Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Henri Wilson on 7 Aug 2005 19:46 On 6 Aug 2005 17:45:02 -0700, "Sue..." <suzysewnshow(a)yahoo.com.au> wrote: > >Henri Wilson wrote: >> On 4 Aug 2005 18:47:53 -0700, "Sue..." <suzysewnshow(a)yahoo.com.au> wrote: >> >> >Henri Wilson wrote: >> >> On Wed, 3 Aug 2005 09:13:20 -0400, "sue jahn" <susysewnshow(a)yahoo.com.au> >> >> wrote: >> >> >> >> >> I'll believe you... >> >> > >> >http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html >> > >> >You have allowed for light to ?ocassionaly? interact >> >with matter? >> > >> >You are using an unrealistic spatial model for propagting >> >light. >> >> I wasn't aware that there was any other model for light propagating in 'empty' >> space. > >I know of no model for less than two charges, so why would you >even consider 'empty' space. >A space with two charges, is not empty space. My theory says there is a threshold matter density, below which 'different things' happen. > >Sue... > 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: bz on 7 Aug 2005 19:54 "George Dishman" <george(a)briar.demon.co.uk> wrote in news:dd616e$v9m$1 @news.freedom2surf.net: > > "bz" <bz+sp(a)ch100-5.chem.lsu.edu> wrote in message > news:Xns96AB8BB274B34WQAHBGMXSZHVspammote(a)130.39.198.139... >> "George Dishman" <george(a)briar.demon.co.uk> wrote in news:dd5g3h$out$1 >> @news.freedom2surf.net: >> >>> >>> "bz" <bz+sp(a)ch100-5.chem.lsu.edu> wrote in message >>> news:Xns96AB76C397DFCWQAHBGMXSZHVspammote(a)130.39.198.139... >>> >>> ... >>> >>>> If we look at femto second pulses, they must consist of bunches of >>>> photons >>>> at different frequencies [the frequencies of the side bands]. >>>> >>>> If we filtered those out, and attenuated the carrier so as to just allow >>>> one photon on the main frequency, would the photons suddenly get longer? >>>> >>>> I don't yet have any reason to think so. >>> >>> If you hit a narrow-band filter with an >>> impulse, how long does it ring ;-) >> >> That depends upon the Q. >> >> Since we only want one photon, we can tolerate a rather low Q. >> >> Aye? > > No, you can get single photons just by reducing the intensity > to a low rate and using quite a wide (long duration) gate. I could, but I don't want to. I want to run through a low pass filter with a fairly shar put off, right above the 'carrier' [the characteristic frequency for the laser in question] and then through a high pass filter with a cut off just below the 'carrier'. I don't mind if I lose most of the carrier also, just as long as I have a few photons left there and have eliminated the 'modulation sidebands'. > You said "one photon on the main frequency". The question is > how close to that frequency is "on"? Low Q means wide band. > The duration of the ringing is of the order of 1 / bandwidth. > Again we come back to the argument from duality, that the > bandwidth is a measure of the uncertainty of the energy of a > particular photon while the "length" is related to the > uncertainty of location and the product is of the order of > Planck's constant. Try to constrain the energy using a filter > and ringing increases the uncertainty in location. High Q filters ring. Low Q filters don't. I know that what I am talking about doing could not work in the electronic signal world because the signals are not quantitized. But light energy is, so I am not convinced it is impossible. The different domain is important. -- 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: the softrat on 7 Aug 2005 22:43 On Sun, 07 Aug 2005 23:46:41 GMT, H@..(Henri Wilson) wrote: >My theory says there is a threshold matter density, below which 'different >things' happen. That theory would apply to the inside of many of the sci.physics posters heads. (Too many 's'es for you, Bucky? Look in a mirror.) the softrat Sometimes I get so tired of the taste of my own toes. mailto:softrat(a)pobox.com -- If at first you don't succeed, destroy all evidence that you tried. -- Steven Wright
