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Natural Science Forum / Physics / Relativity / February 2005Why
Why can the equation for geodesic equation be used to find the period of the pendulum in spite of the fact that the particles do not move along geodesics? Also why the equation can be used even though the centre of the pendulum does not follow a geodesic? > Why can the equation for geodesic equation be used to find the period > of the pendulum in spite of the fact that the particles do not move > along geodesics? Also why the equation can be used even though the > centre of the pendulum does not follow a geodesic? You are intermixing classical and relativistic mechanics. Stick to one or the other. Bill > > Why can the equation for geodesic equation be used to find the period > > of the pendulum in spite of the fact that the particles do not move > > along geodesics? In simple GR the geodesic is modified by ElectroMagnetic force. That is the force imposed by the wire supporting the pendulating bob, modifying the motion of the bob from being in free-fall, to one that undulates, (swings). >>Also why the equation can be used even though the > > centre of the pendulum does not follow a geodesic? Same as above. In unified field theory the effects of EM and gravitation are geodesic. Regards Ken S. Tucker > You are intermixing classical and relativistic mechanics. Stick to one or > the other. > Bill PS: Hobba is a crank, check the sheer quantity of his posts, it's verbal direarya and mental constipation. Thank you Ken. Kalikinkar et al... Conventionally, "geodesical motion" is identical to "free-fall", butt (pun intended), consider the force on your bum while sitting in chair. That force on your bum (cute butt if you're a girl), is attributable to the intermolecular EM-force stressing the chair. Of course your "free-fall" motion is prevented by this force and your geodesical motion is being continuously modified by that force. For ref see R.C. Tolman's "Relativity....Cosmology", Eq.(103.1). Regards Ken S. Tucker > Kalikinkar et al... > Conventionally, "geodesical motion" is identical to "free-fall", [quoted text clipped - 7 lines] > Regards > Ken S. Tucker No, I am not a girl. Thanks for the ref. "No, I am not a girl. Thanks for the ref. " That's ok, likely the chicks will find your butt cute. That's a hobby of mine, see Weinberg's "Grav and Cosmo", Eq.(5.1.11) and the equations immediately following it. Einstein disagrees with Weinberg, see Dover's PoR, GR1916, Eq. (65a), and read down to find "kappa_sigma=0". I happen to agree with Einstein, in GR the Lorentz force vanishes and that becomes the basis of the Quantum Theory, in fact IMO that's one damn good reason for GR in the first place, to place QT on a principled foundation, aside from the original ad hoc foundation Planck set forth. Regards Ken S. Tucker > "No, I am not a girl. > Thanks for the ref. " [quoted text clipped - 6 lines] > PoR, GR1916, Eq. (65a), and read down to find > "kappa_sigma=0". Thanks for the ref. but unfortunately I don't have Weinberg's "Grav and Cosmo" with me. I tried to search it in the net and came accross a few articles by you. Can you please post some relevant ones here too if you have them handy. > I happen to agree with Einstein, in GR the Lorentz > force vanishes and that becomes the basis of the [quoted text clipped - 4 lines] > Regards > Ken S. Tucker You sound great here. Would like to hear more please. > > "No, I am not a girl. > > Thanks for the ref. " [quoted text clipped - 11 lines] > articles by you. Can you please post some relevant ones here too if you > have them handy. I'm afraid my net skills are wanting, I've provided 3 ref's most of the GRist's who read this group have, if I were to mislead, I hope I'd be corrected. Try posting and asking for good net refs. > > I happen to agree with Einstein, in GR the Lorentz > > force vanishes and that becomes the basis of the [quoted text clipped - 6 lines] > > You sound great here. Would like to hear more please. Well of course, the last century of GR developement, has an element of *Peyton Place*, on the surface our small community of GRist's seems rather quaint and happy, but underneath is a seathing rage of intermingled rivalies, tragic suicides (Ehrenfest), insanity (Eddington), and much much more! (bring a sense of humor). As a nosy gossip, I see Einstein (in the 1916 ref) stated the Lorentz Force, f_u = q*F_uv U^v ==> 0 . I'll stop here, because I don't know if your familiar with LF in tensor form. Regards Ken S. Tucker > > > "No, I am not a girl. > > > Thanks for the ref. " [quoted text clipped - 45 lines] > > f_u = q*F_uv U^v ==> 0 . The Lorentzs force-law, can be written as the fourvector equation m*du/d = q*F_u u^ where m is the rest mass of a particle, q its charge, and its proper time. Here F_u are the components of the electromagnetic field tensor. Since an accelerated charge radiates one expects that the electromagnetic field produced by the charge acts upon the charge. This is not taken into account by the Lorentz force-law. Hence one is lead to modify the equation of motion of the charge as m*du/d = q*F_u u^ + (field reaction four-force) > I'll stop here, because I don't know if > your familiar with LF in tensor form. > Regards > Ken S. Tucker Please continue Ken. Read your post several times, glad your familiar with tensors. > > As a nosy gossip, I see Einstein (in the > > 1916 ref) stated the Lorentz Force, > > f_u = q*F_uv U^v ==> 0 . > The Lorentzs force-law, can be written as the fourvector equation > [quoted text clipped - 9 lines] > electromagnetic field produced by the charge acts upon the charge. This > is not taken into account by the Lorentz force-law. True, but LF (Lorentz Force) shouldn't as it doesn't contain derivatives of EM fields, that's what Maxwell's Eqs do, and predict EM waves. >Hence one is lead > to modify the equation of motion of the charge as > m*du/d = q*F_u u^ + (field reaction four-force) I hesitate to embrace that concept, the reason is that the LF I'll denote "f_u", finds 0 = f_u U^u =f , where "f" is invariant, (I use U^u as the 4-velocity) and f_u=0 is a GR requirement. Personally, I regard the f_u=0 to be a "quantum geodesic", but that's not classical GR. > Please continue Ken. I'll stay classical here, about the pendulum problem in your OP, the geodesic is defined by, the absolute derivative, DU^u=0 (classical geodesic) But the wire constraining the bob exerts the LF you know about so that, DU^u = f^u AE's problem with that is the LHS "DU^u" is "absolute acceleration", and in GR that vanishes, i.e. DU^u=0 is the basis for GR. The unification of gravitation and EM-force requires f_u=0, but still account for EM-waves, photons etc. That problem is difficult, my thinking -following AE- is to use nonsymmetrical metrics, but that requires the developement of a new calculus. Regards Ken S. Tucker Im mixing classical .... doing it anyway Then ,,tell us how gravity moves mass. Spill it out once and for all. Make evryone understand what you wrote. ILL tell you ... UP is still a gain in mass. gravity is a push to less mass. Gravity is the energy gain pushing the atom. F is identical to the mass gain pushing the wieght of the atom. Identical as in THE LAW. Re: gravitons are for dumbasses. The gain in mass is F pushing te wieght of the atom. Newton had is right, F = dp/dt is right on! " F is the gain in mass and up is a gain in mass M is the wieght of the atom befor the gain in mass . Gravity is the energy slope across the atom. all the mass of the atom falls twards its center. Atoms change mass at C. So the tom has more mass falling twards its center from one side than the other and the atom pushes its self down the energy slope . V will be the same for evry atom as the mass gain pushes the wieght and the gain is allways proportinal to the mass. SAM wrote F = ma Here F is the applied force, m is the mass of the particle, and a = dv/dt is the particle's acceleration, with v being the particle's velocity. This equation, together with the principle that bodies act symmetrically on one another--so that the force particle A feels from particle B is equal to the force B feels from A--is the basis for understanding particle dynamics". "Newton's law completely describes all the phenomena of classical mechanics...." Hi tj Yeah Newton's good, and your post looks agreeable. I'm kinda of waiting for the results of GP-b and LIGO though, these are subtle effects. Incidentally, I read you post about bouyancy, and I think you were right on. Ken > Read your post several times, glad your > familiar with tensors. [quoted text clipped - 63 lines] > Regards > Ken S. Tucker Thanks Ken for your nice postings. Kalikinkar. Likewise Kalikinkar. About online references, my problem is I'm on Dial-up modem speed right now so hopping about the net is time consuming. What you might want to do in the future is post asking for good net refs on GR. Unfortunately Weinberg's text is not on the net. Though some argue Weinberg's "Grav & Cosmo" is dated (1972), it's a pretty good comprehensive reference book. Regards Ken S. Tucker >> "No, I am not a girl. >> Thanks for the ref. " [quoted text clipped - 6 lines] >> PoR, GR1916, Eq. (65a), and read down to find >> "kappa_sigma=0". >Thanks for the ref. but unfortunately I don't have Weinberg's "Grav and >Cosmo" with me. I tried to search it in the net and came accross a few >articles by you. Can you please post some relevant ones here too if you >have them handy. >> I happen to agree with Einstein, in GR the Lorentz >> force vanishes I note that Ken does not give a reference here to back up his claim that Einstein ever said that the Lorentz force vanishes in GR. >> and that becomes the basis of the >> Quantum Theory, This demonstrates one very good reason why it is not a good idea to take what Ken says too seriously. General Relativity and Quantum Theory were independently constructed as theories, and it is impossible to derive one from the other. I once asked Ken to state the postulates of Quantum Theory (which would be a requirement if you were going to use General Relativity to derive Quantum Theory). He was either unwilling or unable to answer that very simple question. Ken has also never given any satisfactory explanation as to why he considers General Relativity (even with an electromagnetic field) to be a basis for Quantum Theory. The fact is that any attempt like Ken's, here, to link General Relativity to Quantum Theory in such a fundamental manner, is wrong. >> in fact IMO that's one damn good >> reason for GR in the first place, Except that it is not true. >> to place QT on a >> principled foundation, aside from the original ad hoc >> foundation Planck set forth. Quantum Theory was placed on a principled foundation *INDEPENDENTLY* of General Relativity. To reiterate, General Relativity has NOTHING to do with the principles of Quantum Theory. David ----- >>> "No, I am not a girl. >>> Thanks for the ref. " [quoted text clipped - 6 lines] >>> PoR, GR1916, Eq. (65a), and read down to find >>> "kappa_sigma=0". >>Thanks for the ref. but unfortunately I don't have Weinberg's "Grav and >>Cosmo" with me. I tried to search it in the net and came accross a few >>articles by you. Can you please post some relevant ones here too if you >>have them handy. >>> I happen to agree with Einstein, in GR the Lorentz >>> force vanishes >I note that Ken does not give a reference here to back up his claim that >Einstein ever said that the Lorentz force vanishes in GR. I would add here that in General Relativity, it is usual to add the stress-energy tensor for matter and the stress-energy-tensor for the electromagnetic field, to get the full stress-energy tensor which appears in Einstein's Field Equations. The fact that the full stress-energy tensor is divergenceless then implies the Lorentz-type form for the density of the four-force between the matter and the electromagnetic field (you have to use the GR form of Maxwell's Laws in the derivation). That does not mean that the Lorentz four-force is missing. It is embodied in the stress-energy tensor. But that is not what Ken means. For example, let us look at Equation (103.1) in Tolman's "Relativity Thermodynamics and Cosmology" (sic). This is the equation of a charge moving in an electromagnetic field. The equation is given as d^2x^{\mu}/ds^2 + {\alpha\beta,\mu} dx^{\alpha}/ds dx^{\beta}/ds + (e/m_0) F^{\mu}_{\alpha} dx^{\alpha}/ds = 0. The first two terms on the left hand side give the deviation of the particle from inertial motion. The third term is the negative of the Lorentz four-force divided by the mass. Ken, on the other hand, would have us believe that all three terms are inertial. Ken seems to want to see the electromagnetic field as a connection. Interestingly, in the U(1) gauge in Quantum Theory, the electromagnetic *potential* plays the part of a connection, and the electromagnetic field plays the part of a curvature. But that bears no relation to regarding the electromagnetic field as a curvature. One essential difference between gravitational effects and electromagnetic effects is that the behaviour a point particle under gravity is independent of the mass of the particle, but the behaviour of a point charged particle under an electromagnetic field depends on the charge and on the mass. David ----- > I would add here that in General Relativity, it is usual to add the > stress-energy tensor for matter and the stress-energy-tensor for the [quoted text clipped - 5 lines] > does not mean that the Lorentz four-force is missing. It is embodied in > the stress-energy tensor. I argue the EM-field contains no energy, anyway that's a bit removed from OP. > But that is not what Ken means. For example, let us look at Equation > (103.1) in Tolman's "Relativity Thermodynamics and Cosmology" (sic). > This is the equation of a charge moving in an electromagnetic field. [quoted text clipped - 6 lines] > The first two terms on the left hand side give the deviation of the > particle from inertial motion. The third term is the negative of the > Lorentz four-force divided by the mass. Ken, on the other hand, would > have us believe that all three terms are inertial. Ken seems to want to > see the electromagnetic field as a connection. When I wrote DU^u =f^u it was generic with refs, perhaps I should have written, m*DU^u = f_u to complete the dimensionality, but as so often in the brevity of ascii, one presumes a unit mass on the geodesic, and m=1. But I appreciate the correction. Correction: I find the electromagnetic fields are in the metric. Maxwell's Equations are in the connection. > Interestingly, in the U(1) gauge in Quantum Theory, the electromagnetic > *potential* plays the part of a connection, and the electromagnetic field > plays the part of a curvature. >But that bears no relation to regarding > the electromagnetic field as a curvature. That last statement seems to contradict the previous. Curvature in physics is a relation between distinguishable geodesics. You need a particle on the geodesic to measure and define it relatively to another geodesic, beyond an imaginary do-dad, conventional tensor analysis confines us to. >One essential difference > between gravitational effects and electromagnetic effects is that the > behaviour a point particle under gravity is independent of the mass of the > particle, but the behaviour of a point charged particle under an > electromagnetic field depends on the charge and on the mass. Well that's for sure! That's where GR/GC philosophy kicks in. Most GRist's are familiar but I'll try to explain them, it's fun to think about, (IMO)... 1) The physical laws of nature apply to every particle independant of it's substance. 2) A particle is defined by measurements, relatively to some FoR. 3) In a relative relation that particle is a valid FoR as the laws of nature apply to it equally as they do to the measuring FoR. We are therefore entitled to situate a rest FoR on any particle and call that the center of the universe. In that FoR the motion of the particle is described by the absolute derivative DU^u=0, (U^u is 4-velocity = dx^u/ds) since it can't move relatively to itself, that's General Relativity/General Covariance. As detailed by McAnally, an *apparent* problem occurs, due to the idea that LF "f_u =/=0". In fact if I find one FoR where f_u =0 then it generally vanishes, and since I can (must) then f_u =0 is ALWAYS true, it becomes a physical law of nature, (under GR/GC). I will remind the reader that AE's GR1916 paper, past Eq.