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From: Henri Wilson on 17 Jul 2005 19:29 On Sun, 17 Jul 2005 14:00:12 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <H@..> wrote in message >news:2d3jd1dhsv9tht2tn70u4nubj606vq0g47(a)4ax.com... >> On Sat, 16 Jul 2005 09:51:31 +0100, "George Dishman" >> <george(a)briar.demon.co.uk> >> wrote: >>>"Henri Wilson" <H@..> wrote in message >>>>>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. > >In that case you are agreeing that the principle >you call BaT does not apply hence it is refuted >as a general model for light. No I am not George. I say the sagnac effect would occur even if light traveled at c/2 It is not related to light speed. > >> Either a local absolute 'non-rotating' frame exists > >.. which means the light moves at c in that frame >which contradicts BaT. No it doesn't. I said an "absolute 'non-rotating' frame" Rotation is not 'translation'. > >> or photons carry little 'internal gyros'. > >They have "spin" but that is known and dealt with. > >The system is sensitive only to anisotropy in the >time of flight. There should be no anisotropy if >the speed is source-speed dependent. You can believe taht george. I will prefer to not worry about it at this stage. I am only interested in what light does in empty space, after it leaves its source at c. Do you deny that it leaves its source at c, relative to the source? > >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 17 Jul 2005 19:59 On Sun, 17 Jul 2005 20:29:09 GMT, "kenseto" <kenseto(a)erinet.com> wrote: > >"Paul B. Andersen" <paul.b.andersen(a)deletethishia.no> wrote in message >news:dbbu11$cak$1(a)dolly.uninett.no... >> Henri Wilson wrote: >> > THERE IS ONLY ONE CLOCK. >> >> > 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. > >This statement is meaningless. If you are agreeing with me that the GR STATEMENT is meaningles, then I agree. If you claim MY statement is meaningless, then I would like to know why. >All you are saying is that the proper rate of >a clock is according to definition (one second per second) in all frames. >The problem is that each clock has its own proper rate but this proper rate >does not correspond to the proper rate of another clock moving wrt it. IOW, >the passage of a clock second in one frame does not correspond to the >passage of a clock second in another frame. That's why I use the orbit duration as a common time reference. It has only one REAL value..... whatever that value IS does not matter one iota. > >Ken Seto >> >> >> > 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 > 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 17 Jul 2005 20:29 On Sun, 17 Jul 2005 13:18:47 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <H@..> wrote in message >news:ld2jd1plbe5br0bgv0nkcgd9a8vftm01sl(a)4ax.com... >Only if the distribution of the change of >angle is sufficiently small. You haven't >derived that yet. > >>>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. > >You cannot just pick up conventional equations >and tinker with them, you need to do it the >way I outlined: > >>>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. > >Until you do those, you have no starting point. This is good enough, George. "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....." > >>>> 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. > >You again prove me right. If you think >conventional physics is wrong, you cannot >adopt its equations. I don't wan to . I know that light initially moves at c wrt its source. > >> Can you not see that the whole picture of what is going on with light >> becomes >> much clearer under the BaT > >I can see BaT is ruled out by Sagnac and >you have no way to answer that. Come up >with something new. Forget Sagnac. > >....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. > >Don't waste my time with assertions, you >have to show how you derive that result. >I've shown that the mean deflection >angle is 90 degrees, you haven't shown >me anything at all. George, one must always create a 'picture' before on can analyse something mathematically. The mean deflection angle is obviously zero in my model. No need for any maths. >>>> 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 > >No it doesn't. Your theory says nothing >whatsoever on the matter because you >haven't shown how you derive it. 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..... > >> 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. > >ROFL! Sorry Henri, it isn't even a theory >until you publish the derivations. It is >baseless speculation, nothing more. Consider this, George: A laser is pointed at Mars. An optical gate allows a 1ns long pulse of its light to travel in the direction of Mars. The pulse arrives at Mars after a certain time interval, say 10 minutes. Somehow, a 30 cm long 'package of EM' is traversing the space between the two planets, during those ten minutes. In the absence of an aether, what 'form' might that package have during its travel? Nothing in the whole of physics provides ANY information about this. >> >> Well, as pointed out before, the result would be that the photon follows a >> kind >> of random walk path > >That's true for multiple interactions but I meant >compare before and after a single interaction. On >those scales it will look like a