Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Paul B. Andersen on 3 Aug 2005 15:33 Sue... skrev: > 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. Well, Sue? You are not up to it, are you? > > > > 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? Of course you know. This stupid remark of yours was obviously made to avoid having to address the issue. You always make stupid remarks to that end, don't you? Here is another one: > I automatically assume that Andersen believes > the earth's rotation affects gravity. > Others I credit with greater insight. Fleeing the challenge, Sue? You always do, don't you? Paul
From: sue jahn on 3 Aug 2005 15:29 "bz" <bz+sp(a)ch100-5.chem.lsu.edu> wrote in message news:Xns96A78DCA73BA6WQAHBGMXSZHVspammote(a)130.39.198.139... > "Sue..." <suzysewnshow(a)yahoo.com.au> wrote in news:1123091999.580984.298710 > @o13g2000cwo.googlegroups.com: > > > > > bz wrote: > >> "Sue..." <suzysewnshow(a)yahoo.com.au> wrote in > >> news:1123076729.211267.18360(a)o13g2000cwo.googlegroups.com: > >> > >> >> QM is a statistical theory...and stats don't work too well with a > >> >> sample size of ONE. > >> > > >> > Did someone say the *sample* size was one ? > >> > > >> > Santa can deliver 2.3 toys to *ONE* house > >> > with a high degree of certainty that each > >> > child in the house will get a toy. > >> > > BZ: Only if Santa restricts his deliveries to a small portion of the > > globe. > > << I see no reason for a single photon to be longer than one cycle.>> > > > > Sue: Well maybe it has to be at least four cycles in case > > some astromnomer wants to cut it into four pieces. > > http://www.eso.org/outreach/press-rel/pr-2000/phot-26-00.html > > Naw, they just need for 4 photons to arrive from the same source at the > same time. Ya care to show us where that feller is on the periodic table? Sue... > > > GAWD! you don't suppose Santa elves cut up the toys do you? > > Naw, they just stretch them out over millions of cycles. > > > > > > -- > 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
From: Paul B. Andersen on 3 Aug 2005 15:45 sue jahn skrev: > "Paul B. Andersen" <paul.b.andersen(a)hia.no> wrote in message news:1123069101.085286.119210(a)g14g2000cwa.googlegroups.com... > > > > jgreen(a)seol.net.au skrev: > > > Paul B. Andersen wrote: > > > > jgreen(a)seol.net.au wrote: > > > > > 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 know what YOU are discussing, but what I > > was answering in the posting you responded to was > > this particular challenge defined by Sue: > > Take careful notes on all of this. You may be called as > a witness when Andersen sues himself for plagiarism. ;-) I am sure you are a great admirer of your own wittiness, Sue. Do you admire your ability to flee challenges equally much? > > | 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. This challenge goes for you too, Sue. Now you can demonstrate your ability to flee it yet again. > > 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. > > > > Paul Paul
From: sue jahn on 3 Aug 2005 15:46 "Paul B. Andersen" <paul.b.andersen(a)hia.no> wrote in message news:1123097611.454902.209960(a)z14g2000cwz.googlegroups.com... > > Sue... skrev: > > 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. > > Well, Sue? > You are not up to it, are you? I can do it with a bag of marbles and two garden hoses and nobody questions the result. You broke your proof. You can fix your proof. Sue... > > > > > > > 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? > > Of course you know. > This stupid remark of yours was obviously made > to avoid having to address the issue. > > You always make stupid remarks to that end, don't you? > > Here is another one: > > > I automatically assume that Andersen believes > > the earth's rotation affects gravity. > > Others I credit with greater insight. > > Fleeing the challenge, Sue? > > You always do, don't you? > > Paul >
From: Paul B. Andersen on 3 Aug 2005 16:01
Henri Wilson skrev: > 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. ... and thus contradicting the "slowing light" theory. Is it beginning to dawn to you? :-) > >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. Henri, Henri, Henri .. :-) I rest my case. Paul, amused |