance from source. > >>> > >>>So we can conclude that molecules in rare space do NOT tend to unify > >>>the speed of all light travelling in any particular direction, it slows > >>>it down proportionally to the travelled distance. > >> > >> > >> As I have pointed out many times, THERE ARE TWO DISTINCTLY DIFFERENT EFFECTS. > >> Can you not read properly any more? > > > >Henri Wilson wrote: > >| .....and yes, molecules in rare space DO tend to unify the speed of all light > >| traveling in any particular direction. All light is redshifted in the process. > > > >... ARE TWO DISTINCTLY DIFFERENT EFFECTS ? > > > >and these two effects are that all light coming from the same direction > >is going at the same speed, but since light is slowed down as it goes, > >some of the light coming from the same direction may go slower than > >other light from that direction. > > > >Have I got it now? > > No you haven't gotten it. Right. > You are not even trying to get it. I never try the impossible. > You probably don't have the required intelligence to be able to get it. Right. I do however have the required intelligence to see that you are contradicting yourself. How is your foot, Henri? :-) Paul
From: Paul B. Andersen on 2 Aug 2005 14:51 jgreen(a)seol.net.au wrote: > Paul B. Andersen wrote: > >>sue jahn wrote: >> >>>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. >> >>Why do you think I should have any problem with this? >>This is yet another old non 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. >> >>Let A measure the proper duration of one Earth rotation to be T. >>Then, as you now know and have accepted is experimentally >>verified for clocks in GPS orbit, 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." > > > Thanks- I needed a good laugh! Fools laugh at what they don't understand. Paul
From: Henri Wilson on 2 Aug 2005 20:02 On Tue, 2 Aug 2005 10:20:40 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> wrote: >H@..(Henri Wilson) wrote in >news:mc8te1p1jcju71a65e4t4q6nmn3abr170d(a)4ax.com: > >>> >>>> The relationship mentioned above is thus in accordance >>>> with the uncertainty principle. >>>> >>>>> Their theory seems to be falsified by femtosecond laser pulses. >>>>> There are some that would be less than 2 cycles at the frequency of >>>>> the laser. Also, the max keying speed of ELF transmission would seem >>>>> to preclude any requirement for millions of cycled per photon. >>>> >>>> Quite. >>>> If the pulses are very short and determined with high precision >>>> in time (very short wave packet), the uncertainty in energy >>>> must be high (Wave packet with very wide frequency spectrum.) >>> >>>Or there must be some uncertainly in the exact time the photon was >>>emitted by the transmitting antenna. >>> >>>> But I will repeat: >>>> It doesn't make sense to talk about the length of a photon. >>> >>>It would make an interesting experiment, if I had the equipment to do >>>switching at zero crossing and sent 1 cycle at 160 meters [the 1.8 MHz >>>amateur band], I think I should create a bunch of 160 meter photons. >>>They would probably be spread out a bit in time. >>> >>>> Consider this: >>>> We observe the H-alpha line from a very weak astronomical >>>> source. With modern CCDs, we can literally count the photons >>>> as they arrive, maybe only a very few photons per second. >>> >>>yes. >>> >>>> If we use a spectrometer, we can see that the spread in >>>> frequency/energy is very small (it's a spectral line). >>> >>>yes. >>> >>>> But you can say just about nothing about _when_ the next >>>> photon will be detected, they will appear to arrive randomly >>>> (like radioactive radiation - Poisson distribution). >>> >>>yes. >>> >>>> So the "wave packet" of these photons must have a very >>>> narrow frequency spectrum and must be very long in space. >>> >>>No. The packets are spread out in space, but each is very short. >>> >>>> Does it make sense to say that these photons are >>>> light seconds long? :-) >>> >>>No. Only that they are light seconds apart. >>> >>>> Don't think so. >>>> That the wave packet is long only means that we don't >>>> know when the photon will be detected. >>> >>>We know when it is detected. If it is millions of cycles long, does >>>detection take place at the beginning or the end of the millions of >>>cycles? >>> >>>If it is millions of cycles long, what happens if a shutter is dropped >>>into the focal plane in the middle of the wave train? >>> >>>> But yet it is detected at an instant - not gradually. >>> >>>That would indicate that it can NOT be millions of cycles 'long'. >>> >>>I see no reason for it to be more than 1 cycle 'long'. >> >> Ah! but what is a 'cycle'? >> .....a cycle of what? > >A cycle of E <---> M energy transfer. Where the E and M fields exchange >energy. > >A rotation of the energy magnitude vector in EM space. > >A cycle of the AC voltage in my transmitting antenna. >A cycle of the AC voltage induced by the passing M field in my receiving >antenna. No that's not the cycle of a single photon. That involves 'group phasing'. >A cycle of the current in my loop transmitting antenna [which produces an M >field in space] >A cycle of the current induced in my loop receiving antenna by the M field >of the passing radio wave. No bob. Read the question. >... HW. www.users.bigpond.com/hewn/index.htm Sometimes I feel like a complete failure. The most useful thing I have ever done is prove Einstein wrong.
From: Henri Wilson on 2 Aug 2005 20:12 On 2 Aug 2005 08:01:12 -0700, "Paul B. Andersen" <paul.b.andersen(a)hia.no> wrote: > >Henri Wilson skrev: >> On Fri, 29 Jul 2005 12:08:58 +0200, "Paul B. Andersen" >> <paul.b.andersen(a)deletethishia.no> wrote: >> >Right. > >> You are not even trying to get it. > >I never try the impossible. > >> You probably don't have the required intelligence to be able to get it. > >Right. > >I do however have the required intelligence to see >that you are contradicting yourself. > >How is your foot, Henri? :-) To coin a favorite SRians phrase, "Paul you simply don't understand the theory" > > >Paul HW. www.users.bigpond.com/hewn/index.htm Sometimes I feel like a complete failure. The most useful thing I have ever done is prove Einstein wrong.
From: jgreen on 2 Aug 2005 23:46
Paul B. Andersen wrote: > jgreen(a)seol.net.au wrote: > > Paul B. Andersen wrote: > > > >>sue jahn wrote: > >> > >>>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. > >> > >>Why do you think I should have any problem with this? > >>This is yet another old non 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. > >> > >>Let A measure the proper duration of one Earth rotation to be T. > >>Then, as you now know and have accepted is experimentally > >>verified for clocks in GPS orbit, 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." > > > > > > Thanks- I needed a good laugh! > > Fools laugh at what they don't understand. 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 expect you to answer-----you cannot! Jim G c'=c+v > > Paul |