collision regardless >of the details. George, light a fire. Take a garden hose and send a jet of water through some very thin smoke. Does the jet appear to change in any way because of the smoke? >.....but the final deviation at any distance is apparently >> smaller than the resolving power of any telescope. > >I expected to find a maximum mass for the particle >when I did the calculation, the result was billions >of interactions each causing a deflection of the >photon with a mean angle of 90 degrees. Trust me, >when you do your sums properly, the result will be >surprising so just guessing what might happen is >a meaningless exercise. I think you must have made an error. Check your logic. > >Personally, I don't think you are capable of >producing a theory from this speculation because >you need to know the details of the interaction >but I'll be happy if you can surprise me. George, I think your ability to understand experimental physics is sadly lacking. > >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: kenseto on 18 Jul 2005 07:42 "Henri Wilson" <H@..> wrote in message news:32sld19kn75a9laktbfe3srh76rtjj0uch(a)4ax.com... > On Sun, 17 Jul 2005 20:29:09 GMT, "kenseto" <kenseto(a)erinet.com> wrote: > > > > >"Paul B. Andersen" <paul.b.andersen(a)deletethishia.no> wrote in message > >news:dbbu11$cak$1(a)dolly.uninett.no... > >> Henri Wilson wrote: > >> > THERE IS ONLY ONE CLOCK. > >> > >> > 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. > > > >This statement is meaningless. > > If you are agreeing with me that the GR STATEMENT is meaningles, then I agree. > If you claim MY statement is meaningless, then I would like to know why. You were comparing an absolute interval of time (an orbit) with clock time (ticks of a clock). That's why you get the disagreement. When SR says that the proper rate of a clock doesn't it means that the defintion for proper rate doesn't change....ie one second/second at the rest frame of the clock. > > >All you are saying is that the proper rate of > >a clock is according to definition (one second per second) in all frames. > >The problem is that each clock has its own proper rate but this proper rate > >does not correspond to the proper rate of another clock moving wrt it. IOW, > >the passage of a clock second in one frame does not correspond to the > >passage of a clock second in another frame. > > That's why I use the orbit duration as a common time reference. > It has only one REAL value..... whatever that value IS does not matter one > iota. An orbit is common interval of absolute time but it can have different clock time values in different frames. Why? Because the absolute time content for a clock second is different in different frames (different state of absolute motion). BTW that's the reason why the speed of light is a constant math ratio in all frames as follows: Light path length of rod (299,792,458m)/the absolute time content for a clock second co-moving with the rod. Ken Seto
From: Henri Wilson on 18 Jul 2005 18:11
On Mon, 18 Jul 2005 11:42:27 GMT, "kenseto" <kenseto(a)erinet.com> wrote: > >"Henri Wilson" <H@..> wrote in message >news:32sld19kn75a9laktbfe3srh76rtjj0uch(a)4ax.com... >> >> According to GR, there is no physical change of the clock, >> >> it keeps running at its proper rate. >> > >> >This statement is meaningless. >> >> If you are agreeing with me that the GR STATEMENT is meaningles, then I >agree. >> If you claim MY statement is meaningless, then I would like to know why. > >You were comparing an absolute interval of time (an orbit) with clock time >(ticks of a clock). That's why you get the disagreement. When SR says that >the proper rate of a clock doesn't change it means that the defintion for proper >rate doesn't change....ie one second/second at the rest frame of the clock. Yes that's all very clear, Ken. How do you define the 'seconds' in your 'one second/second'? >> >> >All you are saying is that the proper rate of >> >a clock is according to definition (one second per second) in all frames. >> >The problem is that each clock has its own proper rate but this proper >rate >> >does not correspond to the proper rate of another clock moving wrt it. >IOW, >> >the passage of a clock second in one frame does not correspond to the >> >passage of a clock second in another frame. >> >> That's why I use the orbit duration as a common time reference. >> It has only one REAL value..... whatever that value IS does not matter one >> iota. > >An orbit is common interval of absolute time but it can have different clock >time values in different frames. Why? Because the absolute time content for >a clock second is different in different frames (different state of >absolute motion). BTW that's the reason why the speed of light is a constant >math ratio in all frames as follows: >Light path length of rod (299,792,458m)/the absolute time content for a >clock second co-moving with the rod. Ken, there is an observer in the proposed orbit and another on the ground. They both agree that the clock ticks at N ticks/orbit before launch and N+n ticks/orbit after launch. Nothing has changed except the clock. If you cannot see that then there is absolutely no hope for you. You should immediately cease posting to a physics group. > >Ken Seto > 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. |