(65a) quietly specifies this by stating kappa_sigma=0. Regards Ken S. Tucker > > I would add here that in General Relativity, it is usual to add the > > stress-energy tensor for matter and the stress-energy-tensor for the [quoted text clipped - 12 lines] > I argue the EM-field contains no energy, anyway > that's a bit removed from OP. Huh? The energy density of an electromagnetic field is a pretty basic part of E&M. > > I argue the EM-field contains no energy, anyway > > that's a bit removed from OP. > Huh? The energy density of an electromagnetic field is a pretty basic > part of E&M. I agree it's a common concept, but that concept seems to be derived from the system *potential* energy like, e = a*b/r ("a" and "b" are single charges). I would ask, is there a *need* for the concept of electric field energy in that? Ken > > > I argue the EM-field contains no energy, anyway > > > that's a bit removed from OP. [quoted text clipped - 11 lines] > I would ask, is there a *need* for the > concept of electric field energy in that? Potential energy doesn't have an absolute meaning, it's defined between two points. The potential energy also doesn't tell you where that energy *is* for an electromagnetic field, so no I don't think that makes energy density obsolete, especially since in GR what you need for the electromagnetic contribution to the stress-energy is the energy density. > > > > I argue the EM-field contains no energy, anyway > > > > that's a bit removed from OP. [quoted text clipped - 17 lines] > density obsolete, especially since in GR what you need for the > electromagnetic contribution to the stress-energy is the energy density. I think we can discuss this without GR, because about a month ago I provided a solution based on G_uv=T_uv that didn't require the electric field, posted in spr, I'll ref if you're interested. I'll ask a simple question, does one need more information than e = a*b/r to evaluate the energy stored at rest by two charges, no tricky stuff, at least not yet. Ken PS: It's obvious we're talking about static fields so McAnally's post about EM-waves delivering energy from the sun is non-sequitor, and will only confuse. Photons are another subject. > > > > > I argue the EM-field contains no energy, anyway > > > > > that's a bit removed from OP. [quoted text clipped - 27 lines] > require the electric field, posted in spr, > I'll ref if you're interested. I don't want to get too sidetracked from the original statement you made. > > > > > I argue the EM-field contains no energy, anyway > > > > > that's a bit removed from OP. This is what I'm disagreeing with, and my point was not that you couldn't get away with not *using* the energy density of an electromagnetic field, but that it seems quite wrong to say, as you did above, that the EM-field contains no energy. > I'll ask a simple question, does one need > more information than e = a*b/r to evaluate > the energy stored at rest by two charges, > no tricky stuff, at least not yet. No, you wouldn't have to use it to solve this problem. That doesn't really mean much, because I can write a collision problem where you don't have to use forces, just momentum conservation. That doesn't mean forces weren't at work there. > > > > > > I argue the EM-field contains no energy, anyway > > > > > > that's a bit removed from OP. [quoted text clipped - 37 lines] > but that it seems quite wrong to say, as you did above, that the EM-field > contains no energy. Let's be quite careful. That one uses the Electric Field *squared*, to determine energy, is not the same thing as the E-field has energy. Mr. Hogg, let me try an analogy to velocity. You have no problem with KE = (1/2) mv^2, where v is relative. But I doubt you would want to say mv has energy. Perhaps I'm splitting hairs, but if you need a square of E-field founded in more fundamental concepts, to gain energy, then I posed that distinction, the E-field has no more energy than v does, it's relative. Ken > > I'll ask a simple question, does one need > > more information than e = a*b/r to evaluate [quoted text clipped - 5 lines] > have to use forces, just momentum conservation. That doesn't mean forces > weren't at work there. > > This is what I'm disagreeing with, and my point was not that you > couldn't [quoted text clipped - 21 lines] > distinction, the E-field has no more energy > than v does, it's relative. The analogy breaks down because you can find a frame in which an object is not moving, but you cannot choose a reference frame in which there is no electromagnetic field. You might choose a frame in which the electric component goes away or one in which the magnetic component goes away, but you can't find a frame in which it all goes away. >> > This is what I'm disagreeing with, and my point was not that you >> couldn't [quoted text clipped - 30 lines] > but > you can't find a frame in which it all goes away. Remind Tucker to send my grandson the drums and amplifier he was promised, would you please? Androcles. Andro, post your address and I promise i won't tell the local kids which house to throw eggs at. > Andro, post your address and I promise i won't > tell the local kids which house to throw eggs at. Send the drums and 500W amplifier and I won't tell the cops where you get your crack from. :-) Androcles > > > This is what I'm disagreeing with, and my point was not that you > > couldn't [quoted text clipped - 27 lines] > component goes away or one in which the magnetic component goes away, but > you can't find a frame in which it all goes away. Ha...we're getting complicated. Consider the path of an electron in orbit about the nucleus (using Bohr's thinking, and stay at the shallow end of the pool:) Relative to the electron there is no electric field. The evidence of a potential difference requires the application of a photon, or if you prefer an EM wave, to vary "r". The potential difference appears from a*b/r => a*b/r' there is no derivative, at least where QT is concerned, i.e. r + delta r = r' , and that delta is not a differential. Ken S. Tucker Ken S. Tucker: >> > > This is what I'm disagreeing with, and my point was not that you >> > couldn't [quoted text clipped - 34 lines] > >Ha...we're getting complicated. It's not complicated. B^2 - E^2 is an invariant. So is B.E. If those are non-zero in _any_ frame, you can't make those become zero by changing frames. >Consider the path of an electron in orbit >about the nucleus (using Bohr's thinking, >and stay at the shallow end of the pool:) > >Relative to the electron there is no electric field. Huh? You have to be trolling. >The evidence of a potential difference requires the application >of a photon, or if you prefer an EM wave, to vary "r". [quoted text clipped - 4 lines] >there is no derivative, at least where QT is concerned, i.e. >r + delta r = r' , and that delta is not a differential. > Ken S. Tucker: > >> [quoted text clipped - 19 lines] >are non-zero in _any_ frame, you can't make those become zero by changing >frames. This is true. It is also true that if either E or B is nonzero in any frame, then for any frame, at least one of them is nonzero, i.e. if both E and B are zero in one frame, then both are zero in every frame. David ----- >> Ken S. Tucker: >> >> [quoted text clipped - 19 lines] >>are non-zero in _any_ frame, you can't make those become zero by changing >>frames. >This is true. It is also true that if either E or B is nonzero in any >frame, then for any frame, at least one of them is nonzero, i.e. if both >E and B are zero in one frame, then both are zero in every frame. And since the energy density of the electromagnetic field is proportional to E^2 + B^2, it follows that if the energy density of the field is nonzero in one frame, then it is nonzero in all frames. David ----- > Ken S. Tucker: > >Creighton Hogg wrote: [quoted text clipped - 41 lines] > > It's not complicated. Ya, complicated for me. >B^2 - E^2 is an invariant. So is B.E. If those > are non-zero in _any_ frame, you can't make those become zero by changing > frames. Vectors, (ugh), for anyone who's interested see Purcell, pg.268. ((I enjoy the review)), Bilge are you refering to EM-waves? 1)Bilge, we must keep this discussion to static fields, otherwise we'll need to sort out EM-waves from photons. 