OWLS is not equal to c
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From: George Dishman on 16 Jul 2005 08:09 "Henri Wilson" <H@..> wrote in message news:gdegd1p62sv1qtkr0bt444f559kli2ej73(a)4ax.com... > On 15 Jul 2005 05:04:22 -0700, george(a)briar.demon.co.uk wrote: > >> >> >>Henri Wilson wrote: >>> On Sun, 10 Jul 2005 16:13:06 +0100, "George Dishman" >>> <george(a)briar.demon.co.uk> >>> wrote: >>> >>> >> Mine is like moving a bar magnet past a small chip of wood. >>> > >>> >No effect then? >>> >>> How do you know? >> >>Because in a first approximation (in general >>conversation, wood in considered to be non- >>magnetic. Your analogy of "like moving a bar >>magnet past a small chip of wood" herefore >>implies no effect. If that isn't what you >>meant, can you find a better analogy to >>illustrate what you are trying to convey? > > I was trying to convey that the effect is extrememly small. You might try aluminium then, it is not thought of as magnetic in isolation but could be affected by eddy currents as the magnet passes. > Light encountering an atom in space drags that atom along almost > infinitesimally......but enough to cause a redshift over vast distances. > >> >>> >> The atom speeds must be expressed in terms of hte source frame. >>> >> The maths show that the amount of slowing of the light is independent >>> >> of >>> >> atom >>> >> speed. >>> > >>> >Those two statements are contradictory. If the >>> >maths is independent of the speed, you can use >>> >any inertial frame. >>> > >>> >>>Show the calculations that lead you to say it or >>> >>>it is just a worthless assertion. >>> >> >>> >> I did. >>> > >>> >Not in this thread, cite please. >>> >>> If the photon loses momentum p, it loses energy pc-p(c-v), where c-v is >>> the >>> final speed wrt the source. >>> But if an atom with mass m gains this momentum p, it has to gain the >>> kinetic >>> energy p^2/(2m). Conservation of energy implies then pv = p^2/(2m). >>> >>> v=p/2m >>> >>> %redshift = v/c = p/2mc = pc/mc^2 = Eo/mc^2 = h/[m.c.lambda] >>> >>> How interesting.....atom speed doesn't appear in the equation..... >> >>Yes, that rings alarm bells. It's not clear >>to me why but I suspect it is because you have >>taken some shortcuts in working only with >>changes. I think you need to start by showing >>the basic energy and momentum formulae for a >>photon of frequency f moving at speed v. The >>relativistic formulae do not include speed of >>course so I need to se how you take variable >>speed into account. It might then cancel out >>giving the equations above when you take >>differences but I can't check without the base >>equations. > > I think the maths is correct. I got it from Bjoern. ...and he thinks he is > an > expert....I just changed it a little. I would expect Bjoern's maths to be conventional hence be based on an invariant speed of c for the photon. As such it is relevant to your theory. What you need to do is derive the equation or something similar from your own theory. > The atom speed must be equated relative to the light source. In general you have a photon of frequency f_1 moving at speed v_p1 interacting with a particle of mass m moving at speed v_m1. Afterwards the photon has frequency f_2 and speed v_p2 while the particle has speed v_m2. First write down the equations for the energy and momentum of a photon in terms of its frequency and speed, do the same for the massive particle, then solve assuming both total energy and total momentum are conserved. > One would not expect the force imparted on the atom by the photon to be > proportional to the atom speed (relative to the photon), which is c-v. > > PH-->c------------A->v > > But the time that force operates is inversely proportional to c-v. > > So, fair enough. The momentum gain of the atom is independent of atom > speed > (relative to anything) Unfortunately you need to show the proof. Conventional physics doesn't allow for variable speed photons so you cannot just pick it up, sorry. >>It will be interesting to see how you turn the >>equations above into the distribution then, >>that's the important bit that people seem to >>miss when discussing this topic. > > My theory says the angular deflection is negligible. Does it? You'll have to show me how you derived that by solving the equations above. > Intergalactic atom speeds are simply not high enough to deflect a photon > sideways by an appreciable amount. You don't understand, it is the photon that accelerates the particle. A low mass particle usually gets accelerated to high speeds by an energetic photon. > Remember my model assumes that the photon has a much larger cross section > tan > the atom and completely engulfs it. The proces is nothing like a > collision.. Compare the situation a million miles before to a million miles after and it looks like a collision. We aren't worried about the details, just the changes in energy and momentum and most importantly a graph of probability versus deflection angle. It's that final graph which is the key, and probably the hardest part to work out, but without it you cannot work out the overall blurring of a point source which was our original test. George
