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From: Jerry on 30 Jun 2005 11:19 The Ghost In The Machine wrote: > In sci.physics, Jerry > <Cephalobus_alienus(a)comcast.net> > wrote > on 30 Jun 2005 02:20:12 -0700 > <1120123212.664910.186690(a)g43g2000cwa.googlegroups.com>: > > Henri Wilson wrote: > >> Do you have a better model , A? > >> Does anyone? > > > > Sure. Sagnac fits perfectly well within the context of SR. > > > > Jerry > > > > I could see Sagnac fitting into *GR*, but a rotating disc is not > an inertial reference frame. One could, however, say that it > approximates an inertial reference frame for a sufficiently large disc. Don't need GR to explain Sagnac. http://www.mathpages.com/rr/s2-07/2-07.htm Jerry
From: Henri Wilson on 1 Jul 2005 04:39 On Wed, 29 Jun 2005 19:42:46 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <H@..> wrote in message >news:p5u3c1l8gvpr100f69og9d9gao9nl0q1ek(a)4ax.com... >> On Tue, 28 Jun 2005 21:45:58 +0100, "George Dishman" >> <george(a)briar.demon.co.uk> >> wrote: >>>That's a much more sensible suggestion. >>>However, it will suffer from the usual >>>problem that the energy loss must be >>>associated with a momentum change as >>>you said, hence it would cause blurring >>>of the image of distant objects. >> >> >> Not necessarily. Deflection should even out over any significant distacne. > >That's an interesting aspect. Each deflection will give >the photon a small velocity increment perpendicular to >the direction of propagation (say the z axis). However, >the increments will be randomly distributed in the x-y >plane of and have some spread of magnitude. Think of >each deflection looking along the (average) line of >propagation and it will look like a random walk. The >end result is the (angular) deflection will be >proportional to the square root of the number of >deflections and proportional to the mean deflection >angle per encounter. Yes that IS interesting. The effect must be small though because blurring is pretty small even at very large distances. ..nothing like dispersion in a gaseous atmosphere. > >> I imagine a photon as a long pointed cigar shaped object moving rapidly >> through >> space. Every time it passes an atom, that atom is drawn very slightly >> towards >> it. The associated energy can only come from the photon. The photon will >> also >> be deflected but after passing lots of atoms will end up in the same basic >> direction. It will travel in a very slight zig-zag path, losing a little >> energy >> at every direction change. > >You can use the line width to set a minimum on the >number of deflections. The redshift would depend on >the number but for the reason above, spectral >broadening would depend on the square root. There is another and possibly more important factor here that I hadn't considered. Not only would the atoms be deflected, they would also be moved in the direction of photon travel. Hey! George, I think we've got it. As light travels, its 'fields' come into the 'contact sphere' of atoms, which acquire momentum as a result. The photons MUST provide the energy for that momentum. This is 'tired light' in a nutshell. Photons don't have to actually strike an atom to drag it along a little. Energy loss is proportional to distance. Goodbye BB theory! >> >> Note, if two photons pass an atom simultaneously and equidistant but on >> opposite sides, the atom will not move...imagine two electrons passing a >> proton. The electrons are deflected towards each other but the proton does >> not >> move sideways. >> Question, does the proton end up displaced longitudinally? It should not, >> but..... >> >> e- >v >> P >> e- >v > > >It's actually more complex and probably depends on >the speed. I think the answer can be a small >velocity change for the proton in either direction >but parallel to the direction of motion of the >electrons of course, but we are talking of photons, >not electrons. The proton should eventually come to rest exactly where it was.. > >Because EM forces are far stronger than gravitational >other forms of scattering dominate where interactions >with charged particles are concerned. Free neutrons >might give gravitational scattering but of course >they decay so having thought about your suggestion a >bit longer, I don't see how you can construct a >viable model out of it. Well I have now modified it. _____________A->v______p->>>> Whenever a photon goes flying past an atom, the atom is moved a little in the same