From: FrediFizzx on 21 Jun 2007 03:24 "John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message news:kdgi7356het3mddjjcjplvae265lsmbhc7(a)4ax.com... > On Tue, 19 Jun 2007 22:15:43 -0700, "FrediFizzx" > <fredifizzx(a)hotmail.com> wrote: > >>"John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message >>news:l20h7359ctfot73g0qeng6mg9d1h3e28te(a)4ax.com... >>> On Tue, 19 Jun 2007 14:51:53 -0700, "FrediFizzx" >>> <fredifizzx(a)hotmail.com> wrote: >> >>>>I suppose the main argument is whether or not the "vacuum" is >>>>polarizable. SI'ers are convinced that CGS buries "vacuum" >>>>polarization >>>>when it doesn't necessarily do so. If the "vacuum" is polarizable, >>>>then >>>>we will have "vacuum" capacitance and "vacuum" inductance. In order >>>>to >>>>have those, there must be "vacuum" charge. Which can be expressed >>>>as >>>>"Quantum Vacuum Charge" = +,- sqrt(hbar*c) in CGS and which must be >>>>all >>>bound. > Your vacuum charge is defined by hbar and c, whose definitions are not > in dispute in either system of units. > The product hbar*c can be shown in SI as equal to 3.16e-16 meter*volt* > coulombs. We don't want to take the square root of such a thing, and > call it charge. We discussed this before. We get relief by multiplying > by Coulomb's constant, in other words re-introduce eps0, so that now > we try hbar*c/eps0 whose square root is 5.974e-8 volt*meters. Now you > have a cgs representation of charge in esu's voltmeters. > The whole cgs system looks increasingly more synthetic with each > foray. > >>>>SI's statement >>>> >>>>c^2 = 1/(eps0 mu0), (1) >>>> >>>>is merely shorthand for >>>> >>>>c^2 = lambda^2/(4pi^2 Cvac Lvac), (2) > > No, Eq. (1) is not shorthand for Eq. (2). If Quantum Vacuum Charge (QVC) exists, then eq. 1 is indeed shorthand for eq. 2. > Eq. (1) conveys the substantive fact that c depends on two simple > constant vacuum parameters, Their product yields c and their quotient > yields Z. Sure, but eq. 1 doesn't really tell us why they exist where perhaps eq. 2 gives us more insight. What it looks like to me is that eps0 and mu0 can really only exist if QVC exists. > Your Eq. (2) is firstly a simple kinematic declaration that a > broadcast wavelength times the broadcast frequency equals the velocity > of transmission. which can be ascertained by drawing sine waves on > paper. You make this velocity c, which is OK, but in the second place, > you go too far in naming parameters Lvac and Cvac. Are they properties > of the vacuum or of the vagaries of the broadcast transmitter? They (Lvac and Cvac) are probably actually just a different way to look at the electric and magnetic fields of EM radiation. Of course they are not like capacitance and inductance of real caps and inductors. > What is the other constraint (e.g., Z) that allows you to calculate > Cvac and Lvac? I can make Cvac be 1 farad or any value, then solving > for Lvac. These C and L's can't be properties of the vacuum; they must > have the happy values that would satisfy F_brdcst. Cvac and Lvac can't be calculated any more than k_e or k_m can be. However, it is not hard to postulate what their values might be based on eps0 and mu0. Cvac = 2 eps0 lambda Lvac = mu0 lambda/8pi^2 When converted to natural units we obtain, Cvac = Lvac = lambda/2pi Simply the rationalized wavelength of the EM radiation under consideration. And a thing of beauty since they end up equal to each other. > There is no new information in (2) and it does not in any way resemble > (1). Well, I think my simple demonstration above does possibly give us some new information. Best, Fred Diether Moderator sci.physics.foundations
From: Greg Neill on 21 Jun 2007 07:25 "Spirit of Truth" <juneharton(a)prodigy.net> wrote in message news:RQoei.395$K44.183(a)newssvr13.news.prodigy.net... > > "Greg Neill" <gneillREM(a)OVEsympatico.ca> wrote in message > news:466d2fd5$0$21103$9a6e19ea(a)news.newshosting.com... > > "Spirit of Truth" <juneharton(a)prodigy.net> wrote in message > > news:kk3bi.1304$TC1.1254(a)newssvr17.news.prodigy.net... > > > >> Greg, you are coping out again. The lack of simultaneity example in there > >> shows exactly why lack of simultaneity is false. > > > > Simultaneity is an observer dependent thing, and should > > be (to any thinking individual) an obvious consequence of > > the finite speed of light. > > Greg, the example has a reality of a destroyed train in one frame > and no destroyed train in the other frame...both going into the future > ...thus I guess you must believe in a zillion universes? No, it does not. This is another case of your criticizing a theory without understanding it or its mathematics. Once again I feel that we have no basis for a productive discussion, so I will bow out. Have fun.