From: George Dishman on 16 Jul 2005 12:57 Henri, You asked me a specific question at the end and this is getting too big so I'll trim a lot. "Henri Wilson" <H@..> wrote in message news:g9ggd1lun6685iinorj5pbas6i2c75e26t(a)4ax.com... > On Fri, 15 Jul 2005 17:32:01 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >>> George, look at it this way. Both obserevrs use the same clock to >>> measure >>> the >>> time duration of the orbit. >> >>No they don't the observer _is_ the clock, and there >>is one in orbit and one on the ground so they cannot >>possibly be the same clock. > > What the hell are you talking about? .... >>What I can see is that you didn't know it was >>conventional to describe inanimate measuring >>instruments in anthropomorhic terms in science. > > What the hell are you talking about? .... >>Scientifically speaking, your acronym "GO" means >>a clock permanently on the ground throughout the >>experiment while "OO" means a clock permanently >>in orbit. The GPS clock is first compared locally >>against the ground clock, then launched, then >>compared locally against the orbiting clock. > > That is not the experiment I described. I just translated some jargon into more accessible words, otherwise it is the same. If you think of a person being asked to determine the time between two events scientifically, he must measure it using a stopwatch. Done properly, he will get whatever constitutes the events to start and stop the stopwatch automatically, then he just records the numbers. Since all he is doing is noting the result, it is really the stopwatch that determines the result, not the person. However, I'll switch to answering what you say next. > THERE IS ONLY ONE CLOCK. > > Let me repeat: > > There is ONE clock and there are two observers. > > One observer, OO, is already in orbit. OK. > The other, GO, is on the ground, initially with the clock. OK, in that case, the one clock that binds them is also on the ground. > Both observers will measure the duration of the OO's orbit using that one > clock. They will do it firstly when the clock is on the ground then > secondly, > when it has joined the OO in orbit. OK. There is a problem in that you are at first trying to measure the orbit with a clock on the ground so gravitational redshift might confuse the issue. Place a marker at a fixed point in space and let the satellite just miss it and we can use the moment of co-location as a timing reference. > When the clock is on the ground, both observers register N ticks per > orbit. No. When the clock is on the ground, the ground observer is measuring the duration of the orbit. The orbiting chappie is just viewing the result of the ground-based measurement. To state it accurately, both agree that the ground observer measures N ticks for the orbit. > After launch, both observers register N+n ticks per orbit. (we know GPS > clocks DO speed up) Nitpick: if the clock speeds up, it would register N-n ticks per orbit. I'll just call the new value N'. Indeed we do but again you are a bit inaccurate in the way you state it. The clock when orbiting registers N' ticks. That is the duration as measured by the OO if you like. The GO is not making a measurement himself, he is seeing and agreeing that the OO has measured N'. > Now, since neither the orbit nor the states of the two observers have > changed > in any way, the difference in tick counts PER ORBIT can only be attributed > to a > REAL PHYSICAL change in the clock itself. > > If you are going to argue that the 'duration of a tick' is different in > the > orbit than on the ground, you will have to explain why the OO didn't > initially > count N-n ticks per orbit and finally N ticks. Again I shouldn't be arguing this at all, YOU are the one who claimed to understand SR/GR so you should be able to tell me why they get different values even though there has been no physical change. The answer is that a clock is a device that produces ticks equally spaced in time just as a ruler gives equally spaced ticks in space. If you lay two rulers side by side, the ticks match. Lay them at an angle then project lines at right angles to the other and the projected points don't match the marks on the other ruler. The same thing is happening in your experiment. > You lose no matter what. On the contrary, the fact that you had to ask me proves you could not answer yourself so I win, remember the only question here is whether your claim that you understand SR is true. The fact that you don't even recognise the SR answer when I stick it under your nose simply emphasises that fact. >>>>> There exists an 'Absolute Physics' which is NOT observer dependent. >>>>> >>>>> Thus, a rod occupies a length of 'space'. >>>>> An orbit is completed in an 'interval of time'. >>>> >>>>And that religious conviction is why you cannot >>>>grasp the relativistic model. What is absolute >>>>(or better, invariant) is spacetime intervals, >>>>not space or time separately. >>> >>> What the hell is a 'spacetime interval'? >> >>Henri, how can you possibly ask that after claiming >>you understand SR?! In terms of explanation, it is >>the most fundamental concept from which everything >>else follows. > > Rubbish. Again you simply prove you know nothing of SR, and the equivalent of the interval in GR is the metric which is about as fundamental as you can get. Understanding the physical nature of the metric is the key relativity. For example gravitational waves are ripples in that metric so it must be physical in the sense of being able to transport energy. > It is an interval on a graph. > It is not a physical entity. Keep digging Henri, you more you deny it, the more you disprove your claim to understand any of this :-) George