direction. Remember we are discussing deep space here. There are not many photons OR atoms. >> >> I read somewhere authoritative that about half the scientific world does >> not >> agree with the BB theory. >> I don't remember where I read everything I read George. > >I think you'll find it was just a crank site or >publication. There's a lot of wishful thinkers >out there. It's certainly untrue. Well it wont be when my latest theory comes out. Even YOU must agree it is credible. > >>>> >>>> What do you gain by accepting the BB idea? Nothing. >>> >>>True, the universe isn't there for my gain. >>>The measurements tell me that's the way it >>>is whether I want to accept it or not. >> >> the measurments tell us that light loses energy apparently in proportion >> to the >> distance it travels.....although that is by no means certain. > >No, the SNe measurements rule that out conclusively >though it had been pretty well discounted by more >complex evidence long before then. I like my latest theory. Every time a photon whizzes past an atom, the atom gains a minute amount of energy and the photon loses a little. ....it slows slightly. I am willing to bet that redshifted light is moving at <c. Blue shifts result from genuine c+v doppler.....but is not quite as blueshifted as it should be. > >George > > 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 Jul 2005 04:41 On 30 Jun 2005 02:20:12 -0700, "Jerry" <Cephalobus_alienus(a)comcast.net> wrote: >Henri Wilson wrote: >> On 28 Jun 2005 20:21:27 -0700, "Arthur Dent" <jp006t2227(a)blueyonder.co.uk> >> wrote: >> >> > >> > >> >Henri Wilson wrote: >> >> On 27 Jun 2005 18:52:28 -0700, "Jerry" <Cephalobus_alienus(a)comcast.net> wrote: >> >> >> >> >> >In your dreams, Henri... >> >> > >> >> >Jerry >> >> >> >> I suppose you are one of those people who believe that all photons are little, >> >> round, perfectly elastic and identical dimensionless points. >> >> >> >> They are not. If that were true there would be no way they would differ from >> >> 'nothing'. They a long, pointed, cigar shaped aggregations of standing EM >> >> waves. >> > >> > >> >You are bullshitting, H. Sorry to have to say that, but you have no >> >evidence >> >and lose credibility when you make such assertions. >> >> Do you have a better model , A? >> Does anyone? > >Sure. Sagnac fits perfectly well within the context of SR. And SR fits perfectly well within the context of a waste collection site. > >Jerry 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: George Dishman on 1 Jul 2005 15:37 "Henri Wilson" <H@..> wrote in message news:u2v9c11n8ue1gqurdds45t90v4bea1lpmc(a)4ax.com... > On Wed, 29 Jun 2005 19:42:46 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <H@..> wrote in message >>news:p5u3c1l8gvpr100f69og9d9gao9nl0q1ek(a)4ax.com... >>> On Tue, 28 Jun 2005 21:45:58 +0100, "George Dishman" >>> <george(a)briar.demon.co.uk> >>> wrote: > >>>>That's a much more sensible suggestion. >>>>However, it will suffer from the usual >>>>problem that the energy loss must be >>>>associated with a momentum change as >>>>you said, hence it would cause blurring >>>>of the image of distant objects. >>> >>> Not necessarily. Deflection should even out over any significant >>> distacne. >> >>That's an interesting aspect. Each deflection will give >>the photon a small velocity increment perpendicular to >>the direction of propagation (say the z axis). However, >>the increments will be randomly distributed in the x-y >>plane of and have some spread of magnitude. Think of >>each deflection looking along the (average) line of >>propagation and it will look like a random walk. The >>end result is the (angular) deflection will be >>proportional to the square root of the number of >>deflections and proportional to the mean deflection >>angle per encounter. > > Yes that IS interesting. > > The effect must be small though because blurring is pretty small even at > very > large distances. ..nothing like dispersion in a gaseous atmosphere. Indeed, that's where the problem lies for that approach to tired light, the predicted blurring is far to great with usual models. >>> I imagine a photon as a long pointed cigar shaped object moving rapidly >>> through >>> space. Every time it passes an atom, that atom is drawn very slightly >>> towards >>> it. The associated energy can only come from the photon. The photon will >>> also >>> be deflected but after passing lots of atoms will end up in the same >>> basic >>> direction. It will travel