Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Paul B. Andersen on 3 Aug 2005 08:00 sue jahn skrev: > "Paul B. Andersen" <paul.b.andersen(a)deletethishia.no> wrote in message news:dclsna$41n$1(a)dolly.uninett.no... > > 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." Sue, I have responded to your challenge and explained "what Einstein's relativity say's the shaft should do." Now I challenge you to point out an inconsitency in the above. This response of yours: > <<geosynchronous satellite.>> > Neither will see the earth rotate. > So your experssion: > > <<N times per Earth rotation.>> > > Reduces to division by zero. > > << Why do you think I should have any problem with this?>> > > Heal thyself. ... is just too silly. Even you know that the Earth rotates once per sidereal day, even if you don't see it. Or don't you? Paul
From: Sue... on 3 Aug 2005 08:13 Paul B. Andersen wrote: > sue jahn skrev: > > "Paul B. Andersen" <paul.b.andersen(a)deletethishia.no> wrote in message news:dclsna$41n$1(a)dolly.uninett.no... > > > 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." > > Sue, I have responded to your challenge and explained > "what Einstein's relativity say's the shaft should do." > > Now I challenge you to point out an inconsitency in the above. > > This response of yours: > > > <<geosynchronous satellite.>> > > Neither will see the earth rotate. > > So your experssion: > > > > <<N times per Earth rotation.>> > > > > Reduces to division by zero. > > > > << Why do you think I should have any problem with this?>> > > > > Heal thyself. > > .. is just too silly. > > Even you know that the Earth rotates once > per sidereal day, even if you don't see it. > > Or don't you? I automatically assume that Andersen believes the earth's rotation affects gravity. Others I credit with greater insight. Sue... > > > Paul
From: Henri Wilson on 3 Aug 2005 08:41 On Wed, 3 Aug 2005 07:16:11 -0400, "sue jahn" <susysewnshow(a)yahoo.com.au> wrote: > >"bz" <bz+sp(a)ch100-5.chem.lsu.edu> wrote in message news:Xns96A72FEAEF37WQAHBGMXSZHVspammote(a)130.39.198.139... >> H@..(Henri Wilson) wrote in >> news:0920f1pfofa1omcth88bbp598mt4riv84a(a)4ax.com: >> >> >>>>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'. >> >> Depends on the transmitter's power. >> >> >>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. >> >> >>> Ah! but what is a 'cycle'? >> >>> .....a cycle of what? >> >> I answered the question with respect to the topic under discussion. >> >> I see no reason for a single photon to be longer than one cycle. > >Since the rules of Quantum Mechanics are already written, and you >AFAIK are not being consulted on a rewrite, the point is rather moot. Eh ? QM is a statistical theory...and stats don't work too well with a sample size of ONE. > >Sue... > >> >> >> >> >> -- >> bz >> >> please pardon my infinite ignorance, the set-of-things-I-do-not-know is an >> infinite set. >> >> bz+sp(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap > 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 3 Aug 2005 08:56 On 3 Aug 2005 03:35:20 -0700, "Paul B. Andersen" <paul.b.andersen(a)hia.no> wrote: > >Henri Wilson skrev: >> On 2 Aug 2005 08:01:12 -0700, "Paul B. Andersen" <paul.b.andersen(a)hia.no> >> wrote: >> >> >How is your foot, Henri? :-) >> >> To coin a favorite SRians phrase, >> "Paul you simply don't understand the theory" > >And which theory are you referring to? > >Is it this theory: >| light loses a minute amount of momentum every time 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. That is ONE very sound theory >or is it this theory: >| molecules in rare space DO tend to unify the speed of all light >| traveling in any particular direction. > >I have no problem understanding either of them. >But I also understand that they contradict each other. They do not contradict. ...the slow and fast light from a star merely slows at different rates. Im not adamant about the unification theory becasue I cannot see how Earth observers could detect doppler shifts in light from faraway stars. > >I will give you a hint: >All the light "going in the same direction" >do not have to come from the same source. I think I told YOU that....and I suggested that this fact might be instrumental in unifying the speed of all light in any particular direction. > >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: sue jahn on 3 Aug 2005 09:13
"Henri Wilson" <H@..> wrote in message news:pcf1f1hlqkj1548ov7apeqnvgm6i1m15go(a)4ax.com... > On 3 Aug 2005 03:35:20 -0700, "Paul B. Andersen" <paul.b.andersen(a)hia.no> > wrote: > > > > >Henri Wilson skrev: > >> On 2 Aug 2005 08:01:12 -0700, "Paul B. Andersen" <paul.b.andersen(a)hia.no> > >> wrote: > >> > > >> >How is your foot, Henri? :-) > >> > >> To coin a favorite SRians phrase, > >> "Paul you simply don't understand the theory" > > > >And which theory are you referring to? > > > >Is it this theory: > >| light loses a minute amount of momentum every time 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. > > That is ONE very sound theory > > >or is it this theory: > >| molecules in rare space DO tend to unify the speed of all light > >| traveling in any particular direction. > > > >I have no problem understanding either of them. > >But I also understand that they contradict each other. > > They do not contradict. ...the slow and fast light from a star merely slows at > different rates. > Im not adamant about the unification theory becasue I cannot see how Earth > observers could detect doppler shifts in light from faraway stars. You can't see it because you deny the Coulomb coupling that exist between the EM coupling structures and the matter in their local region of space. The only proof you can offer it that is mathematically convenient for folks with an ultra violet catastrophe. << There is in particular one problem whose exhaustive solution could provide considerable elucidation. What becomes of the energy of a photon after complete emission? Does it spread out in all directions with further propagation in the sense of Huygens' wave theory, so constantly taking up more space, in boundless progressive attenuation? Or does it fly out like a projectile in one direction in the sense of Newton's emanation theory? In the first case, the quantum would no longer be in the position to concentrate energy upon a single point in space in such a way as to release an electron from its atomic bond, and in the second case, the main triumph of the Maxwell theory - the continuity between the static and the dynamic fields and, with it, the complete understanding we have enjoyed, until now, of the fully investigated interference phenomena - would have to be sacrificed, both being very unhappy consequences for today's theoreticians. >> --Max Planck From Nobel Lectures, Physics 1901-1921, Elsevier Publishing Company, Amsterdam, 1967 http://nobelprize.org/physics/laureates/1918/planck-lecture.html Sue... http://www.eso.org/projects/vlti/images/vlti-array-smallsize.jpg > > > > > >I will give you a hint: > >All the light "going in the same direction" > >do not have to come from the same source. > > I think I told YOU that....and I suggested that this fact might be instrumental > in unifying the speed of all light in any particular direction. > > > > >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. |