Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Jeff Root on 31 Jul 2005 22:24 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, in Minneapolis
From: bz on 1 Aug 2005 10:29 "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. 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. 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. -- 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 1 Aug 2005 15:21 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
From: Paul B. Andersen on 1 Aug 2005 16:38 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. > 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. 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. 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). 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.) But I will repeat: It doesn't make sense to talk about the length of a photon. 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. If we use a spectrometer, we can see that the spread in frequency/energy is very small (it's a spectral line). 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). So the "wave packet" of these photons must have a very narrow frequency spectrum and must be very long in space. Does it make sense to say that these photons are light seconds long? :-) Don't think so. That the wave packet is long only means that we don't know when the photon will be detected. But yet it is detected at an instant - not gradually. Paul
From: sue jahn on 1 Aug 2005 16:38
"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&btnG=Search Sue... |