in a very slight zig-zag path, losing a little >>> energy >>> at every direction change. >> >>You can use the line width to set a minimum on the >>number of deflections. The redshift would depend on >>the number but for the reason above, spectral >>broadening would depend on the square root. > > There is another and possibly more important factor here that I hadn't > considered. > Not only would the atoms be deflected, they would also be moved in the > direction of photon travel. > > Hey! George, I think we've got it. Don't get carried away, both energy and momentum are conserved within the small region over which the interaction occurs. > As light travels, its 'fields' come into the 'contact sphere' of atoms, > which > acquire momentum as a result. > > The photons MUST provide the energy for that momentum. > > This is 'tired light' in a nutshell. Photons don't have to actually strike > an > atom to drag it along a little. That's correct so far. > Energy loss is proportional to distance. Whoa! You mean a high frequency photon would lose the same energy as a low frequency one? The redshift _factor_ has been shown to be independent of frequency which means the photon should lose a fixed _fraction_ of its energy per interaction. That's a fairly well known result even from tired light supporters. > Goodbye BB theory! > > >>> >>> Note, if two photons pass an atom simultaneously and equidistant but on >>> opposite sides, the atom will not move...imagine two electrons passing a >>> proton. The electrons are deflected towards each other but the proton >>> does >>> not >>> move sideways. >>> Question, does the proton end up displaced longitudinally? It should >>> not, >>> but..... >>> >>> e- >v >>> P >>> e- >v >> >> >>It's actually more complex and probably depends on >>the speed. I think the answer can be a small >>velocity change for the proton in either direction >>but parallel to the direction of motion of the >>electrons of course, but we are talking of photons, >>not electrons. > > The proton should eventually come to rest exactly where it was.. Nope, no friction in a vacuum, well mostly none :-o In the bigger picture, the proton would have occasional interactions with lots of CMBR photons. If it was moving in some direction (relative to the mean motion of the CMBR source), CMBR photons moving the same way would be reddened by Doppler while those moving the other way would be blueshifted. Since blueshifted photons would transfer more momentum, gradually the proton's speed would be adjusted so that there was no mean dipole component to the CMBR anisotropy. That is almost like a drag term. That could take a very long time though! I'll leave you to do the sums ;-) >>Because EM forces are far stronger than gravitational >>other forms of scattering dominate where interactions >>with charged particles are concerned. Free neutrons >>might give gravitational scattering but of course >>they decay so having thought about your suggestion a >>bit longer, I don't see how you can construct a >>viable model out of it. > > Well I have now modified it. > > _____________A->v______p->>>> > > Whenever a photon goes flying past an atom, the atom is moved a little in > the > same direction. > Remember we are discussing deep space here. There are not many photons OR > atoms. Check if you can conserve both energy and momentum (you'll find the answer on many web pages). >>> I read somewhere authoritative that about half the scientific world does >>> not >>> agree with the BB theory. >>> I don't remember where I read everything I read George. >> >>I think you'll find it was just a crank site or >>publication. There's a lot of wishful thinkers >>out there. It's certainly untrue. > > Well it wont be when my latest theory comes out. > Even YOU must agree it is credible. It appears credible at first glance which is why I looked at it some time ago. The sphere of influence within which the photon is significantly affected can be given a cross-sectional area, i.e. the area of a surface perpendicular to the line of flight of the photon, and you can split up that area like a bullseye target. If the photon hits the centre, it gets bounced back the way it came. Next ring out deflects it by sligtly less and so on. By integrating the whole, you get the total cross section but by considering a narrow ring you get the probability that the photon will be deflected by an angle between say a and a+da where da is a small increment. All this is