From: Paul B. Andersen on 16 Jul 2005 17:24 Henri Wilson wrote: > THERE IS ONLY ONE CLOCK. > > Let me repeat: > > There is ONE clock and there are two observers. > > One observer, OO, is already in orbit. > The other, GO, is on the ground, initially with the clock. > > Both observers will measure the duration of the OO's orbit using that one > clock. They will do it firstly when the clock is on the ground then secondly, > when it has joined the OO in orbit. > > When the clock is on the ground, both observers register N ticks per orbit. > After launch, both observers register N+n ticks per orbit. (we know GPS clocks > DO speed up) OK. Good to see that you accept that clocks behave as predicted by GR. > Now, since neither the orbit nor the states of the two observers have changed > in any way, the difference in tick counts PER ORBIT can only be attributed to a > REAL PHYSICAL change in the clock itself. Why is that? The proper duration of the orbit is N+n ticks. While on ground the clock will measure the duration to be N ticks, but this is not proper time. It proves that time is not absolute. > If you are going to argue that the 'duration of a tick' is different in the > orbit than on the ground, YOU are the one who insists that "the 'duration of a tick' is different in the orbit than on the ground" because you claim that "the difference in tick counts PER ORBIT can only be attributed to a REAL PHYSICAL change in the clock itself." According to GR, there is no physical change of the clock, it keeps running at its proper rate. > you will have to explain why the OO didn't initially > count N-n ticks per orbit and finally N ticks. Uh? :-) Your Wonderland fairies at work again? :-) >> Henri Wilson wrote: >>>What the hell is a 'spacetime interval'? >George Dishman wrote: >>Henri, how can you possibly ask that after claiming >>you understand SR?! In terms of explanation, it is >>the most fundamental concept from which everything >>else follows. > > > Rubbish. > It is an interval on a graph. > It is not a physical entity. This is kind of right for the time measured by the clock on ground. In orbit, the clock measures the proper duration of the orbit, that is the "length" of the path through space-time the satellite has travelled during one orbit. You could call that time "a physical entity" if you like. On ground, the clock is measuring the space time interval of a geometric projection of the orbit. A projection has not necessarily the same length as the projected entity. In this case, it's shorter. I will not quarrel about whether a projection is "a physical entity" or not, but I think we can agree that the proper length of a rod is more "physical" than the length of its shadow. Paul
From: Henri Wilson on 16 Jul 2005 18:42 On Sat, 16 Jul 2005 13:09:54 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <H@..> wrote in message >news:gdegd1p62sv1qtkr0bt444f559kli2ej73(a)4ax.com... >> On 15 Jul 2005 05:04:22 -0700, george(a)briar.demon.co.uk wrote: >>>> How do you know? >>> >>>Because in a first approximation (in general >>>conversation, wood in considered to be non- >>>magnetic. Your analogy of "like moving a bar >>>magnet past a small chip of wood" herefore >>>implies no effect. If that isn't what you >>>meant, can you find a better analogy to >>>illustrate what you are trying to convey? >> >> I was trying to convey that the effect is extrememly small. > >You might try aluminium then, it is not thought >of as magnetic in isolation but could be affected >by eddy currents as the magnet passes. That's exactly the type of process I had in mind. When a photon engulfs an atom, it drags that atom ever so slightly. The atom gains a little momentum in the directio of photon travel and the photon loses a little. This would completely explain the cosmic redshift. >>> >>>Yes, that rings alarm bells. It's not clear >>>to me why but I suspect it is because you have >>>taken some shortcuts in working only with >>>changes. I think you need to start by showing >>>the basic energy and momentum formulae for a >>>photon of frequency f moving at speed v. The >>>relativistic formulae do not include speed of >>>course so I need to se how you take variable >>>speed into account. It might then cancel out >>>giving the equations above when you take >>>differences but I can't check without the base >>>equations. >> >> I think the maths is correct. I got it from Bjoern. ...and he thinks he is >> an >> expert....I just changed it a little. > >I would expect Bjoern's maths to be conventional >hence be based on an invariant speed of c for >the photon. As such it is relevant to your theory. >What you need to do is derive the equation or >something similar from your own theory. I did. I simply altered light speed after the interaction event. > >> The atom speed must be equated relative to the light source. > >In general you have a photon of frequency f_1 moving >at speed v_p1 interacting with a particle of mass m >moving at speed v_m1. Afterwards the photon has >frequency