2)It is obvious "B" will always vanish, as it results from relative motion. > >Consider the path of an electron in orbit > >about the nucleus (using Bohr's thinking, [quoted text clipped - 3 lines] > > Huh? You have to be trolling. Say what? qE.v=0 that's QT, no spiralling allowed. IOW's f_0 = q*F_0i U^i =0, no absolute force in the direction of time. Now q>0, U^i>0 so what term is zero...hmmm ...let me thinks...I thunked...I vote F_0i, (==E_i), where's the error? > >The evidence of a potential difference requires the application > >of a photon, or if you prefer an EM wave, to vary "r". [quoted text clipped - 4 lines] > >there is no derivative, at least where QT is concerned, i.e. > >r + delta r = r' , and that delta is not a differential. Regards Ken S. Tucker >> Ken S. Tucker: >> >Creighton Hogg wrote: [quoted text clipped - 45 lines] >> >> It's not complicated. >Ya, complicated for me. >>B^2 - E^2 is an invariant. So is B.E. If those >> are non-zero in _any_ frame, you can't make those become zero by >changing >> frames. >Vectors, (ugh), for anyone who's interested >see Purcell, pg.268. ((I enjoy the review)), >Bilge are you refering to EM-waves? No, Bilge is not referring to EM-waves. B^2 - E^2 and B.E are both invariants in Special Relativity, independently of whether EM-waves are involved or not. >1)Bilge, we must keep this discussion to >static fields, otherwise we'll need to >sort out EM-waves from photons. Bilge never mentioned EM-waves. The fact is that under the transformation law for the EM-field, B^2 - E^2 and B.E are invariants. >2)It is obvious "B" will always vanish, >as it results from relative motion. This presupposes that all charges are stationary. Since there can be a charge moving in any given frame of reference, then B need not be zero in any frame of reference. >> >Consider the path of an electron in orbit >> >about the nucleus (using Bohr's thinking, [quoted text clipped - 3 lines] >> >> Huh? You have to be trolling. >Say what? qE.v=0 Why? >that's QT, I presume that by "QT", you mean "Quantum Theory". In that case, qE.v= 0 has nothing to do with Quantum Theory. What is the Hamiltonian for the charged body? What does its wave function look like? >no spiralling allowed. There are no trajectories in Quantum Theory. >IOW's >f_0 = q*F_0i U^i =0, >no absolute force in the direction of time. You mean that there is no component of the 4-force in the direction of time. Why not? >Now q>0, U^i>0 For which i? >so what term is zero...hmmm >...let me thinks...I thunked...I vote F_0i, >(==E_i), where's the error? It is possible for two nonzero vetors to have dot product equal to zero. >> >The evidence of a potential difference requires the application >> >of a photon, or if you prefer an EM wave, to vary "r". [quoted text clipped - 4 lines] >> >there is no derivative, at least where QT is concerned, i.e. >> >r + delta r = r' , and that delta is not a differential. >Regards >Ken S. Tucker > >> Ken S. Tucker: > >> >Creighton Hogg wrote: [quoted text clipped - 60 lines] > invariants in Special Relativity, independently of whether EM-waves are > involved or not. You've imported a problem, you've specified I quote, "invariants in Special Relativity". SR is a specialized CS, (uses R^a_bcd=0). Wouldn't be easier if we put invariants in a Generally Covariant (GC) form consistent with GR. I my previous post, I translated Bilges vectors into a GC LF form. > >1)Bilge, we must keep this discussion to > >static fields, otherwise we'll need to [quoted text clipped - 9 lines] > charge moving in any given frame of reference, then B need not be zero in > any frame of reference. Careful, are you saying we're dealing with charges varing in relative position, that would require an energy potential diff to be actualized, either by emission or absorption of EM radiation, (photons), or a set of relatively stationary charges that are regarded from a moving FoR. I do wish you would be more specific. No offense is intended but, care must be taken. > >> >Consider the path of an electron in orbit > >> >about the nucleus (using Bohr's thinking, [quoted text clipped - 7 lines] > > Why? > >that's QT, > > I presume that by "QT", you mean "Quantum Theory". In that case, qE.v= 0 > has nothing to do with Quantum Theory. Excuse me, if qE.v >0 then E and v are not perpendicular. This means a force occurs from E to v, and energy will *continuously radiant*, that's not what happens, although it was a classical calculation. QT and GR support qE.v=0. >What is the Hamiltonian for the > charged body? There is no such thing. Your "charged" body would need a test charge to realize your body is charged, and that in turn is a solution to a system with the charged body inclusive. For example Mr. Non Ame, the General Theory of Relativity insists you may attach a stationary FoR to Mercury and arrive at the solution the Sun has a perihelion shift relative to Mercury. How would you do that? >What does its wave function look like? I think real good, GR's G_uv=T_uv is damn close to a standing EM-wave. One way to that is to consider how to move from one geodesic to another for a particle in free-fall. We need a photon to do that, hence the relation of two distinct geodesics is physically varied and thus measured with photons. > >no spiralling allowed. > [quoted text clipped - 8 lines] > You mean that there is no component of the 4-force in the direction of > time. Why not? Because DU^0 =/=0 violates QT. > >Now q>0, U^i>0 > [quoted text clipped - 5 lines] > > It is possible for two nonzero vetors to have dot product equal to zero. And the point is...? > >> >The evidence of a potential difference requires the application > >> >of a photon, or if you prefer an EM wave, to vary "r". [quoted text clipped - 7 lines] > >Regards > >Ken S. Tucker Ken >> >> Ken S. Tucker: >> >> >Creighton Hogg wrote: [quoted text clipped - 65 lines] >are >> involved or not. >You've imported a problem, you've specified >I quote, "invariants in Special Relativity". >SR is a specialized CS, (uses R^a_bcd=0). >Wouldn't be easier if we put invariants in >a Generally Covariant (GC) form consistent >with GR. The General Relativistic invariant, F^{uv} F_{uv} = g^{ab} g^{cd} F_{ac} F_{bd}, is proportional to B^2 - E^2. The General Relativistic scalar density, e^{abcd} F_{ab} F_{cd}, where e^{abcd} is the Levi-Civita tensor density, is proportional to B.E. This answers your question about the General Relativistic objects which correspond to B^2 - E^2 and to B.E. > I my previous post, I translated Bilges >vectors into a GC LF form. >> >1)Bilge, we must keep this discussion to >> >static fields, otherwise we'll need to [quoted text clipped - 12 lines] >zero in >> any frame of reference. >Careful, are you saying we're dealing with >charges varing in relative position, that would >require an energy potential diff to be actualized, No, it doesn't. It merely requires them to be in motion relative to each other. >either by emission or absorption of EM radiation, >(photons), or a set of relatively stationary >charges that are regarded from a moving FoR. What possible reason could you give for expecting all charges to be stationary in the one frame of reference? >I do wish you would be more specific. >No offense is intended but, care must be taken. The situation in which charged bodies are in motion relative to each other (even without the influence of an electromagnetic potential) is far more natural than the artificial situation that you are suggesting, in which all the charges are stationary with respect to the same frame of reference. >> >> >Consider the path of an electron in orbit >> >> >about the nucleus (using Bohr's thinking, [quoted text clipped - 7 lines] >> >> Why? >> >that's QT, >> >> I presume that by "QT", you mean "Quantum Theory". In that case, >qE.v= 0 >> has nothing to do with Quantum Theory. >Excuse me, if qE.v >0 then E and v are not >perpendicular. This is true. >This means a force occurs >from E to v, What does "from E to v" mean? >and energy will *continuously radiant*, "Radiant" is not a verb. There is nothing to stop qE.v from being positive at one event, negative at another event, and zero at some event in between. >that's not what happens, although it was a classical >calculation. QT and GR support qE.v=0. No, they don't. I haven't seen any real explanation about why they would. >>What is the Hamiltonian for the >> charged body? >There is no such thing. Of course there is a Hamiltonian for the body. >Your "charged" >body would need a test charge to realize >your body is charged, No. The interaction of the charge with the electromagnetic field would be sufficient. In fact, in Relativity (both Special and General), the interaction with the electromagnetic field is the full extent to which an electric charge can interact electrically with the rest of the universe. The Hamiltonian for a single free charged body in an electromagnetic field is: H = (p - q A)^2/(2m) + q phi in the non-relativistic case, H = sqrt[(p - q A)^2 + m^2] + q phi in the relativistic case, and H = alpha.