important for X-ray work and you'll find stuff on it on medical sites. The key formula is called "Klein-Nishina" and you should find my analysis if you search for that in the sci.astro group early last year. There is a significant factor which is the ratio of the photon energy E = h * nu to the particle energy given by m c^2 of course. Things are different depending on which is larger. Think of playing pool but with with one bowling ball and one table tennis ball. >>>>> What do you gain by accepting the BB idea? Nothing. >>>> >>>>True, the universe isn't there for my gain. >>>>The measurements tell me that's the way it >>>>is whether I want to accept it or not. >>> >>> the measurments tell us that light loses energy apparently in proportion >>> to the >>> distance it travels.....although that is by no means certain. >> >>No, the SNe measurements rule that out conclusively >>though it had been pretty well discounted by more >>complex evidence long before then. > > I like my latest theory. > > Every time a photon whizzes past an atom, the atom gains a minute amount > of > energy and the photon loses a little. ....it slows slightly. > I am willing to bet that redshifted light is moving at <c. Blue shifts > result > from genuine c+v doppler.....but is not quite as blueshifted as it should > be. It sounds good in principle but I found the problem appears when you do the sums. Typically the minimum number of interactions for is in the billions and the average deflection of the photon turns out to be close to 90 degrees. It surprised me when I calculated it but nobody has been able to show a fault in my analysis yet. Of course it means every star should light up the whole sky as if we were embedded in translucent plastic. George
From: Henri Wilson on 1 Jul 2005 19:43
On Fri, 1 Jul 2005 20:37:34 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <H@..> wrote in message >news:u2v9c11n8ue1gqurdds45t90v4bea1lpmc(a)4ax.com... >> On Wed, 29 Jun 2005 19:42:46 +0100, "George Dishman" >> <george(a)briar.demon.co.uk> >> wrote: >> There is another and possibly more important factor here that I hadn't >> considered. >> Not only would the atoms be deflected, they would also be moved in the >> direction of photon travel. >> >> Hey! George, I think we've got it. > >Don't get carried away, both energy and >momentum are conserved within the small >region over which the interaction occurs. > >> As light travels, its 'fields' come into the 'contact sphere' of atoms, >> which >> acquire momentum as a result. >> >> The photons MUST provide the energy for that momentum. >> >> This is 'tired light' in a nutshell. Photons don't have to actually strike >> an >> atom to drag it along a little. > >That's correct so far. > >> Energy loss is proportional to distance. > >Whoa! You mean a high frequency photon would >lose the same energy as a low frequency one? That is quite likely...a proportional loss. > > >The redshift _factor_ has been shown to be >independent of frequency which means the >photon should lose a fixed _fraction_ of >its energy per interaction. That's a fairly >well known result even from tired light >supporters. This theory could easily conform with that rule. The amount of energy gained by an atom through 'photon drag' should be proportional to the photon's energy. >> Goodbye BB theory! >> >> >>>> >>>> Note, if two photons pass an atom simultaneously and equidistant but on >>>> opposite sides, the atom will not move...imagine two electrons passing a >>>> proton. The electrons are deflected towards each other but the proton >>>> does >>>> not >>>> move sideways. >>>> Question, does the proton end up displaced longitudinally? It should >>>> not, >>>> but..... >>>> >>>> e- >v >>>> P >>>> e- >v >>> >>> >>>It's actually more complex and probably depends on >>>the speed. I think the answer can be a small >>>velocity change for the proton in either direction >>>but parallel to the direction of motion of the >>>electrons of course, but we are talking of photons, >>>not electrons. >> >> The proton should eventually come to rest exactly where it was.. > >Nope, no friction in a vacuum, well mostly >none :-o > >In the bigger picture, the proton would have >occasional interactions with lots of CMBR >photons. If it was moving in some direction >(relative to the mean motion of the CMBR >source), CMBR photons moving the same way >would be reddened by Doppler while those >moving the other way would be blueshifted. >Since blueshifted photons