Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: bz on 1 Aug 2005 17:42 "Paul B. Andersen" <paul.b.andersen(a)deletethishia.no> wrote in news:dcm17g$9u5$1(a)dolly.uninett.no: > bz wrote: >> "Jeff Root" <jeff5(a)freemars.org> wrote in news:1122862205.456748.163740 >> @g44g2000cwa.googlegroups.com: >> >> >>>George replied to Bob: >>> >>> >>>>>The spectra seem to indicate that the single photons >>>>>have very narrow bandwidth [as I would expect]. >>>> >>>>If a photon has a specific energy and that is >>>>proportional to frequency, then a single photon >>>>has a unique frequency hence zero bandwidth and >>>>infinite duration >:-( >>>> >>>>At least it does with a semi-classical view. >>>> >>>>If you include Heisenberg, then the uncertainty >>>>in the measured energy relates to the uncertainty >>>>in the frequency which depends on the time over >>>>which the frequency is measured, hence the >>>>bandwidth is related to the method of measurement, >>>>and I don't need to point out the crucial role of >>>>measurement methods in QM. >>>> >>>>As a result, I don't think a photon has a specific >>>>length or duration, but the idea of it as a single >>>>cycle with hard ends at the zero crossings can be >>>>ruled out as too simplistic. >>> >>>The notion that a single photon can have a bandwidth seems >>>absurd to me. For reasons analogous to the notion that the >>>North Pole can have a longitude. It seems obvious that the >>>property of bandwidth only applies to statistical collections >>>of photons, or waves. >>> >>>I'm not sure it makes sense to talk about the frequency of a >>>single photon, either. Any photon has a particular, measurable >>>energy which is associated with a particular frequency, but >>>there is no way to measure that frequency. Only by analyzing >>>the behavior of a collection of photons does the property of >>>frequency become evident. The analysis shows that f = E/h, >>>so you can *know* the frequency of a single photon within the >>>limits of uncertainty, but you can't actually measure it. >>> >>>If you have enough photons to directly measure their frequency, >>>then you have enough photons to detect their bandwidth. >>> >> >> >> The question is 'how big'[long] is a photon. >> >> I can't see any reason that it should be more than one cycle in length. > > It doesn't make much sense to talk about the "length" of a photon. > A photon is a point particle. Something can be a point when seen from one direction and still have duration (length) along another axis. >> There is a paper >> http://jchemed.chem.wisc.edu/JCEWWW/Articles/DynaPub/DynaPub.html#ref16 >> That I disagree with. They appear to believe that each photon consist >> of a wavetrain that is millions of cycles long. > > Remember that a "photon" is the particle aspect of the wave-particle > duality. A photon is the name we have given to the smallest possible 'bundle' of EM energy. In some experiments it looks like a wave, in some, it looks like a particle. Limitiations of our testing equipment. Some equations treat it like a wave, some like a probability function, some like a particle. None [that I know of] encompass all the photons observed properties. > It is true that you can make a classical (according to Maxwell) > "EM wave packet" which is limited in time and space, and which has > the energy of one photon. According to Fourier this "wave packet" > must have a certain relationship between the width of the spectrum > (spread of frequencies) and the extension in space. > The narrower the spectrum, the longer in space, and vice versa. I understand FT and FFT, modulation and what information theory says 'must' happen. > But you cannot equate this wave packet to a photon - a particle. > You must consider it to be a probability function. > The width of the spectrum is then the uncertainty in the frequency > and thus the energy of the photon, and the extension in space > is the uncertainty in time (since the wave packet is propagating > the spatial extension can be interpreted as an uncertainty in time). I agree. I think there is an uncertainly in time for single photons. A small fraction of the time of one wavelength, a small phase uncertainty, should suffice. > 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'. -- 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: bz on 1 Aug 2005 17:51 "sue jahn" <susysewnshow(a)yahoo.com.au> wrote in news:42ee88e9$0$18637$14726298(a)news.sunsite.dk: > > "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." >> >> Paul > > <<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. > http://www.google.com/search?hl=en&lr=&safe=off&q=gear+ratio+calculator&b > tnG=Search They each (satellite and earth) make 1 rotation about their common barycenter per day. N=1, not zero. -- 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: Henri Wilson on 1 Aug 2005 18:27 On Mon, 1 Aug 2005 21:42:57 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> wrote: >"Paul B. Andersen" <paul.b.andersen(a)deletethishia.no> wrote in >news:dcm17g$9u5$1(a)dolly.uninett.no: > >> bz wrote: >>> "Jeff Root" <jeff5(a)freemars.org> wrote in news:1122862205.456748.163740 >>> @g44g2000cwa.googlegroups.com: >>> >>> >>>>George replied to Bob: >>>> >>>> >>>>>>The spectra seem to indicate that the single photons >>>>>>have very narrow bandwidth [as I would expect]. >>>>> >>>>>If a photon has a specific energy and that is >>>>>proportional to frequency, then a single photon >>>>>has a unique frequency hence zero bandwidth and >>>>>infinite duration >:-( >>>>> >>>>>At least it does with a semi-classical view. >>>>> >>>>>If you include Heisenberg, then the uncertainty >>>>>in the measured energy relates to the uncertainty >>>>>in the frequency which depends on the time