From: John C. Polasek on 21 Jun 2007 10:48 On Thu, 21 Jun 2007 00:24:00 -0700, "FrediFizzx" <fredifizzx(a)hotmail.com> wrote: >"John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message >news:kdgi7356het3mddjjcjplvae265lsmbhc7(a)4ax.com... >> On Tue, 19 Jun 2007 22:15:43 -0700, "FrediFizzx" >> <fredifizzx(a)hotmail.com> wrote: >> >>>"John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message >>>news:l20h7359ctfot73g0qeng6mg9d1h3e28te(a)4ax.com... >>>> On Tue, 19 Jun 2007 14:51:53 -0700, "FrediFizzx" >>>> <fredifizzx(a)hotmail.com> wrote: >>> >>>>>I suppose the main argument is whether or not the "vacuum" is >>>>>polarizable. SI'ers are convinced that CGS buries "vacuum" >>>>>polarization >>>>>when it doesn't necessarily do so. If the "vacuum" is polarizable, >>>>>then >>>>>we will have "vacuum" capacitance and "vacuum" inductance. In order >>>>>to >>>>>have those, there must be "vacuum" charge. Which can be expressed >>>>>as >>>>>"Quantum Vacuum Charge" = +,- sqrt(hbar*c) in CGS and which must be >>>>>all >>>bound. >> Your vacuum charge is defined by hbar and c, whose definitions are not >> in dispute in either system of units. >> The product hbar*c can be shown in SI as equal to 3.16e-16 meter*volt* >> coulombs. We don't want to take the square root of such a thing, and >> call it charge. We discussed this before. We get relief by multiplying >> by Coulomb's constant, in other words re-introduce eps0, so that now >> we try hbar*c/eps0 whose square root is 5.974e-8 volt*meters. Now you >> have a cgs representation of charge in esu's voltmeters. >> The whole cgs system looks increasingly more synthetic with each >> foray. >> >>>>>SI's statement >>>>> >>>>>c^2 = 1/(eps0 mu0), (1) >>>>> >>>>>is merely shorthand for >>>>> >>>>>c^2 = lambda^2/(4pi^2 Cvac Lvac), (2) >> >> No, Eq. (1) is not shorthand for Eq. (2). > >If Quantum Vacuum Charge (QVC) exists, then eq. 1 is indeed shorthand >for eq. 2. It's always a red flag for me if equations can't have units affixed. Please say what the units are for QVC, Lvac and Cvac. You also would do us a favor by rewriting, in cgs, my SI equations in the other note for a field of 1 million volts/meter. It should be simple arithmetic. What is D and what is H? >> Eq. (1) conveys the substantive fact that c depends on two simple >> constant vacuum parameters, Their product yields c and their quotient >> yields Z. > >Sure, but eq. 1 doesn't really tell us why they exist where perhaps eq. >2 gives us more insight. What it looks like to me is that eps0 and mu0 >can really only exist if QVC exists. No, of course one equation cannot tell what eps0 is. For that you have the discoveries in my Dual Space book, or the permittivity paper at http://www.dualspace.net. It demonstrates that, in order to reproduce eps0, you need an electron-positron pair elastically restrained in a cubic cell of size L = 3.514x10^-14m with spring constant of K = 2.612x10^14 N/m. 2e/L^3 makes a charge density P of 7.2x10^21 coul/m^3. Then Eq. 14 for eps0 gives the whole picture: eps0 = 2e*P/K = 8.854x10^-12 farad/meter It makes the very sensible statement that permittivity is proportional to e (for Ee = force), times charge density, and inverse to the aforesaid spring constant K. It all works and every one of a dozen parameters of pairspace are "forced" by logical laws without hypothesis. Unfortunately, the paper is not something to be speed-read nor is it, in retrospect, very lucid. > >> Your Eq. (2) is firstly a simple kinematic declaration that a >> broadcast wavelength times the broadcast frequency equals the velocity >> of transmission. which can be ascertained by drawing sine waves on >> paper. You make this velocity c, which is OK, but in the second place, >> you go too far in naming parameters Lvac and Cvac. Are they properties >> of the vacuum