f_2 and speed v_p2 while the particle has >speed v_m2. > >First write down the equations for the energy and >momentum of a photon in terms of its frequency and >speed, do the same for the massive particle, then >solve assuming both total energy and total momentum >are conserved. > >> One would not expect the force imparted on the atom by the photon to be >> proportional to the atom speed (relative to the photon), which is c-v. >> >> PH-->c------------A->v >> >> But the time that force operates is inversely proportional to c-v. >> >> So, fair enough. The momentum gain of the atom is independent of atom >> speed >> (relative to anything) > >Unfortunately you need to show the proof. Conventional >physics doesn't allow for variable speed photons so >you cannot just pick it up, sorry. George, conventional physics is so obviously wrong about light speed that what you just said is really quite amusing. Can you not see that the whole picture of what is going on with light becomes much clearer under the BaT....and there is absolutely no evidence that light retains a speed c relative to its source over long distances. >>>It will be interesting to see how you turn the >>>equations above into the distribution then, >>>that's the important bit that people seem to >>>miss when discussing this topic. >> >> My theory says the angular deflection is negligible. > >Does it? You'll have to show me how you derived >that by solving the equations above. George, don't regard the photon/atom interaction as a collision. The atom gains momentum almost entirely in the direction of photon travel. Therefore the photon loses momentum almost entirely in that same direction. >> Intergalactic atom speeds are simply not high enough to deflect a photon >> sideways by an appreciable amount. > >You don't understand, it is the photon that >accelerates the particle. A low mass particle >usually gets accelerated to high speeds by an >energetic photon. My theory says that the cross section of a photon is roughly inversely proportional to its energy. Hence gamma particles are tiny but possess enormous 'internal energy'. When they interact with an atom, I agree, the result is more like that of a standard collision. ...but for long wavelength EM, the cross section is much greater than atomic radii..hence the small drag... You have to admit this appears to be a plausible theory. >> Remember my model assumes that the photon has a much larger cross section >> tan >> the atom and completely engulfs it. The proces is nothing like a >> collision.. > >Compare the situation a million miles before to a >million miles after and it looks like a collision. >We aren't worried about the details, just the >changes in energy and momentum and most importantly >a graph of probability versus deflection angle. >It's that final graph which is the key, and probably >the hardest part to work out, but without it you >cannot work out the overall blurring of a point >source which was our original test. Well, as pointed out before, the result would be that the photon follows a kind of random walk path.....but the final deviation at any distance is apparently smaller than the resolving power of any telescope. > >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 16 Jul 2005 18:45
On Sat, 16 Jul 2005 09:51:31 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <H@..> wrote in message >news:nqfgd1lna457fb8tmqluso3r8ldu5p0i1e(a)4ax.com... >> On Fri, 15 Jul 2005 17:02:23 +0100, "George Dishman" >> <george(a)briar.demon.co.uk> >> wrote: >... >>>Again let me point out that I only queried your >>>apparent claim that aberration didn't exist at all >... >>>I never suggested it did. The path of a photon >>>which is "vertical" in the barycentric frame is >>>"diagonal" to the extent of about 19 arc seconds >>>in the Earth frame which you seemed to deny in >>>your earlier posting. I am content to accept it >>>was just a misunderstanding. >> >> Even though the path of one infintesimal element of the vertical beam >> follows a >> diaginal path in a telescope, if you plot the positions of successive such >> elements, you will find they remain vertically aligned. > >Of course, but again nobody suggested anything >other than that. Abberration describes the path >of individual photons which becomes angled on >a change of frame, even in Ritzian theory. It >is also illustrated by the family of 'diagonal' >lines in your animation. > >Actually, in the case of the path of starlight >plotted in the Earth frame, it becomes a sine >wave with a period of a year ;-) > >> this is shown in my program www.users.bigpond.com/hewn/movingframe.exe >> >>> >>>The historical significance as the page says was >>>for certain types of aether theory as it was hard >>>for them to reconcile with the MMX. >> >> true. >> Not surprising when one considers that light speed is source dependent. > >Sorry Henri, Sagnac refutes that. We have >resolved all the queries you raised on that and >the result remains that source dependency would >produce a null result from Sagnac. I am happy to accept that the sagnac effect is based on an entirely different principle than the nonrotating case. Either a local absolute 'non-rotating' frame exists or photons carry little 'internal gyros'. > >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. |