(p - q A) + m beta + q phi in the case of the Dirac electron. Here, A is the vector potential, phi is the scalar potential, q is the charge, p is the canonical momentum, m is the mass, and alpha and beta are 4x4 matrices such that alpha_i alpha_j + alpha_j alpha_i = 2 delta_{ij}, i, j = 1, 2, 3, alpha_i beta + beta alpha_i = 0, i = 1, 2, 3, beta^2 = 1. Note in particular that in the non-relativistic non-quantum case, the canonical momentum is given by p = m v + q A. >and that in turn is >a solution to a system with the charged >body inclusive. Your vision seems unduly restrictive. >For example Mr. Non Ame, the General Theory >of Relativity insists you may attach a >stationary FoR to Mercury and arrive at >the solution the Sun has a perihelion shift >relative to Mercury. How would you do that? It's already been done, using the Schwarzchild metric, and it is the orbit of Mercury which has the precession of the perihelion, not the Sun. >>What does its wave function look like? >I think real good, GR's G_uv=T_uv is damn >close to a standing EM-wave. Do you even know what a wave function is? Look it up in a book on quantum mechanics some time. > One way to that is to consider how to >move from one geodesic to another for a >particle in free-fall. Why? >We need a photon to >do that, hence the relation of two distinct >geodesics is physically varied and thus >measured with photons. Why? Why not something else? >> >no spiralling allowed. >> [quoted text clipped - 9 lines] >of >> time. Why not? >Because DU^0 =/=0 violates QT. You haven't shown this yet. >> >Now q>0, U^i>0 >> [quoted text clipped - 6 lines] >> It is possible for two nonzero vetors to have dot product equal to >zero. >And the point is...? You had two vectors, and that their dot product was equal to zero. You then invoked the assertion that one of the vectors is nonzero in order to conclude that the other vector was equal to zero. The point is that my observation invalidates that conclusion. Mr. Ame, Read your post. Evidentally you wish to remain anonymous, wow give a MENSA point for that one, and knowledgeable. > >> >> Ken S. Tucker: > >> >> >Creighton Hogg wrote: [quoted text clipped - 78 lines] > > is proportional to B^2 - E^2. Good and thanks, obviously F^uv F_uv = invariant (I'll denote that F^2). 1)Would you characterise F^2 as an energy density? 2)If so, is it non-zero? > The General Relativistic scalar density, e^{abcd} F_{ab} F_{cd}, > where e^{abcd} is the Levi-Civita tensor density, is proportional to > B.E. Sure, as I understand it that will always vanish in some CS, because "B" is relative, I'm with you. > This answers your question about the General Relativistic objects which > correspond to B^2 - E^2 and to B.E. Yes thank you, I'm ok with those. > > I my previous post, I translated Bilges > >vectors into a GC LF form. [quoted text clipped - 22 lines] > No, it doesn't. It merely requires them to be in motion relative to each > other. I'll repeat, IOW's if you push or pull charges together you require force*distance. Coulomb's F = q1*q2/r^2 , (energy) = Force*radius = q1*q2/r You can't get around that. your "relative motion" implies a dr, and that needs energy. > >either by emission or absorption of EM radiation, > >(photons), or a set of relatively stationary > >charges that are regarded from a moving FoR. > > What possible reason could you give for expecting all charges to be > stationary in the one frame of reference? What's all charges we only have two, in my gedanken. > >I do wish you would be more specific. > >No offense is intended but, care must be taken. [quoted text clipped - 4 lines] > which all the charges are stationary with respect to the same frame of > reference. Opinion noted. > >> >> >Consider the path of an electron in orbit > >> >> >about the nucleus (using Bohr's thinking, [quoted text clipped - 23 lines] > > What does "from E to v" mean? That was succintly worded, it means an effective force will result on a charge "q" following vector "v" in an E-field "E". > >and energy will *continuously radiate*, > There is nothing to stop qE.v from being > positive at one event, negative at another event, and zero at some event > in between. If you want to write a theory of physics supporting that notion you go for it, I argue for GR, not against you, if I can see anyway to help your theory I'll likely contribute if it's worthwhile for you. I'm arguing also for Old AE, see GR1916 Eq.(65a) where he *suggests* kappa_sigma=0. > >that's not what happens, although it was a classical > >calculation. QT and GR support qE.v=0. > > No, they don't. I haven't seen any real explanation about why they would. Be patient, it'll sink in. > >>What is the Hamiltonian for the > >> charged body? [quoted text clipped - 9 lines] > No. The interaction of the charge with the electromagnetic field would be > sufficient. Gee I wonder if said "electromagnetic field" results from a a relative body? >In fact, in Relativity (both Special and General), the > interaction with the electromagnetic field is the full extent to which an > electric charge can interact electrically with the rest of the universe. Ok > The Hamiltonian for a single free charged body in an electromagnetic field > is: [quoted text clipped - 22 lines] > > Note in particular that in the non-relativistic non-quantum case, the > canonical momentum is given by p = m v + q A. p^u = q*A^u when v=0? > >and that in turn is > >a solution to a system with the charged > >body inclusive. > > Your vision seems unduly restrictive. On the contrary. > >For example Mr. Non Ame, the General Theory > >of Relativity insists you may attach a [quoted text clipped - 4 lines] > It's already been done, using the Schwarzchild metric, and it is the orbit > of Mercury which has the precession of the perihelion, not the Sun. Careful, an observer standing on Mercury has an right to claim that to be the centre of the universe as a solar centric one, is that a good definition of General Covariance? > >>What does its wave function look like? > [quoted text clipped - 3 lines] > Do you even know what a wave function is? Look it up in a book on quantum > mechanics some time. Check out that wave mechanic probabilty density and GR's field have the same units, ENERGY DENSITY. Confirm please, I may be wrong. > > One way to that is to consider how to > >move from one geodesic to another for a [quoted text clipped - 8 lines] > > Why? Why not something else? To date physically, surveying with light-rays, "radar, lasers" is the standard. > >> >no spiralling allowed. > >> [quoted text clipped - 13 lines] > > You haven't shown this yet. > >> >Now q>0, U^i>0 > >> [quoted text clipped - 13 lines] > conclude that the other vector was equal to zero. The point is that my > observation invalidates that conclusion. Yes, I see your point, the argument I intended was to show that the "E" field is a geodesical to a charged particle, like the g-field is to a massive particle. Doing that gets DU^u=0 in EM-fields, i.e. eliminates Lorentz. I can put Maxwell's Eq's into the Christoffel if you want to see that. Regards Ken S. Tucker > > >>What does its wave function look like? > > [quoted text clipped - 9 lines] > > Confirm please, I may be wrong. Probability density can't have the same units as energy density. Probability carries no units. No connection between quantum mechanics and GR. In fact, they're downright incompatible. | > > >>What does its wave function look like? | > > [quoted text clipped - 14 lines] | No connection between quantum mechanics and GR. In fact, they're | downright incompatible. Hmm... I think Mr. Hogg needs more unit lessons here maybe. Say we have an E-field and it is proportional to, E_0 ~ sqrt(hbar*c*n)/lambda^2 Where n is the number of hbar*c "cells". Square that and we have energy density, u ~ hbar*c*n/lambda^4 Now in natural units of hbar = c = 1, we get, u = n/lambda^4 Hmm... the numerator has no units now. How about that? Well, you can do this with e^2 = alpha also. You end up with, alpha/length = energy alpha/length^4 = energy density What is alpha? It is a probability according to Feynman. FrediFizzx http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf or postscript http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps Thanks Freddi. We're not here to solve all the worlds problems. Nor to say we're unifying QT and GR. But for Mr. Hogg and ourselves, it's a reasonable excersize to compare the probability disturbutions of wave mechanics, with the ENERGY DENSITY in GR... G_uv = T_uv. The first test is compatibility of units. I could rant big time on the equivalence of a particle *probably* being somewhere and the smear the G_uv=T_uv does that. The units of ENERGY DENSITY seem common. Barring error suppose that's true. Call me crazy, but I figure the wave function for the Sun, and solution G_uv=T_uv is the same. One