or of the vagaries of the broadcast transmitter? > >They (Lvac and Cvac) are probably actually just a different way to look >at the electric and magnetic fields of EM radiation. Of course they are >not like capacitance and inductance of real caps and inductors. > >> What is the other constraint (e.g., Z) that allows you to calculate >> Cvac and Lvac? I can make Cvac be 1 farad or any value, then solving >> for Lvac. These C and L's can't be properties of the vacuum; they must >> have the happy values that would satisfy F_brdcst. > >Cvac and Lvac can't be calculated any more than k_e or k_m can be. I just calculated ke (eps0?) above. >However, it is not hard to postulate what their values might be based on >eps0 and mu0. > >Cvac = 2 eps0 lambda Farads, no? But you don't have eps0. You just have capacity in centimeters. > >Lvac = mu0 lambda/8pi^2 Got to be henries. But you don't have mu0. >When converted to natural units we obtain, > >Cvac = Lvac = lambda/2pi Natural units; don't make me sick. It says farads = henries = cm's. >Simply the rationalized wavelength of the EM radiation under >consideration. And a thing of beauty since they end up equal to each >other. You should be revulsed. >> There is no new information in (2) and it does not in any way resemble >> (1). > >Well, I think my simple demonstration above does possibly give us some >new information. > >Best, > >Fred Diether >Moderator sci.physics.foundations John Polasek
From: FrediFizzx on 22 Jun 2007 01:25 "John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message news:5u1l7350jqkok44stf7mp6argpjiqavmdt(a)4ax.com... > On Thu, 21 Jun 2007 00:24:00 -0700, "FrediFizzx" > <fredifizzx(a)hotmail.com> wrote: > >>"John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message >>news:kdgi7356het3mddjjcjplvae265lsmbhc7(a)4ax.com... >>> On Tue, 19 Jun 2007 22:15:43 -0700, "FrediFizzx" >>> <fredifizzx(a)hotmail.com> wrote: >>> >>>>"John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message >>>>news:l20h7359ctfot73g0qeng6mg9d1h3e28te(a)4ax.com... >>>>> On Tue, 19 Jun 2007 14:51:53 -0700, "FrediFizzx" >>>>> <fredifizzx(a)hotmail.com> wrote: >>>> >>>>>>I suppose the main argument is whether or not the "vacuum" is >>>>>>polarizable. SI'ers are convinced that CGS buries "vacuum" >>>>>>polarization >>>>>>when it doesn't necessarily do so. If the "vacuum" is >>>>>>polarizable, >>>>>>then >>>>>>we will have "vacuum" capacitance and "vacuum" inductance. In >>>>>>order >>>>>>to >>>>>>have those, there must be "vacuum" charge. Which can be expressed >>>>>>as >>>>>>"Quantum Vacuum Charge" = +,- sqrt(hbar*c) in CGS and which must >>>>>>be >>>>>>all >>>bound. >>> Your vacuum charge is defined by hbar and c, whose definitions are >>> not >>> in dispute in either system of units. >>> The product hbar*c can be shown in SI as equal to 3.16e-16 >>> meter*volt* >>> coulombs. We don't want to take the square root of such a thing, and >>> call it charge. We discussed this before. We get relief by >>> multiplying >>> by Coulomb's constant, in other words re-introduce eps0, so that now >>> we try hbar*c/eps0 whose square root is 5.974e-8 volt*meters. Now >>> you >>> have a cgs representation of charge in esu's voltmeters. >>> The whole cgs system looks increasingly more synthetic with each >>> foray. >>> >>>>>>SI's statement >>>>>> >>>>>>c^2 = 1/(eps0 mu0), (1) >>>>>> >>>>>>is merely shorthand for >>>>>> >>>>>>c^2 = lambda^2/(4pi^2 Cvac Lvac), (2) >>> >>> No, Eq. (1) is not shorthand for Eq. (2). >> >>If Quantum Vacuum Charge (QVC) exists, then eq. 1 is indeed shorthand >>for eq. 2. > > It's always a red flag for me if equations can't have units affixed. > Please say what the units are for QVC, Lvac and Cvac. Depends on what system of units you want to use. For SI it is, coulomb, henry, and farad. > You also would do us a