Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Paul B. Andersen on 3 Aug 2005 06:35 Henri Wilson skrev: > 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" 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. 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. I will give you a hint: All the light "going in the same direction" do not have to come from the same source. Paul
From: sue jahn on 3 Aug 2005 07:10 <jgreen(a)seol.net.au> wrote in message news:1123055654.402098.166360(a)g44g2000cwa.googlegroups.com... > > Sue... wrote: > > jgreen(a)seol.net.au wrote: > > > bz wrote: > > > > jgreen(a)seol.net.au wrote in news:1122960081.759862.165870 > > > > @g14g2000cwa.googlegroups.com: > > > > > > > > > Jim Greenfield > > > > > c'=c+v > > > > > > > > > > > > > Jim, what about the c'=c-v photons from sources going away from us? > > > > > > Light is emitted from source at c. > > > > << Light is emitted from source at c. >> > > > > Are you sure of this? > > > > What if light is "dragged" out of the source > > by all the charges in the universe ? > > Would that alter the way you visualize light's propagation? > > I take on board, and consider it likely, that field propagation is > instantaneous (infinitely fast). Yes... that is a good notion. Attach a bungee cord to the Sphinx's nose (look near at your feet) Stretch the cord to the Eiffel tower. Then release it so it pops the Sphinx's rump. What is the delay between the time you release the end of the cord and the time it begin's moving towared the Sphinx? [rhetorical] > On the other hand, given the inverse > square (for attraction in this case) for distance/force, two isolated > atoms colliding at high speed far from significant gravity inspiring > material, are unlikely to be significantly influenced by your > proposition. You are unlikey to find such a condition. << It [ISM] fills interstellar space. This mixture is usually extremely tenuous, with typical gas densities ranging from a few single to a few hundred particles per cubic centimeter. As a result of primordial nucleosynthesis, the gas is roughly 90% hydrogen and 10% helium, with additional elements ("metals" in astronomical parlance) present in trace amounts. >> http://www.answers.com/topic/interstellar-medium-2 Sue... > I suspect photons will be emitted from the crash site at c (in whatever > direction, which may need further thought), and the debris left at the > scene will have reduced mass (be different elements, or changed > energies/hotter) > > Push is only a pull in the opposite direction. > > Bye > Jim G > c'=c+v > > > > Sue... > > > > snip >
From: sue jahn on 3 Aug 2005 07:16 "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 ? 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
From: Paul B. Andersen on 3 Aug 2005 07:38 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: | 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. 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
From: sue jahn on 3 Aug 2005 07:47
"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. ;-) Sue... http://news.google.com/news?hl=en&lr=&tab=wn&ie=UTF-8&q=rove+novak+plame&btnG=Search+News > | 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. > 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 > |