experiment I'd suggest theoretically is to impact a photon, on the wave function and simultaneuosly on AE's G=T. Suppose the result was gauge indifferent. (Meaning an adjust between results may be accomplished by the relativity of the socalled constants)). Better stop here, pending in-bound spitballs. Ken S. Tucker > Thanks Freddi. > We're not here to solve all the worlds [quoted text clipped - 7 lines] > > The first test is compatibility of units. Which it fails. > I could rant big time on the equivalence > of a particle *probably* being somewhere > and the smear the G_uv=T_uv does that. > The units of ENERGY DENSITY seem > common. Barring error suppose that's > true. The Einstein field equations don't "smear" anything. They're amazingly, perfectly deterministic and point particles most certainly remain point particles in case that was what you meant. They are also completely incompatible with quantum mechanics. > Call me crazy, but I figure the wave function > for the Sun, and solution G_uv=T_uv is the > same. Except that the solution of Einstein's field equations is deterministic. The wave function for the entire sun as a system would most certainly not be deterministic. > > Thanks Freddi. > > We're not here to solve all the worlds [quoted text clipped - 9 lines] > > Which it fails. Then why reply, do you expect people to spend an afternoon figuring out how to explain some- thing to you. > > I could rant big time on the equivalence > > of a particle *probably* being somewhere [quoted text clipped - 6 lines] > perfectly deterministic and point particles most certainly remain point > particles in case that was what you meant. It's interpretation, the stronger the g-field the more probable the mass density. Please do not confuse that with the electromagnetic forces, specifically, locating a massive body using EM-forces, like a chemically constructed surface. Disregard entirely EM-force and think in g-field only. Become a phantom that can move through solid objects and your only sense is G_uv. >They are also completely > incompatible with quantum mechanics. If you want. > > Call me crazy, but I figure the wave function > > for the Sun, and solution G_uv=T_uv is the > > same. > > Except that the solution of Einstein's field equations is deterministic. No, regard the retardation. It's not instanteous. Objects in a g-system can only communicate their g-field presence to other objects at c. It's certainly not deterministic. Try photo- graphing the solar system with radar, then convert that image into a "true" picture of the system, by true I mean determined. Because "c" (radar) is finite, you'll get differing images dependant on location. In addition, by the time you have received your reflection, a planet could have exploded or something. If you ask me, you want to be very careful about importing Newtonian ideas into a universe that functions using GR. I empathize, it's very easy to adopt a lazy minded approach, but it will bite your bum-bum, (I have pictures)! > The wave function for the entire sun as a system would most certainly not > be deterministic. Yup, I figure it's probablity field to be very near the ENERGY DENSITY T_uv, measured gravitationally. Regards Ken S. Tucker > > > Thanks Freddi. > > > We're not here to solve all the worlds [quoted text clipped - 13 lines] > an afternoon figuring out how to explain some- > thing to you. Because I hope that someday you'll take stock and realize that when everyone who works in physics here tells you that you've got it wrong, that maybe you're wrong. You don't bother listening to anyone when they explain the mistakes you make. You don't even admit you've made them when they've been pointed out. You never admitted that you didn't understand Bjoern's thesis, hell you haven't even admitted you were wrong about him not having a phd. You never admitted you were wrong about heavy ion physics and the connection to exotic quark matter. You never admit you're wrong when Steve Carlip, who has more experience with GR than everyone else on the newsgroup, including you, combined, tells you that you're wrong. > > > I could rant big time on the equivalence > > > of a particle *probably* being somewhere [quoted text clipped - 11 lines] > It's interpretation, the stronger the g-field > the more probable the mass density. There's nothing to do with probabilities in GR. It's completely deterministic. The exact positions and velocities of your system are the input you feed into the Einstein field equations. > Please do not confuse that with the electromagnetic > forces, specifically, locating a massive body [quoted text clipped - 8 lines] > > If you want. Not if I want Ken, that's just how it is. > > > > Thanks Freddi. > > > > We're not here to solve all the worlds [quoted text clipped - 17 lines] > everyone who works in physics here tells you that you've got it wrong, > that maybe you're wrong. Right and wrong is for god, and making pronouncements of truth is to use refs. What a lot of people do in this NG is consider alternatives. >You don't bother listening to anyone when they > explain the mistakes you make. Oh-Oh a flame, next you'll invoke peer pressure and insist I conform... [yup snip gossip] > > > > I could rant big time on the equivalence > > > > of a particle *probably* being somewhere [quoted text clipped - 15 lines] > deterministic. The exact positions and velocities of your system are the > input you feed into the Einstein field equations. That's a reasonable classical interpretation. > > Please do not confuse that with the electromagnetic > > forces, specifically, locating a massive body [quoted text clipped - 10 lines] > > Not if I want Ken, that's just how it is. Ok, Mr. Hogg, you have a hardened position that's intolerant of alternatives. Ken Ken S. Tucker: >> > > > Thanks Freddi. >> > > > We're not here to solve all the worlds [quoted text clipped - 20 lines] > >Right and wrong is for god, and making This is god speaking. You're wrong.  Signaturegod who will soon be sending an email to reserve a special place in hell for ken. On Tue, 22 Feb 2005, God wrote: > > This is god speaking. You're wrong. I always thought you'd be taller. > > > > Thanks Freddi. > > > > We're not here to solve all the worlds [quoted text clipped - 17 lines] > everyone who works in physics here tells you that you've got it wrong, > that maybe you're wrong. Right and wrong is for god, and making pronouncements of truth is to use refs. What a lot of people do in this NG is consider alternatives. >You don't bother listening to anyone when they > explain the mistakes you make. Oh-Oh a flame, next you'll invoke peer pressure and insist I conform... [yup snip gossip] > > > > I could rant big time on the equivalence > > > > of a particle *probably* being somewhere [quoted text clipped - 15 lines] > deterministic. The exact positions and velocities of your system are the > input you feed into the Einstein field equations. That's a reasonable classical interpretation. > > Please do not confuse that with the electromagnetic > > forces, specifically, locating a massive body [quoted text clipped - 10 lines] > > Not if I want Ken, that's just how it is. Ok, Mr. Hogg, you have a hardened position that's intolerant of alternatives. Ken > > > > > Thanks Freddi. > > > > > We're not here to solve all the worlds [quoted text clipped - 23 lines] > What a lot of people do in this NG is > consider alternatives. Right and wrong is determined by mathematical consistency and empirical correctness. Alternatives are constrained by these things very tightly. > >You don't bother listening to anyone when they > > explain the mistakes you make. [quoted text clipped - 3 lines] > > [yup snip gossip] It wasn't gossip. I pointed out a few times you've said blatantly false things and never admitted it when it was pointed out to you. It's not asking you to conform, it's asking you stop insisting 0 = 1. > > > > > I could rant big time on the equivalence > > > > > of a particle *probably* being somewhere [quoted text clipped - 36 lines] > Ok, Mr. Hogg, you have a hardened position > that's intolerant of alternatives. That isn't true. I am open to just about anything if it can help us understand nature just a little better. I'm opposed to alot of the things you say because they are false, empirically or mathematically. In this case, GR and quantum mechanics have a horrible time getting along. A quantum theory of gravity requires changing GR, changing QM, or changing both in a way no one has thought of. String theory takes the first route, LQG