favor by rewriting, in cgs, my SI equations in > the other note for a field of 1 million volts/meter. It should be > simple arithmetic. What is D and what is H? 10^6 volt/m ~= 33.3564095198 statvolt/cm in CGS. Is this for free space EM radiation? >>> Eq. (1) conveys the substantive fact that c depends on two simple >>> constant vacuum parameters, Their product yields c and their >>> quotient >>> yields Z. >> >>Sure, but eq. 1 doesn't really tell us why they exist where perhaps >>eq. >>2 gives us more insight. What it looks like to me is that eps0 and >>mu0 >>can really only exist if QVC exists. > > No, of course one equation cannot tell what eps0 is. For that you have > the discoveries in my Dual Space book, or the permittivity paper at > http://www.dualspace.net. > It demonstrates that, in order to reproduce eps0, you need an > electron-positron pair elastically restrained in a cubic cell of size What is "restraining" the electron-positron pair? Why don't we see them or detect them? Is it all the other e+e- "vacuum" pairs that are doing the restraining? If so, don't you think there is some kind of BOUND "vacuum" charge happening? Of course there would be; Quantum Vacuum Charge. > L = 3.514x10^-14m with spring constant of > K = 2.612x10^14 N/m. > 2e/L^3 makes a charge density P of 7.2x10^21 coul/m^3. > Then Eq. 14 for eps0 gives the whole picture: > eps0 = 2e*P/K = 8.854x10^-12 farad/meter > It makes the very sensible statement that permittivity is proportional > to e (for Ee = force), times charge density, and inverse to the > aforesaid spring constant K. It all works and every one of a dozen > parameters of pairspace are "forced" by logical laws without > hypothesis. > Unfortunately, the paper is not something to be speed-read nor is it, > in retrospect, very lucid. Well, it would be much better if you didn't ignore the rest of the quantum objects of the Standard Model for your "vacuum". >>> Your Eq. (2) is firstly a simple kinematic declaration that a >>> broadcast wavelength times the broadcast frequency equals the >>> velocity >>> of transmission. which can be ascertained by drawing sine waves on >>> paper. You make this velocity c, which is OK, but in the second >>> place, >>> you go too far in naming parameters Lvac and Cvac. Are they >>> properties >>> of the vacuum or of the vagaries of the broadcast transmitter? >> >>They (Lvac and Cvac) are probably actually just a different way to >>look >>at the electric and magnetic fields of EM radiation. Of course they >>are >>not like capacitance and inductance of real caps and inductors. >> >>> What is the other constraint (e.g., Z) that allows you to calculate >>> Cvac and Lvac? I can make Cvac be 1 farad or any value, then solving >>> for Lvac. These C and L's can't be properties of the vacuum; they >>> must >>> have the happy values that would satisfy F_brdcst. >> >>Cvac and Lvac can't be calculated any more than k_e or k_m can be. > I just calculated ke (eps0?) above. k_e is not eps0. In SI, k_e = 1/(4pi eps0) >>However, it is not hard to postulate what their values might be based >>on >>eps0 and mu0. >> >>Cvac = 2 eps0 lambda > Farads, no? But you don't have eps0. You just have capacity in > centimeters. ??? Figure it out if the wavelength of the EM radiation is 1 meter. Cvac ~= 1.7708376 E-11 farad in SI units. >>Lvac = mu0 lambda/8pi^2 > Got to be henries. But you don't have mu0. I have no idea what you mean by "But you don't have mu0 or eps0". >>When converted to natural units we obtain, >> >>Cvac = Lvac = lambda/2pi > Natural units; don't make me sick. It says farads = henries = cm's. I can asure you that it is only you that is making yourself sick. ;-) Most people know how to deal with this. >>Simply the rationalized wavelength of the EM radiation under >>consideration. And a thing of beauty since they end up equal to each >>other. > You should be revulsed. It is not my fault that you can't see physics from a different perspective. Many people have no problem with looking at physics from different perspectives. Best, Fred Diether Moderator sci.physics.foundations http://www.vacuum-physics.com