over >>>>>which the frequency is measured, hence the >>>>>bandwidth is related to the method of measurement, >>>>>and I don't need to point out the crucial role of >>>>>measurement methods in QM. >>>>> >>>>>As a result, I don't think a photon has a specific >>>>>length or duration, but the idea of it as a single >>>>>cycle with hard ends at the zero crossings can be >>>>>ruled out as too simplistic. >>>> >>>>The notion that a single photon can have a bandwidth seems >>>>absurd to me. For reasons analogous to the notion that the >>>>North Pole can have a longitude. It seems obvious that the >>>>property of bandwidth only applies to statistical collections >>>>of photons, or waves. >>>> >>>>I'm not sure it makes sense to talk about the frequency of a >>>>single photon, either. Any photon has a particular, measurable >>>>energy which is associated with a particular frequency, but >>>>there is no way to measure that frequency. Only by analyzing >>>>the behavior of a collection of photons does the property of >>>>frequency become evident. The analysis shows that f = E/h, >>>>so you can *know* the frequency of a single photon within the >>>>limits of uncertainty, but you can't actually measure it. >>>> >>>>If you have enough photons to directly measure their frequency, >>>>then you have enough photons to detect their bandwidth. >>>> >>> >>> >>> The question is 'how big'[long] is a photon. >>> >>> I can't see any reason that it should be more than one cycle in length. >> >> It doesn't make much sense to talk about the "length" of a photon. >> A photon is a point particle. > >Something can be a point when seen from one direction and still have >duration (length) along another axis. > >>> There is a paper >>> http://jchemed.chem.wisc.edu/JCEWWW/Articles/DynaPub/DynaPub.html#ref16 >>> That I disagree with. They appear to believe that each photon consist >>> of a wavetrain that is millions of cycles long. >> >> Remember that a "photon" is the particle aspect of the wave-particle >> duality. > >A photon is the name we have given to the smallest possible 'bundle' of EM >energy. In some experiments it looks like a wave, in some, it looks like a >particle. Limitiations of our testing equipment. > >Some equations treat it like a wave, some like a probability function, some >like a particle. > >None [that I know of] encompass all the photons observed properties. > >> It is true that you can make a classical (according to Maxwell) >> "EM wave packet" which is limited in time and space, and which has >> the energy of one photon. According to Fourier this "wave packet" >> must have a certain relationship between the width of the spectrum >> (spread of frequencies) and the extension in space. >> The narrower the spectrum, the longer in space, and vice versa. > >I understand FT and FFT, modulation and what information theory says >'must' happen. > >> But you cannot equate this wave packet to a photon - a particle. >> You must consider it to be a probability function. >> The width of the spectrum is then the uncertainty in the frequency >> and thus the energy of the photon, and the extension in space >> is the uncertainty in time (since the wave packet is propagating >> the spatial extension can be interpreted as an uncertainty in time). > >I agree. I think there is an uncertainly in time for single photons. A >small fraction of the time of one wavelength, a small phase uncertainty, >should suffice. > >> 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? 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 1 Aug 2005 18:34 On 31 Jul 2005 19:24:14 -0700, "Jeff Root" <jeff5(a)freemars.org> wrote: >On July 29, 2005, Henri Wilson wrote: > >> I am waiting to have it explained. >> What exactly defines 'proper time'. > >My understanding is that 'proper time' is the time experienced >where the observer is, as opposed to the time somewhere else. >My proper time is the time I experience where I am, and your >proper time is the time you experience where you are. > >That's a very simple explanation, but 'proper time' seems to >be a very simple thing. > >On the other hand, if you want to know what 'time' in general >is, I can't explain it. That is something a person can only >learn through experience. Virtually everyone learns what time >is before they learn to talk, anyway, so it isn't a problem. > >As Aurelius Augustinus wrote (1,600 years ago), "What is time? >If someone asks me, I know. If I wish to explain it to someone >who asks, I know not." Jeff you have come in late on these threads. You will have to read about my experiment detailed in the thread "GPS GR Correction myth" As for TIME. It is a fundamental dimension like space. Time also has three subdimesnions. Time "instant" is a point on the absolute time axis (t2) Time "interval" is a length on a time axis. Time 'flow' is the ratio of one to another (dt1/dt2) Time normally flows at 1 second(t1) per second(t2) That is equivalent to defining the slope of a hill as "10 metres/metre" I'm not sure what t3 does yet but I think it might be associated with consciousness. > > -- Jeff, in Minneapolis 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 01:21
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! None of your above attends the issue: there are THREE clocks involved, two of which are physically connected to the axle, and one near the top which was synchronised with the ground clock pre experiment. What I, Sue, Henri, and all other thinkers wish to know IIISSSSSSS Relativity claiming the two top clocks will show the same??????? And if so, how is a torque on the axle avoided? (oops! now DHR's have the top two showing the same, but not the top and bottom connected ones.........) Jim Greenfield c'=c+v > Paul |