the second, and I wish to God someone had a workable theory that took the third. [snip] | > Ok, Mr. Hogg, you have a hardened position | > that's intolerant of alternatives. [quoted text clipped - 9 lines] | LQG the second, and I wish to God someone had a workable theory that took | the third. So what is wrong with figuring out how to change (improve) GR or QT? They both fit nature fairly well so they have to meet somewhere. I get the impression that Ken is going down that path. Mistakes are bound to be made on the journey. What exactly is it that you think is the most "horrible time of getting along"? I think we would like to know what you think that is. Math can describe more than nature probably uses physically. Don't you think humans have the capability of using math to cheat a little bit? Go ahead and cheat a little bit. It makes physics more fun if it can still get the right answers. FrediFizzx http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf or postscript http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps Fredi, Creighton and all... I'd like to compare QM's "probability density" with GR's energy density very simply. I have 10 basketballs each with mass 1, and I throw them in a box 10x10x10 metres. (1)What is the energy density of that box? (2)What is the probability density I will find a basketball in that box at some point? Let's solve that problem this way, if we throw in another 10 balls, (3) would it be true that energy density doubles, and probablity density doubles? (4) If (3) is true then we have an arbituary proportionality constant relating (1) and (2). If (4) is true, then the conceptual relation, (5) "probability density" == "energy density". is true. In conventional QM, texts show "probablity density" as (PSI)*(PSI) dx dy dz and in GR "energy density" as Tuv . Regards Ken S. Tucker >Fredi, Creighton and all... > I'd like to compare QM's "probability density" with GR's >energy density very simply. >I have 10 basketballs each with mass 1, and I throw them >in a box 10x10x10 metres. >(1)What is the energy density of that box? More details are needed before that question can be answered. >(2)What is the probability density I will find a basketball >in that box at some point? The answer to that question is not a probability density, but a probability. Or do you mean a spatial point in the box, rather than a point in time? If you mean a spatial point, and the balls are not moving under the action of gravity, then the probability density is very close to 10 times the volume of a basketball divided by the volume of the box. If gravity is present, then the probability distribution will have a marked increase as you move towards the bottom of the box, and above a certain height, the probability density will be negligible. >Let's solve that problem this way, if we throw in another >10 balls, >(3) would it be true that energy density doubles, >and probablity density doubles? The probability density will double approximately. The doubling is not exact. What happens to the energy density is anybody's guess. (3) is true for neither, although it is quite close to the truth for the probability density. Of course, this does not double the probability of finding the balls in the box. This is because the simultaneous presence of balls at distinct points in the box is not forbidden, and the presence of a ball at P and the presence of a ball at a distinct point Q are not mutually exclusive. It follows that the nett probability is strictly less than the integral of the probability density over the whole box (in fact, it is MUCH less than the integral). >(4) If (3) is true then we have an arbituary proportionality >constant relating (1) and (2). There is no guarantee that (3) is true. >If (4) is true, then the conceptual relation, (4) need not be true. >(5) "probability density" == "energy density". >is true. (5) need not be true. >In conventional QM, texts show "probablity density" as >(PSI)*(PSI) dx dy dz That is the probability per unit volume (if you are measuring coordinates). If you are measuring a completely different observable, then the probability density looks nothing like that. For example, the probability density with respect to the momentum is (2 pi hbar)^(-3) [int psi(x,y,z) exp(- i (p_x x + p_y y + p_z z)/hbar) dx dy dz]* [int psi(x,y,z) exp(- i (p_x x + p_y y + p_z z)/hbar) dx dy dz] = (2 pi hbar)^(-3) int dx dy dz dx' dy' dz' psi(x,y,z)* psi(x',y'z') exp(i [p_x (x - x') + p_y (y - y') + p_z (z - z')]/hbar). >and in GR "energy density" as >Tuv . Energy density is T^{00}. >Regards >Ken S. Tucker Cheers! > >Fredi, Creighton and all... > > I'd like to compare QM's "probability density" with GR's [quoted text clipped - 13 lines] > a probability. Or do you mean a spatial point in the box, > rather than a point in time? ... > For example, the probability density with respect to the momentum is [snip, mcanally, you're not even wrong] Poincare transformations do apply, I snipped. Do'not waste time. Ken >> >Fredi, Creighton and all... >> > I'd like to compare QM's "probability density" with GR's [quoted text clipped - 14 lines] >> rather than a point in time? >... >> For example, the probability density with respect to the momentum is >[snip, mcanally, you're not even wrong] You are an idiot. My name is not McAnally, but thank you for the compliment. He seems to be a *lot* more intelligent than you are. Your exposition are full of false statements, and complete misunderstandings of physics. On the other hand, I thought that his comments on Hamilton's principle for a charge (and its generalization to General Relativity) were interesting. You would seem to be basing your claim on David Bowman's innuendo that David McAnally and I have the same internet provider. More fool you. Surely, you are not so limited in intelligence as to think that a commercial provider only has *one* user. Do you, for example, think that you are the only user of vianet.on.ca? If you are so limited in intelligence as to think that, then you are too stupid to be worth worrying about, any more. Like many others, I have tried to enlighten you. And, like many others, I have come to the inevitable conclusion that to attempt to enlighten one such as you is a waste of time, because you are either unable or unwilling to open your mind and to learn that of which you are ignorant. I personally think that it is because you are too stupid to be able to learn. I note that Creighton Hogg came to exactly the same conclusion about you. But I presume that you are upset by anybody shooting down your obsession that energy density is the same as probability density, and you refuse to accept any argument against your prejudices, no matter how rigourous and waterproof. >Poincare transformations Fourier transforms. Get the terminology right. And if you can't even understand the importance of the Fourier transforms where I introduced them, then you are ignorant of quantum mechanics. Of course, your ignorance of quantum mechanics comes as no surprise. > >> For example, the probability density with respect to the momentum is > > >[snip, mcanally, you're not even wrong] > > You are an idiot. My name is not McAnally, I didn't say it is, why are you so defensive, it was an honest web error. (David Bowman, I would say you're 99% probably right, that high intelligence you have is as good as a blood hound chasing a burgler from a fish store...koala too, Detective Dick Dirky missed that one, of course if you're wrong about Ms. Non Ame, I may have to provide make-up sex, I'll stop here, after all McAnally might be posting from his mom's computer). This has got to be the most worst post i've done in the last hour or so... Sorry ken | Fredi, Creighton and all... | I'd like to compare QM's "probability density" with GR's [quoted text clipped - 28 lines] | | Tuv . You don't have to convince me that there is or ought to be a connection between them. But, I think I liked my alpha/length^4 = e^2/length^4 equals energy density better. But we have an ambiguous 4D volume here. Let's take the Dirac Lagrangian for a spinor field which has units of energy density with cgs units; L = i(hbar*c)psibar*gamma^u*@_u*psi - (mc^2)*psibar*psi Where @_u is the partial. This can be written as; L = (hbar*c)[i*psibar*gamma^u*@_u*psi - (2pi/lambda_m)psibar*psi] The expression in the brackets, [ ], is simply a 1/(4D volume). There is a probability of ONE that the spinor field is in that 4D volume. So maybe this is a better example to use for going to GR. Quantum field theory is all