From: Jeff Root on 8 Aug 2005 18:57 Jim Greenfield replied to Jeff Root: > Jeff Root wrote: >> Jim Greenfield replied to Paul B. Andersen: >> >> >> > 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. >> > >> > After all your pathetic attempts to duck the question, it STILL >> > remains!! >> > You claim here that the axle does NOT twist under GR. Since two of the >> > clocks are ATTACHED to the axle, they MUST read the SAME time. But GR >> > says the high unconnected (different gravity) one runs differentlty. >> > This is a CONTRADICTION!!!!!!!!!!!!!!!!!!! of the "theory" >> >> Jim, >> >> How can a prediction made by a theory contradict the theory? Didn't you want to answer that question? You claimed that the prediction made by GR contradicts GR. Obviously it is impossible for a prediction of a theory to contradict the theory which made it. Clearly, what you said was not what you meant. >> Why do you say that because the two clocks are attached to the >> axle, they must read the same? > > Because of the axle in my car. Really? Your car has only one axle? Assuming that you are referring to the axles on which the wheels turn, a car with four wheels has four axles so each can turn at a different speed. See http://auto.howstuffworks.com/differential.htm I'd think a simpler example would be better. Though I don't know why you needed an example at all. An axle is an axle, and an axle by itself is an extremely simple thing. All it is in this case is a single, solid rotating body. > A clock driven at each end mechanically will read the same > for eternity That is what you need to show. You haven't shown it, merely asserted it. It seems obvious that the two clocks would read the same, but what seems obvious isn't always true. GR predicts that in certain circumstances, the two clocks would not read the same. If you think that is wrong, you need to show why. > (sans diff slippage). Slippage? The thought experiment doesn't permit any slippage. If you are thinking of your car wheels slipping on the pavement, you are thinking of something utterly irrelevant. > Gr is trying to tell me, that if I lay the old girl on the side, > time is passing differently at each end due to gravity disparity. Yes. > So how are the rotations per time at each end the same, > if a clock at a higher altitude reads differently, due to > gravity difference? They aren't. GR predicts that the rotational speed of the bottom of the shaft, measured with a clock at the bottom of the shaft, is greater than the rotational speed of the top of the shaft, as measured by a clock at the top of the shaft. You think that means the shaft must become twisted. Paul showed that it doesn't mean that. > For Paul to maintain the claim of time passing differently due > to gravity, he must disconnect! > The free (call it a GPS based clock) and the one connected to > the (very high) axle, CANNOT read differently, WITHOUT a torque > on the axle. Paul showed that they do read differently without a torque. > THIS is the contradiction. : The "Theory" says they will, and > the axle proves they won't. The entire body of the axle is rotating at constant speed. What that speed is measured to be depends on where in the gravity field the measurement is made. That idea is very surprising to almost everyone, including me. To some people, it is incomprehensible. They reject the idea as being absurd, even though the time difference is directly observed every day in careful measurements. Those people sometimes reject the observations because they contradict their own beliefs. -- Jeff, in Minneapolis
From: George Dishman on 9 Aug 2005 14:28
"bz" <bz+sp(a)ch100-5.chem.lsu.edu> wrote in message news:Xns96ABC086CA9AWQAHBGMXSZHVspammote(a)130.39.198.139... > "George Dishman" <george(a)briar.demon.co.uk> wrote in news:dd616e$v9m$1 > @news.freedom2surf.net: .... > > I want to run through a low pass filter with a fairly shar put off, right > above the 'carrier' [the characteristic frequency for the laser in > question] and then through a high pass filter with a cut off just below > the > 'carrier'. I don't mind if I lose most of the carrier also, just as long > as > I have a few photons left there and have eliminated the 'modulation > sidebands'. > >> You said "one photon on the main frequency". The question is >> how close to that frequency is "on"? Low Q means wide band. >> The duration of the ringing is of the order of 1 / bandwidth. >> Again we come back to the argument from duality, that the >> bandwidth is a measure of the uncertainty of the energy of a >> particular photon while the "length" is related to the >> uncertainty of location and the product is of the order of >> Planck's constant. Try to constrain the energy using a filter >> and ringing increases the uncertainty in location. > > High Q filters ring. Low Q filters don't. Anything narrower than critical damping rings. > I know that what I am talking about doing could not work in the electronic > signal world because the signals are not quantitized. Yes they are, the quanta are just too small to detect. > But light energy is, so I am not convinced it is impossible. > > The different domain is important. I'll be interested to see what you get if you ever have the chance to try it. Thanks for a thought-provoking discussion. Best regards George |