would transfer >more momentum, gradually the proton's speed >would be adjusted so that there was no mean >dipole component to the CMBR anisotropy. >That is almost like a drag term. I don't quite understand your logic here George. there would be protons moving in all directions....so why should the CMBR be affected? > >That could take a very long time though! >I'll leave you to do the sums ;-) > >>>Because EM forces are far stronger than gravitational >>>other forms of scattering dominate where interactions >>>with charged particles are concerned. Free neutrons >>>might give gravitational scattering but of course >>>they decay so having thought about your suggestion a >>>bit longer, I don't see how you can construct a >>>viable model out of it. >> >> Well I have now modified it. >> >> _____________A->v______p->>>> >> >> Whenever a photon goes flying past an atom, the atom is moved a little in >> the >> same direction. >> Remember we are discussing deep space here. There are not many photons OR >> atoms. > >Check if you can conserve both energy and >momentum (you'll find the answer on many >web pages). George, of course you can. It is the same as any collision....but I will check because it might reveal something interesting. >>>> I read somewhere authoritative that about half the scientific world does >>>> not >>>> agree with the BB theory. >>>> I don't remember where I read everything I read George. >>> >>>I think you'll find it was just a crank site or >>>publication. There's a lot of wishful thinkers >>>out there. It's certainly untrue. >> >> Well it wont be when my latest theory comes out. >> Even YOU must agree it is credible. > >It appears credible at first glance which >is why I looked at it some time ago. The >sphere of influence within which the photon >is significantly affected can be given a >cross-sectional area, i.e. the area of a >surface perpendicular to the line of flight >of the photon, and you can split up that >area like a bullseye target. If the photon >hits the centre, it gets bounced back the >way it came. Next ring out deflects it by >sligtly less and so on. By integrating the >whole, you get the total cross section but >by considering a narrow ring you get the >probability that the photon will be >deflected by an angle between say a and >a+da where da is a small increment. > >All this is important for X-ray work and >you'll find stuff on it on medical sites. >The key formula is called "Klein-Nishina" >and you should find my analysis if you >search for that in the sci.astro group >early last year. > >There is a significant factor which is >the ratio of the photon energy E = h * nu >to the particle energy given by m c^2 of >course. Things are different depending >on which is larger. Think of playing pool >but with with one bowling ball and one >table tennis ball. OK > >>>>>> What do you gain by accepting the BB idea? Nothing. >>>>> >>>>>True, the universe isn't there for my gain. >>>>>The measurements tell me that's the way it >>>>>is whether I want to accept it or not. >>>> >>>> the measurments tell us that light loses energy apparently in proportion >>>> to the >>>> distance it travels.....although that is by no means certain. >>> >>>No, the SNe measurements rule that out conclusively >>>though it had been pretty well discounted by more >>>complex evidence long before then. >> >> I like my latest theory. >> >> Every time a photon whizzes past an atom, the atom gains a minute amount >> of >> energy and the photon loses a little. ....it slows slightly. >> I am willing to bet that redshifted light is moving at <c. Blue shifts >> result >> from genuine c+v doppler.....but is not quite as blueshifted as it should >> be. > >It sounds good in principle but I found >the problem appears when you do the sums. >Typically the minimum number of >interactions for is in the billions >and the average deflection of the photon >turns out to be close to 90 degrees. It >surprised me when I calculated it but >nobody has been able to show a fault in >my analysis yet. Of course it means every >star should light up the whole sky as if >we were embedded in translucent plastic. Your 90 degrees is the problem. I say photon 'cross-sections are large compared to atoms and there is practically NO sideways deflection. Remember atoms are mainly 'nothing'. But the atom can easily be dragged along by the 'photon fields' (for want of a better term) .. > >George > 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. |