based on the Lagrangian density which has units of energy density and it pretty much can successfully describe elementary particle interactions in the absence of gravity since gravity is so weak. But to be fully compliant with nature, gravity ought to be included in there. The problem is that there is no way to measure it until well after the correspondence principle kicks in. Plus I have a suspicion that the "Partial Inter-dimensional Transformations" are at work here. ;-) I think we might be able to blame "dual space" for that besides just blaming you. LOL FrediFizzx >> >> >>B^2 - E^2 is an invariant. So is B.E. If those >> >> >> are non-zero in _any_ frame, you can't make those become zero [quoted text clipped - 25 lines] >> >> is proportional to B^2 - E^2. >Good and thanks, obviously >F^uv F_uv = invariant (I'll denote that F^2). >1)Would you characterise F^2 as an energy density? Why would I? >2)If so, is it non-zero? F^{uv} F_{uv} can certainly be nonzero. >> The General Relativistic scalar density, e^{abcd} F_{ab} F_{cd}, >> where e^{abcd} is the Levi-Civita tensor density, is proportional to >> B.E. >Sure, as I understand it that will always vanish >in some CS, because "B" is relative, I'm with you. No, you are most certainly not "with me". I never stated that e^{abcd} F_{ab} F_{cd} would vanish in *any* frame of reference. In fact, if e^{abcd} F_{ab} F_{cd} vanishes in one frame of reference, then it vanishes in all frames of reference. More generally, if a scalar density is nonzero in one frame of reference, then it is nonzero in all frames of reference, and if a scalar density is zero in one frame of reference, then it is zero in all frames of reference. It seems that you have not comprehended this important point. >> This answers your question about the General Relativistic objects >which >> correspond to B^2 - E^2 and to B.E. >Yes thank you, I'm ok with those. >> > I my previous post, I translated Bilges >> >vectors into a GC LF form. [quoted text clipped - 24 lines] >each >> other. >I'll repeat, IOW's if you push or pull charges >together you require force*distance. Coulomb's What does this have to do with the fact that the charges may be moving relative to each other? Why does any relative motion between the charges have to be caused by a force between them? You have not justified this assertion. >F = q1*q2/r^2 , >(energy) = Force*radius = q1*q2/r That is not the definition of potential energy. That what you have on the far right-hand side is a formula for the potential energy is a consequence of the fact that the force is inverse square in r, and therefore the two formulae agree by accident. >You can't get around that. your "relative >motion" implies a dr, and that needs energy. Why does this forbid any possibility of relative motion? >> >either by emission or absorption of EM radiation, >> >(photons), or a set of relatively stationary >> >charges that are regarded from a moving FoR. >> >> What possible reason could you give for expecting all charges to be >> stationary in the one frame of reference? >What's all charges we only have two, in my >gedanken. What possible reason could you give for expecting the two charges to be stationary in the one frame of reference? >> >I do wish you would be more specific. >> >No offense is intended but, care must be taken. [quoted text clipped - 7 lines] >of >> reference. >Opinion noted. Why is it an opinion? You have not demonstrated that the situation that you are suggesting is any more natural than the situation in which the charges are in relative motion. You seem to think that you don't need to provide a proof that the charges must be stationary in one frame of reference. >> >> >> >Consider the path of an electron in orbit >> >> >> >about the nucleus (using Bohr's thinking, [quoted text clipped - 23 lines] >> >> What does "from E to v" mean? >That was succintly worded, it means an effective >force will result on a charge "q" following >vector "v" in an E-field "E". That has nothing to do with what qE.v signifies. Why can't we use standard terminology? You aren't even following standard terminology here. >> >and energy will *continuously radiate*, "Radiant" is not a verb. The verb is "radiate". >> There is nothing to stop qE.v from being >> positive at one event, negative at another event, and zero at some >event >> in between. >If you want to write a theory of physics supporting >that notion you go for it, I argue for GR, GR does not contradict the assertion that I made above. You seem to have very unusual ideas about what physics theories assert and what they don't assert. >not against >you, if I can see anyway to help your theory Seeing that my assertion does not contradict GR, then I don't see any need to create any new theory. I don't know how you could possibly think that my assertion above contradicts GR. I don't need any help with a new theory since I am not developing a new theory. And if I were developing a new theory, why would I want your help? >I'll >likely contribute if it's worthwhile for you. > I'm arguing also for Old AE, see GR1916 Eq.(65a) >where he *suggests* kappa_sigma=0. I will check this bit out in detail, to see if I can discern Einstein's actual intention with what he wrote. Have you considered the possibility that something may have been missed in the translation from German to English? >> >that's not what happens, although it was a classical >> >calculation. QT and GR support qE.v=0. >> >> No, they don't. I haven't seen any real explanation about why they >would. >Be patient, it'll sink in. Not without your giving a valid explanation for this particular prejudice. So far, I have seen no such explanation. >> >>What is the Hamiltonian for the >> >> charged body? [quoted text clipped - 10 lines] >would be >> sufficient. >Gee I wonder if said "electromagnetic field" >results from a a relative body? It is possible to have a nonzero electromagnetic field without any sources. Also, your observation was not relevant to what I wrote. The electromagnetic field takes time to transfer the interaction from one charge to another. >>In fact, in Relativity (both Special and General), the >> interaction with the electromagnetic field is the full extent to >which an >> electric charge can interact electrically with the rest of the >universe. >Ok >> The Hamiltonian for a single free charged body in an electromagnetic >field [quoted text clipped - 25 lines] >> >> Note in particular that in the non-relativistic non-quantum case, the >> canonical momentum is given by p = m v + q A. >p^u = q*A^u when v=0? The canonical conjugate momentum is equal to q A when v = 0. >> >and that in turn is >> >a solution to a system with the charged >> >body inclusive. >> >> Your vision seems unduly restrictive. >On the contrary. As I remarked earlier, a charge only needs an electromagnetic field with which to interact. >> >For example Mr. Non Ame, the General Theory >> >of Relativity insists you may attach a [quoted text clipped - 5 lines] >orbit >> of Mercury which has the precession of the perihelion, not the Sun. >Careful, an observer standing on Mercury >has an right to claim that to be the centre >of the universe as a solar centric one, >is that a good definition of General Covariance? If you want to attach a frame of reference to Mercury, then you do not have a set of coordinates with which to solve Einstein's field equations (in fact, the problem is almost certainly intractable), except in a neighbourhood of Mercury which certainly excludes the Sun. >> >>What does its wave function look like? >> [quoted text clipped - 4 lines] >quantum >> mechanics some time. >Check out that wave mechanic probabilty density >and GR's field have the same units, ENERGY DENSITY. Probability density has the units of inverse volume. Energy density has the units of energy per unit volume. Even after setting universal constants appropriate to GR to unity, energy density has the units of inverse area. >Confirm please, I may be wrong. You are. One is inverse volume. The closest that the other can get is inverse area. And even if they had the same units, that would not mean that they were related in any way. >> > One way to that is to consider how to >> >move from one geodesic to another for a [quoted text clipped - 8 lines] >> >> Why? Why not something else? >To date physically, surveying with light-rays, >"radar, lasers" is the standard. But you have not shown that other possibilities are excluded. >> >> >no spiralling allowed. >> >> |
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