From: anandsr21 on 22 Jun 2007 09:47
On Jun 20, 1:04 am, "Greg Neill" <gneill...(a)OVEsympatico.ca> wrote: > "Florian" <firstn...(a)lastname.net> wrote in message > > news:1hzyjoi.thrw70pivls0N%firstname(a)lastname.net... > > > Greg Neill <gneill...(a)OVEsympatico.ca> wrote: > > > I know. we need to invent more trick to make it work. At some point > > there would be so many tricks (MONDtheory, dark matter, dark energy) > > that it will certainly be time to find a better model. Right? > > I wonder if the physicist who will find a better model is born yet? > > It's not "inventing more trick" if the mathematics is > already in place in the current theory. MONDis an > attempt to ignore theory and employ a curve-fit model > instead. Dark matter is not a "trick" in that it is > an empirical necessity. > > MOND is not an attempt to ignore theory. It is an attempt to explain the galactic data by thinking out of the box. We don't yet know the reason why MOND works so well at the galactic scales. And any theory including GR will need to come up with a reason why MOND phenomenology works. DM would be very credible if MOND did not exist, but since it does, GR needs to explain it. Just to debunk some myths here: 1) MOND has exactly ONE UNIVERSAL free parameter. This parameter has already been fixed to 1.2E-10m/s^2. 2) It has fit more than a hundred galaxies. You can search for "mond fits" to find them. There are peer reviewed papers explaining each of those fits. MOND is not favored by cranks, they have their own theory. http://www.journals.uchicago.edu/ApJ/journal/issues/ApJ/v508n1/38018/sc0.html 3) MOND had predicted the properties of LSB galaxies, before they could be studied, so it has made more successful predictions compared to DM theory (which cannot really predict anything). 4) Trying to fit a galaxy with DM is a purely curve fitting exercise because there are multiple free parameters per galaxy. These parameters have no constraints, provided by previous fits. 5) DM is only required if you don't know what is causing the higher gravitational force. All this is regarding galactic scales. Don't tell me about the Bullet Cluster. I know about it and it doesn't give anything new compared to other galaxy clusters. MOND also doesn't fit them properly. The real problem is that we don't know what space is. How mass deforms it? And how mass traverses space. Once we have an explanations for these we may find the reason for MOND effects. I do hope that LQG, which is based on a quantum transformation of GR, will provide answers to these questions. I am sure it will take a long time ;-). I think that MOND effect happens due to the meeting of gravitation effects due to a massive body and the cosmic gravitational effects. I think that Gravitational force will never truly reduce to zero. So that any particle will not be able to execute a straight line trajectory. It will always execute a circle, whose radius depends on its speed, in absence of massive objects in its vicinity. Light and other particles moving at the speed of light will move in a circle equal to the size of the universe and we will observe them moving in a straight line. Near a massive object the newtonian force will completely dominate the cosmic force and we will see the normal newtonian drop off. Further away the cosmic force will have more of an effect. At the MOND scales the two forces will become equal and we will see a MONDian drop off. Incidently MONDs force is a geometrical mean of a0 and GM/R^2. regards, -anandsr |